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Qtn No.	1	2	3	4	5	6	Total
Marks	10	7	10	10	10	10	57
Score

Question 1 Code: 9709/31/M/J/14/8, Topic: Numerical solutions of equations

$\text{(i)}$ By sketching each of the graphs $y=\operatorname{cosec} x$ and $y=x(\pi-x)$ for $0 < x < \pi$, show that the equation

$$ \operatorname{cosec} x=x(\pi-x) $$

has exactly two real roots in the interval $0 < x < \pi$. $[2]$

$\text{(ii)}$ Show that the equation $\operatorname{cosec} x=x(\pi-x)$ can be written in the form $\displaystyle x=\frac{1+x^{2} \sin x}{\pi \sin x}$. $[2]$

$\text{(iii)}$ The two real roots of the equation $\operatorname{cosec} x=x(\pi-x)$ in the interval $0 < x < \pi$ are denoted by $\alpha$ and $\beta$, where $\alpha < \beta$.

$\text{(a)}$ Use the iterative formula

$$ \displaystyle x_{n+1}=\frac{1+x_{n}^{2} \sin x_{n}}{\pi \sin x_{n}} $$

to find $\alpha$ correct to 2 decimal places. Give the result of each iteration to 4 decimal places. $[3]$

$\text{(b)}$ Deduce the value of $\beta$ correct to 2 decimal places. $[1]$

Question 2 Code: 9709/32/M/J/14/8, Topic: Differentiation

 

The diagram shows the curve $y=x \cos \frac{1}{2} x$ for $0 \leqslant x \leqslant \pi$.

$\text{(i)}$ Find $\displaystyle\frac{\mathrm{d} y}{\mathrm{~d} x}$ and show that $4 \displaystyle\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}+y+4 \sin \frac{1}{2} x=0$. $[5]$

$\text{(ii)}$ Find the exact value of the area of the region enclosed by this part of the curve and the $x$-axis. $[5]$

Question 3 Code: 9709/33/M/J/14/8, Topic: Algebra, Integration

Let $\displaystyle\mathrm{f}(x)=\frac{6+6 x}{(2-x)\left(2+x^{2}\right)}$.

$\text{(i)}$ Express $\mathrm{f}(x)$ in the form $\displaystyle\frac{A}{2-x}+\frac{B x+C}{2+x^{2}}$. $[4]$

$\text{(ii)}$ Show that $\displaystyle\int_{-1}^{1} \mathrm{f}(x) \mathrm{d} x=3 \ln 3$. $[5]$

Question 4 Code: 9709/31/O/N/14/8, Topic: Trigonometry

$\text{(i)}$ By first expanding $\sin (2 \theta+\theta)$, show that $[4]$

$$ \sin 3 \theta=3 \sin \theta-4 \sin ^{3} \theta. $$

$\text{(ii)}$ Show that, after making the substitution $\displaystyle x=\frac{2 \sin \theta}{\sqrt{3}}$, the equation $x^{3}-x+\frac{1}{6} \sqrt{3}=0$ can be written in the form $\sin 3 \theta=\frac{3}{4}$. $[1]$

$\text{(iii)}$ Hence solve the equation

$$ x^{3}-x+\frac{1}{6} \sqrt{3}=0, $$

giving your answers correct to 3 significant figures. $[4]$

Question 5 Code: 9709/32/O/N/14/8, Topic: Trigonometry

Question 6 Code: 9709/33/O/N/14/8, Topic: Differentiation

The variables $x$ and $y$ are related by the differential equation

$$ \displaystyle\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{5} x y^{\frac{1}{2}} \sin \left(\frac{1}{3} x\right) \text {. } $$

$\text{(i)}$ Find the general solution, giving $y$ in terms of $x$. $[6]$

$\text{(ii)}$ Given that $y=100$ when $x=0$, find the value of $y$ when $x=25$. $[3]$