HSC Mathematics Exam 2U 2016

Question 1: Answer (B)
Noting that sine is positive in Quadrants 1 & 2 and cosine is negative in quadrants 2 & 3.
Hence $\theta$ must lie in quadrant 2. 

Question 2: Answer (C)

$p={6\over 30}={1\over 5}$

Question 3: Answer (B)

The first clue is the minus sign which indicates the coefficient of $x^2$ is negative, hence its a concave down parabola, thus B or C. If $y=0$, $3-(x-2)^2=0\implies x=2\pm \sqrt{3}$ so it has two zeros, so it must be (B) since (C) has no zeros as it doesn't cross the $x$-axis.

Question 4: Answer (A)

The only graph that satisfies $f(-x)=-f(x)$ is (A). Or the only graph to go onto itself under rotation f $180^o$ about the origin is (A).

Question 5: Answer (B)

${d\over dx} \ln(\cos x)={ -\sin x\over \cos x}=-\tan x$

Question 6: Answer (A)

The normal period of $y=\tan x$ is $\pi$. So in this case the period is ${\pi\over 3}$.

Question 7: Answer (A)

With $\theta$ radians, arc length is $l=r\times \theta$.

$\theta={l\over r}$

Area is $A={1\over 2}r^2\theta={1\over 2}.5^2.{7\over 5}={32\over 5}$

Question 8: Answer (D)

From the graph of $y=\cos 2x$ it is seen there are $5$ solutions to the equation $|\cos (2x)|=1$.

Question 9: Answer (C)

Drawing a picture we see two triangles whose combined area is $A=0.5\times 2\times 2+0.5\times 3\times 3={13\over 2}$.

Or if you happen to know the formula for integral of absolute value we have

$\int^2_{-3} |x+1| dx =  \left.{ (x+1) |x+1|\over 2}\right|^{x=2}_{x=-3}=9/2+2={13\over 2}$.

Question 10: Answer (D)
$4+\log_2x=4\log_22+\log_2x=\log_22^4+\log_2x=\log_2 16x$

Question 11(a) ====================================




Question 11(b)
$y={x+2\over 3x-4}$

$y'={vu'-uv'\over v^2}$

$y'={ (3x-4)\times 1-(x+2)\times 3\over (3x-4)^2}$

$y'=-{ 10\over (3x-4)^2}$


Question 11(c)  Solve $|x-2|\le 3$

$-3\le x-2 \le 3$

$-3+2\le x-2+2 \le 3+2$

$-1\le x \le 5$


Question 11(d)

$$\int_0^1(2x+1)^3 dx=\left. {(2x+1)^4\over 8}\right|^1_0$$
$$3^4/8-1/8=10$$

Question 11(e)

Solve the equations
$$\begin{array}{rl} y= & -5-4x\\ y= & 3-2x-x^2\\ \therefore -5-4x & =3-2x-x^2\\x^2-2x- 8 & =0\\ (x-4)(x+2) & =0\\ x=4  & \& \quad x=-2 \end{array}$$

When $x=-2$, $y=3$ and when $x=4$, $y=-21$
Points of intersection are: $$(-2,3)\quad , \quad (4,-21)$$

Question 11(f)

$y=\tan x$

$$y'=\left. \sec^2x\right|_{x=\pi/8}=1.17$$

Question 11(g)

Solve $\sin {x\over 2}={1\over 2}$.

$$0\le {x\over 2}\le \pi$$

$${x\over 2}={\pi\over 6}, \pi-\pi/6={\pi\over 6}, {5\pi\over 6}$$

$$x={\pi\over 3}, {5\pi\over 3}$$

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(from BOS notes)

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