Polynomials - Polynomial Long Division - Example5

Polynomials: Factor and Remainder Theorem

Polynomials: Factor and Remainder Theorem

In Algebra, we regularly come across the polynomials. The skill in solving algebraic operations of polynomials is very important in mathematics.

If you need to revise on the basic of algebra, you can look at it here.

The definition

\( {\small P(x) \ = \ a_{n} x^{n} + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + … + a_{0} }\)

\(\\[5pt] {\small n \ }\) is a non-negative integer. It is the degree or the order of the polynomial.

\(\\[5pt] {\small x \ }\) is the variable of the polynomial.

\(\\[5pt] {\small a_{n}, a_{n-1}, a_{n-2}, … , a_{0} \ }\) are the coefficients of each term in \({ \small x \ }\) from the highest power to the lowest power.

Here are some of the examples:

1. \(\\[5pt]\) Linear polynomial or polynomial of degree 1
\( \hspace{1em} {\small P(x) \ = \ ax \ + \ b }\)
e.g.: \({\small P(x) \ = \ 2x \ + \ 3 }\)

2. \(\\[5pt]\) Quadratic polynomial or polynomial of degree 2
\( \hspace{1em} {\small P(x) \ = \ ax^{2} + bx + c }\)
e.g.: \({\small P(x) \ = \ 3x^{2} + 2x + 3 }\)

3. \(\\[5pt]\) Cubic polynomial or polynomial of degree 3
\( \hspace{1em} {\small P(x) \ = \ ax^{3} + bx^{2} + cx + d }\)
e.g.: \({\small P(x) \ = \ 3x^{3} + 2x^2 + 3x + 5 }\)

4. \(\\[5pt]\) Quartic polynomial or polynomial of degree 4
\( \hspace{1em} {\small P(x) \ = \ ax^{4} + bx^{3} + cx^{2} + dx + e }\)
e.g.: \({\small P(x) \ = \ 5x^{4} + 2x^{3} + 3x^2 + x + 6 }\)

The addition and subtraction of polynomials

Collect all the like terms and simply the expression by adding or subtracting the like terms.

e.g.:

\( \hspace{1em} {\small P(x) \ = \ 2x^{2} + 3x + 4 }\)
\( \hspace{1em} {\small Q(x) \ = \ x^{2} \ – \ 2x + 3 }\)

\( \hspace{1em} {\small P(x) \ + \ Q(x) }\)
\( \hspace{1.5em} {\small = \ (2x^{2} + 3x + 4) + (x^{2} \ – \ 2x + 3) }\)
\( \hspace{1.5em} {\small = \ (2x^{2} + x^{2}) + (3x \ – \ 2x) + (4 + 3) }\)
\( \hspace{1.5em} {\small = \ 3x^{2} + x + 7 }\)

\( \hspace{1em} {\small P(x) \ – \ Q(x) }\)
\( \hspace{1.5em} {\small = \ (2x^{2} + 3x + 4) \ – \ (x^{2} \ – \ 2x + 3) }\)
\( \hspace{1.5em} {\small = \ (2x^{2} \ – \ x^{2}) + (3x \ + \ 2x) + (4 \ – \ 3) }\)
\( \hspace{1.5em} {\small = \ x^{2} + 5x + 1 }\)

The multiplication of polynomials

Use the distributive law by multiplying each term.

e.g.:

\( \hspace{1em} {\small P(x) \ = \ 2x^{2} + 3x + 4 }\)
\( \hspace{1em} {\small Q(x) \ = \ x^{2} \ – \ 2x + 3 }\)

Polynomials - Multiplication of Polynomials

\( \hspace{1em} {\small P(x) \ \times \ Q(x) }\)
\( \hspace{1.5em} {\small = \ 2x^{4} + (3x^{3} \ – \ 4x^{3}) + (6x^{2} \ – \ 6x^{2} + 4x^{2}) }\)
\(\\[8pt] \hspace{2.5em} {\small + \ (9x \ – \ 8x) + 12 }\)
\( \hspace{1.5em} {\small = \ 2x^{4} \ – \ x^{3} + 4x^{2} + x + 12 }\)

The division of polynomials

We’ll use a simple analogy of long division of a number to illustrate the polynomial division.

Just as an example, what is the result of 7 divided by 3?

Polynomials - Simple Long Division

So, \(\hspace{1em} {\large\frac{7}{3}} \ = \ 2 \ + \ {\large\frac{1}{3}} \)

Or, we can conveniently write the above result as,

\(\hspace{1.5em} 7 \ = \ 2 \times 3 \ + \ 1 \ \)

Then, in terms of polynomial division,

\( \hspace{1em} {\large\frac{P(x)}{D(x)}} \ = \ Q(x) \ + {\large\frac{R(x)}{D(x)}} \)

or in its general form:

\( \hspace{1em} P(x) \ = \ D(x) \times Q(x) + R(x) \)

with,
P(x) is the polynomial to be divided with or the dividend,
D(x) is the divisor,
Q(x) is the quotient and
R(x) is the remainder.

The process of finding the quotient Q(x) and the remainder R(x) can be done similarly with polynomial long division.

Here is an example,

\(\hspace{1em} {\Large\frac{x^{2} \ – \ 4x \ – \ 3}{x \ + \ 1}} \ = \ ? \)

Step 1. Divide the highest power of the polynomial P(x) with the highest power of the divisor D(x).

Polynomials - Long Division - Step 1

Step 2. Multiply the result we get in the quotient Q(x) from Step 1 with the divisor D(x).

Polynomials - Long Division - Step 2

Step 3. Subtract the polynomial P(x) from the result in Step 2.

Polynomials - Long Division - Step 3

Step 4. Repeat the process from Step 1 to the next available highest power of polynomial P(x) until we are left with only the remainder.

Polynomials - Long Division - Step 4

Now, one important thing to remember is that the degree of the remainder R(x) is at most one fewer than the order of the divisor D(X).

For example, if we have a linear divisor \({\small \ ax \ + \ b \ }\) then the remainder is a constant.

If we have a quadratic divisor \({\small \ ax^{2} \ + \ bx \ + \ c \ }\) then the remainder is a linear polynomial and so on.

By knowing that, we’ll know when to stop the process of multiplying back the quotient.

Factor Theorem

If \({\small \ (ax \ + \ b) \ }\) is a linear factor of a polynomial P(x) then \({\small \ P({\large\frac{-b}{a}}) \ = \ 0 }\).

e.g.: Show that \({\small \ (x \ – \ 2) \ }\) is a linear factor of a polynomial \({\small P(x) \ = \ x^{4} \ – \ 5x^{2} + 2x }\).

Solution: If \({\small \ (x \ – \ 2) \ }\) is a linear factor, then according to the factor theorem \({\small \ P(2) \ }\) must be equal to zero.

We’ll try by substituting the value of \({\small \ x \ = \ 2 }\),

\(\hspace{1.5em} {\small P(2) \ = \ ({2})^{4} \ – \ 5({2})^{2} + 2(2) }\)
\(\hspace{3.2em} {\small \ = \ 16 \ – \ 20 + 4 }\)
\(\hspace{3.2em} {\small \ = \ 0 }\)

Therefore, \({\small \ (x \ – \ 2) \ }\) is a linear factor of \({\small \ P(x) \ = \ x^{4} \ – \ 5x^{2} + 2x }\).

Another great use of factor theorem is when we want to factorise a higher order polynomial or the roots of a higher order polynomial equation.

We can use the constant term of the higher order polynomial and quickly substituting the value of each factors of the constant term.

If any of the results equals to zero then by factor theorem we can find its linear factor(s).

The other factors can then be found by applying polynomial long division.

You can study some of the examples below to fully understand the concept.

Remainder Theorem

If a polynomial P(x) is divided by a linear divisor \({\small \ (ax \ + \ b) \ }\) then \({\small \ P({\large\frac{-b}{a}}) \ }\) is the remainder.

e.g.: Find the remainder when a polynomial \({\small \ P(x) \ = \ x^{4} \ – \ 2x^{3} + 3x^{2} \ – \ 8 \ }\) is divided by \({\small \ (x \ – \ 1) }\).

Solution: According to the remainder theorem, the remainder is \({\small \ P(1) \ }\),

\(\hspace{1.5em} {\small P(1) \ = \ (1)^{4} \ – \ 2(1)^{3} + 3(1)^{2} \ – \ 8 }\)
\(\hspace{3.2em} {\small \ = \ 1 \ – \ 2 + 3 \ – \ 8 }\)
\(\hspace{3.2em} {\small \ = \ -6 }\)

Therefore, the remainder of \({\small \ P(x) \ = \ x^{4} \ – \ 2x^{3} + 3x^{2} \ – \ 8 \ }\) is \({\small \ -6 \ }\) when it is divided by \({\small \ (x \ – \ 1) }\).

Try some of the examples below and if you need any help, just look at the solution I have written. Cheers ! =) .
\(\\[1pt]\)


EXAMPLE:

\({\small 1.\enspace}\) Find the remainder of \({\small \ P(x) \ = \ 2x^{3} + 4x^{2} \ – \ 6x + 7 \ }\) when it is divided by \({\small \ (2x \ – \ 1) }\).

\(\\[1pt]\)
\({\small 2.\enspace}\) Find the value of \({\small \ k \ }\) if \({\small \ (x \ – \ 1) \ }\) is a factor of \({\small \ {k}^{2}{x}^{4} \ – \ 3k{x}^{2} + 2 }\).

\(\\[1pt]\)
\({\small 3.\enspace}\) This is an example of the aplication of the factor theorem. We would try to find the roots of higher order polynomials.
\(\\[1pt]\)
Factorise \({\small \ 2{π‘₯}^{3} + 9{π‘₯}^{2} + 7{π‘₯} \ βˆ’ \ 6 \ }\) completely and hence find all the roots of the polynomial.

\(\\[1pt]\)
\({\small 4.\enspace}\) Given that a polynomial \({\small \ p(x) \ = \ {π‘₯}^{3} + a{π‘₯}^{2} + b{π‘₯} \ βˆ’ \ 3 \ }\) when divided by \({\small \ (x + 1) \ }\) and \({\small \ (x \ – \ 1) }\), the remainders are -9 and 1 respectively. Find the values of a and b.

\(\\[1pt]\)
\({\small 5.\enspace}\) Continuing from the same question in example 4, find the reminder when the polynomial \({\small \ p(x) \ }\) is divided by \({\small \ ({x}^{2} \ – \ 1) }\).

\(\\[1pt]\)
\({\small 6.\enspace}\) 9709/21/w19
The polynomial p(x) is defined by:
\(\\[1pt]\)
\(\hspace{3em} {\small p(x) \ = \ a{x}^{3} + a{x}^{2} \ βˆ’ \ 15x \ βˆ’ \ 18}\),
\(\\[1pt]\)
where a is a constant. It is given that (x βˆ’ 2) is a factor of p(x).
\({\small (\textrm{i}).\hspace{0.7em}}\) Find the value of a.
\({\small (\textrm{ii}).\hspace{0.6em}}\) Using this value of a, factorise p(x) completely.
\({\small (\textrm{iii}).\hspace{0.5em}}\) Hence solve the equation \({\small \ p(e^{\sqrt{y}}) \ = \ 0}\), giving the answer correct to 2 significant figures.

\(\\[1pt]\)


PRACTICE MORE WITH THESE QUESTIONS BELOW!

\({\small 1.\enspace}\) The polynomial p(x) is defined by

\( \hspace{3em} {\small \ p(x) \ = \ 6{x}^{3} + a{x}^{2} + 9{x} + b }\),

where a and b are constants. It is given that \({\small \ (x \ – \ 2) \ }\) and \({\small \ (2x + 1) }\) are factors of p(x). Find the values of a and b.

\({\small 2. \enspace}\) The polynomial p(x) is defined by

\( \hspace{3em} {\small \ p(x) \ = \ 6{x}^{3} + a{x}^{2} \ – \ 4{x} \ – \ 3 }\),

where a is a constant. It is given that \({\small \ (x + 3) \ }\) is a factor of p(x).
\({\small\hspace{1.2em} (\textrm{a}).\hspace{0.8em}}\) Find the value of a.
\({\small\hspace{1.2em} (\textrm{b}).\hspace{0.8em}}\) Using this value of a, factorise p(x) completely.

\({\small 3. \enspace}\) Find the quotient and remainder when \({\small \ 6{x}^{4} + {x}^{3} \ βˆ’ \ {x}^{2} + 5x \ βˆ’ \ 6 \ }\) is divided by \( {\small \ (2{x}^{2} \ βˆ’ \ x + 1) }\).

\({\small 4. \enspace}\) The polynomial f(x) is defined by

\( \hspace{3em} {\small \ f(x) \ = \ {x}^{4} \ – \ 3{x}^{3} + 5{x}^{2} \ – \ 6{x} + 11 }\),

Find the quotient and remainder when f(x) is divided by \({\small \ ({x}^2 + 2) }\).

\({\small 5. \enspace}\) The polynomial \({\small \ 6{x}^{3} + a{x}^{2} + b{x} \ βˆ’ \ 2 }\), where a and b are constants, is denoted by p(x). It is given that \({\small \ (2x + 1) }\) is a factor of p(x) and that when p(x) is divided by \({\small \ (x + 2) \ }\) the remainder is βˆ’24. Find the values of a and b.

\({\small 6. \enspace}\) The polynomial \({\small \ {x}^{4} + 3{x}^{3} + ax + b }\), where a and b are constants, is denoted by p(x). When p(x) is divided by \({\small \ {x}^{2} + x \ βˆ’ \ 1 \ }\) the remainder is \({\small \ 2x + 3}\). Find the values of a and b.

\({\small 7. \enspace}\) The cubic polynomial f(x) is defined by \({\small \ f(x) \ = \ {x}^{3} + a{x}^{2} + 14x + a + 1 }\), where a is a constant. It is given that \({\small \ (x + 2) \ }\) is a factor of f(x). Use the factor theorem to find the value of a and hence factorise f(x) completely.

\({\small 8. \enspace}\) \({\small (\textrm{i}).\hspace{0.7em}}\) Find the quotient when \({\small \ {x}^{4} \ βˆ’ \ 2{x}^{3} + 8{x}^{2} \ βˆ’ \ 12x + 13 \ }\) is divided by \({\small \ {x}^{2} + 6 \ }\) and show that the remainder is 1.
\({\small\hspace{1.3em}(\textrm{ii}).\hspace{0.7em}}\) Show that the equation \({\small \ {x}^{4} \ βˆ’ \ 2{x}^{3} + 8{x}^{2} \ βˆ’ \ 12x + 12 \ = \ 0 \ }\) has no real roots.

\({\small 9. \enspace}\) The polynomial \({\small \ {x}^{4} + {x}^{3} + ax + b}\), where a and b are constants, is divisible by \({\small \ {x}^{2} \ βˆ’ \ x + 1}\). Find the values of a and b.

\({\small 10.\enspace}\) The polynomial p(x) is defined by \({\small \ p(x) \ = \ 4{x}^{3} + 4{x}^{2} \ βˆ’ \ 29x \ βˆ’ \ 15 }\).
\({\small\hspace{2.8em}(\textrm{i}).\hspace{0.7em}}\) Use the factor theorem to show that \({\small \ x + 3 \ }\) is a factor of p(x).
\({\small\hspace{2.8em}(\textrm{ii}).\hspace{0.7em}}\) Factorise p(x) completely.


As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .

9709 Statistics Paper 6 Question 4 June 2006

Permutations and Combinations

Permutations and Combinations

When given a number of objects and we would to find the possible different arrangements or selections , we may use permutations or combinations.

Permutations deal with the arrangement of objects. The order of the objects is important. The number of different arrangements of r objects from n distinct object is,

\(\\[20pt]\hspace{2em} ^{n}P_{r } \ = \ \displaystyle \frac{n!}{(n-r)!} \)

Combinations deal with the selection of objects. The order of the objects is not important. The number of different selections of r objects from n distinct object is,

\(\\[20pt]\hspace{2em} ^{n}C_{r } \ = \ \displaystyle \binom{n}{r} \ = \ \frac{n!}{r!(n-r)!} \)

Notice that it is also used as the binomial coefficients in Binomial Expansion and Binomial Series.

Let’s see a simple example below to illustrate the difference between permutations and combinations.

We have 3 letters, A, B and C and we would like to find the permutations and combinations of 2 letters from the 3 letters.

For permutations, we have the following arrangements: AB, BA, AC, CA, BC and CB.

For combinations, we have the following selections: AB, AC and BC.

As you can see, AB and BA, AC and CA or BC and CB are considered as the same selection since the order of the objects is not important.

In other words, combinations is actually the permutations of r identical objects from n distinct object.

Try the exercises below and if you need any help, just look at the solution I have written. Cheers ! =) .
\(\\[1pt]\)


EXERCISE 5A:

\({\small 1.\enspace}\) Seven different cars are to be loaded on to a transporter truck. In how many different ways can the cars be arranged?

\(\\[1pt]\)
\({\small 2.\enspace}\) How many numbers are there between 1245 and 5421 inclusive which contain each of the digits 1, 2, 4 and 5 once and once only?

\(\\[1pt]\)
\({\small 3.\enspace}\) An artist is going to arrange five paintings in a row on a wall. In how many ways can this be done?

\(\\[1pt]\)
\({\small 4.\enspace}\) Ten athletes are running in a 100-metre race. In how many different ways can the first three places be filled?

\(\\[1pt]\)
\({\small 5.\enspace}\) By writing out all the possible arrangements of \({\small {D}_{1}{E}_{1}{E}_{2}{D}_{2} }\), show that there are \( \frac{4!}{2!(2)!} = {\small 6 \ }\) different arrangements of the letters of the word DEED.

\(\\[1pt]\)
\({\small 6.\enspace}\) A typist has five letters and five addressed envelopes. In how many different ways can the letters be placed in each envelope without getting every letter in the right envelope? If the letters are placed in the envelopes at random what is the probability that each letter is in its correct envelope?

\(\\[1pt]\)
\({\small 7.\enspace}\) How many different arrangements can be made of the letters in the word STATISTICS?

\(\\[1pt]\)
\({\small 8.\enspace \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) Calculate the number of arrangements of the letters in the word NUMBER.
\(\\[1pt]\)
\({\small \hspace{2.3em}\textrm{(b)}.\hspace{0.8em}}\) How many of the arrrangements in part (a) begin and end with a vowel?

\(\\[1pt]\)
\({\small 9.\enspace}\) How many different numbers can be formed by taking one, two, three and four digits from the digits 1, 2, 7 and 8, if repetitions are not allowed?
One of these numbers is chosen at random. What is the probability that it is greater than 200?

