SPM Additional Mathematics 2017, Paper 2 (Question 10 & 11)

Question 10 (10 marks):
(a) Prove that 2 tan x cos² x = sin 2x.

(b) Hence, solve the equation 4 tan x cos² x = 1 for 0 ≤ x ≤ 2π.

(c)(i) Sketch the graph of y = sin 2x for 0 ≤ x ≤ 2π.

(c)(ii) Hence, using the same axes, sketch a suitable straight line to find the number of solutions for the equation 4π tan x cos² x = x – 2π for 0 ≤ x ≤ 2π.
State the number of solutions.

Solution: 
(a)
$\begin{array}{l}2\tan x{\cos }^{2}x=\sin 2x\\ \text{Left hand side}\\ =2\tan x{\cos }^{2}x\\ =2\times \frac{\sin x}{\cos x}\times {\cos }^{2}x\\ =2\sin x\cos x\\ =\sin 2x\\ =\text{ Right hand side }\left(\text{Proven}\right)\end{array}$


(b)
$\begin{array}{l}4\tan x{\cos }^{2}x=1,0\le x\le 2\pi \\ 2\left(2\tan x{\cos }^{2}x\right)=1\\ 2\sin 2x=1\\ \sin 2x=\frac{1}{2}\\ \text{Basic angle}=\frac{\pi }{6}\\ 2x=\frac{\pi }{6},\left(\pi -\frac{\pi }{6}\right),\left(2\pi +\frac{\pi }{6}\right),\left(3\pi -\frac{\pi }{6}\right)\\ 2x=\frac{\pi }{6},\frac{5\pi }{6},\frac{13\pi }{6},\frac{17\pi }{6}\\ x=\frac{\pi }{12},\frac{5\pi }{12},\frac{13\pi }{12},\frac{17\pi }{12}\end{array}$



(c)(i)
y = sin 2x, 0 ≤ x ≤ 2π.




(c)(ii)
$\begin{array}{l}4\pi \tan x{\cos }^{2}x=x-2\pi \\ 2\pi \left(2\tan x{\cos }^{2}x\right)=x-2\pi \\ 2\pi \sin 2x=x-2\pi \\ \sin 2x=\frac{x}{2\pi }-\frac{2\pi }{2\pi }\\ \sin 2x=\frac{x}{2\pi }-1\\ y=\frac{x}{2\pi }-1\end{array}$


Number of solutions = 4

Question 11 (10 marks):
Diagram 6 shows a curve y = 2x² – 18 and the straight line PQ which is a tangent to the curve at point K.

It is given that the gradient of the straight line PQ is 4.
(a) Find the coordinates of point K
(b) Calculate the area of the shaded region.
(c) When the region bounded by the curve, the x-axis and the straight line y = h is rotated through 180^o about the y-axis, the volume generated is 65π unit³.
Find the value of h.


Solution: 
(a)
$\begin{array}{l}y=2x^2-18\\ \frac{dy}{dx}=4x\\ \\ \text{Gradient of straight line }PQ=4\\ 4x=4\\ x=1\\ \\ \text{When }x=1,\\ y=2{\left(1\right)}^2-18=-16\\ \\ \text{Coordinates of }K=\left(1,-16\right).\end{array}$


(b)
$\begin{array}{l}\text{At }x\text{-axis, }y=0\\ \therefore 2x^2-18=0\\ 2x^2=18\\ x^2=9\\ x=\pm 3\\ \text{The curve cuts the }x\text{-axis at }\left(-3,0\right)\text{ and }\left(3,0\right)\text{.}\\ \\ \text{Area of shaded region}\\ \text{= Area of triangle}-\text{Area under the curve}\\ =\frac{1}{2}\left(5-1\right)\left(16\right)-{\displaystyle {\int }_1^{3}ydx}\\ =32-{\displaystyle {\int }_1^3\left(2x^2-18\right)dx}\\ =32-\left|{\left[\frac{2x^3}{3}-18x\right]}_1^3\right|\\ =32-\left|\left(\frac{2{\left(3\right)}^3}{3}-18\left(3\right)\right)-\left(\frac{2{\left(1\right)}^3}{3}-18\left(1\right)\right)\right|\\ =32-\left|\left(18-54-\frac{2}{3}+18\right)\right|\\ =32-\left|-18\frac{2}{3}\right|\\ =32-18\frac{2}{3}\\ =13\frac{1}{3}{\text{ units}}^2\end{array}$


(c)
$\begin{array}{l}\text{Volume generated}=65\pi \\ \pi {\displaystyle {\int }_h^0{x}^{2}dy}=65\pi \\ \pi {\displaystyle {\int }_h^0\left(\frac{y}{2}+9\right)dx}=65\pi \leftarrow \boxed{\begin{array}{l}y=2x^2-18\\ x^2=\frac{y}{2}+9\end{array}}\\ {\left[\frac{y^2}{4}+9y\right]}_h^0=65\\ 0-\left(\frac{h^2}{4}+9h\right)=65\\ -\frac{h^2}{4}-9h=65\\ h^2+36h+260=0\\ \left(h+10\right)\left(h+26\right)=0\\ h=-10\text{ or }h=-26\left(\text{rejected}\right)\\ \text{Thus, }h=-10\end{array}$