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 Pure Maths 3 No 10
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9709/32/F/M/17 – Paper 32 Feb March 2017 No 10
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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.
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\({\small\hspace{1.2em}(\textrm{i}).\hspace{0.7em}}\) Find the x-coordinate of Q.
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\({\small\hspace{1.2em}(\textrm{ii}).\hspace{0.7em}}\) Show that \({\small \displaystyle \int \ln x \ \mathrm{d}x \ = \ x \ln x \ – \ x \ + \ c }\), where c is a constant.
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\({\small\hspace{1.2em}(\textrm{iii}).\hspace{0.5em}}\) 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.

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\({\small 2.\enspace}\) 9709/32/F/M/19 – Paper 32 Feb March 2019 Pure Maths 3 No 10
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9709/32/F/M/19 – Paper 32 Feb March 2019 No 10
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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.
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\({\small\hspace{1.2em}(\textrm{i}).\hspace{0.7em}}\) 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.
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\({\small\hspace{1.2em}(\textrm{ii}).\hspace{0.7em}}\) Showing all your working, find the x-coordinate of M, giving your answer correct to 3 decimal places.

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\({\small 3.\enspace}\) 9709/32/M/J/20 – Paper 32 May June 2020 Pure Maths 3 No 9
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9709/32/M/J/20 – Paper 32 May June 2020 No 9
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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.
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\({\small\hspace{1.2em}(\textrm{a}).\hspace{0.8em}}\) Show that p satisfies the equation \( {\small \ \tan p \ = \ \large{ \frac{1}{1 \ + \ p} } }\).

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

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

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\({\small 6. \hspace{0.8em}(i).\hspace{0.8em}}\) 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.
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\({\small\hspace{1.2em}\left(ii\right).\hspace{0.8em}}\) 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).

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\({\small 7.\enspace}\) Solve these integrations.
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}} \displaystyle \int_{0}^{\infty} \frac{1}{{x}^{2} \ + \ 4} \ \mathrm{d}x\)
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}} \displaystyle \int_{0}^{3} \frac{1}{\sqrt{9 \ – \ {x}^{2}}} \ \mathrm{d}x\)
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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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\({\small 9.\enspace}\) 9709/33/M/J/20 – Paper 33 June 2020 Pure Maths 3 No 2
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Find the exact value of \(\displaystyle \int_{0}^{1} (2 \ – \ x) \mathrm{e}^{-2x} \ \mathrm{d}x \).

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\({\small 10.\enspace}\) 9709/33/M/J/20 – Paper 33 June 2020 Pure Maths 3 No 7(a), (b), (c)
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Let \({\small f(x) \ = \ {\large \frac{ 2 }{(2x \ – \ 1)( 2x \ + \ 1 ) }} }\)
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Express \({\small f(x) }\) in partial fractions.
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Using your answer to part (a), show that
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\({\scriptsize {\Big( f(x) \Big)}^{2} \ = \ {\large \frac{ 1 }{ {(2x \ – \ 1)}^{2} }} \ – \ {\large \frac{ 1 }{ (2x \ – \ 1) }} }\)
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\({\hspace{3em} \scriptsize \ + \ {\large \frac{ 1 }{ (2x \ + \ 1)}} \ + \ {\large \frac{ 1 }{ {(2x \ + \ 1)}^{2} }} . }\)
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\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}\) Hence show that \(\displaystyle \int_{1}^{2} {\Big( f(x) \Big)}^{2} \ \mathrm{d}x = \frac{2}{5} + \frac{1}{2} \ln \frac{5}{9}\).

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\({\small 11.\enspace}\) 9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 3
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Find the exact value of \(\displaystyle \int_{1}^{4} x^{\frac{3}{2}} \ln x \ \mathrm{d}x \).

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\({\small 12.\enspace}\) 9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 4
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A curve has equation \( y = \cos x \sin 2x \).
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Find the \(x\)-coordinate of the stationary point in the interval \(0 \lt x \lt \frac{1}{2}\pi\), giving your answer correct to 3 significant figures.

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\({\small 13.\enspace}\) 9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 6(a), (b)
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 Question no 6 Paper 32 June 2020
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The diagram shows the curve \(y = {\large\frac{x}{1+{3x}^{4}} } \), for \(x \geq 0\), and its maximum point \(M\).
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the \(x\)-coordinate of \(M\), giving your answer correct to 3 decimal places.
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Using the substitution \(u = \sqrt{3}{x}^{2}\), find by integration the exact area of the shaded region bounded by the curve, the \(x\)-axis and the line \(x = 1\).