\(\\[1pt]\)


EXERCISE 5B:

\({\small 1.\enspace}\) How many three-card hands can be dealt from a pack of 52 cards?

\(\\[1pt]\)
\({\small 2.\enspace}\) From a group of 30 boys and 32 girls, two girls and two boys are to be chosen to represent their school. How many possible selections are there?

\(\\[1pt]\)
\({\small 3.\enspace}\) A history exam paper contains eight questions, four in Part A and four in Part B. Candidates are required to attempt five questions. In how many ways can this be done if
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) there are no restrictions,
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) at least two questions from Part A and at least two questions from Part B must be attempted?

\(\\[1pt]\)
\({\small 4.\enspace}\) A committee of three people is to be selected from four women and five men. The rules state that there must be at least one man and one woman on the committee. In how many different ways can the committee be chosen?
Subsequently one of the men and one of the women marry each other. The rules also state that a married couple may not both serve on the committee. In how many ways can the committee be chosen now?

\(\\[1pt]\)
\({\small 5.\enspace}\) A box of one dozen eggs contains one that is bad. If three eggs are chosen at random, what is the probability that one of them will be bad?

\(\\[1pt]\)
\({\small 6.\enspace}\) In a game of bridge the pack of 52 cards is shared equally between all four players. What is the probability that one particular player has no hearts?

\(\\[1pt]\)
\({\small 7.\enspace}\) A bag contains 20 chocolates, 15 toffees and 12 peppermints. If three sweets are chosen at random, what is the probability that they are
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) all different,
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) all chocolates,
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em}}\) all the same,
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(d)}.\hspace{0.8em}}\) all not chocolates?

\(\\[1pt]\)
\({\small 8.\enspace}\) Show that \( \displaystyle \binom{n}{r} \ = \ \binom{n}{n \ – \ r} \).

\(\\[1pt]\)
\({\small 9.\enspace}\) Show that the number of permutations of n objects of which r are of one kind and n – r are of another kind is \( \displaystyle \binom{n}{r} \).

\(\\[1pt]\)


EXERCISE 5C:

\({\small 1.\enspace}\) The letters of the word CONSTANTINOPLE are written on 14 cards, one on each card. The cards are shuffled and then arranged in a straight line.
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) How many different possible arrangements are there?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) How many arrangements begin with P?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em}}\) How many arrangements start and end with O?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(d)}.\hspace{0.8em}}\) How many arrangements are there where no two vowels are next to each other?

\(\\[1pt]\)
\({\small 2.\enspace}\) A coin is tossed 10 times.
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) How many different sequences of heads and tails are possible?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) How many different sequences containing six heads and four tails are possible?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em}}\) What is the probability of getting six heads and four tails?

\(\\[1pt]\)
\({\small 3.\enspace}\) Eight cards are selected with replacement from a standard pack of 52 playing cards, with 12 picture cards, 20 odd cards and 20 even cards.
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) How many different sequences of eight cards are possible?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) How many of the sequences in part (a) will contain three picture cards, three odd-numbered cards and two even-numbered cards?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em}}\) Use parts (a) and (b) to determine the probability of getting three picture cards, three odd-numbered cards and two even-numbered cards if eight cards are selected with replacement from a standard pack of 52 playing cards.

\(\\[1pt]\)
\({\small 4.\enspace}\) Eight women and five men are standing in a line.
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) How many arrangements are possible if any individual can stand in any position?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) In how many arrangements will all five men be standing next to one another?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em}}\) In how many arrangements will no two men be standing next to one another?

\(\\[1pt]\)
\({\small 5.\enspace}\) Each of the digits 1, 1, 2, 3, 3, 4, 6 is written on a separate card. The seven cards are then laid out in a row to form a 7-digit number.
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) How many distinct 7-digit numbers are there?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) How many of these 7-digit numbers are even?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em}}\) How many of these 7-digit numbers are divisible by 4?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(d)}.\hspace{0.8em}}\) How many of these 7-digit numbers start and end with the same digit?

\(\\[1pt]\)
\({\small 6.\enspace}\) Three families, the Mehtas, the Mupondas and the Lams, go to the cinema together to watch a film. Mr and Mrs Mehta take their daughter Indira, Mr and Mrs Muponda take their sons Paul and John, and Mrs Lam takes her children Susi, Kim and Lee. The families occupy a single row with eleven seats.
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) In how many ways could the eleven people be seated if there were no restriction?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) In how many ways could the eleven people sit down so that the members of each family are all sitting together?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em}}\) In how many of the arrangements will no two adults be sitting next to one another?

\(\\[1pt]\)
\({\small 7.\enspace}\) The letters of the word POSSESSES are written on nine cards, one on each card. The cards are shuffled and four of them are selected and arranged in a straight line.
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) How many possible selections are there of four letters?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) How many arrangements are there of four letters?

\(\\[1pt]\)


MISCELLANEOUS EXERCISE 5:

\({\small 1.\enspace}\) The judges in a ‘Beautiful Baby’ competition have to arrange 10 babies in order of merit. In how many different ways could this be done? Two babies are to be selected to be photographed. In how many ways can this selection be made?

\(\\[1pt]\)
\({\small 2.\enspace}\) In how many ways can a committee of four men and four women be seated in a row if
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) they can sit in any position,
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) no one is seated next to a person of the same sex?

\(\\[1pt]\)
\({\small 3.\enspace}\) How many distinct arrangements are there of the letters in the word ABRACADABRA?

\(\\[1pt]\)
\({\small 4.\enspace}\) Six people are going to travel in a six-seater minibus but only three of them can drive. In how many different ways can they seat themselves?

\(\\[1pt]\)
\({\small 5.\enspace}\) There are eight different books on a bookshelf: three of them are hardbacks and the rest are paperbacks.
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) In how many different ways can the books be arranged if all the paperbacks are together and all the hardbacks are together?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) In how many different ways can the books be arranged if all the paperbacks are together?

\(\\[1pt]\)
\({\small 6.\enspace}\) Four boys and two girls sit in a line on stools in front of a coffee bar.
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) In how many ways can they arrange themselves so that the two girls are together?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) In how many ways can they sit if the two girls are not together?

\(\\[1pt]\)
\({\small 7.\enspace}\) Ten people travel in two cars, a saloon and a Mini. If the saloon has seats for six and the Mini has seats for four, find the number of different ways in which the party can travel, assuming that the order of seating in each car does not matter and all the people can drive.

\(\\[1pt]\)
\({\small 8.\enspace}\) Giving a brief explanation of your method, calculate the number of different ways in which the letters of the word TRIANGLES can be arranged if no two vowels may come together.

\(\\[1pt]\)
\({\small 9.\enspace}\) I have seven fruit bars to last the week. Two are apricot, three fig and two peach. I select one bar each day. In how many different orders can I eat the bars?
If I select a fruit bar at random each day, what is the probability that I eat the two appricot ones on consecutive days?

\(\\[1pt]\)
\({\small 10.\enspace}\) A class contains 30 children, 18 girls and 12 boys. Four complimentary theatre tickets are distributed at random to the children in the class. What is the probability that
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) all four tickets go to the girls,
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) two boys and two girls receive tickets?

\(\\[1pt]\)
\({\small 11.\enspace \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) How many different 7-digit numbers can be formed from the digits 0, 1, 2, 2, 3, 3, 3 assuming that a number cannot start with 0?
\(\\[1pt]\)
\({\small \hspace{2.8em}\textrm{(b)}.\hspace{0.8em}}\) How many of these numbers will end in 0?

\(\\[1pt]\)
\({\small 12.\enspace}\) Calculate the number of ways in which three girls and four boys can be seated on a row of seven chairs if each arrangement is to be symmetrical.

\(\\[1pt]\)
\({\small 13.\enspace}\) Find the number of ways in which
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) 3 people can be arranged in 4 seats,
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) 5 people can be arranged in 5 seats.
\(\\[1pt]\)
In a block of 8 seats, 4 are in row A and 4 are in row B. Find the number of ways of arranging 8 people in the 8 seats given that 3 specified people must be in row A.

\(\\[1pt]\)
\({\small 14.\enspace}\) Eight different cards, of which four are red and four are black, are dealt to two players so that each receives a hand of four cards. Calculate
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) the total number of different hands which a given player could receive,
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) the probability that each player receives a hand consisting of four cards all of the same colour.

\(\\[1pt]\)
\({\small 15.\enspace}\) A piece of wood of length of 10 cm is to be divided into 3 pieces so that the length of each piece is a whole number of cm, for example, 2 cm, 3 cm and 5 cm.
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) List all the different sets of lengths which could be obtained.
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) If one of these sets is selected at random, what is the probability that the lengths of the pieces could be lengths of the sides of a triangle?

\(\\[1pt]\)
\({\small 16.\enspace}\) Nine persons are to be seated at three tables holding 2, 3 and 4 persons respectively. In how many ways can the groups sitting at the tables be selected, assuming that the order of sitting at the tables does not matter?

\(\\[1pt]\)
\({\small 17.\enspace \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) Calculate the number of different arrangements which can be made using all the letters of the word BANANA.
\(\\[1pt]\)
\({\small \hspace{2.8em}\textrm{(b)}.\hspace{0.8em}}\) The number of combinations of 2 objects from n is equal to the number of combinations of 3 objects from n. Determine n.

\(\\[1pt]\)
\({\small 18.\enspace}\) A ‘hand’ of 5 cards is dealt from an ordinary pack of 52 playing cards. Show that there are nearly 2.6 million distinct hands and that, of these, 575 757 contain no card from the heart suit.
On three successive occasions a card player is dealt a hand containing no heart. What is the probability of this happening? What conclusion might the player be justifiably reach?

\(\\[1pt]\)


PRACTICE MORE WITH THESE QUESTIONS BELOW!

\({\small 1.\enspace}\) In a 60-metre hurdles race there are five runners, one from each of the nations Austria, Belgium, Canada, Denmark and England.

\({\small\hspace{1.2em} (\textrm{i}).\hspace{0.8em}}\) How many different finishing orders are there?

\({\small\hspace{1.2em} (\textrm{ii}).\hspace{0.7em}}\) What is the probability of predicting the finishing order by choosing first, second, third, fourth and fifth at random?

\({\small 2. \enspace}\) In a ‘Goal of the season’ competition, participants are asked to rank ten goals in order of quality.

The organisers select their ‘correct’ order at random. Anybody who matches their order will be invited to join the television commentary team for the next international match.

\({\small\hspace{1.2em} (\textrm{i}).\hspace{0.8em}}\) What is the probability of a participant’s order being the same as that of the organisers?

\({\small\hspace{1.2em} (\textrm{ii}).\hspace{0.7em}}\) Five million people enter the competition. How many people would be expected to join the commentary team?

\({\small 3. \enspace}\) How many arrangements of the word ACHIEVE are there if

\({\small\hspace{1.2em} (\textrm{i}).\hspace{0.8em}}\) there are no restrictions on the order the letters are to be in

\({\small\hspace{1.2em} (\textrm{ii}).\hspace{0.7em}}\) the first letter is an A

\({\small\hspace{1.2em} (\textrm{iii}).\hspace{0.5em}}\) the letters A and I are to be together.

\({\small\hspace{1.2em} (\textrm{iv}).\hspace{0.6em}}\)the letters C and H are to be apart.

\({\small 4. \enspace (\textrm{i}).\hspace{0.7em}}\) A football team consists of 3 players who play in a defence position, 3 players who play in a midfield position and 5 players who play in a forward position. Three players are chosen to collect a gold medal for the team. Find in how many ways this can be done
\({\small\hspace{2.8em}(\textrm{a}).\hspace{0.7em}}\) if the captain, who is a midfield player, must be included, together with one defence and one forward player.
\({\small\hspace{2.8em}(\textrm{b}).\hspace{0.7em}}\) if exactly one forward player must be included, together with any two others.

\({\small \hspace{1.2em} (\textrm{ii}).\hspace{0.6em}}\) Find how many different arrangements there are of the nine letters in the words GOLD MEDAL
\({\small\hspace{2.8em}(\textrm{a}).\hspace{0.7em}}\) if there are no restrictions on the order of the letters,
\({\small\hspace{2.8em}(\textrm{b}).\hspace{0.7em}}\) if the two letters D come first and the letters L come last.

\({\small 5. \enspace}\) The diagram shows the seating plan for passengers in a minibus, which has 17 seats arranged in 4 rows. The back row has 5 seats and the other 3 rows have 2 seats on each side. 11 passengers get on the minibus.
\(\\[1pt]\)
9709 Statistics Paper 6 Question 4 June 2006
\(\\[1pt]\)
\({\small\hspace{1.2em} (\textrm{i}).\hspace{0.7em}}\) How many possible seating arrangements are there for the 11 passengers?
\({\small\hspace{1.2em} (\textrm{ii}).\hspace{0.6em}}\) How many possible seating arrangements are there if 5 particular people sit in the back row?

Of the 11 passengers, 5 are unmarried and the other 6 consist of 3 married couples.

\({\small\hspace{1.2em} (\textrm{iii}).\hspace{0.5em}}\) In how many ways can 5 of the 11 passengers on the bus be chosen if there must be 2 married couples and 1 other person, who may or may not be married?

\({\small 6. \enspace}\) A choir consists of 13 sopranos, 12 altos, 6 tenors and 7 basses. A group consisting of 10 sopranos, 9 altos, 4 tenors and 4 basses is to be chosen from the choir.
\({\small\hspace{1.2em} (\textrm{i}).\hspace{0.7em}}\) In how many different ways can the group be chosen?
\({\small\hspace{1.2em} (\textrm{ii}).\hspace{0.6em}}\) In how many ways can the 10 chosen sopranos be arranged in a line if the 6 tallest stand next to each other?
\({\small\hspace{1.2em} (\textrm{iii}).\hspace{0.5em}}\) The 4 tenors and the 4 basses in the group stand in a single line with all the tenors next to each other and all the basses next to each other. How many possible arrangements are there if three of the tenors refuse to stand next to any of the basses?

\({\small 7. \enspace (\textrm{i}).\hspace{0.7em}}\) Find how many numbers between 5000 and 6000 can be formed from the digits 1, 2, 3, 4, 5 and 6
\({\small\hspace{2.8em}(\textrm{a}).\hspace{0.7em}}\) if no digits are repeated,
\({\small\hspace{2.8em}(\textrm{b}).\hspace{0.7em}}\) if repeated digits are allowed.

\({\small \hspace{1.2em} (\textrm{ii}).\hspace{0.6em}}\) Find the number of ways of choosing a school team of 5 pupils from 6 boys and 8 girls
\({\small\hspace{2.8em}(\textrm{a}).\hspace{0.7em}}\) if there are more girls than boys in the team,
\({\small\hspace{2.8em}(\textrm{b}).\hspace{0.7em}}\) if three of the boys are cousins and are either all in the team or all not in the team.

\({\small 8. \enspace (\textrm{i}).\hspace{0.7em}}\) Find the number of different ways in which all 12 letters of the word STEEPLECHASE can be arranged so that all four Es are together.

\({\small \hspace{1.2em} (\textrm{ii}).\hspace{0.6em}}\) Find the number of different ways in which all 12 letters of the word STEEPLECHASE can be arranged so that the Ss are not next to each other.

Four letters are selected from the 12 letters of the word STEEPLECHASE.

\({\small \hspace{1.2em} (\textrm{iii}).\hspace{0.4em}}\) Find the number of different selections if the four letters include exactly one S.

\({\small 9. \enspace (\textrm{i}).\hspace{0.7em}}\) Find the number of different ways in which the 9 letters of the word TOADSTOOL can be arranged so that all three Os are together and both Ts are together.

\({\small \hspace{1.2em} (\textrm{ii}).\hspace{0.6em}}\) Find the number of different ways in which the 9 letters of the word TOADSTOOL can be arranged so that the Ts are not together.

\({\small \hspace{1.2em} (\textrm{iii}).\hspace{0.4em}}\) Find the probability that a randomly chosen arrangement of the 9 letters of the word TOADSTOOL has a T at the beginning and a T at the end.

\({\small \hspace{1.2em} (\textrm{iv}).\hspace{0.6em}}\) Five letters are selected from the 9 letters of the word TOADSTOOL. Find the number of different selections if the five letters include at least 2 Os and at least 1 T.

\({\small 10. \enspace (\textrm{a}).\hspace{0.7em}}\) A group of 6 teenagers go boating. There are three boats available. One boat has room for 3 people, one has room for 2 people and one has room for 1 person. Find the number of different ways the group of 6 teenagers can be divided between the three boats.

\({\small \hspace{1.6em} (\textrm{b}).\hspace{0.6em}}\) Find the number of different 7-digit numbers which can be formed from the seven digits 2, 2, 3, 7, 7, 7, 8 in each of the following cases.

\({\small \hspace{2.8em} (\textrm{i}).\hspace{0.7em}}\) The odd digits are together and the even digits are together.

\({\small \hspace{2.8em} (\textrm{ii}).\hspace{0.6em}}\) The 2s are not together.


As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .

Sequence and Series - Arithmetric Progression, Arithmetic Series, Geometric Progression, Geometric Series

Sequences and Series

Sequences and Series

Sequence is a list of numbers with a predefined rule or formula.

The numbers are called as terms and we can find the pattern in the sequence.

There are quite a limitless number of possible sequences can be made, we however will mainly focus on two main categories of sequences, the arithmetic sequence and the geometric sequence.

Arithmetic Sequence

In arithmetic sequence or progression any two consecutive terms always have the same difference.

\(\\[12pt]\hspace{2em}{\small {u}_{n} \ = \ a \ + \ (n \ – \ 1) d }\)
\(\\[7pt]{\small {u}_{n} }\) is the \({\small \boldsymbol{{n}^{\textrm{th}}} }\) term
\(\\[7pt]{\small a }\) is the first term or \({\small \boldsymbol{{u}_{1}} }\)
\({\small d }\) is the common difference.

Example:

Example of Arithmetic Sequence

\(\\[7pt]{\small a \ \ \ = \ {u}_{1} }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 10 }\)
\(\\[7pt]{\small {u}_{2} \ = \ 15, \ {u}_{3} \ = \ 20, \ {u}_{4} \ = \ 25, \ \textrm{etc} }\)
\(\\[7pt]{\small d \ \ \ = \ 5}\)
\(\\[7pt]{\small {u}_{n} \ = \ a \ + \ (n \ – \ 1) d }\)
\(\\[7pt]{\small {u}_{n} \ = \ 10 \ + \ (n \ – \ 1) \ \times \ 5 }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 10 \ + \ 5n \ – \ 5 }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 5n \ + \ 5 }\)

So, for example, if we want to find the 10th term of the sequence above,

\(\\[7pt]{\small {u}_{10} \ = \ 5(10) \ + \ 5 }\)
\(\hspace{1.6em}{\small = \ 55 }\)

Geometric Sequence

In geometric sequence or progression any two consecutive terms always have the same ratio.

