## 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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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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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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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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$${\small \hspace{1.2em}(i). \enspace }$$ 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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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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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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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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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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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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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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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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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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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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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).

$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ 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

### 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 ! =) .
$$\\[1pt]$$

EXAMPLE:

$${\small 1.\enspace}$$ 9709/03/SP/20 – Specimen Paper 2020 Pure Maths 3 No 10
$$\\[1pt]$$
In a chemical reaction, a compound X is formed from two compounds Y and Z.
$$\\[1pt]$$
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 $${\small \ {\large\frac{\textrm{d}x}{\textrm{d}t}} \ = \ 2}$$.
$$\\[1pt]$$
$${\small\hspace{1.2em}\left(\textrm{a}\right).\hspace{0.8em}}$$ Show that x and t satisfy the differential equation
$$\\[1pt]$$
$$\hspace{3em} {\large \frac{\mathrm{d}x}{\mathrm{d}t }} \ = \ 0.01(10 \ – \ x)(20 \ – \ x)$$ .
$$\\[1pt]$$
$${\small\hspace{1.2em}\left(\textrm{b}\right).\hspace{0.8em}}$$ Solve this differential equation and obtain an expression for x in terms of t.

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

$$\\[1pt]$$
$${\small 3.\enspace}$$ 9709/03/SP/17 – Specimen Paper 03 2017 No 8
$$\\[1pt]$$
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.

$$\\[1pt]$$
$${\small 4.\enspace}$$ 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
$$\\[1pt]$$
$$\hspace{3em} {\large\frac{\textrm{d}P}{\textrm{d}t}} = -k \ \sqrt{P \ – \ A}$$,
$$\\[1pt]$$
$$\\[7pt]$$ where k is a positive constant.
$$\\[1pt]$$
$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Find the general solution of this differential equation.
$$\\[1pt]$$
$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Given that $${\small P \ = \ 5A}$$ when $${\small t \ = \ 0}$$ and that $${\small P \ = \ 2A}$$ when $${\small t \ = \ 2}$$, show that $${\small k \ = \ \sqrt{A}}$$.
$$\\[1pt]$$
$${\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}$$ Find the value of t when $${\small P \ = \ A}$$.
$$\\[1pt]$$
$${\small\hspace{1.2em}\left(d\right).\hspace{0.8em}}$$ Obtain an expression for P in terms of A and t.

$$\\[1pt]$$
$${\small 5.\enspace}$$ 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
$$\\[1pt]$$
$$\hspace{3em} {\large\frac{\textrm{d}\theta}{\textrm{d}t}} = -k \ (\theta \ – \ A)$$,
$$\\[1pt]$$
$$\\[7pt]$$ where k is a positive constant.
$$\\[1pt]$$
$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ Find the solution of this differential equation, given that $${\small \ \theta \ = \ 4A \ }$$ when $${\small t \ = \ 0}$$.
$$\\[1pt]$$
$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Given also that $${\small \theta \ = \ 3A}$$ when $${\small t \ = \ 1}$$, show that $${\small k \ = \ \ln{\large\frac{3}{2}}}$$.
$$\\[1pt]$$
$${\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}$$ Find $${\small \theta}$$ in terms of A when $${\small t \ = \ 2}$$, expressing your answer in its simplest form.

$$\\[1pt]$$
$${\small 6.\enspace}$$ The variables x and y are related by the differential equation
$$\\[1pt]$$
$$\hspace{3em} {\large\frac{\textrm{d}y}{\textrm{d}x}} = {\large\frac{6x \ {\textrm{e}}^{3x}}{{y}^{2}}} \ .$$
$$\\[1pt]$$
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.

$$\\[1pt]$$
$${\small 7.\enspace}$$ 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.
$$\\[1pt]$$

$$\\[1pt]$$
$${\small\hspace{1.2em}\left(i\right).\hspace{0.8em}}$$ Using the fact that the gradient of the curve at P is $${\small {\large \frac{PN}{TN}}}$$, show that
$$\\[1pt]$$
$$\hspace{3em} {\large\frac{\textrm{d}y}{\textrm{d}x}} = {\large \frac{1}{2}} {y}^{2} \cot x \ .$$
$$\\[1pt]$$
$${\small\hspace{1.2em}\left(ii\right).\hspace{0.8em}}$$ 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
$$\\[1pt]$$
$$\hspace{3em} x \tan \theta {\large\frac{\textrm{d}x}{\textrm{d}\theta}} \ + \ \textrm{cosec}^{2} \theta = 0,$$
$$\\[1pt]$$
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 $${\small x \ = \ 0}$$.
$$\\[1pt]$$
$${\small\hspace{1.2em}\textrm{i}.\hspace{0.8em}}$$ Solve the differential equation, obtaining $${\small y }$$ in terms of $${\small x }$$.
$$\\[1pt]$$
$${\small\hspace{1.2em}\textrm{ii}.\hspace{0.8em}}$$ Explain why $${\small x }$$ can only take values that are less than 1.

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

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

$$\\[1pt]$$
$${\small 12.\enspace}$$ 9709/31/M/J/20 – Paper 31 June 2020 Pure Maths 3 No 8(a), (b)
$$\\[1pt]$$
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 $$(4, \mathrm{e})$$.
$$\\[1pt]$$
$${\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}$$ By setting up and solving a differential equation, find the equation of the curve, expressing $$y$$ in terms of $$x$$.
$$\\[1pt]$$
$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ Describe what happens to $$y$$ as $$x$$ tends to infinity.

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

$$\\[1pt]$$
$${\small 14.\enspace}$$ 9709/33/M/J/21 – Paper 33 June 2021 Pure Maths 3 No 7(a), (b)
$$\\[1pt]$$

$$\\[1pt]$$
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.
$$\\[1pt]$$
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 $$\tan x$$.
$$\\[1pt]$$
$${\small\hspace{1.2em}\left(a\right)\hspace{0.4em} (i). \hspace{0.6em} }$$ Show that $${\large \frac{MN}{y} } \ = \ {\large \frac{\mathrm{d}y}{\mathrm{d}x} }$$.
$$\\[1pt]$$
$${\small\hspace{2.7em} (ii). \hspace{0.4em} }$$ Hence show that $$x$$ and $$y$$ satisfy the differential equation $${\large \frac{1}{2} } {y}^{2} {\large \frac{\mathrm{d}y}{\mathrm{d}x} } \ = \ \tan x$$.
$$\\[1pt]$$
$${\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}$$ 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
$$\\[1pt]$$
A curve is such that the gradient at a general point with coordinates $$(x, y)$$ is proportional to $${\large \frac{y}{\sqrt{x \ + \ 1}} }$$.
$$\\[1pt]$$
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
$$\\[1pt]$$
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$$.
$$\\[1pt]$$
Using partial fractions, solve the differential equation, obtaining an expression for $$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 $${\small V = \ {\large \frac{4}{3}}{h}^{3} }$$.

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 =) .