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\({\small 14.\enspace}\) 9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 9(a)
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Question 9 (a) Paper 32 June 2020
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The diagram shows the curves \( y = \cos x \) and \( y = \frac{k}{1 \ + \ x} \), where \(k\) is a constant for \(0 \leq x \leq \frac{1}{2\pi}\). The curves touch at the point where \(x = p\).
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Show that \(p\) satisfies the equation \( \tan p = \frac{1}{1+p} \).

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\({\small 15.\enspace}\) 9709/31/M/J/20 – Paper 31 June 2020 Pure Maths 3 No 4(a), (b)
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The curve with equation \(y = { \mathrm{e} }^{2x} (\sin x + 3 \cos x) \) has a stationary point in the interval \(0 \leq x \leq \pi \).
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the \(x\)-coordinate of this point, giving your answer correct to 2 decimal places.
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Determine whether the stationary point is a maximum or a minimum.

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\({\small 16.\enspace}\) 9709/31/M/J/20 – Paper 31 June 2020 Pure Maths 3 No 5(a), (b)
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the quotient and remainder when \(2x^3 − x^2 + 6x + 3\) is divided by \(x^2 + 3\).
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Using your answer to part (a), find the exact value of \(\displaystyle \int_{1}^{3} \frac{2x^3 \ – \ x^2 \ + \ 6x \ + \ 3}{x^2 \ + \ 3} \mathrm{d}x \).

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\({\small 17.\enspace}\) 9709/31/M/J/20 – Paper 31 June 2020 Pure Maths 3 No 7(a), (b)
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Let \( f(x) = { \large \frac{\cos x}{1 \ + \ \sin x} } \).
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Show that \(f'(x) \lt 0 \) for all \(x\) in the interval \(-\frac{1}{2} \pi \lt x \lt \frac{3}{2} \pi\).
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Find \(\displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{2} } f(x) \ \mathrm{d}x \). Give your answer in a simplified exact form.

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\({\small 18.\enspace}\) 9709/32/F/M/21 – Paper 32 March 2021 Pure Maths 3 No 6(a), (b)
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Let \({\small f(x) \ = \ {\large \frac{ 5a }{(2x \ – \ a)( 3a \ – \ x ) }} }\)
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Express \({\small f(x) }\) in partial fractions.
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Hence show that \(\displaystyle \int_{a}^{ 2a } f(x) \ \mathrm{d}x = \ln 6\).

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\({\small 19.\enspace}\) 9709/32/F/M/21 – Paper 32 March 2021 Pure Maths 3 No 9(c)
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Let \({\small f(x) \ = \ {\large \frac{ { \mathrm{e} }^{2x} \ + \ 1 }{{ \mathrm{e} }^{2x} \ – \ 1 }} }\), for \(x \gt 0\).
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Find \(f'(x)\). Hence find the exact value of x for which \( f'(x) = -8 \).

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\({\small 20.\enspace}\) 9709/32/F/M/21 – Paper 32 March 2021 Pure Maths 3 No 10(a), (b)
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Question 10 Paper 32 March 2021
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The diagram shows the curve \( y = \sin 2x \ {\cos}^{2} x \) for \(0 \leq x \leq \frac{1}{2} \pi \), and its maximum point \(M\).
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Using the substitution \(u = \sin x\), find the exact area of the region bounded by the curve and the \(x\)-axis.
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Find the exact \(x\)-coordinate of \(M\).

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\({\small 21.\enspace}\) 9709/33/M/J/21 – Paper 33 June 2021 Pure Maths 3 No 4(a),(b)
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Let \({\small f(x) \ = \ {\large \frac{ 15 \ – \ 6x }{(1 \ + \ 2x)( 4 \ – \ x ) }} }\)
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Express \({\small f(x) }\) in partial fractions.
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Hence find \(\displaystyle \int_{1}^{ 2 } f(x) \ \mathrm{d}x \), giving your answer in the form \( \ln ( \frac{a}{b} ) \), where \(a\) and \(b\) are integers.

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\({\small 22.\enspace}\) 9709/33/M/J/21 – Paper 33 June 2021 Pure Maths 3 No 8(a),(b)
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Question 8 Paper 33 June 2021
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The diagram shows the curve \( y = {\large \frac{ \ln x }{ x^4 } } \) and its maximum point \(M\).
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the exact \(x\)-coordinate of \(M\).
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) By using integration by parts, show that for all \( a \gt 1, \displaystyle \int_{1}^{ a } \frac{ \ln x }{ x^4 } \ \mathrm{d}x \lt \frac{1}{9} \).