\(\\[12pt]\hspace{2em}{\small {u}_{n} \ = \ a {r}^{ (n \ – \ 1) } }\)
\(\\[7pt]{\small {u}_{n} }\) is the \({\small \boldsymbol{{n}^{\textrm{th}}} }\) term
\(\\[7pt]{\small a }\) is the first term or \({\small \boldsymbol{{u}_{1}} }\)
\({\small r }\) is the common ratio.

Example:

Example of Geometric Sequence

\(\\[7pt]{\small a \ \ \ = \ {u}_{1} }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 100 }\)
\(\\[7pt]{\small {u}_{2} \ = \ 50, \ {u}_{3} \ = \ 25, \ {u}_{4} \ = \ 12.5, \ \textrm{etc} }\)
\(\\[12pt]{\small r \ \ \ = \ {\large\frac{1}{2}} }\)
\(\\[7pt]{\small {u}_{n} \ = \ a {r}^{ (n \ – \ 1) } }\)
\(\\[12pt]{\small {u}_{n} \ = \ 100 \ \times \ { \big({\large\frac{1}{2}} \big) }^{ (n \ – \ 1) } }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 100 \ \times \ {2}^{(-n+1)} }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 200 \ \times \ {2}^{-n} }\)

So, for example, if we want to find the 10th term of the sequence above,

\(\\[7pt]{\small {u}_{10} \ = \ 200 \ \times \ {2}^{-10} }\)
\(\hspace{1.6em}{\small = \ 0.1953125 }\)

Meanwhile, we can also calculate the sum of n terms, that is, the addition from the first term of the sequence up to the \({\small {n}^{\textrm{th}} }\) term.

The addition of the sequence is called series.

The formula for both arithmetic series and geometric series are as follows:

Arithmetic Series

\(\\[18pt]\hspace{1.5em}{\small {S}_{n} \ = \ {\large\frac{n}{2}} \ ( a \ + \ {u}_{n} ) \ = \ {\large\frac{n}{2}} \ \big[ 2a \ + \ (n \ – \ 1)d \big] }\)
\(\\[7pt]{\small {S}_{n} }\) is the sum up to \({\small \boldsymbol{{n}^{\textrm{th}}} }\) term.

Example:

Example of Arithmetic Series

\(\\[7pt]{\small {S}_{1} \ = \ {u}_{1} }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 10 }\)
\(\\[7pt]{\small {S}_{2} \ = \ {u}_{1} \ + \ {u}_{2} }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 10 \ + \ 15 }\)
\(\\[7pt]{\small {S}_{3} \ = \ {u}_{1} \ + \ {u}_{2} \ + \ {u}_{3} }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 10 \ + \ 15 \ + \ 20 }\)
\(\\[7pt]{\small {S}_{4} \ = \ {u}_{1} \ + \ {u}_{2} \ + \ {u}_{3} \ + \ {u}_{4} }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 10 \ + \ 15 \ + \ 20 \ + \ 25}\)
\(\\[12pt]\), etc
\(\\[7pt]{\small d \ \ \ = \ 5}\)
\(\\[12pt]{\small {S}_{n} \ = \ {\large\frac{n}{2}} \ \big[ 2a \ + \ (n \ – \ 1)d \big] } \)
\(\\[12pt]{\small {S}_{n} \ = \ {\large\frac{n}{2}} \ \big[ 2(10) \ + \ (n \ – \ 1) \ \times \ 5 \big] } \)
\(\\[12pt]\hspace{1.4em}{\small = \ {\large\frac{n}{2}} \ ( 20 \ + \ 5n \ – \ 5 ) } \)
\(\\[7pt]\hspace{1.4em}{\small = \ {\large\frac{5}{2}}{n}^2 \ + \ {\large\frac{15}{2}}n }\)

So, for example, if we want to find the sum of the first 10 terms of the sequence above,

\(\\[12pt]{\small {S}_{10} \ = \ {\large\frac{5}{2}}{(10)}^2 \ + \ {\large\frac{15}{2}}n }\)
\(\hspace{1.6em}{\small = \ 325 }\)

Geometric Series

\(\\[12pt]\hspace{2em}{\small {S}_{n} \ = \ {\large\frac{a \ ( {r}^{ n } \ – \ 1 ) }{r \ – \ 1} } }\) for \({\small |r| \ \gt \ 1 }\)
\(\\[7pt]\) or,
\(\\[20pt]\hspace{2em}{\small {S}_{n} \ = \ {\large\frac{a \ ( 1 \ – \ {r}^{ n } )}{1 \ – \ r } } }\) for \({\small |r| \ \lt \ 1 }\)
\(\\[7pt]{\small {S}_{n} }\) is the sum up to \({\small \boldsymbol{{n}^{\textrm{th}}} }\) term.

Example:

Example of Geometric Series

\(\\[7pt]{\small {S}_{1} \ = \ {u}_{1} }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 100 }\)
\(\\[7pt]{\small {S}_{2} \ = \ {u}_{1} \ + \ {u}_{2} }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 100 \ + \ 50 }\)
\(\\[7pt]{\small {S}_{3} \ = \ {u}_{1} \ + \ {u}_{2} \ + \ {u}_{3} }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 100 \ + \ 50 \ + \ 25 }\)
\(\\[7pt]{\small {S}_{4} \ = \ {u}_{1} \ + \ {u}_{2} \ + \ {u}_{3} \ + \ {u}_{4} }\)
\(\\[7pt]\hspace{1.4em}{\small = \ 100 \ + \ 50 \ + \ 25 \ + \ 12.5}\)
\(\\[12pt]\), etc
\(\\[12pt]{\small r \ \ \ = \ {\large\frac{1}{2}} }\)
\(\\[12pt]{\small {S}_{n} \ = \ {\large\frac{a \ ( 1 \ – \ {r}^{ n } )}{1 \ – \ r } } }\)
\(\\[18pt]{\small {S}_{n} \ = \ {\large\frac{100 \ \big( 1 \ – \ {{\large (\frac{1}{2} )} }^{ n } \big)}{1 \ – \ {\large\frac{1}{2}} } } }\)
\(\\[18pt]{\small {S}_{n} \ = \ {\large\frac{100 \ \big( 1 \ – \ {2} ^{ -n } \big)}{ {\large\frac{1}{2}} } } }\)
\(\\[10pt]{\small {S}_{n} \ = \ 200 \ \times \ \big( 1 \ – \ {2 }^{ -n } \big) }\)

So, for example, if we want to find the sum of the first 10 terms of the sequence above,

\(\\[10pt]{\small {S}_{n} \ = \ 200 \ \times \ \big( 1 \ – \ {(2) }^{ -10 } \big) }\)
\(\hspace{1.4em}{\small = \ 199.8046875 }\)

In geometric series, we can also find the sum to infinity given that the series is convergent \({\small (|r| \ \lt \ 1) }\).

Since \({\small \displaystyle \lim_{n \to \infty} 1 \ – \ {r}^{ n } \ = \ 1 \ }\) for \({\small \ (|r| \ \lt \ 1) }\),

\(\\[12pt]\hspace{2em}{\small {S}_{\infty} \ = \ {\large\frac{a}{1 \ – \ r } } }\)
\(\\[7pt]{\small {S}_{\infty} }\) is the sum to infinity.

Example:

Example of Infinite Geometric Series

\(\\[18pt]{\small {S}_{\infty} \ = \ {\large\frac{ a }{1 \ – \ r } } }\)
\(\\[18pt]{\small {S}_{\infty} \ = \ {\large\frac{ 100 }{1 \ – \ {\large\frac{1}{2}} } } }\)
\(\hspace{1.6em}{\small = \ 200 }\)

Try some of the examples below and if you need any help, just look at the solution I have written. Cheers ! =) .
\(\\[1pt]\)


EXAMPLE:

\({\small 1.\enspace}\) A runner who is training for a long-distance race plans to run increasing distances each day for 21 days. She will run x km on day 1, and on each subsequent day she will increase the distance by 10% of the previous day’s distance. On day 21 she will run 20 km.
\({\small\hspace{2.8em}(\textrm{i}).\hspace{0.7em}}\) Find the distance she must run on day 1 in order to achieve this. Give your answer in km correct to 1 decimal place.
\({\small\hspace{2.8em}(\textrm{ii}).\hspace{0.7em}}\) Find the total distance she runs over the 21 days.

\(\\[1pt]\)
\({\small 2.\enspace}\) The first, second and third terms of a geometric progression are x, x – 3 and x – 5 respectively.
\({\small\hspace{2.8em}(\textrm{i}).\hspace{0.7em}}\) Find the value of x.
\({\small\hspace{2.8em}(\textrm{ii}).\hspace{0.7em}}\) Find the fourth term of the progression.
\({\small\hspace{2.8em}(\textrm{iii}).\hspace{0.5em}}\) Find the sum to infinity of the progression.

\(\\[1pt]\)
\({\small 3.\enspace}\) In an arithmetic progression, the sum of the first ten terms is equal to the sum of the next five terms. The first term is a.
\({\small\hspace{2.8em}(\textrm{i}).\hspace{0.7em}}\) Show that the common difference of the progression is \({\small {\large \frac{1}{3} } a}\).
\({\small\hspace{2.8em}(\textrm{ii}).\hspace{0.7em}}\) Given that the tenth term is 36 more than the fourth term, find the value of a.

\(\\[1pt]\)
\({\small 4.\enspace}\) The sum to infinity of a geometric progression is 9 times the sum of the first four terms. Given that the first term is 12, find the value of the fifth term.

\(\\[1pt]\)
\({\small 5.\enspace}\) Two heavyweight boxers decide that they would be more successful if they competed in a lower weight class. For each boxer this would require a total weight loss of 13 kg. At the end of week 1 they have each recorded a weight loss of 1 kg and they both find that in each of the following weeks their weight loss is slightly less than the week before.
\(\\[1pt]\)
Boxer A’s weight loss in week 2 is 0.98 kg. It is given that his weekly weight loss follows an arithmetic progression.
\({\small\hspace{2.8em}(\textrm{i}).\hspace{0.7em}}\) Write down an expression for his total weight loss after x weeks.
\({\small\hspace{2.8em}(\textrm{ii}).\hspace{0.7em}}\) He reaches his 13 kg target during week n. Use your answer to part (i) to find the value of n.
\(\\[1pt]\)
Boxer B’s weight loss in week 2 is 0.92 kg and it is given that his weekly weight loss follows a geometric progression.
\({\small\hspace{2.8em}(\textrm{iii}).\hspace{0.5em}}\) Calculate his total weight loss after 20 weeks and show that he can never reach his target.

\(\\[1pt]\)
\({\small 6.\enspace}\) The sum of the first twenty terms of an arithmetic progression is 50, and the sum of the next twenty terms is -50. Find the sum of the first hundred terms of the progression.

\(\\[1pt]\)
\({\small 7.\enspace}\) In 1971 a newly-built flat was sold with a 999-year lease. The terms of the sale included a requirement to pay ‘ground rent’ yearly. The ground rent was set at Β£28 per year for the first 21 years of the lease, increasing by Β£14 to Β£42 per year for the next 21 years, and then increasing again by Β£14 at the end of each subsequent period of 21 years.
\({\small\hspace{1.2em} (\textrm{a}).\hspace{0.8em}}\) Find how many complete 21-year periods there would be if the lease ran for the full 999 years, and how many years there would be left over.
\({\small\hspace{1.2em} (\textrm{b}).\hspace{0.8em}}\) Find the total amount of ground rent that would be paid in all of the complete 21-year periods of the lease.

\(\\[1pt]\)
\({\small 8.\enspace}\) Jayesh invests $100 in a savings acoount on the first day of each month for one complete year. The account pays interest at \( {\small \frac{1}{2} \% \ }\) for each complete month. How much does Jayesh have invested at the end of the year (but before making a thirteenth payment)?

\(\\[1pt]\)
\({\small 9.\enspace}\) Neeta takes out a 25-year mortgage of $40 000 to buy her house. Compound interest is charged on the loan at a rate of 8% per annum. She has to pay off the mortgage with 25 equal payments, the first of which is to be one year after the loan is taken out. Continue the following argument to calculate the value of each annual payment.
\(\\[1pt]\)
\({\small \hspace{1.2em} \bullet \hspace{1.2em} }\) After 1 year she owes \({\small $(40 000 \times 1.08) }\) (loan plus interest) less the payment made, $P, that is, she owes \({\small $(40 000 \times 1.08 \ – P)}\).
\(\\[1pt]\)
\( {\small \hspace{1.2em} \bullet \hspace{1.2em} }\) After 2 years she owes \({\small $\big((40 000 \times 1.08 \ – P) \times 1.08 \ – P \big) }\).
\(\\[1pt]\)
\({\small \hspace{1.2em} \bullet \hspace{1.2em} }\) After 3 years she owes \({\small $\Big(\big((40 000 \times 1.08 \ – P) \times 1.08 \ – P \big) \times 1.08 \ – P \Big)}\).
\(\\[1pt]\)
At the end of 25 years this (continued) expression must be zero. Form an equation in P and solve it.

\(\\[1pt]\)
\({\small 10.\enspace}\) The Bank of Utopia offers an interest rate of 100% per annum with various options as to how the interest may be added. Gopal invests $1000 and considers the following options.
\(\\[1pt]\)
\( {\small \hspace{1.2em} \textrm{Option A} \hspace{1em} }\) Interest added annually at the end of the year.
\(\\[1pt]\)
\( {\small \hspace{1.2em} \textrm{Option B} \hspace{1em} }\) Interest of 50% credited at the end of each half-year.
\(\\[1pt]\)
\( {\small \hspace{1.2em} \textrm{Option C, D, E, …} \hspace{1em} }\) The Bank is willing to add interest as often as required, subject to (interest rate) \( {\small \times }\) (number of credits per year) = 100.
\(\\[1pt]\)
Investigate to find the maximum possible amount in Gopal’s account after one year.

\(\\[1pt]\)


PRACTICE MORE WITH THESE QUESTIONS BELOW!

\({\small 1. \hspace{0.6em}(\textrm{i}).\hspace{0.7em}}\) The first and second terms of a geometric progression are p and 2p respectively, where p is a positive constant. The sum of the first n terms is greater than 1000p. Show that \({\small {2}^{n} \ \gt \ 1001}\).

\({\small\hspace{1.2em}(\textrm{ii}).\hspace{0.7em}}\) In another case, p and 2p are the first and second terms respectively of an arithmetic progression. The nth term is 336 and the sum of the first n terms is 7224. Write down two equations in n and p and hence find the values of n and p.

\({\small 2. \enspace}\) The first three terms of an arithmetic progression are 4, x and y respectively. The first three terms of a geometric progression are x, y and 18 respectively. It is given that both x and y are positive.
\({\small\hspace{1.2em}(\textrm{i}).\hspace{0.7em}}\) Find the value of x and the value of y.
\({\small\hspace{1.2em}(\textrm{ii}).\hspace{0.7em}}\)Find the fourth term of each progression.

\({\small 3. \enspace}\) In an arithmetic progression the first term is a and the common difference is 3. The nth term is 94 and the sum of the first n terms is 1420. Find n and a.

\({\small 4. \enspace}\) A person wants to buy a pension which will provide her with an income of $10 000 at the end of each of the next n years. Show that, with a steady interest rate of 5% per year, the pension should cost her
\(\\[1pt]\)
\({\small \$10 \ 000 } \big(\frac{1}{1.05} + \frac{1}{{1.05}^{2}} + \frac{1}{{1.05}^{3}} + \cdots + \frac{1}{{1.05}^{n}} \big) . \)
\(\\[1pt]\)
Find a simple formula for calculating this sum, and find its value when n = 10, 20, 30, 40, 50.

\({\small 5. \enspace}\) As part of a fund-raising campaign, I have been given some books of raffle tickets to sell. Each book has the same number of tickets and all the tickets I have been given are numbered in sequence. The number of the ticket on the front of the 5th book is 205 and that on the front of the 19th book is 373.
\({\small\hspace{1.2em} (\textrm{i}).\hspace{0.8em}}\) By writing the number of the ticket on the front of the first book as a and the number of tickets in each book as d, write down two equations involving a and d.
\({\small\hspace{1.2em} (\textrm{ii}).\hspace{0.7em}}\) From these two equations find how many tickets are in each book and the number on the front of the first book I have been given.
\({\small\hspace{1.2em} (\textrm{iii}).\hspace{0.6em}}\) The last ticket I have been given is numbered 492. How many books have I been given?

\({\small 6. \enspace}\) A tank is filled with 20 litres of water. Half the water is removed and replaced with anti-freeze and thoroughly mixed. Half this mixture is then removed and replaced with anti-freeze. This process continues.
\({\small\hspace{1.2em} (\textrm{i}).\hspace{0.8em}}\) Find the first five terms in the sequence of amounts of water in the tank at each stage.
\({\small\hspace{1.2em} (\textrm{ii}).\hspace{0.6em}}\) Find the first five terms in the sequence of amounts of anti-freeze in the tank at each stage.
\({\small\hspace{1.2em} (\textrm{iii}).\hspace{0.4em}}\) Is either of these sequences geometric? Explain.

\({\small 7. \enspace}\) A company producing salt from sea water changed to a new process. The amount of salt obtained each week increased by 2% of the amount obtained in the preceding week. It is given that in the first week after the change the company obtained 8000 kg of salt.
\({\small\hspace{1.2em} (\textrm{i}).\hspace{0.8em}}\) Find the amount of salt obtained in the 12th week after the change.
\({\small\hspace{1.2em} (\textrm{ii}).\hspace{0.6em}}\) Find the total amount of salt obtained in the first 12 weeks after the change.

\({\small 8. \enspace}\) The common ratio of a geometric progression is 0.99. Express the sum of the first 100 terms as a percentage of the sum to infinity, giving your answer correct to 2 significant figures.

\({\small 9. \hspace{0.6em}(\textrm{a}).\hspace{0.7em}}\) Each year, the value of a certain rare stamp increases by 5% of its value at the beginning of the year. A collector bought the stamp for $10 000 at the beginning of 2005. Find its value at the beginning of 2015 correct to the nearest $100.

\({\small\hspace{1.2em}(\textrm{b}).\hspace{0.8em}}\) The sum of the first n terms of an arithmetic progression is \({\small \frac{1}{2}n(3n+7)}\). Find the 1st term and the common difference of the progression.

