$\require{\cancel}$ $\require{\stix[upint]}$

Qtn No.	1	2	3	4	5	6	7	Total
Marks	4	6	7	8	8	11	10	54
Score

Question 1 Code: 9709/31/O/N/18/2, Topic: Logarithmic and exponential functions

Showing all necessary working, solve the equation $\displaystyle\frac{2 \mathrm{e}^{x}+\mathrm{e}^{-x}}{\mathrm{e}^{x}-\mathrm{e}^{-x}}=4$, giving your answer correct to 2 decimal places. $[4]$

Question 2 Code: 9709/31/O/N/14/4, Topic: Differentiation

The parametric equations of a curve are

$$ x=\frac{1}{\cos ^{3} t}, \quad y=\tan ^{3} t, $$

where $0 \leqslant t < \frac{1}{2} \pi$.

$\text{(i)}$ Show that $\displaystyle\frac{\mathrm{d} y}{\mathrm{~d} x}=\sin t$. $[4]$

$\text{(ii)}$ Hence show that the equation of the tangent to the curve at the point with parameter $t$ is $y=x \sin t-\tan t$. $[3]$

Question 3 Code: 9709/31/O/N/13/5, Topic: Trigonometry, Integration

$\text{(i)}$ Prove that $\cot \theta+\tan \theta \equiv 2 \operatorname{cosec} 2 \theta$. $[3]$

$\text{(ii)}$ Hence show that $\displaystyle\int_{\frac{1}{6} \pi}^{\frac{1}{3} \pi} \operatorname{cosec} 2 \theta \mathrm{d} \theta=\frac{1}{2} \ln 3.$ $[4]$

Question 4 Code: 9709/31/O/N/11/6, Topic: Trigonometry

$\text{(i)}$ Express $\cos x+3 \sin x$ in the form $R \cos (x-\alpha)$, where $R>0$ and $0^{\circ}< \alpha <90^{\circ}$, giving the exact value of $R$ and the value of $\alpha$ correct to 2 decimal places. $[3]$

$\text{(ii)}$ Hence solve the equation $\cos 2 \theta+3 \sin 2 \theta=2$, for $0^{\circ}< \theta <90^{\circ}$. $[5]$

Question 5 Code: 9709/31/O/N/19/6, Topic: Differentiation, Integration

$\text{(i)}$ By differentiating $\displaystyle\frac{\cos x}{\sin x}$, show that if $\displaystyle y=\cot x$ then $\displaystyle\frac{\mathrm{d} y}{\mathrm{~d} x}=-\operatorname{cosec}^{2} x$. $[2]$

$\text{(ii)}$ Show that $\displaystyle\int_{\frac{1}{4} \pi}^{\frac{1}{2} \pi} x \operatorname{cosec}^{2} x \mathrm{~d} x=\frac{1}{4}(\pi+\ln 4)$. $[6]$

Question 6 Code: 9709/31/O/N/13/9, Topic: Vectors

 

The diagram shows three points $A, B$ and $C$ whose position vectors with respect to the origin $O$ are given by $\overrightarrow{O A}=\left(\begin{array}{r}2 \\ -1 \\ 2\end{array}\right), \overrightarrow{O B}=\left(\begin{array}{l}0 \\ 3 \\ 1\end{array}\right)$ and $\overrightarrow{O C}=\left(\begin{array}{l}3 \\ 0 \\ 4\end{array}\right).$ The point $D$ lies on $B C$, between $B$ and $C$, and is such that $C D=2 D B$.

$\text{(i)}$ Find the equation of the plane $A B C$, giving your answer in the form $a x+b y+c z=d$. $[6]$

$\text{(ii)}$ Find the position vector of $D$. $[1]$

$\text{(iii)}$ Show that the length of the perpendicular from $A$ to $O D$ is $\frac{1}{3} \sqrt{(65)}$. $[4]$

Question 7 Code: 9709/31/O/N/18/9, Topic: Algebra, Integration

Let $\displaystyle\mathrm{f}(x)=\frac{6 x^{2}+8 x+9}{(2-x)(3+2 x)^{2}}$

$\text{(i)}$ Express $\mathrm{f}(x)$ in partial fractions. $[5]$

$\text{(ii)}$ Hence, showing all necessary working, show that $\displaystyle\int_{-1}^{0} \mathrm{f}(x) \, \mathrm{d} x=1+\frac{1}{2} \ln \left(\frac{3}{4}\right)$. $[5]$