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\({\small 23.\enspace}\) 9709/32/M/J/21 – Paper 32 June 2021 Pure Maths 3 No 4
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Using integration by parts, find the exact value of \( \displaystyle \int_{0}^{ 2 } {\tan}^{-1} \big( \frac{1}{2} x \big) \ \mathrm{d}x \).

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\({\small 24.\enspace}\) 9709/32/M/J/21 – Paper 32 June 2021 Pure Maths 3 No 6(a), (b)
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Prove that \( \mathrm{cosec} 2\theta − \cot 2\theta \equiv tan \theta \).
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Hence show that \( \displaystyle \int_{\frac{1}{4}\pi}^{ \frac{1}{3}\pi } (\mathrm{cosec} 2\theta − \cot 2\theta) \ \mathrm{d}\theta = \frac{1}{2} \ln 2 \).

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\({\small 25.\enspace}\) 9709/32/M/J/21 – Paper 32 June 2021 Pure Maths 3 No 8
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The equation of a curve is \( y = { \mathrm{e} }^{-5x} \ {\tan}^{2} x \) for \( -\frac{1}{2}\pi \lt x \lt \frac{1}{2}\pi \).
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Find the \(x\)-coordinates of the stationary points of the curve. Give your answers correct to 3 decimal places where appropriate.

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\({\small 26.\enspace}\) 9709/31/M/J/21 – Paper 31 June 2021 Pure Maths 3 No 7(a)
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Question 7 Paper 31 June 2021
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The diagram shows the curve \( y = {\large \frac{ {\tan}^{-1} x }{ \sqrt{x} } } \) and its maximum point \(M\) where \(x = a\).
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Show that a satisfies the equation \( a = \tan \Big( {\large \frac{2a}{1 \ + \ a^2} }\Big) \).

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\({\small 27.\enspace}\) 9709/31/M/J/21 – Paper 31 June 2021 Pure Maths 3 No 9(a), (b)
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The equation of a curve is \( y = { x^{-\frac{2}{3}}} \ \ln x \ \) for \( x \gt 0 \). The curve has one stationary point.
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the exact coordinates of the stationary point.
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Show that \( \displaystyle \int_{1}^{ 8 } y \ \mathrm{d}x = 18 \ln 2 \ – \ 9 \).

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\({\small 28.\enspace}\) 9709/32/F/M/22 – Paper 32 March 2022 Pure Maths 3 No 8(a), (b)
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the quotient and remainder when \({\small 8x^3 + 4x^2 + 2x + 7}\) is divided by \({\small 4x^2 + 1}\).
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Hence find the exact value of \( \displaystyle \int_{0}^{ \frac{1}{2} } \frac{8x^3 + 4x^2 + 2x + 7}{4x^2 + 1} \ \mathrm{d}x \).

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\({\small 29.\enspace}\) 9709/32/F/M/22 – Paper 32 March 2022 Pure Maths 3 No 11
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Integration - Direct substitution & Differentiation - Maximum point - Paper 32 March 2022
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The diagram shows the curve \( \ y \ = \ \sin x \ \cos 2x \ \) for \({\small \ 0 \le x \le {\large\frac{1}{2}}\pi}\), and its maximum point \(M\).
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the \(x\)-coordinate of \(M\), giving your answer correct to 3 significant figures.
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Using the substitution \( \ u = \cos x \), find the area of the shaded region enclosed by the curve and the \(x\)-axis in the first quadrant, giving your answer in a simplified exact form.

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\({\small 30.\enspace}\) 9709/13/O/N/21 – Paper 13 November 2021 Pure Maths 1 No 8
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Integration - Finding Area Between Two Curves - Paper 13 Nov 2021 No 8
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The diagram shows the curves with equations \( \ y \ = \ {x}^{ { \large – \frac{1}{2} } } \ \) and \( \ y \ = \ \frac{5}{2} \ – \ {x}^{ { \large \frac{1}{2} } } \). The curves intersect at the points \( \ A( {\large\frac{1}{4}},2) \ \) and \( \ B( 4 , {\large\frac{1}{2}}) \).
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the area of the region between the two curves.
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) The normal to the curve \( \ y \ = \ {x}^{ { \large – \frac{1}{2} } } \ \) at the point \( (1, 1) \) intersects the \(y\)-axis at the point \( (0, p) \). Find the value of \(p\).