\({\small 10. \hspace{0.4em}(\textrm{a}).\hspace{0.7em}}\) An organ has 8 tubes. The lengths of these tubes are a geometrical progression, and the length of the shortest tube is exactly half that of the longest which is 66 cm long. What is the total length of all the tubes in the organ to the nearest cm?

\({\small\hspace{1.2em}(\textrm{b}).\hspace{0.8em}}\) Determine how long it will take to repay an interest free loan $6 625 by monthly payments initially of $10 and increasing by $5 each month. How much is the final payment?

If the loan is $10 000, find the smallest number of months that would be needed to repay it.

\(\\[1pt]\)


As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .

Vector Line Equation

Vector Line and Plane Equation

Vector Line and Plane Equation

We have learned the cartesian form of a line equation: \( \boxed{ \ y \ = \ mx \ + \ c \ }\).

While it comes in handy in solving one-dimensional (1D) and two-dimensional (2D) problems, it may not be as such when solving three-dimensional problems.

Using the vector form of a line equation and a plane equation helps us to solve 3D problems much easier than using its cartesian form.

Given any two points, A and B, we can draw the vector \({\small \vec{a}}\) and \({\small \vec{b}}\) from the origin. Then, the line equation of line AB in the vector form can be written as follows:

\( \hspace{3em} \vec{r} \ = \ \vec{a} \ + \ \lambda(\vec{b} \ – \ \vec{a}) \)

\(\\[5pt] {\small \vec{r} \ = \ }\) position vector
: it represents any points along the line AB

\(\\[5pt] {\small \vec{a} \ \ \textrm{or} \ \ \vec{b} \ = \ }\) location vector
: it shows the location vector of any one point along the line AB which can be represented by \({\small \vec{a}}\) or \({\small \vec{b}}\)

\(\\[5pt] {\small \vec{b} \ – \ \vec{a} \ = \ }\) direction vector
: it gives the direction vector of the line AB

\(\\[5pt] {\small \lambda \ }\) is a constant value.

Vector Line Equation

For the 2D shape, the vector form of a plane equation is shown below:

\( \hspace{3em} (\vec{r} \ – \ \vec{a}) \cdot \vec{n} \ = \ 0 \)

\(\\[5pt] {\small \vec{r} \ = \ }\) position vector
: it represents any points on the plane

\(\\[5pt] {\small \vec{a} \ = \ }\) location vector
: it shows the location of a point on the plane which is represented by \({\small \vec{a}}\)

\(\\[5pt] {\small \vec{n} \ = \ }\) normal vector
: it is the vector that gives perpendicular direction to the plane.

Vector Plane Equation

Note that we can find the cartesian form of a plane equation from its vector form,

\(\\[8pt] \hspace{3em} (\vec{r} \ – \ \vec{a}) \cdot \vec{n} \ \hspace{0.7em} \ = \ 0 \)
\(\\[8pt] \hspace{3em} \vec{r} \ \cdot \vec{n} \ – \ \vec{a} \ \cdot \vec{n} \ = \ 0 \)
\(\\[8pt] \hspace{3em} \vec{r} \ \cdot \vec{n} \hspace{3.3em} \ = \ \vec{a} \ \cdot \vec{n} \)

Let \( \ {\small \vec{r}}=\begin{pmatrix}
x \\[1pt]
y \\[1pt]
z
\end{pmatrix}
\ \) and \( \ {\small \vec{n}}=\begin{pmatrix}
a \\[1pt]
b \\[1pt]
c
\end{pmatrix} \)

with \({\small \ (x,y,z) \ } \) are the cartesian coordinates of any points on the plane and \({\small \ (a,b,c)} \ \) are the cartesian components of the normal vector \( \ {\small \vec{n} \ }\). Then the cartesian form is:

\( \hspace{3em} ax \ + \ by \ + \ cz \ = \ d \)

with \( \ {\small d \ = \ \vec{a} \ \cdot \vec{n} \ }\) (the dot product of vector \( \ {\small \vec{a} \ }\) and \( \ {\small \vec{n} \ }\) ).

I have put together some of the questions I received in the comment section below. You can try these questions also to further your understanding on this topic.

To check your answer, you can look through the solutions that I have posted either in Youtube videos or Instagram posts.

You can subscribe, like or follow my youtube channel and IG account. I will keep updating my IG daily post, preferably.

Furthermore, you can find some examples and more practices below! =).

Try some of the examples below and if you need any help, just look at the solution I have written. Cheers ! =) .
\(\\[1pt]\)


EXAMPLE:

\({\small 1.\enspace}\) 9709/03/SP/20 – Specimen Paper 2020 Pure Maths 3 No 8
\(\\[1pt]\)
9709/03/SP/20 – Specimen Paper 2020 Pure Maths 3 No 8
\(\\[1pt]\)
In the diagram, OABC is a pyramid in which OA = 2 units, OB = 4 units and OC = 2 units. The edge OC is vertical, the base OAB is horizontal and angle \({\small \ AOB \ = \ 90^{\large{\circ}}}\). Unit vectors i, j and k are parallel to OA, OB and OC respectively. The midpoints of AB and BC are M and N respectively.
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{a}).\hspace{0.8em}}\) Express the vectors \({\small \ \overrightarrow{ON} \ }\) and \({\small \ \overrightarrow{CM} \ }\) in terms of i, j and k.
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{b}).\hspace{0.8em}}\) Calculate the angle between the directions of \({\small \ \overrightarrow{ON} \ }\) and \({\small \ \overrightarrow{CM} \ }\).
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{c}).\hspace{0.8em}}\) Show that the length of the perpendicular from M to ON is \( {\small \ {\large \frac{3}{5} } \sqrt 5 }\).

\(\\[1pt]\)
\({\small 2.\enspace}\) 9709/11/O/N/16 – Paper 11 Oct Nov 2020 Pure Maths 1 No 9
\(\\[1pt]\)
9709/11/O/N/16 – Paper 11 Oct Nov 2020 Pure Maths 1 No 9
\(\\[1pt]\)
The diagram shows a cuboid OABCDEFG with a horizontal base OABC in which OA = 4 cm and AB = 15 cm. The height OD of the cuboid is 2 cm. The point X on AB is such that AX = 5 cm and the point P on DG is such that DP = p cm, where p is a constant. Unit vectors i, j and k are parallel to OA, OC and OD respectively.
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{i}).\hspace{0.7em}}\) Find the possible values of p such that angle \({\small \ OPX \ = \ 90^{\large{\circ}}}\).
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{ii}).\hspace{0.7em}}\) For the case where p = 9, find the unit vector in the direction of \({\small \ \overrightarrow{XP} }\).
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{iii}).\hspace{0.5em}}\) A point Q lies on the face CBFG and is such that XQ is parallel to AG. Find \({\small \ \overrightarrow{XQ} }\).

\(\\[1pt]\)
\({\small 3.\enspace}\) The points A and B have position vectors i + 2jk and 3i + j + k respectively. The line l has equation r = 2i + j + k + \({\small \mu}\)(i + j + 2k).
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}\) Show that l does not intersect the line passing through A and B.
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(ii)}.\hspace{0.8em}}\) The plane m is perpendicular to AB and passes through the mid-point of AB. The plane m intersects the line l at the point P. Find the equation of m and the position vector of P.

\(\\[1pt]\)
\({\small 4.\enspace}\)
Vector Line and Plane Equation Example 2
\(\\[1pt]\)
The diagram shows a set of rectangular axes Ox, Oy and Oz, and four points A, B, C and D with position vectors \({\small \overrightarrow{OA} \ = \ 3\textbf{i} }\), \({\small \overrightarrow{OB} \ = \ 3\textbf{i} \ + \ 4\textbf{j} \ }\), \({\small \overrightarrow{OC} \ = \ \textbf{i} \ + \ 3\textbf{j} \ }\) and \({\small \overrightarrow{OD} \ = \ 2\textbf{i} \ + \ 3\textbf{j} \ + \ 5\textbf{k} \ }\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}\) Find the equation of the plane BCD, giving your answer in the form \({\small ax + by + cz = d}\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(ii)}.\hspace{0.8em}}\) Calculate the acute angle between the planes BCD and OABC .

\(\\[1pt]\)
\({\small 5.\enspace}\) The line l has equation r = i + 2j + 3k + \({\small \mu}\)(2ij – 2k).
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}\) The point P has position vector 4i + 2j – 3k. Find the length of the perpendicular from P to l.
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(ii)}.\hspace{0.8em}}\) It is given that l lies in the plane \({\small ax + by + 2z = 13}\), where a and b are constants. Find the values of a and b.

\(\\[1pt]\)
\({\small 6.\enspace}\) Two planes have equations \({\small 2x + 3y \ – \ z \ = \ 1 \ }\) and \( \ {\small x \ – \ 2y + z \ = \ 3}\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}\) Find the acute angle between the planes.
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(ii)}.\hspace{0.8em}}\) Find a vector equation for the line of intersection of the planes.

\(\\[1pt]\)
\({\small 7.\enspace}\) The planes m and n have equations \({\small 3x + y \ – \ 2z \ = \ 10}\) and \({\small x \ – \ 2y + 2z \ = \ 5}\) respectively. The line l has equation r = 4i + 2j + k + \({\small \lambda}\)(i + j + 2k).
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}\) Show that l is parallel to m.
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(ii)}.\hspace{0.8em}}\) Calculate the acute angle between the planes m and n.
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(iii)}.\hspace{0.8em}}\) A point P lies on the line l. The perpendicular distance P from the plane l is equal to 2. Find the position vectors of the two possible positions of P.

\(\\[1pt]\)
\({\small 8.\enspace}\) The line l has equation r = 5i – 3jk + \({\small \lambda}\)(i – 2j + k). The plane p has equation (ri – 2j) . (3i + j + k) = 0. The line l intersects the plane p at the point A.
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}\) Find the position vector of A.
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(ii)}.\hspace{0.8em}}\) Calculate the acute angle between l and p.
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(iii)}.\hspace{0.8em}}\) Find the equation of the line which lies in p and intersects l at right angles.

\(\\[1pt]\)


PRACTICE MORE WITH THESE QUESTIONS BELOW!

\({\small 1.\enspace}\) The points A and B have position vectors, relative to the origin O, given by \({\small \overrightarrow{OA} \ = \ \textbf{i} \ + \ \textbf{j} \ + \ \textbf{k} }\) and \({\small \overrightarrow{OB} \ = \ 2\textbf{i} \ + \ 3\ \textbf{k} }\). The line l has vector equation \( {\small \vec{r} = 2\textbf{i} \ – \ 2\textbf{j} \ – \ \textbf{k} + \mu(-\textbf{i} + 2\textbf{j} + \textbf{k} ) }\).
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}\) Show that the line passing through A and B does not intersect l.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}\) Show that the length of the perpendicular from A to l is \({\small {\large\frac{1}{\sqrt{2}} }}\).
\(\\[1pt]\)

\({\small 2. \enspace}\) The point P has position vector \({\small 3\textbf{i} \ – \ 2\textbf{j} \ + \ \textbf{k} }\). The line l has equation \( {\small \vec{r} = 4\textbf{i} + 2\textbf{j} + 5\textbf{k} + \mu(\textbf{i} + 2\textbf{j} + 3\textbf{k} ) }\).
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}\) Find the length of the perpendicular from P to l, giving your answer correct to 3 significant figures.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}\) Find the equation of the plane containing l and P, giving your answer in the form \({\small ax + by + cz = d}\).
\(\\[1pt]\)

\({\small 3. \enspace}\) Two lines l and m have equations \( {\small \vec{r} = 2\textbf{i} \ – \textbf{j} + \textbf{k} + s(2\textbf{i} + 3\textbf{j} \ – \textbf{k} ) }\) and \( {\small \vec{r} = \textbf{i} + 3\textbf{j} + 4\textbf{k} + t(\textbf{i} + 2\textbf{j} + \textbf{k} ) }\) respectively.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}\) Show that the lines are skew.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}\) A plane p is parallel to the lines l and m. Find a vector that is normal to p.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(iii)}.\hspace{0.5em}}\) Given that p is equidistant from the lines l and m, find the equation of p. Give your answer in
the form \({\small ax + by + cz = d}\).
\(\\[1pt]\)

\({\small 4. \enspace}\) The line l has equation \( {\small \vec{r} = 4\textbf{i} + 3\textbf{j} \ – \textbf{k} + \mu(\textbf{i} + 2\textbf{j} \ – 2\textbf{k} ) }\). The plane p has equation \({\small 2x \ – 3y \ – z = 4}\).
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}\) Find the position vector of the point of intersection of l and p.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}\) Find the acute angle between l and p.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(iii)}.\hspace{0.5em}}\) A second plane q is parallel to l, perpendicular to p and contains the point with position vector
4j βˆ’ k. Find the equation of q, giving your answer in the form \({\small ax + by + cz = d}\).
\(\\[1pt]\)

\({\small 5. \enspace}\) Two planes p and q have equations \({\small x + y + 3z = 8}\) and \({\small 2x \ – 2y + z = 3}\) respectively.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}\) Calculate the acute angle between the planes p and q.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}\) The point A on the line of intersection of p and q has y-coordinate equal to 2. Find the equation of the plane which contains the point A and is perpendicular to both the planes p and q. Give your answer in the form \({\small ax + by + cz = d}\).
\(\\[1pt]\)

\({\small 6. \enspace}\) The equations of two lines l and m are \( {\small \vec{r} = 3\textbf{i} \ – \textbf{j} \ – 2\textbf{k} + \lambda(\ -\textbf{i} + \textbf{j} + 4\textbf{k} ) }\) and \( {\small \vec{r} = 4\textbf{i} + 4\textbf{j} \ – 3\textbf{k} + \mu(2\textbf{i} + \textbf{j} \ – 2\textbf{k} ) }\) respectively.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}\) Show that the lines do not intersect.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}\) ) Calculate the acute angle between the directions of the lines.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(iii)}.\hspace{0.5em}}\) Find the equation of the plane which passes through the point (3, βˆ’2, βˆ’1) and which is parallel to both l and m. Give your answer in the form \({\small ax + by + cz = d}\).
\(\\[1pt]\)

\({\small 7. \enspace}\) The points A and B have position vectors, relative to the origin O, given by \({\small \overrightarrow{OA} \ = \ \textbf{i} \ + \ 2\textbf{j} \ + \ 3\textbf{k} }\) and \({\small \overrightarrow{OB} \ = \ 2\textbf{i} \ + \ \textbf{j} \ + \ 3\ \textbf{k} }\). The line l has vector equation \( {\small \vec{r} = (1 \ – \ 2t)\textbf{i} + (5 \ + \ t)\textbf{j} \ + (2 \ – \ t)\textbf{k} }\).
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}\) Show that l does not intersect the line passing through A and B.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}\) ) The point P lies on l and is such that angle PAB is equal to \({\small 60^{\large{\circ}} }\). Given that the position vector of P is \({\small (1 \ – \ 2t)\textbf{i} + (5 \ + \ t)\textbf{j} \ + (2 \ – \ t)\textbf{k} }\), show that \({\small 3t^2 \ + \ 7t \ + \ 2 \ = \ 0 }\). Hence find the only possible position vector of P.
\(\\[1pt]\)

\({\small 8. \enspace}\) The plane p has equation \({\small 3x + 2y + 4z = 13}\). A second plane q is perpendicular to p and has equation \({\small ax + y + z = 4}\), where a is a constant.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}\) Find the value of a.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}\) The line with equation \( {\small \vec{r} = \textbf{j} \ – \textbf{k} + \lambda( \textbf{i} + 2\textbf{j} + 2\textbf{k} ) }\) meets the plane p at the point A and the plane q at the point B. Find the length of AB.
\(\\[1pt]\)

\({\small 9. \enspace}\) The lines \({\small l_{1}}\) and \({\small l_{2}}\) have equations \( {\small \vec{r} = \textbf{i} + 2\textbf{j} + 3\textbf{k} + \lambda(a\textbf{i} + 4\textbf{j} + 3 \textbf{k} ) }\) and \( {\small \vec{r} = 4\textbf{i} \ – \textbf{k} + \mu(2\textbf{i} + 4\textbf{j} + b\textbf{k} ) }\) respectively. Given that \({\small l_{1}}\) and \({\small l_{2}}\) are parallel:
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}\) Write down the values of a and b.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}\) Find the shortest distance d between \({\small l_{1}}\) and \({\small l_{2}}\).
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(iii)}.\hspace{0.5em}}\) Find a vector equation of the plane p containing \({\small l_{1}}\) and \({\small l_{2}}\).
\(\\[1pt]\)

\({\small 10.\enspace}\) Two lines \({\small l_{1}}\) and \({\small l_{2}}\) have equations \( {\small \vec{r} = \ -8\textbf{i} + 12\textbf{j} + 16\textbf{k} + \lambda(\textbf{i} + 7\textbf{j} \ – 2 \textbf{k} ) }\) and \( {\small \vec{r} = 4\textbf{i} + 6 \textbf{j} \ – 8\textbf{k} + \mu(2\textbf{i} \ – \textbf{j} + 2\textbf{k} ) }\) respectively.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}\) Show that \({\small l_{1}}\) and \({\small l_{2}}\) are skew lines.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}\) The points P and Q lie on \({\small l_{1}}\) and \({\small l_{2}}\) respectively such that PQ is perpendicular to both \({\small l_{1}}\) and \({\small l_{2}}\). Show that \({\small \overrightarrow{PQ} \ = \ 16\textbf{i} \ – \ 8\textbf{j} \ – \ 20\textbf{k} }\).
\(\\[1pt]\)


As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .

Complex Numbers - Argand diagram

Complex Numbers

Complex Numbers

Just as we need negative integers to represent what positive integers can’t, we need imaginary numbers to represent what the real numbers can’t.

Complex numbers represent a number system that combines both real and imaginary numbers. While it may be tempting to ask what is the practical use of complex numbers, their importance is undeniable especially in the field of signal processing.

The usage of complex numbers is an essential part in designing “the poles and zeroes” or simply put, the stability of a system.

“The complex power” of an electrical signal is also an important design consideration in power lines and generators.

We frequently encountered imaginary numbers before, take for example, the solution of this quadratic equation:

\(\\[5pt] \hspace{2em} {x}^{2} \ + \ 1 \ = \ 0 \)
\(\\[5pt] \hspace{2em} {x}^{2} \hspace{2.4em} = \ -1 \)
\(\\[5pt] \hspace{2em} x \hspace{2.8em} = \ \pm \sqrt{-1} \)

\( {\small \sqrt{-1} \ }\) is an example of an imaginary number. Whenever you met an even n-th root of a negative number, you will need to know the complex number theory to simplify or solve it further.

We will start with the two basic forms to represent complex numbers, the cartesian form and the polar form.

Complex numbers in cartesian form can be shown as below:

\( \hspace{2em} {\large z \ = \ a \ + \ b \textrm{i} }\)

\(\\[5pt] {\small a \ = \ \textrm{Re(} z \textrm{)} }\)
: the real part of a complex number z

\(\\[5pt] {\small b \ = \ \textrm{Im(} z \textrm{)} }\)
: the imaginary part of a complex number z

\( {\small \textrm{i} \ = \ \sqrt{-1} }\)

Argand diagram is usually used to show complex number graphically. The x-axis represents the “real part” and the y-axis represents the “imaginary part”.

The Argand diagram representation of the complex number z above can be seen as follows:

Argand diagram in cartesian form

The cartesian form is especially handy in dealing with addition and subtraction of complex numbers.

The result of addition and subtraction of complex numbers can be found by adding or subtracting each of the real parts and each of the imaginary parts separately.

Example:

\( \hspace{2em} {z}_{1} \ = \ a \ + \ b \textrm{i} \)
\( \hspace{2em} {z}_{2} \ = \ c \ + \ d \textrm{i} \)

\( \hspace{2em} {z}_{1} \ + \ {z}_{2} \ = \ (a + c) \ + \ (b + d) \textrm{i} \)
\( \hspace{2em} {z}_{1} \ – \ {z}_{2} \ = \ (a \ – \ c) \ + \ (b \ – \ d) \textrm{i} \)

The second form of complex numbers is the polar form or the modulus-argument form.

\(\\[7pt] \hspace{2em} {\large z \ = \ r \cos \theta \ + \ \textrm{i} \ r \sin \theta }\)
\( \hspace{3.1em} {\large = \ r \ (\cos \theta \ + \ \textrm{i} \ \sin \theta) }\)

\(\\[5pt] {\small r \ = \ |z| }\)
: the modulus of a complex number z

\(\\[5pt] {\small \theta \ = \ \textrm{arg(} z \textrm{)} }\)
: the principal argument of a complex number z, \( \enspace {\small -\pi \lt \theta \le \pi } \)

The modulus \({\small r }\) and argument \({\small \theta}\) can be used to show the sets of points or regions of complex numbers in Argand diagram.

The Argand diagram of a complex number z in its polar form can be seen below:

Argand diagram in polar form

Alternatively, the polar form can also be expressed in euler notation and phase-angle notation.

In euler notation,

\(\\[10pt] \hspace{2.6em} {\large {e}^{\textrm{i} \theta} \ = \ \cos \theta \ + \ \textrm{i} \ \sin \theta }\)
Since, \(\\[7pt] \hspace{1em} {\large z \ = \ r \ (\cos \theta \ + \ \textrm{i} \ \sin \theta) }\)
then \( \hspace{1.5em} {\large z \ = \ r \ {e}^{\textrm{i} \theta} }\)

In phase-angle notation,

\(\\[10pt] \hspace{2.6em} {\large \angle \theta \ = \ {e}^{\textrm{i} \theta} }\)
Since, \(\\[7pt] \hspace{1em} {\large z \ = \ r \ {e}^{\textrm{i} \theta} }\)
then \( \hspace{1.5em} {\large z \ = \ r \ \angle \theta }\)

The euler and phase-angle notation are merely a simpler way of writing the polar form.

The polar form, especially the euler notation and phase-angle notation are useful in multiplication and division of complex numbers.

The result of multiplication of complex numbers can be found by multiplying the moduli and adding the arguments.

The result of division of complex numbers can be found by dividing the moduli and subtracting the arguments.

Example:

\( \hspace{2em} {z}_{1} \ = \ {r}_{1} \ {e}^{\textrm{i} {\theta}_{1}} \)
\( \hspace{2em} {z}_{2} \ = \ {r}_{2} \ {e}^{\textrm{i} {\theta}_{2}} \)

\(\\[10pt] \hspace{2em} {z}_{1} \times {z}_{2} \ = \ ({r}_{1} \times {r}_{2}) \ {e}^{\textrm{i} ({\theta}_{1} \ + \ {\theta}_{2})} \)
\( \hspace{2em} {\large \frac{{z}_{1}}{{z}_{2}}} \hspace{1.9em} = \ {\large \frac{{r}_{1}}{{r}_{2}}} \ {e}^{\textrm{i} ({\theta}_{1} \ – \ {\theta}_{2})} \)

By comparing the complex numbers in both cartesian & polar form, they can be converted from one form to another.

To find the cartesian form from the polar form, \( \enspace z \ = \ a \ + \ b \textrm{i} \)

\(\\[5pt] \enspace a \ = \ r \ \cos \theta \)
\( \enspace b \ = \ r \ \sin \theta \)

To find the polar form from the cartesian form, \( \enspace z \ = \ r \ (\cos \theta \ + \ \textrm{i} \sin \theta )\)

\(\\[7pt] \enspace r \ = \ \sqrt{{a}^{2} \ + \ {b}^{2}} \)
\( \enspace \theta \ = \ {\tan}^{-1} \Big({\large \frac{b}{a} } \Big) \)

The complex conjugate of a complex number has the same real part and opposite sign for its imaginary part. That is, to say, if \( z \ = \ a \ + \ b \textrm{i} \ \) then the complex conjugate is \( {z}^{*} \ = \ a \ – \ b \textrm{i} \).

One such example of usage is when dealing with polynomial equations with real coefficients (non-complex number coefficients).

If a polynomial equation has a complex number z as one of its roots, then the complex conjugate \({\small {z}^{*} }\) is also a root of the polynomial equation.

To find the n-th power of a complex number, De Moivre’s Theorem can be used. It can be shown as follows:

If \( \enspace z \ = \ r \ {e}^{\textrm{i} \theta} \)

then \( \enspace {z}^{n} \ = \ {r}^{n} \ {e}^{ \textrm{i} (n\theta) } \)

I have put together some of the questions I received in the comment section below. You can try these questions also to further your understanding on this topic.

To check your answer, you can look through the solutions that I have posted either in Youtube videos or Instagram posts.

You can subscribe, like or follow my youtube channel and IG account. I will keep updating my IG daily post, preferably.

Furthermore, you can find some examples and more practices below! =).

Try some of the examples below and if you need any help, just look at the solution I have written. Cheers ! =) .
\(\\[1pt]\)


EXAMPLE:

\({\small 1.\enspace}\) 9709/03/SP/17 – Specimen Paper 2017 Pure Maths 3 No 9
\(\\[1pt]\)
The complex number \({\small \ 3 \ βˆ’ \ \textrm{i} \ }\) is denoted by u. Its complex conjugate is denoted by u*.
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{i}).\hspace{0.7em}}\) On an Argand diagram with origin O, show the points A, B and C representing the complex numbers u, u* and u*u respectively. What type of quadrilateral is OABC?
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{ii}).\hspace{0.7em}}\) Showing your working and without using a calculator, express \( {\small \ {\large\frac{u}{u*}} }\) in the form of x + iy, where x and y are real.
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{iii}).\hspace{0.5em}}\) By considering the argument of \( {\small \ {\large\frac{u}{u*}} }\), prove that
\({\small\hspace{3em} \ {\tan}^{-1} \big(\frac{3}{4}\big) \ = \ 2 {\tan}^{-1} \big(\frac{1}{3}\big) }\)

\(\\[1pt]\)
\({\small 2.\enspace}\) 9709/03/SP/20 – Specimen Paper 2020 Pure Maths 3 No 6
\(\\[1pt]\)
The complex number \({\small \ 1 \ + \ 3\textrm{i} \ }\) and \({\small \ 4 \ + \ 2\textrm{i} \ }\) are denoted by u and v respectively.
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{a}).\hspace{0.8em}}\) Find \( {\small \ {\large \frac{u}{v}} \ }\) in the form of x + iy, where x and y are real.
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{b}).\hspace{0.8em}}\) State the argument of \( {\small \ \large { \frac{u}{v} }}\).
\(\\[1pt]\)
In an Argand diagram, with origin O, the points A, B and C represent the complex numbers u, v and u – v respectively.
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{c}).\hspace{0.8em}}\) State fully the geometrical relationship between OC and BA.
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{d}).\hspace{0.8em}}\) Show that angle \({\small AOB = \frac{1}{4} \pi \small }\) radians.

\(\\[1pt]\)
\({\small 3.\enspace}\) The complex number w is given by \({\small w = {\large -\frac {1}{2}} \ + \ {\large \textrm{i}\frac {\sqrt{3}}{2}} }\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(i\right).\hspace{0.8em}}\) Find the modulus and argument of w.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(ii\right).\hspace{0.8em}}\) The complex number z has a modulus R and argument \({\small \theta}\), where \( \ {\small – {\large \frac{1}{3}}\pi \lt \theta \lt {\large \frac{1}{3}}\pi } \). State the modulus and argument of wz and the modulus and argument of \({\small {\large \frac{z}{w}} }\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(iii\right).\hspace{0.8em}}\) Hence explain why, in Argand diagram, the points representing z, wz and \({\small {\large \frac{z}{w}} }\) are the vertices of an equilateral triangle.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(iv\right).\hspace{0.8em}}\) In an Argand diagram, the vertices of an equilateral triangle lie on a circle with centre at the origin. One of the vertices represents the complex number 4 + 2i. Find the complex numbers represented by the other two vertices. Give your answers in the form x + iy, where x and y are real and exact.

\(\\[1pt]\)
\({\small 4.\enspace}\) The complex number -2 + i is denoted by u.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(i\right).\hspace{0.8em}}\) Given that u is a root of the equation \({\small {x}^{3} \ – \ 11x \ – \ k \ = \ 0}\), where k is real, find the value of k.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(ii\right).\hspace{0.8em}}\) Write down the other complex root of this equation.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(iii\right).\hspace{0.8em}}\) Find the modulus and argument of u.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(iv\right).\hspace{0.8em}}\) Sketch an Argand diagram showing the point representing u. Shade the region whose points represent the complex numbers z satisfying both the inequalities
\({\small \quad |z| \ \lt \ |z \ – \ 2| \ }\) and \( \ {\small 0 \ \lt \ \textrm{arg}(z \ – \ u) \ \lt \ {\large \frac{1}{4} }\pi }\).

\(\\[1pt]\)
\({\small 5.\enspace}\) The complex number w is defined by \({\small w = -1 + \textrm{i} }\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(i\right).\hspace{0.8em}}\) Find the modulus and argument of \( {\small {w}^{2} }\) and \({\small {w}^{3}}\), showing your working.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(ii\right).\hspace{0.8em}}\) The points in an Argand diagram representing \({\small w }\) and \({\small {w}^{2}}\) are the ends of a diameter of a circle. Find the equation of the circle, giving your answer in the form \({\small |z \ – \ (a \ + \ b\textrm{i})| \ = \ k}\).

\(\\[1pt]\)
\({\small 6.\enspace (a).\hspace{0.8em} }\) Without using a calculator, solve the equation
\(\\[1pt]\)
\({\small \quad 3w \ + \ 2\textrm{i}{w}^{*} \ = \ 17 \ + \ 8\textrm{i}, }\)
\(\\[1pt]\)
where \({\small {w}^{*} }\) denotes the complex conjugate of \({\small w }\). Give your answer in the form \({\small (a \ + \ b\textrm{i}). }\)
\(\\[1pt]\)
\({\small \hspace{1.3em} (b).\hspace{0.8em} }\) In an Argand diagram, the loci
\(\\[1pt]\)
\({\small \quad \textrm{arg}(z \ – \ 2\textrm{i}) \ = \ {\large \frac{1}{6} }\pi \ }\) and \( \ {\small |z \ – \ 3| \ = \ |z \ – \ 3\textrm{i}| }\)
\(\\[1pt]\)
intersect at the point P. Express the complex number represented by P in the form of \({\small r {e}^{\textrm{i}\theta}}\), giving the exact value of \({\small \theta }\) and the value of r correct to 3 significant figures.

\(\\[1pt]\)
\({\small 7.\enspace}\) The complex number z is defined by \({\small \ z \ = \ {\large \frac{9\sqrt{3} \ + \ 9\textrm{i}}{\sqrt{3} \ – \ \textrm{i}} } }\). Find, showing all your working,
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(i\right).\hspace{0.8em}}\) an expression for z in the form \({\small r {e}^{\textrm{i}\theta}}\), where \({\small r \ \gt \ 0 \ }\) and \( \ {\small -\pi \lt \theta \le \pi } \),
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(ii\right).\hspace{0.8em}}\) the two square roots of z, giving your answers in the form \({\small r {e}^{\textrm{i}\theta}}\), where \({\small r \ \gt \ 0 \ }\) and \( \ {\small -\pi \lt \theta \le \pi } \).

\(\\[1pt]\)
\({\small 8.\enspace (a).\hspace{0.8em} }\) Showing all working and without using a calculator, solve the equation
\(\\[1pt]\)
\( \quad {\small (1 \ + \ \textrm{i}){z}^{2} \ – \ (4 \ + \ 3\textrm{i})z \ + \ 5 \ + \ \textrm{i} \ = \ 0}\).
\(\\[1pt]\)
Give your answers in the form x + iy, where x and y are real.
\(\\[1pt]\)
\({\small \hspace{1.3em} (b).\hspace{0.8em} }\) The complex number u is given by
\( \quad {\small u \ = \ -1 \ – \ \textrm{i} }\).
\(\\[1pt]\)
On a sketch of an Argand diagram show the point representing u. Shade the region whose points represent complex numbers satisfying the inequalities
\({\small \ |z| \ \lt \ |z \ – \ 2\textrm{i}| }\) and \( {\small {\large \frac{1}{4} }\pi \ \lt \ \textrm{arg}(z \ – \ u) \ \lt \ {\large \frac{1}{2} }\pi }\).

\(\\[1pt]\)


PRACTICE MAKES PERFECT!

\({\small 1.\enspace(a).\hspace{0.8em}}\) Showing all necessary working, express the complex number \({\small {\large \frac{2 \ + \ 3\textrm{i}}{1 \ – \ 2\textrm{i}} } }\) in the form \({\small r {e}^{\textrm{i}\theta}}\), where \({\small r \ \gt \ 0 \ }\) and \( \ {\small -\pi \lt \theta \le \pi } \). Give the values of r and \({\small \theta}\) correct to 3 significant figures.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) On an Argand diagram sketch the locus of points representing complex numbers z satisfying the equation \({\small |z \ – \ 3 \ + \ 2\textrm{i}| \ = \ 1 }\). Find the least value of |z| for points on this locus, giving your answer in an exact form.
\(\\[1pt]\)

\({\small 2.\enspace(a) \ (i).\hspace{0.5em}}\) Without using a calculator, express the complex number \({\small {\large \frac{2 \ + \ 6\textrm{i}}{1 \ – \ 2\textrm{i}} } }\) in the form x + iy, where x and y are real.
\(\\[1pt]\)
\({\small\hspace{2.4em}\left(ii\right).\hspace{0.5em}}\) Hence, without using a calculator, express \({\small {\large \frac{2 \ + \ 6\textrm{i}}{1 \ – \ 2\textrm{i}} } }\) in the form \({\small r(\cos \theta \ + \ \textrm{i} \sin \theta)}\), where \({\small r \ \gt \ 0 \ }\) and \( \ {\small -\pi \lt \theta \le \pi } \), giving the exact values of r and \({\small \theta}\).
\(\\[1pt]\)
\({\small\hspace{1.5em}\left(b\right).\hspace{0.8em}}\) On a sketch of an Argand diagram, shade the region whose points represent complex numbers z satisfying both the inequalities \({\small |z \ – \ 3 \textrm{i}| \ \le \ 1 }\) and Re \({\small z \ \le \ 0}\), where Re z denotes the real part of z. Find the greatest value of arg z for points in this region, giving your answer in radians correct to 2 decimal places.

\({\small 3. \enspace}\) The complex number \({\small 1 \ βˆ’ \ (\sqrt{3})\textrm{i} }\) is denoted by u.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(i\right).\hspace{0.8em}}\) Find the modulus and argument of u.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(ii\right).\hspace{0.8em}}\) Show that \({\small \ {u}^{3} \ + \ 8 \ = \ 0}\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(iii\right).\hspace{0.8em}}\) On a sketch of an Argand diagram, shade the region whose points represent complex numbers z satisfying both the inequalities \({\small |z \ – \ u| \ \le \ 2 }\) and Re \({\small z \ \ge \ 2}\), where Re z denotes the real part of z.

\({\small 4. \enspace}\) The polynomial \({\small {z}^{4} \ + \ 3{z}^{2} \ + \ 6z \ + \ 10 }\) is denoted by p(z). The complex number \({\small -1 \ + \ \textrm{i} }\) is denoted by u.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(i\right).\hspace{0.8em}}\) Showing all your working, verify that u is a root of the equation p(z) \({\small = \ 0}\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(ii\right).\hspace{0.8em}}\) Find the other three roots of the equation p(z) \({\small = \ 0}\).

\({\small 5.\enspace(a).\hspace{0.8em}}\) Showing all necessary working, solve the equation \({\small \textrm{i}{z}^{2} \ + \ 2z \ βˆ’ \ 3\textrm{i} \ = \ 0 }\), giving your answers in the form x + iy, where x and y are real and exact.
\(\\[1pt]\)
\({\small \quad(b) \ (i).\hspace{0.8em}}\) On a sketch of an Argand diagram, show the locus representing complex numbers satisfying the equation \({\small |z| \ = \ |z \ βˆ’ \ 4 \ βˆ’ \ 3\textrm{i}| }\).
\(\\[1pt]\)
\({\small \hspace{2em} (ii).\hspace{0.7em}}\) Find the complex number represented by the point on the locus where |z| is least. Find the modulus and argument of this complex number, giving the argument correct to 2 decimal places.

\({\small 6. \enspace}\) Sketch, on an Argand diagram, the locus of the points representing the complex number z such that \({\small |\textrm{i}z \ – \ 2| \ = \ |2 \ – \ z|}\). Hence find the least value of \({\small |z \ + \ 2|}\).

\({\small 7. \enspace}\) Solve the equation \({\small \textrm{i}{z}^{4} \ = \ -81 }\), expressing the roots in the form \({\small r {e}^{\textrm{i}\theta}}\), where \({\small r \ \gt \ 0 \ }\) and \( \ {\small -\pi \lt \theta \le \pi } \).

\({\small 8. \enspace}\) In an Argand diagram, the point P represents the complex number z such that:
\(\\[1pt]\)
\( {\small |z \ – \ 2 \ + \ 4\textrm{i}| \le 4 \ }\) and \({\small \ {\large \frac{\pi}{4} } \le \textrm{arg}(z \ + \ 2\textrm{i}) \lt 0 }\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(i\right).\hspace{0.8em}}\) Sketch the locus of P.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(ii\right).\hspace{0.8em}}\) Hence, find the exact range of values of \({\small |z \ + \ 2|}\).