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\({\small 31.\enspace}\) 9709/13/O/N/21 – Paper 13 November 2021 Pure Maths 1 No 10
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A curve has equation \( \ y = \mathrm{f}(x) \ \) and it is given that
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\( \hspace{2em} \mathrm{f}^{\prime}(x) = { \big( \frac{1}{2}x \ + \ k \big) }^{-2} \ – \ { ( 1 \ + \ k ) }^{-2} \),
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where \({\small \ k \ }\) is a constant. The curve has a minimum point at \({\small \ x \ = \ 2 }\).
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find \( \ \mathrm{f}^{\prime\prime}(x) \) in terms of \(k\) and \(x\), and hence find the set of possible values of \(k\).
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It is now given that \( {\small \ k \ = \ −3 \ }\) and the minimum point is at \({\small \ (2, \ 3\frac{1}{2}) }\).
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Find \( \ \mathrm{f}(x) \).

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\({\small 32.\enspace}\) 9709/12/M/J/21 – Paper 12 June 2021 Pure Maths 1 No 9
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Volume Revolution Integration - Paper 12 June 2021 No 9
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The diagram shows part of the curve with equation \( \ {\small y^2 \ = \ x \ − \ 2 } \ \) and the lines \( \ {\small x \ = \ 5 } \ \) and \( \ {\small y \ = \ 1 } \). The shaded region enclosed by the curve and the lines is rotated through \( \ {\small 360^{\circ} \ }\) about the \(x\)-axis.
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Find the volume obtained.

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\({\small 33.\enspace}\) 9709/12/M/J/21 – Paper 12 June 2021 Pure Maths 1 No 11
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The gradient of a curve is given by
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\( \hspace{2em} {\small {\large \frac{\mathrm{d}y}{\mathrm{d}x}} \ = \ 6{(3x \ – \ 5)}^{3} \ – \ k{x}^{2}}\),
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where \({\small \ k \ }\) is a constant. The curve has a stationary point at \( \ {\small (2, -3.5) }\).
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the value of \( \ k\).
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Find the equation of the curve.
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\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}\) Find \( {\small {\large \frac{ {\mathrm{d}}^{2} y }{ \mathrm{d}{x}^{2}} } }\).
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\({\small\hspace{1.2em}\left(d\right).\hspace{0.8em}}\) Determine the nature of the stationary point at \( \ {\small (2, -3.5) }\).

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\({\small 34.\enspace}\) 9709/12/M/J/20 – Paper 12 June 2020 Pure Maths 1 No 8
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Volume Revolution - Paper 12 June 2020 No 8
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The diagram shows part of the curve with equation \( \ {\small y \ = \ {\large\frac{6}{x}} } \). The points \( \ {\small (1, 6) \ }\) and \( \ {\small (3, 2) \ }\) lie on the curve. The shaded region is bounded by the curve and the lines \( \ {\small y \ = \ 2 } \ \) and \( \ {\small x \ = \ 1 } \).
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the volume generated when the shaded region is rotated through \( \ {\small 360^{\circ} \ }\) about the \(y\)-axis.
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) The tangent to the curve at a point \(X\) is parallel to the line \( \ {\small y \ + \ 2x \ = \ 0 } \). Show that \(X\) lies on the line \( \ {\small y \ = \ 2x } \).

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\({\small 35.\enspace}\) 9709/12/M/J/20 – Paper 12 June 2020 Pure Maths 1 No 10
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The equation of a curve is
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\( \hspace{2em} y \ = 54x \ – \ {(2x \ – \ 7)}^{3} \).
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find \( \ {\small {\large \frac{\mathrm{d}y}{\mathrm{d}x}} \ } \) and \( {\small \ {\large \frac{ {\mathrm{d}}^{2} y }{ \mathrm{d}{x}^{2}} } }\).
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Find the coordinates of each of the stationary points on the curve.
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\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}\) Determine the nature of each of the stationary points.