\({\small 9. \enspace}\) In an Argand diagram, the point P represents the complex number z. Draw on a single clearly labelled diagram to show the locus of P when z satisfies
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(i\right).\hspace{0.8em}}\) \( {\small |z \ – \ 3 \ – \ 4\textrm{i}| = 5 \ }\)
\({\small\hspace{1.2em}\left(ii\right).\hspace{0.8em}}\) \( {\small |z \ – \ 2 \ + \ 3\textrm{i}| = |z \ – \ 10 \ – \ 3\textrm{i}| \ }\)
\(\\[1pt]\)
Hence find the greatest and least possible values of |z| when z satisfies both
\(\\[1pt]\)
\( {\small |z – 3 – 4\textrm{i}| \le 5 }\) and \( {\small |z – 2 + 3\textrm{i}| \ge |z – 10 – 3\textrm{i}| }\)

\({\small 10.\enspace}\) Given that \({\small {z}^{*} \ = \ \frac{{(2 \ – \ 2\textrm{i})}^{3}}{{(-1 \ + \ \sqrt{3}\textrm{i})}^{4}}}\), find the exact value of |z| and arg(z). Hence, state the smallest positive integer n such that \({\small {z}^{n}}\) is a purely real number.


As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .

Binomial expansion formula

Binomial Expansion and Binomial Series

Binomial Expansion and Binomial Series

In algebra, we all have learnt the following basic algebraic expansion:

\( \hspace{3em} {(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2} \).

We can keep multiplying the expression \( { \small (a + b) } \) by itself to find the expression for higher index value. For example:

\(\\[12pt] {(a + b)}^{3} \ = \ {(a + b)}^{2} \times (a + b) \)
\(\\[12pt] \hspace{1.5em} \ = \ ({a}^{2} + 2ab + {b}^{2}) \times (a + b) \)
\(\\[12pt] \hspace{1.5em} \ = \ {a}^{3} + {a}^{2}b + 2{a}^2b + 2a{b}^2 + a{b}^2 + {b}^{3} \)
\( \hspace{1.5em} \ = \ {a}^{3} + 3{a}^{2}b + 3a{b}^2 + {b}^{3} \)

Of course this is a really tedious way for a very large index/power/exponent number.

Instead of doing it manually, we could use a formula called the Binomial Theorem which is shown below:

\( {(a + b)}^{n} \ = \ \displaystyle \sum_{k=0}^{n} \binom{n}{k} {a}^{n \ – \ k} \ {b}^{k} \)

with \( \displaystyle \binom{n}{k} \ = \ \frac{n!}{k!(n-k)!} \)

\(\hspace{4.4em} = \ {\large \frac{n (n \ – \ 1) (n \ – \ 2) … (n \ – \ k \ + \ 1)}{k(k \ – \ 1)(k \ – \ 2) \ … \ 1} } \)

for \( k \geq 1 \) and \( \displaystyle \binom{n}{0} \ = \ 1 \)

\(\\[10pt]\) Example:
\(\\[12pt] {(a + b)}^{10} \)
\(\\[15pt] = \binom{10}{0} {a}^{10}{b}^{0} + \binom{10}{1} {a}^{9}{b}^{1}+…+ \binom{10}{10} {a}^{0}{b}^{10} \)
\(\\[15pt] = {a}^{10} + 10 {a}^{9}b+ 45 {a}^{8}{b}^{2} + … + 10{a}{b}^{9} + {b}^{10} \)

\(\\[10pt]\) Note that,
\(\\[20pt]\quad \ {\displaystyle \binom{10}{0} } \ = \ 1 \)
\(\\[20pt]\quad \ {\displaystyle \binom{10}{1} } \ = \ {\large \frac{ 10 }{ 1 } } \ = \ 10 \)
\(\\[20pt]\quad \ {\displaystyle \binom{10}{2} } \ = \ {\large \frac{ 10 \ \times \ 9 }{ 2 \ \times \ 1 } } \ = \ 45 \)
\(\\[20pt]\quad \ {\displaystyle \binom{10}{3} } \ = \ {\large \frac{ 10 \ \times \ 9 \ \times \ 8 }{ 3 \ \times \ 2 \ \times \ 1 } } \ = \ 120 \)
\(\\[20pt]\quad \ {\displaystyle \binom{10}{4} } \ = \ {\large \frac{ 10 \ \times \ 9 \ \times \ 8 \ \times \ 7 }{4 \ \times \ 3 \ \times \ 2 \ \times \ 1 } } \ = \ 210 \)
\(\\[20pt]\quad \ {\displaystyle \binom{10}{5} } \ = \ {\large \frac{10 \ \times \ 9 \ \times \ 8 \ \times \ 7 \ \times \ 6 }{5 \ \times \ 4 \ \times \ 3 \ \times \ 2 \ \times \ 1 } } \ = \ 252 \)
, etc.

\(\binom{n}{k} \) is read as “n choose k” or sometimes referred to as the binomial coefficients.

Notice that this binomial expansion has a finite number of terms with the k values take the non-negative numbers from 0, 1, 2, … , n.

Then the next question would be: Can we still use the binomial theorem for the expansion with negative number or fractional number for the index value?

Thankfully, Sir Isaac Newton has shown that the binomial theorem can be generalized to take in any numbers for the index value including the negative and fractional numbers as long as it is within a convergence rule.

Using this result, we have the Binomial Series which can be expressed as follows:

\(\\[25pt]{(1 + x)}^{n} = \displaystyle \sum_{k=0}^{n} \binom{n}{k} {x}^{k} \)
\(\\[10pt]{\small = \ 1 + nx + \frac{(n)(n-1)}{2!}{x}^{2} + \frac{(n)(n-1)(n-2)}{3!}{x}^{3} + … }\)

with \( {\small k } \) is any real number and \( {\small -1 \lt x \lt 1 } \).

\(\\[10pt]\) Example:
\(\\[12pt] { {\large \frac{1}{1 + x}} \ = \ (1 + x)}^{-1} \)
\(\\[15pt]{\small = \ 1 + (-1)x + \frac{(-1)(-2)}{2!}{x}^{2} + \frac{(-1)(-2)(-3)}{3!}{x}^{3} + … }\)
\(\\[15pt]{\small = \ 1 \ – \ x \ + \ {x}^{2} \ – \ {x}^{3} \ + \ … \ – \ … }\)
\(\\[15pt]\) The expansion is valid when \({\small -1 \lt x \lt 1 } \).

\(\\[10pt]\) Another example:
\(\\[12pt] { \frac{1}{\sqrt{(1 + x)}} \ = \ (1 + x)}^{\frac{1}{2}} \)
\(\\[15pt]{\small = \ 1 + \frac{1}{2}x + \frac{(\frac{1}{2})(- \frac{1}{2})}{2!}{x}^{2} + \frac{(\frac{1}{2})(- \frac{1}{2})( – \frac{3}{2})}{3!}{x}^{3} + … }\)
\(\\[15pt]{ = \ 1 + \frac{x}{2} – \frac{{x}^{2}}{8} + \frac{{x}^{3}}{16} \ – \ … + … }\)
\(\\[15pt]\) The expansion is valid when \({\small -1 \lt x \lt 1 } \).

Try some of the examples below and if you need any help, just look at the solution I have written. Cheers ! =) .
\(\\[1pt]\)


EXAMPLE:

\({\small 1.\enspace}\) Let \({\small f(x) \ = \ {\large \frac{7{x}^{2} \ – \ 15x \ + \ 8}{(1 \ – \ 2x){(2 \ – \ x)}^{2} }} }\)
\(\\[1pt]\)
Express \({\small f(x) }\) in partial fractions and hence obtain the expansion of \({\small f(x) }\) in ascending powers of x, up to and including the term in \({\small {x}^{2}}\).

\(\\[1pt]\)
\({\small 2.\enspace}\) Let \({\small f(x) \ = \ {\large \frac{ x \ – \ 4{x}^{2} }{(3 \ – \ x)(2 \ + \ {x}^{2}) }} }\)
\(\\[1pt]\)
Express \({\small f(x) }\) in partial fractions and hence obtain the expansion of \({\small f(x) }\) in ascending powers of x, up to and including the term in \({\small {x}^{3}}\).

\(\\[1pt]\)
\({\small 3.\enspace}\) Find the value of \( {\small {(2 \ + \ x)}^{6} \ – \ {(2 \ – \ x)}^{6} }\) in ascending powers of x, up to and including the term in \({\small {x}^{3}}\). Hence, find the value of \( {\small {(1.99)}^{6} \ – \ {(2.01)}^{6} }\).

\(\\[1pt]\)
\({\small 4.\enspace}\) Find the last four terms in the expansion in ascending powers of x of \( (2x – {x}^{2})^{10}\).

\(\\[1pt]\)
\({\small 5.\enspace}\) Find the value of \( {\small \frac{a}{b} }\) in \({\small {(a + bx)}^{12}}\) given that the coefficient of \({\small {x}^{2} }\) is 11 times the coefficient of \({\small x }\).

\(\\[1pt]\)
\({\small 6.\enspace}\) Find the value of a, b and n in the following expansion:
\(\\[1pt]\)
\({\small {(a + bx)}^{n} \ = \ … + {\large\frac{20}{3}}{x}^{2} + 20 {x}^{3} + … }\).
\(\\[1pt]\)
given that \( {\small {\large \frac{a}{b}} \ = \ {\large \frac{4}{9}} }\).

\(\\[1pt]\)
\({\small 7.\enspace}\) Consider the binomial expansion of \({\small {\large \frac{1}{\sqrt{4-x}} } }\).
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{a}).\hspace{0.8em}}\) Write down the first 4 terms.
\({\small\hspace{1.2em}(\textrm{b}).\hspace{0.8em}}\) State the interval of convergence for the complete expression.
\({\small\hspace{1.2em}(\textrm{c}).\hspace{0.8em}}\) Use the expansion to estimate \({\small {\large \frac{1}{\sqrt{3.6}} } }\). Check your answer by direct calculation.

\(\\[1pt]\)
\({\small 8.\hspace{0.4em}(\textrm{i}).\hspace{0.4em}}\) Find the first 3 terms in the expansion of \({\small {(2x \ – \ {\large \frac{1}{16x}})}^{8} }\) in descending powers of x.
\(\\[1pt]\)
\({\small\hspace{1em}(\textrm{ii}).\hspace{0.7em}}\) Hence find the coefficient of \({\small {x}^{4}}\) in the expansion of \({\small {(2x \ – \ {\large \frac{1}{16x}})}^{8} {( {\large \frac{1}{{x}^{2}}} \ + \ 1 )}^{2} }\).

\(\\[1pt]\)


PRACTICE MORE WITH THESE QUESTIONS BELOW!

\({\small 1.\enspace}\) Expand \(\frac{1}{\sqrt[3]{(1 \ + \ 6x)}}\) in ascending powers of x up to and including the term in \({\small x^3 }\), simplifying the coefficients.

\({\small 2. \enspace}\) Expand \(\frac{4}{\sqrt{(4 \ – \ 3x)}}\) in ascending powers of x up to and including the term in \({\small x^2 }\), simplifying the coefficients.

\({\small 3. \enspace}\) Let \({\small f(x) \ = \ {\large \frac{3{x}^{2} \ + \ x \ + \ 6}{(x \ + \ 2)({x}^{2} \ + \ 4) }} }\)

(i). Express \({\small f(x) }\) in partial fractions.
(ii). Hence obtain the expansion of \({\small f(x) }\) in ascending powers of x, up to and including the term in \({\small {x}^{2}}\).

\({\small 4. \enspace}\) Express \( {\small {(3 \ + \ 4x)}^{\frac{1}{2}} }\) as a series of descending powers of x up to and including the third non-zero coefficient. State the set of values of x for which the series expansion is valid.

\({\small 5. \enspace}\) (i). Given that the first three terms in the expansion of \( {\small {(1 \ + \ ax)}^{b}}\) in ascending powers of x are \({\small 1 \ + \ x \ + \ \frac{3}{2}{x}^{2}}\), where a and b are constants, find the values of a and b.
(ii). Hence, with the values of a and b found in (i), expand \(\frac{ {(1 \ + \ ax)}^{b} }{1 \ – \ x}\) as a series in ascending powers of x up to and including the term in \({\small x^2 }\).

\({\small 6. \enspace}\) Expand \({\small {(1-2p)}^{8} }\) up to and including the term in \({\small p^3 }\). Hence, find the first four terms in the expansion in ascending powers of x of \({\small {(1 \ – \ 4x \ + \ \frac{2}{x})}^{8} }\).

\({\small 7. \enspace}\) Find the coefficient of \({\small {x}^{2}}\) in the expansion of \({\small {(1 \ + \ 3x)}^{2}{(1 \ – \ 3x)}^{10} }\).

\({\small 8. \enspace}\) Find the coefficient of \({ \small x }\) in the expansion of \({\small {(3x \ – \ \frac{2}{x})}^{9} }\).

\({\small 9. \enspace}\) Find the value of a and b in the following expansion:

\({\small \hspace{1.5em} {(a + bx)}^{9} \ = \ … + 672{x}^{3} + 252 {x}^{4} + … }\)

\({\small 10.\enspace}\) Find the value of a and b in the expansion of \({\small {(1 \ + \ ax)}^{5}{(1 \ – \ bx)}^{7} }\) if the coefficients of \({ \small x }\) and \({\small {x}^{2}}\) are 3 and -9 respectively.


As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .

Partial Fractions - Summary of Forms

Partial Fractions

Partial Fractions

In solving algebra related problems and questions, we may sometimes deal with rational functions. A rational function is basically an algebraic polynomial fraction, in which we have polynomials on both the numerator and denominator.

Partial fractions is one of the simplest and most effective method in solving algebra related problems regarding rational functions.

In partial fractions, we separate the polynomials in our rational function into simpler form of polynomials.

Some of the applications of partial fractions include the solving of integration problems with rational functions, the binomial expansion and also the arithmetic series and sequences.

There are a few basic forms we need to memorize in partial fractions:

1.\(\enspace\) The linear form:
\(\hspace{6em} {\small (ax + b)}\)
Example:

\({\large\frac{3x \ + \ 5}{(x \ + \ 1)(2x \ + \ 7)} \ \equiv \ \frac{A}{(x \ + \ 1)} + \frac{B}{(2x \ + \ 7)} }\)

2.\(\enspace\) The quadratic form of a linear factor:
\(\hspace{6em} {\small (cx \ + \ d)^{2}} \)
Example:

\(\frac{3x + 5}{(x + 1){(2x + 7)}^{2}} \equiv \frac{A}{(x + 1)} + {\small\boxed{\frac{B}{(2x + 7)} + \frac{C}{{(2x + 7)}^{2}}}} \)

3.\(\enspace\) The quadratic form that cannot be factorized:
\(\hspace{6em} {\small (c{x}^{2} \ + \ d) }\)
Example:

\({\large\frac{3x \ + \ 5}{(x \ + \ 1)(2{x}^{2} \ + \ 7)} \ \equiv \ \frac{A}{(x \ + \ 1)} + {\small\boxed{\frac{Bx \ + \ C}{(2{x}^{2} \ + \ 7)}}}}\)

After the polynomials in the denominator of the rational function is separated, make the denominators of the simpler terms to be the same. This is typically done by multiplying the denominators together .

To find each of the coefficients in the numerator (A, B or C), we can use substitution method or equating the coefficient method.

In the substitution method, we substitute a value of x that we freely choose in the left hand side numerator and the right hand side numerator and then find the coefficients one by one.

While in the equating the coefficient method, we expand the right hand side numerator and then compare each of the coefficients in the right hand side numerator with the left hand side numerator.

Both methods will be shown in the solution of the examples below. Give it a try and if you need any help, just look at the solution I have written. Cheers ! =) .
\(\\[1pt]\)


EXAMPLE:

\({\small 1.\enspace}\) Let \({\small f(x) \ = \ {\large \frac{7{x}^{2} \ – \ 15x \ + \ 8}{(1 \ – \ 2x){(2 \ – \ x)}^{2} }} }\)
\(\\[1pt]\)
Express \({\small f(x) }\) in partial fractions.
\(\\[1pt]\)

\(\\[1pt]\)
\({\small 2.\enspace}\) Let \({\small f(x) \ = \ {\large \frac{ x \ – \ 4{x}^{2} }{(3 \ – \ x)(2 \ + \ {x}^{2}) }} }\)
\(\\[1pt]\)
Express \({\small f(x) }\) in partial fractions.
\(\\[1pt]\)

\(\\[1pt]\)
\({\small 3.\enspace}\) Let \({\small f(x) \ = \ {\large \frac{ 5{x}^{2} \ + \ x \ + \ 27 }{(2x \ + \ 1)( {x}^{2} \ + \ 9 ) }} }\)
\(\\[1pt]\)
Express \({\small f(x) }\) in partial fractions.
\(\\[1pt]\)

\(\\[1pt]\)
\({\small 4.\enspace}\) Let \({\small f(x) \ = \ {\large \frac{ 10 x \ + \ 9 }{(2x \ + \ 1){( 2x \ + \ 3 )}^{2} }} }\)
\(\\[1pt]\)
Express \({\small f(x) }\) in partial fractions.
\(\\[1pt]\)

\(\\[1pt]\)
\({\small 5.\enspace}\) Let \({\small f(x) \ = \ {\large \frac{ 2x(5 \ – \ x) }{(3 \ + \ x){( 1 \ – \ x )}^{2} }} }\)
\(\\[1pt]\)
Express \({\small f(x) }\) in partial fractions.
\(\\[1pt]\)

\(\\[1pt]\)


PRACTICE MORE WITH THESE QUESTIONS BELOW!

\({\small 1.\enspace}\) Express \({\small {\large \frac{7{x}^{2} \ – \ 3x \ + \ 2}{x({x}^{2} \ + \ 1) }} }\) in partial fractions.

\({\small 2. \enspace}\) Let \({\small f(x) \ = \ {\large \frac{ 5{x}^{2} \ + \ x \ + \ 6 }{(3 \ – \ 2x)({x}^{2} \ + \ 4 )}} }\)
Express \({\small f(x) }\) in partial fractions.

\({\small 3. \enspace}\) Express \({\small {\large \frac{2 \ – \ x \ + \ 8{x}^{2}}{(1 \ – \ x)(1 \ + \ 2x)(2 \ + \ x) }} }\) in partial fractions.

\({\small 4. \enspace}\) Let \({\small f(x) \ = \ {\large \frac{ {x}^{2} \ + \ 3x \ + \ 3 }{(x \ + \ 1)(x \ + \ 3 )}} }\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Express f(x) in partial fractions.
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Hence show that,
\(\hspace{3em} {\small \displaystyle \int_{0}^{3} f(x) \ \mathrm{d}x = 3 \ – \ \frac{1}{2} \ln 2.}\)

\({\small 5. \enspace}\) Let \({\small f(x) \ = \ {\large \frac{ {x}^{2} \ – \ 8x \ + \ 9 }{(1 \ – \ x){(2 \ – \ x )}^{2}}} }\)
Express \({\small f(x) }\) in partial fractions.

\({\small 6. \enspace}\) Let \({\small f(x) \ = \ {\large \frac{ 2{x}^{2} \ – \ 7x \ – \ 1 }{(x \ – \ 2)({x}^{2} \ + \ 3 )}} }\)
Express \({\small f(x) }\) in partial fractions.

\(\\[12pt]{\small 7. \enspace}\) Express \({\small {\large \frac{ x \ + \ 5 }{(x \ + \ 1)({x}^{2} \ + \ 3 )}} }\) in the form:
\(\hspace{2em} {\large \frac{A}{( x \ + \ 1)} + \frac{Bx \ + \ C}{ ( {x}^{2} \ + \ 3) } } \)

\({\small 8. \enspace}\) Express in partial fractions \({\small {\large \frac{{x}^{4}}{ {x}^{4} \ – \ 1 }} }\)

\({\small 9. \enspace}\) Express in partial fractions \({\small {\large \frac{{x}^{3} \ + \ x \ – \ 1}{ {x}^{2} \ + \ {x}^{4} }} }\)

\(\\[10pt]{\small 10.\enspace}\) Express in partial fractions
\(\hspace{2em}{\small {\large \frac{x \ + \ 5}{ {x}^{3} \ + \ 5{x}^{2} \ + \ 7x \ + \ 3 }} }\)


As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .

Parametric equations-feature image

Parametric Equations

Parametric Equations

The traditional representation of y as a function of x or \({\small y = f(x)}\) is inadequate to represent a curve or a surface.

To be able to draw a curve or a surface, we need to separate the x and y and write them in terms of an independent variable.

Parametric equations are used to express the Cartesian coordinates (x and y) in terms of another independent variable, usually named as t.

The typical procedure in this topic is to find the gradient of the curve or \({ \large\frac{\mathrm{d}y}{\mathrm{d}x} }\). We can do this by finding each derivative of x and y with respect to t and then divide them both.

Let’s dig into some of the examples to show you what I mean. Cheers ! =) .
\(\\[1pt]\)


EXAMPLE:

\({\small 1.\enspace}\) The parametric equations of a curve are
\(\\[1pt]\)
\(\hspace{3em} x = {\large \frac{1}{{\cos}^{3}t}}, \quad y = {\tan}^{3}t\).
\(\\[1pt]\)
where \(0 \ \le \ t \ \le \ \frac{\pi}{2}\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right). \enspace }\) Show that \( {\large\frac{\mathrm{d}y}{\mathrm{d}x} } = \sin \ t\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right). \enspace }\) Hence, show that the equation of the tangent to the curve at the point with parameter t is \( y = x \sin t \ – \ \tan t\).

\(\\[1pt]\)
\({\small 2.\enspace}\) The parametric equations of a curve are
\(\\[1pt]\)
\(\hspace{3em} x = e^{-t} \cos t, \quad y = e^{-t} \sin t\).
\(\\[1pt]\)
Show that \( {\large\frac{\mathrm{d}y}{\mathrm{d}x} } = \tan \big(t \ – \ {\large\frac{\pi}{4}}\big)\)

\(\\[1pt]\)
\({\small 3.\enspace}\) The parametric equations of a curve are
\(\\[1pt]\)
\(\hspace{3em} x = \ln (2t \ + \ 3), \quad y = { \large\frac{3t \ + \ 2}{2t \ + \ 3} }\).
\(\\[1pt]\)
Find the gradient of the curve at the point where it crosses the y-axis.

\(\\[1pt]\)
\({\small 4.\enspace}\) The parametric equations of a curve are
\(\\[1pt]\)
\( {\small x = \ 2\sin \theta \ + \ \sin 2\theta, \enspace y = \ 2\cos \theta \ + \ \cos 2\theta }\),
\(\\[1pt]\)
where \(0 \ \lt \ \theta \ \lt \ \pi\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(\textrm{i}\right). \enspace }\) Obtain an expression for \( {\small {\large\frac{\mathrm{d}y}{\mathrm{d}x} } }\) in terms of \({\small \theta }\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(\textrm{ii}\right). \enspace }\) Hence find the exact coordinates of the point on the curve at which the tangent is parallel to the y-axis.

\(\\[1pt]\)
\({\small 5.\enspace}\) The parametric equations of a curve are
\(\\[1pt]\)
\( x = \ 2 t \ + \sin 2t, \enspace y = \ 1 \ – \ 2\cos 2t \),
\(\\[1pt]\)
where \(-\frac{1}{2}\pi \ \lt \ t \ \lt \ \frac{1}{2}\pi\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(\textrm{i}\right). \enspace }\) Show that \( {\small {\large\frac{\mathrm{d}y}{\mathrm{d}x} } \ = \ 2 \tan t }\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(\textrm{ii}\right). \enspace }\) Hence find the x-coordinate of the point on the curve at which the gradient of the normal is 2. Give your answer correct to 3 significant figures.

\(\\[1pt]\)


PRACTICE MORE WITH THESE QUESTIONS BELOW!

\({\small 1.\enspace}\) The parametric equations of a curve are

\( x = \ \sin t \ + \cos t, \enspace y = \ {\sin}^{3}t \ + \ {\cos}^{3}t\),

where \(\frac{\pi}{4} \ \le \ t \ \le \ \frac{5\pi}{4}\)

\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Show that \({ \large\frac{\mathrm{d}y}{\mathrm{d}x}} = \ -3 \ \sin t \ \cos t \).
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Find the gradient of the curve at the origin.
\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}\) Find the values of t for which the gradient of the curve is 1, giving your answers correct to 2 significant figures.

\({\small 2. \enspace}\) The parametric equations of a curve are

\( x = \ a(2\theta \ – \ \sin 2\theta), \enspace y = \ a(1 \ – \ \cos 2\theta)\).

Show that \({ \large\frac{\mathrm{d}y}{\mathrm{d}x}} = \ \cot \theta \).

\({\small 3.\enspace}\) The parametric equations of a curve are

\( x = \ \ln \cos \theta, \enspace y = \ 3\theta \ – \ \tan \theta\),

where \(0 \ \le \ \theta \ \le \ \frac{1}{2}\pi\).

\({\small\hspace{1.2em}\left(\textrm{i}\right).\hspace{0.8em}}\) Express \({\small { \large\frac{\mathrm{d}y}{\mathrm{d}x}} }\) in terms of \({\small \tan \theta}\).
\({\small\hspace{1.2em}\left(\textrm{ii}\right).\hspace{0.6em}}\) Find the exact y-coordinate of the point on the curve at which the gradient of the normal is equal to 1.

\({\small 4.\enspace}\) The parametric equations of a curve are

\( x = \ {t}^{2} \ + 1, \enspace y = \ 4t \ + \ \ln \ (2t \ – \ 1)\).

\({\small\hspace{1.2em}\left(\textrm{i}\right).\hspace{0.8em}}\) Express \({\small { \large\frac{\mathrm{d}y}{\mathrm{d}x}} }\) in terms of \({\small t}\).
\({\small\hspace{1.2em}\left(\textrm{ii}\right).\hspace{0.6em}}\) Find the equation of the normal to the curve at the point where \({\small t \ = \ 1}\). Give your answer in the form ax + by + cz = 0.

\({\small 5.\enspace}\) The parametric equations of a curve are

\( x = \ t \ + \cos t, \enspace y = \ \ln \ (1 + \sin t)\),

where \(-\frac{1}{2}\pi \ \lt \ t \ \lt \ \frac{1}{2}\pi\).

\({\small\hspace{1.2em}\left(\textrm{i}\right).\hspace{0.8em}}\) Show that \({\small { \large\frac{\mathrm{d}y}{\mathrm{d}x}} \ = \ \sec t}\).
\({\small\hspace{1.2em}\left(\textrm{ii}\right).\hspace{0.6em}}\) Hence find the x-coordinates of the points on the curve at which the gradient is equal to 3. Give your answer correct to 3 significant figures.

\({\small 6.\enspace}\) A curve has parametric equations

\( x = \ {t}^{2} \ + \ 3t \ + \ 1, \enspace y = \ {t}^{4} \ + \ 1\).

The point P on the curve has parameter p. It is given that the gradient of the curve at P is 4. Show that \({\small p \ = \ \sqrt[3]{(2p \ + \ 3)} }\).

\({\small 7.\enspace}\) A curve has parametric equations

\( x = \ 1 \ + \ 2 \ \sin \theta \ \) and \( \ y = \ 4 \ + \ \sqrt{3} \ \cos \theta\).

\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the equations of the tangent and normal at the point P where \({\small \theta \ = \ {\large\frac{\pi}{6}} }\). Hence, find the area A of the triangle bounded by the tangent and normal at P, and the y-axis.
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Determine the rate of change of xy at \({\small \theta \ = \ {\large\frac{\pi}{6}} }\) if x increases at a constant rate of 0.1 units/s.

\({\small 8.\enspace}\) A curve is defined parametrically by \( x \ = \ \frac{2t}{t + 1} \) and \( y \ = \ \frac{{t}^{2}}{t + 1} \).

\({\small\hspace{1.2em}\left(\textrm{i}\right).\hspace{0.8em}}\) Find the equation of the normal to the curve at the point P(1, 1/2).
\({\small\hspace{1.2em}\left(\textrm{ii}\right).\hspace{0.6em}}\) The normal at P meets the curve again at Q. Find the exact coordinates of Q.

\({\small 9.\enspace}\) A curve has parametric equations:

\( x = \ \sec (\frac{\theta}{2}), \enspace y = \ \ln \ \sec (\frac{\theta}{2})\),

where \(-\pi \ \lt \ \theta \ \lt \ \pi \).

\({\small\hspace{1.2em}\left(\textrm{i}\right).\hspace{0.8em}}\) Find the equation of the normal to the curve at the point at \({\small \theta \ = \ {\large\frac{\pi}{3}} }\).
\({\small\hspace{1.2em}\left(\textrm{ii}\right).\hspace{0.6em}}\) Determine the rate of change of x if the gradient of the curve at \({\small \theta \ = \ {\large\frac{\pi}{2}} }\) is decreasing at a rate of 0.4 units per second.

\({\small 10.\enspace}\) The parametric equations of a curve are

\( x = \ t \ + \ \ln t, \enspace y = \ t \ + \ {\textrm{e}}^{t}\) for \( t \gt 0\).

\({\small\hspace{1.2em}\left(\textrm{i}\right).\hspace{0.8em}}\) Sketch the curve, indicating clearly all intercepts and asymptotes.
\({\small\hspace{1.2em}\left(\textrm{ii}\right).\hspace{0.6em}}\) Show that, for all the points on the curve, \({\small { \large\frac{\mathrm{d}y}{\mathrm{d}x}} \ = \ { \large\frac{t(1 \ + \ {\textrm{e}}^{t})}{t \ + \ 1}} }\).
Hence, deduce that the curve does not have any turning points.
\({\small\hspace{1.2em}\left(\textrm{iii}\right).\hspace{0.4em}}\) Find, in exact form, the equation of the normal of the curve at the point where \({\small t \ = \ 1}\).


As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .

9709/32/F/M/19 – Paper 32 Feb March 2019 No 10

Integration and Differentiation

Integration and Differentiation

Integration is an essential part of basic calculus. Algebra plays a very important part to become proficient in this topic.

I have compiled some of the questions that I have encountered during my Math tutoring classes. Do take your time to try the questions and learn from the solutions I have provided below. Cheers ! =) .

More Integration Exercises can be found here.


EXAMPLE:

\({\small 1.\enspace}\) 9709/32/F/M/17 – Paper 32 Feb March 2017 No 10
\(\\[1pt]\)
9709/32/F/M/17 – Paper 32 Feb March 2017 No 10
\(\\[1pt]\)
The diagram shows the curve \( \ {\small y \ = \ {(\ln x)}^{2} }\). The x-coordinate of the point P is equal to e, and the normal to the curve at P meets the x-axis at Q.
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{i}).\hspace{0.7em}}\) Find the x-coordinate of Q.
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{ii}).\hspace{0.7em}}\) Show that \({\small \displaystyle \int \ln x \ \mathrm{d}x \ = \ x \ln x \ – \ x \ + \ c }\), where c is a constant.
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{iii}).\hspace{0.5em}}\) Using integration by parts, or otherwise, find the exact value of the area of the shaded region between the curve, the x-axis and the normal PQ.

\(\\[1pt]\)
\({\small 2.\enspace}\) 9709/32/F/M/19 – Paper 32 Feb March 2019 No 10
\(\\[1pt]\)
9709/32/F/M/19 – Paper 32 Feb March 2019 No 10
\(\\[1pt]\)
The diagram shows the curve \( \ {\small y \ = \ {\sin}^{3} x \sqrt{(\cos x)} \ }\) for \( \ {\small 0 \leq x \leq \large{ \frac{1}{2}} \pi } \), and its maximum point M.
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{i}).\hspace{0.7em}}\) Using the substitution \({\small \ u \ = \ \cos x }\), find by integration the exact area of the shaded region bounded by the curve and the x-axis.
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{ii}).\hspace{0.7em}}\) Showing all your working, find the x-coordinate of M, giving your answer correct to 3 decimal places.

\(\\[1pt]\)
\({\small 3.\enspace}\) 9709/32/M/J/20 – Paper 32 May June 2020 No 9
\(\\[1pt]\)
9709/32/M/J/20 – Paper 32 May June 2020 No 9
\(\\[1pt]\)
The diagram shows the curves \( \ {\small \ y \ = \ \cos x \ }\) and \( \ {\small \ y \ = \ \large{ \frac{k}{1 \ + \ x} } }\), where k is a constant, for \( \ {\small \ 0 \leq x \leq \large{ \frac{1}{2}} \pi } \). The curves touch at the point where x = p.
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{a}).\hspace{0.8em}}\) Show that p satisfies the equation \( {\small \ \tan p \ = \ \large{ \frac{1}{1 \ + \ p} } }\).

\(\\[1pt]\)

\({\small 4.\enspace} \displaystyle \int_{1}^{a} \ln 2x \ \mathrm{d}x = 1.\) Find \({\small a} \).


\(\\[1pt]\)
\({\small 5.\enspace}\) Use the substitution \(u = \sin 4x\) to find the exact value of \(\displaystyle \int_{0}^{{\Large\frac{\pi}{24}}} \cos^{3} 4x \ \mathrm{d}x.\)

\(\\[1pt]\)
\({\small 6. \hspace{0.8em}(i).\hspace{0.8em}}\) Use the trapezium rule with 3 intervals to estimate the value of: \(\displaystyle \int_{{\Large\frac{\pi}{9}}}^{{\Large\frac{2\pi}{3}}} \csc x \ \mathrm{d}x\) giving your answer correct to 2 decimal places.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(ii\right).\hspace{0.8em}}\) Using a sketch of the graph of \(y = \csc x\), explain whether the trapezium rule gives an overestimate or an underestimate of the true value of the integral in part (i).

\(\\[1pt]\)
\({\small 7.\enspace}\) Solve these integrations.
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}} \displaystyle \int_{0}^{\infty} \frac{1}{{x}^{2} \ + \ 4} \ \mathrm{d}x\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}} \displaystyle \int_{0}^{3} \frac{1}{\sqrt{9 \ – \ {x}^{2}}} \ \mathrm{d}x\)
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\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}} \displaystyle \int_{-\infty}^{\infty} \frac{1}{9{x}^{2} \ + \ 4} \ \mathrm{d}x\)
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\({\small\hspace{1.2em}\left(d\right).\hspace{0.8em}} \displaystyle \int_{0}^{1} \frac{1}{\sqrt{x(1 \ – \ x)}} \ \mathrm{d}x\)
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\({\small\hspace{1.2em}\left(e\right).\hspace{0.8em}} \displaystyle \int_{1}^{\infty} \frac{1}{{(1 \ + \ x^2)}^{{\large\frac{3}{2}}}} \ \mathrm{d}x\)
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\({\small\hspace{1.2em}\left(f\right).\hspace{0.8em}} \displaystyle \int_{1}^{\infty} \frac{1}{x \sqrt{{x}^{2} \ – \ 1}} \ \mathrm{d}x\)

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\({\small 8.\enspace}\) The diagram shows the curve \({\small y = {e}^{{\large – \frac{1}{2}x}} \ \sqrt{(1 \ + \ 2x)}}\) and its maximum point M. The shaded region between the curve and the axes is denoted by R.
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Integration Example 5
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\({\small \hspace{1.2em}(i). \enspace }\) Find the x-coordinate of M.
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\({\small \hspace{1.2em}(ii). \enspace }\) Find by integration the volume of the solid obtained when R is rotated completely about the x-axis. Give your answer in terms of \({\small \pi}\) and e.

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PRACTICE MORE WITH THESE QUESTIONS BELOW!

\({\small 1.\enspace}\) Find \(\displaystyle \int \frac{1}{x^2\sqrt{x^2 \ – \ 4}} \ \mathrm{d}x\) using the substitution \({\small x \ = \ 2 \sec \theta }\).

\({\small 2. \enspace}\) Find the exact value of \(\displaystyle \int_{1}^{e} x^4 \ \ln \ x \ \mathrm{d}x \).

\({\small 3. \enspace}\) Find the exact value of \(\displaystyle \int_{4}^{10} \frac{2x \ + \ 1}{(x \ – \ 3)^2} \ \mathrm{d}x \), giving your answer in the form of \({\small a \ + \ b \ \ln \ c}\), where a, b and c are integers.

\({\small 4. \enspace}\) Find the exact value of \(\displaystyle \int_{1}^{4} \frac{\ln \ x}{\sqrt{x}} \ \mathrm{d}x \).

\({\small 5. \enspace}\) Find the exact value of

\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}} \displaystyle \int_{0}^{\infty} {e}^{1 \ – \ 2x} \ \mathrm{d}x\)

\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}} \displaystyle \int_{-1}^{0} \big(
2 \ + \ \frac{1}{x \ – \ 1} \big) \ \mathrm{d}x\)

\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}} \displaystyle \int_{{\large\frac{\pi}{6}}}^{{\large \frac{\pi}{4}}} \cot x \ \mathrm{d}x\)

\({\small\hspace{1.2em}\left(d\right).\hspace{0.8em}}\) Using your result in (c), find also the exact value of \(\displaystyle \int_{{\large\frac{\pi}{6}}}^{{\large \frac{\pi}{4}}} \csc 2x \ \mathrm{d}x\) by using the identity \(\cot x \ – \ \cot 2x \ \equiv \ \csc 2x\).