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\({\small 36.\enspace}\) 9709/12/O/N/19 – Paper 12 June 2019 Pure Maths 1 No 10
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Integration - Paper 12 Nov 2019 No 10
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The diagram shows part of the curve \( \ {\small y \ = \ 1 \ – \ {\large\frac{4}{ {(2x \ + \ 1)}^{2} }} } \). The curve intersects the \(x\)-axis at \(A\). The normal to the curve at A intersects the \(y\)-axis at \(B\).
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\({\small\hspace{1.2em}\left(i\right).\hspace{0.8em}}\) Obtain expressions for \( \ {\small {\large \frac{\mathrm{d}y}{\mathrm{d}x}} \ } \) and \(\displaystyle \int y \ \mathrm{d}x\).
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\({\small\hspace{1.2em}\left(ii\right).\hspace{0.7em}}\) Find the coordinates of \(B\).
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\({\small\hspace{1.2em}\left(iii\right).\hspace{0.6em}}\) Find, showing all necessary working, the area of the shaded region.

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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 Pure Maths 3 No 6
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The variables \({\small x}\) and \({\small \theta}\) satisfy the differential equation
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\(\hspace{3em} (3 \ + \ \cos2\theta) \ {\large \frac{\mathrm{d}x}{\mathrm{d} \theta }} \ = \ x \sin 2\theta \),
\(\\[1pt]\)
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 \({\small x }\).

\(\\[1pt]\)
\({\small 3.\enspace}\) 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
\(\\[1pt]\)
\(\hspace{3em} {\large \frac{\mathrm{d}x}{\mathrm{d} \theta }} \ = \ (x \ + \ 2) \ {\sin}^{2} 2\theta \),
\(\\[1pt]\)
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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\(\hspace{3em} {\large\frac{\textrm{d}P}{\textrm{d}t}} = -k \ \sqrt{P \ – \ A}\),
\(\\[1pt]\)
\(\\[7pt]\) 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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\(\hspace{3em} {\large\frac{\textrm{d}\theta}{\textrm{d}t}} = -k \ (\theta \ – \ A) \),
\(\\[1pt]\)
\(\\[7pt]\) 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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\(\hspace{3em} x \tan \theta {\large\frac{\textrm{d}x}{\textrm{d}\theta}} \ + \ \textrm{cosec}^{2} \theta = 0,\)
\(\\[1pt]\)
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 \({\small \theta }\).

\(\\[1pt]\)
\({\small 9.\enspace}\) The variables x and y satisfy differential equation
\(\\[1pt]\)
\(\hspace{3em} {\large\frac{\textrm{d}y}{\textrm{d}x}} = x \ \textrm{e}^{x+y}\).
\(\\[1pt]\)
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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\({\small 10.\enspace}\) 9709/33/M/J/20 – Paper 33 June 2020 Pure Maths 3 No 4(a), (b)
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The equation of a curve is \(y = x \ {\tan}^{-1} \big(\frac{1}{2}x\big) \).
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find \( {\large \frac{\mathrm{d}y}{\mathrm{d}x}}\).
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) The tangent to the curve at the point where \(x = 2\) meets the \(y\)-axis at the point with coordinates (0, \(p\)).
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Find \(p\).

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\({\small 11.\enspace}\) 9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 7
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The variables \(x\) and \(y\) satisfy the differential equation
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\( {\large \frac{\mathrm{d}y}{\mathrm{d}x}} = { \large\frac{y \ – \ 1}{(x \ + \ 1)(x \ + \ 3)} } \).
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It is given that \(y = 2\) when \(x = 0\).
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Solve the differential equation, obtaining an expression for \(y\) in terms of \(x\).

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\({\small 12.\enspace}\) 9709/31/M/J/20 – Paper 31 June 2020 Pure Maths 3 No 8(a), (b)
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A certain curve is such that its gradient at a point \( (x, y) \) is proportional to \( {\large \frac{y}{x\sqrt{x}} }\). The curve passes through the points with coordinates \( (1, 1) \) and \( (4, \mathrm{e}) \).
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) By setting up and solving a differential equation, find the equation of the curve, expressing \(y\) in terms of \(x\).
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Describe what happens to \(y\) as \(x\) tends to infinity.

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\({\small 13.\enspace}\) 9709/32/F/M/20 – Paper 32 March 2021 Pure Maths 3 No 4(a), (b)
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The variables \(x\) and \(y\) satisfy the differential equation
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\(\hspace{1.2em} (1 \ – \ \cos x){\large\frac{\mathrm{d}y}{\mathrm{d}x}} \ = \ y \sin x \).
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It is given that \(y = 4\) when \(x = \pi \).
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\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Solve the differential equation, obtaining an expression for \(y\) in terms of \(x\).
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Sketch the graph of \(y\) against \(x\) for \( 0 \lt x \lt 2 \pi \).

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\({\small 14.\enspace}\) 9709/33/M/J/21 – Paper 33 June 2021 Pure Maths 3 No 7(a), (b)
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Question 7 Paper 33 June 2021
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For the curve shown in the diagram, the normal to the curve at the point \(P\) with coordinates \( (x, y) \) meets the \(x\)-axis at \(N\). The point \(M\) is the foot of the perpendicular from \(P\) to the \(x\)-axis.
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The curve is such that for all values of \(x\) in the interval \(0 \leq x \lt {\large \frac{1}{2} }\pi \), the area of triangle \(PMN\) is equal to \( \tan x \).
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\({\small\hspace{1.2em}\left(a\right)\hspace{0.4em} (i). \hspace{0.6em} }\) Show that \( {\large \frac{MN}{y} } \ = \ {\large \frac{\mathrm{d}y}{\mathrm{d}x} } \).
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\({\small\hspace{2.7em} (ii). \hspace{0.4em} }\) Hence show that \(x\) and \(y\) satisfy the differential equation \( {\large \frac{1}{2} } {y}^{2} {\large \frac{\mathrm{d}y}{\mathrm{d}x} } \ = \ \tan x \).
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\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Given that \(y = 1 \ \) when \(x = 0\), solve this differential equation to find the equation of the curve, expressing \(y\) in terms of \(x\).

\(\\[1pt]\)
\({\small 15.\enspace}\) 9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 7
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A curve is such that the gradient at a general point with coordinates \( (x, y) \) is proportional to \( {\large \frac{y}{\sqrt{x \ + \ 1}} }\).
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The curve passes through the points with coordinates \( (0, 1) \) and \( (3, \mathrm{e}) \).
\(\\[1pt]\)
By setting up and solving a differential equation, find the equation of the curve, expressing \(y\) in terms of \(x\).

\(\\[1pt]\)
\({\small 16.\enspace}\) 9709/31/M/J/21 – Paper 31 June 2021 Pure Maths 3 No 10
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The variables \(x\) and \(t\) satisfy the differential equation
\(\\[1pt]\)
\(\hspace{1.2em} {\large \frac{\mathrm{d}x}{\mathrm{d}t} } \ = \ x^{2} (1 \ + \ 2x) \),
\(\\[1pt]\)
and \( x = 1 \ \) when \( t \ = \ 0 \).
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Using partial fractions, solve the differential equation, obtaining an expression for \(t\) in terms of \(x\).

\(\\[1pt]\)
\({\small 17.\enspace}\) 9709/32/F/M/22 – Paper 32 March 2022 Pure Maths 3 No 9
\(\\[1pt]\)
The variables \(x\) and \(y\) satisfy the differential equation
\(\\[1pt]\)
\(\hspace{1.2em} (x \ + \ 1)(3x \ + \ 1){\large \frac{\mathrm{d}y}{\mathrm{d}x} } \ = \ y \),
\(\\[1pt]\)
and it is given that \( y = 1 \ \) when \( x \ = \ 1 \).
\(\\[1pt]\)
Solve the differential equation and find the exact value of \(y\) when \(x \ = \ 3\), giving your answer in a simplified form.

\(\\[1pt]\)
\({\small 18.\enspace}\) 9709/32/F/M/22 – Paper 32 March 2022 Pure Maths 3 No 9
\(\\[1pt]\)
The variables \(x\) and \(y\) satisfy the differential equation
\(\\[1pt]\)
\(\hspace{1.2em} (x \ + \ 1)(3x \ + \ 1){\large \frac{\mathrm{d}y}{\mathrm{d}x} } \ = \ y \),
\(\\[1pt]\)
and it is given that \( \ y \ = \ 1 \ \) when \( x \ = \ 1 \).
\(\\[1pt]\)
Solve the differential equation and find the exact value of \(y\) when \( \ x \ = \ 3 \), giving your answer in a simplified form.

\(\\[1pt]\)
\({\small 19.\enspace}\) 9709/12/O/N/19 – Paper 12 November 2019 Pure Maths 1 No 3
\(\\[1pt]\)
A curve is such that
\(\\[1pt]\)
\(\hspace{1.2em} {\large \frac{\mathrm{d}y}{\mathrm{d}x} } \ = \ {\large \frac{k}{\sqrt{x} } } \),
\(\\[1pt]\)
where \( \ k \ \) is a constant.
\(\\[1pt]\)
The points \( \ P \ (1, −1) \ \) and \( \ Q \ (4, 4) \ \) lie on the curve. Find the equation of the curve.

\(\\[1pt]\)


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 =) .