\({\small 6. \enspace}\) The diagram shows the part of the curve \({\small y \ = \ f(x)}\), where \({\small f(x) \ = \ p \ – \ {e}^{x} }\) and p is a constant. The curve crosses the y-axis at (0, 2).

Integration Practice 6

\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the value of p.

\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Find the coordinates of the point where the curve crosses the x-axis.

\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}\) What is the area of the shaded region R?

\({\small 7. \enspace}\) Integrate the following:

\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}} \displaystyle \int \frac{x^2}{1 \ + \ {x}^{3}} \ \mathrm{d}x\)

\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}} \displaystyle \int x^4 \ \sin (x^5 \ + \ 2) \ \mathrm{d}x\)

\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}} \displaystyle \int e^{x} \ \sin x \ \mathrm{d}x\)

\({\small 8. \enspace}\) Let \(I \ = \ \displaystyle \int_{0}^{1} {\large \frac{\sqrt{x}}{2 \ – \ \sqrt{x}}} \ \mathrm{d}x\).

\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Using the substitution \({\small u = \ 2 \ – \ \sqrt{x}}\), show that \(I \ = \ \displaystyle \int_{1}^{2} {\large \frac{2 {(2 \ – \ u)}^{2}}{u}} \ \mathrm{d}u\).

\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Hence show that \(I \ = \ 8 \ \ln 2 \ – \ 5 \).

\({\small 9. \enspace}\) The constant a is such that

\({\small\hspace{3em}} \displaystyle \int_{0}^{a} x{e}^{{\large \frac{1}{2}x}} \mathrm{d}x \ = \ 6 \).

Show that a satisfies the equation

\({\small\hspace{3em}} a \ = \ 2 \ + \ {e}^{{\large -\frac{1}{2}a}}\).

\({\small 10. \enspace}\) Use the substitution \({\small u \ = \ 1 \ + \ 3 \ \tan x }\) to find the exact value of

\({\small\hspace{3em}} \ \displaystyle \int_{0}^{{\large\frac{\pi}{4}}} {\large \frac{\sqrt{1 \ + \ 3 \ \tan x}}{{\cos}^{2}x}} \ \mathrm{d}x\).


As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .

Differential Equations-Practice 5

Differential Equations

Differential Equations

In this topic, we are looking into transforming some of the real-life physical quantities into mathematical models. The goal is to understand “how fast” a certain quantity change with respect to time.

The typical procedure is to create a mathematical model relating the phenomenon in question. By separating the variables and then perform the integration, we will arrive at the general solution to the problem.

Given a set of initial conditions, we can then create a specific equation relating to the problem, namely the particular solution.

I have put together some of the questions I received in the comment section below. You can try these questions also to further your understanding on this topic.

To check your answer, you can look through the solutions that I have posted either in Youtube videos or Instagram posts.

You can subscribe, like or follow my youtube channel and IG account. I will keep updating my IG daily post, preferably.

Furthermore, you can find some examples and more practices below! =).

Try some of the examples below. You can look at the solution I have written below to study and understand the topic. Cheers ! =) .
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EXAMPLE:

\({\small 1.\enspace}\) 9709/03/SP/20 – Specimen Paper 2020 Pure Maths 3 No 10
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In a chemical reaction, a compound X is formed from two compounds Y and Z.
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The masses in grams of X, Y and Z present at time t seconds after the start of the reaction are x, 10 – x, 20 – x respectively. At any time the rate of formation of X is proportional to the product of the masses of Y and Z present at the time. When t = 0, x = 0 and \({\small \ {\large\frac{\textrm{d}x}{\textrm{d}t}} \ = \ 2}\).
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\({\small\hspace{1.2em}\left(\textrm{a}\right).\hspace{0.8em}}\) Show that x and t satisfy the differential equation
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\(\hspace{3em} {\large \frac{\mathrm{d}x}{\mathrm{d}t }} \ = \ 0.01(10 \ – \ x)(20 \ – \ x) \) .
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\({\small\hspace{1.2em}\left(\textrm{b}\right).\hspace{0.8em}}\) Solve this differential equation and obtain an expression for x in terms of t.

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\({\small 2.\enspace}\) 9709/32/M/J/16 – Paper 32 May June 2016 No 6
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The variables \({\small x}\) and \({\small \theta}\) satisfy the differential equation
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\(\hspace{3em} (3 \ + \ \cos2\theta) \ {\large \frac{\mathrm{d}x}{\mathrm{d} \theta }} \ = \ x \sin 2\theta \),
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and its given that \({\small x \ = \ 3}\) when \({\small \theta \ = \ {\large \frac{1}{4}} \pi}\).
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\({\small\hspace{1.2em}(\textrm{i}).\hspace{0.7em}}\) Solve the differential equation and obtain an expression for \({\small x}\) in terms of \({\small \theta}\).
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\({\small\hspace{1.2em}(\textrm{ii}).\hspace{0.7em}}\) State the least value taken by \({\small x }\).

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\({\small 3.\enspace}\) 9709/03/SP/17 – Specimen Paper 03 2017 No 8
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The variables \({\small x}\) and \({\small \theta}\) satisfy the differential equation
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\(\hspace{3em} {\large \frac{\mathrm{d}x}{\mathrm{d} \theta }} \ = \ (x \ + \ 2) \ {\sin}^{2} 2\theta \),
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and its given that \({\small x \ = \ 0}\) when \({\small \theta \ = \ 0}\). Solve the differential equation and calculate the value of \({\small x}\) when \({\small \ \theta \ = \ {\large \frac{1}{4}} \pi }\), giving your answer correct to 3 significant figures.

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\({\small 4.\enspace}\) Compressed air is escaping from a container. The pressure of the air in the container at time t is P, and the constant atmospheric pressure of the air outside the container is A. The rate of decrease of P is proportional to the square root of the pressure difference (PA). Thus the differential equation connecting P and t is
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\(\hspace{3em} {\large\frac{\textrm{d}P}{\textrm{d}t}} = -k \ \sqrt{P \ – \ A}\),
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\(\\[7pt]\) where k is a positive constant.
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the general solution of this differential equation.
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Given that \({\small P \ = \ 5A}\) when \({\small t \ = \ 0}\) and that \({\small P \ = \ 2A}\) when \({\small t \ = \ 2}\), show that \({\small k \ = \ \sqrt{A}}\).
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\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}\) Find the value of t when \({\small P \ = \ A}\).
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\({\small\hspace{1.2em}\left(d\right).\hspace{0.8em}}\) Obtain an expression for P in terms of A and t.

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\({\small 5.\enspace}\) The temperature of a quantity of liquid at time t is \({\small \theta}\). The liquid is cooling an atmosphere whose temperature is constant and equal to A. The rate of decrease of \({\small \theta}\) is proportional to the temperature difference \({\small (\theta \ – \ A)}\). Thus \({\small \theta}\) and t satisfy the differential equation
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\(\hspace{3em} {\large\frac{\textrm{d}\theta}{\textrm{d}t}} = -k \ (\theta \ – \ A) \),
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\(\\[7pt]\) where k is a positive constant.
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the solution of this differential equation, given that \({\small \ \theta \ = \ 4A \ }\) when \({\small t \ = \ 0}\).
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Given also that \({\small \theta \ = \ 3A}\) when \({\small t \ = \ 1}\), show that \({\small k \ = \ \ln{\large\frac{3}{2}}}\).
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\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}\) Find \({\small \theta}\) in terms of A when \({\small t \ = \ 2}\), expressing your answer in its simplest form.

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\({\small 6.\enspace}\) The variables x and y are related by the differential equation
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\(\hspace{3em} {\large\frac{\textrm{d}y}{\textrm{d}x}} = {\large\frac{6x \ {\textrm{e}}^{3x}}{{y}^{2}}} \ .\)
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It is given that \({\small y \ = \ 2}\) when \({\small x \ = \ 0}\). Solve the differential equation and hence find the value of y when \({\small x \ = \ 0.5}\), giving your answer correct to 2 decimal places.

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\({\small 7.\enspace}\) In the diagram the tangent to a curve at a general point P with coordinates (x, y) meets the x-axis at T. The point N on the x-axis such that PN is perpendicular to the x-axis. The curve is such that, for all values of x in the interval \({\small 0 \lt x \lt {\large \frac{1}{2}} \pi}\), the area of triangle PTN is equal to \({\small \tan x}\), where x is in radians.
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Differential Equations Exercise 4
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\({\small\hspace{1.2em}\left(i\right).\hspace{0.8em}}\) Using the fact that the gradient of the curve at P is \({\small {\large \frac{PN}{TN}}}\), show that
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\(\hspace{3em} {\large\frac{\textrm{d}y}{\textrm{d}x}} = {\large \frac{1}{2}} {y}^{2} \cot x \ .\)
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\({\small\hspace{1.2em}\left(ii\right).\hspace{0.8em}}\) Given that \({\small y = 2}\) when \({\small x = {\large \frac{1}{6} } \pi}\), solve this differential equation to find the equation of the curve, expressing y in terms of x.

\({\small 8.\enspace}\) The variables x and \({\small \theta}\) satisfy the differential equation
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\(\hspace{3em} x \tan \theta {\large\frac{\textrm{d}x}{\textrm{d}\theta}} \ + \ \textrm{cosec}^{2} \theta = 0,\)
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for \({\small 0 \lt \theta \lt {\large \frac{1}{2}} \pi} \ \) and \( \ {\small x \gt 0}\). It is given that \({\small x \ = \ 4}\) when \({\small \theta \ = \ {\large \frac{1}{6}} \pi}\). Solve the differential equation, obtaining an expression for \({\small x }\) in terms of \({\small \theta }\).

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\({\small 9.\enspace}\) The variables x and y satisfy differential equation
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\(\hspace{3em} {\large\frac{\textrm{d}y}{\textrm{d}x}} = x \ \textrm{e}^{x+y}\).
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It is given that \({\small y \ = \ 0}\) when \({\small x \ = \ 0}\).
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\({\small\hspace{1.2em}\textrm{i}.\hspace{0.8em}}\) Solve the differential equation, obtaining \({\small y }\) in terms of \({\small x }\).
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\({\small\hspace{1.2em}\textrm{ii}.\hspace{0.8em}}\) Explain why \({\small x }\) can only take values that are less than 1.

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PRACTICE MORE WITH THESE QUESTIONS BELOW!

\({\small 1.\enspace (\textrm{i}) \enspace}\) The number of bacteria in a colony is increasing at a rate that is proportional to the square root of the number of bacteria present. Form a differential equation relating number of bacteria, x, to the time t.

\({\small \quad (\textrm{ii}) \enspace}\) In another colony the number of bacteria, y, after time t minutes is modelled by the differential equation \({\large \frac{\mathrm{d}y}{\mathrm{d}t} \ = \ \frac{10000}{\sqrt{y}}} \). Find y in terms of t, given that \({\small y = 900 }\) when \({\small t = 0 }\). Hence, find the number of bacteria after 10 minutes.

\({\small 2. \enspace}\) A skydiver drops from a helicopter. Before she opens her parachute, her speed v m/s after time t seconds is modelled by the differential equation

\(\hspace{3em}{\large \frac{\mathrm{d}v}{\mathrm{d}t}} \ = \ 10 {\large {\textrm{e}}^{-\frac{1}{2}t}}\)

when \({\small t = 0 }\), \({\small v = 0 }\).

\({\small \quad (\textrm{i}) \hspace{0.7em}}\) Find v in terms of t.

\({\small \quad (\textrm{ii}) \enspace}\) According to this model, what is the speed of the skydiver in the long term?

She opens her parachute when her speed is 10 m/s. Her speed t seconds after this is w m/s and is modelled by the differential equation \( {\large\frac{\mathrm{d}w}{\mathrm{d}t}} \ = \ -\frac{1}{2}(w \ – \ 4)(w \ + \ 5) \).

\({\small \quad (\textrm{iii}) \hspace{0.3em}}\) Express \({\large\frac{1}{(w \ – \ 4)(w \ + \ 5)}} \) in partial fractions.

\({\small \quad (\textrm{iv}) \enspace}\) Using this result show that

\(\hspace{3em}{\large \frac{w \ – \ 4}{w \ + \ 5}} \ = \ 0.4 {\large {\textrm{e}}^{-4.5t}} \).

\({\small \quad (\textrm{v}) \hspace{0.7em}}\) According to this model, what is the speed of the skydiver in the long term?

\({\small 3. \enspace}\) The number of organisms in a population at time t is denoted by x. Treating x as a continuous variable, the differential equation satisfied by x and t is

\(\hspace{3em}{\large \frac{\mathrm{d}x}{\mathrm{d}t}} \ = \ {\large \frac{x {\textrm{e}}^{-t}}{k \ + \ {\textrm{e}}^{-t} } }\),

where \({\small k}\) is a positive constant.

\({\small \quad (\textrm{i}) \hspace{0.7em}}\) Given that \({\small x = 10 }\) when \({\small t = 0 }\), solve the differential equation, obtaining a relation between x, k and t.

\({\small \quad (\textrm{ii}) \enspace}\) Given also that \({\small x = 20 }\) when \({\small t = 1 }\), show that \( k = 1 \ – \ {\large \frac{2}{\textrm{e}}}\).

\({\small \quad (\textrm{iii}) \hspace{0.3em}}\) Show that the number of organisms never reaches 48, however large t becomes.

\({\small 4. \enspace}\) In a model of the expansion of a sphere of radius r cm, it is assumed that, at time t seconds after the start, the rate of increase of the surface area of the sphere is proportional to its volume. When \({\small t = 0 }\), \({\small r = 5 }\) and \({\large \frac{\mathrm{d}r}{\mathrm{d}t}} = \ 2\).

\({\small \quad (\textrm{i}) \hspace{0.7em}}\) Show that r satisfies the differential equation

\(\hspace{3em}{\large \frac{\mathrm{d}r}{\mathrm{d}t}} \ = \ 0.08{r}^{2}\).

\({\small \quad (\textrm{ii}) \enspace}\) Solve this differential equation, obtaining an expression for r in terms of t.

\({\small \quad (\textrm{iii}) \hspace{0.3em}}\) Deduce from your answer to part (ii) the set of values that t can take, according to this model.

\({\small 5. \enspace}\) An underground storage tank is being filled with liquid as shown in the diagram. Initially the tank is empty. At time t hours after filling begins, the volume of liquid is V \({\small \mathrm{{m}^{3}}}\) and the depth of liquid is h m. It is given that \({\small V = \ {\large \frac{4}{3}}{h}^{3} }\).

Differential Equations-Practice 5

The liquid is poured in at a rate of 20 \({\small \mathrm{{m}^{3}}}\) per hour, but owing to leakage, liquid is lost at a rate proportional to \({\small {h}^{2} }\). When \({\small h = 1 }\), \({\large \frac{\mathrm{d}h}{\mathrm{d}t}} = \ 4.95\).

\({\small \quad (\textrm{i}) \hspace{0.7em}}\) Show that h satisfies the differential equation

\(\hspace{3em}{\large \frac{\mathrm{d}h}{\mathrm{d}t}} \ = \ {\large \frac{5}{{h}^{2}} } \ – \ {\large \frac{1}{20} }\).

\({\small \quad (\textrm{ii}) \enspace}\) Verify that

\(\hspace{3em} {\large \frac{20 {h}^{2}}{100 \ – \ {h}^{2}} } \ \equiv \ -20 \ + \ {\large \frac{2000}{(10 \ – \ h)(10 \ + \ h)} }\).

\({\small \quad (\textrm{iii}) \hspace{0.3em}}\) Hence, solve the differential equation in part (i), obtaining an expression for t in terms of h.

\({\small 6. \enspace}\) The variables x and y are related by the differential equation

\(\hspace{3em}{\large \frac{\mathrm{d}y}{\mathrm{d}x}} \ = \ {\large \frac{6y {\textrm{e}}^{3x}}{2 \ + \ {\textrm{e}}^{3x} } }\).

Given that \({\small y = 36 }\) when \({\small x = 0 }\), find an expression for y in terms of x.

\({\small 7. \enspace}\) The variables x and y satisfy the differential equation

\(\hspace{3em} (x + 1) \ y \ {\large \frac{\mathrm{d}y}{\mathrm{d}x}} \ = \ {y}^{2} \ + \ 5 \).

It is given that \({\small y = 2 }\) when \({\small x = 0 }\). Solve the differential equation obtaining an expression for \({\small {y}^{2} }\) in terms of \({\small x }\).

\({\small 8. \enspace}\) The coordinates (x, y) of a general point on a curve satisfy the differential equation

\(\hspace{3em} x \ {\large \frac{\mathrm{d}y}{\mathrm{d}x}} \ = \ {(2 \ – \ x)}^{2} \ y \).

The curve passes through the point (1, 1). Find the equation of the curve, obtaining an expression for \({\small y }\) in terms of \({\small x }\).

\({\small 9. \enspace}\) The variables x and y satisfy the differential equation

\(\hspace{3em} x \ {\large \frac{\mathrm{d}y}{\mathrm{d}x}} \ = \ {(4 \ – \ y)}^{2} \),

and \({\small y = 1 }\) when \({\small x = 1 }\). Solve the differential equation, obtaining an expression for \( {\small {y}^{2} }\) in terms of \({\small x }\).

\({\small 10. \enspace}\) The variables \({\small x}\) and \({\small \theta}\) satisfy the differential equation

\(\hspace{3em} x \ {\cos}^{2} \theta \ {\large \frac{\mathrm{d}x}{\mathrm{d} \theta }} \ = \ 2 \tan \theta \ + \ 1 \),

for \({\small 0 \leq \theta \lt {\large \frac{1}{2}} \pi} \ \) and \( \ {\small x \gt 0}\). It is given that \({\small x \ = \ 1}\) when \({\small \theta \ = \ {\large \frac{1}{4}} \pi}\).

\({\small \quad (\textrm{i}) \hspace{0.7em}}\) Show that

\(\hspace{3em} {\large \frac{\mathrm{d}}{\mathrm{d} \theta }}({\tan}^{2} \theta) \ = \ {\large \frac{2 \tan \theta}{{\cos}^{2} \theta} } \).

\({\small \quad (\textrm{ii}) \enspace}\) Solve the differential equation and calculate the value of \({\small x }\) when \({\small \theta \ = \ {\large \frac{1}{3}} \pi}\), giving your answer correct to 3 significant figures.


As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .