Solutions to HSC2022 Math Advanced

 

 

Exam papers can be found here (and official solutions too. Only answers to MC questions (no explanations)

https://educationstandards.nsw.edu.au/wps/portal/nesa/11-12/resources/hsc-exam-papers

Integral involving exponential function

 

Integrals with logarithmic functions,

 

Integrals with logarithmic functions,

  

Integration with trigonometric substitutions - Several Examples

 

Integration - Trigonometric - Several Examples

 

HSC Mathematics Ext1 Exam 3U 2016

Question 1: ===========================

 

Question 11(a,b,c): =============================



Question 11(d,e): =============================



Question 11(f), 12(a) i ii: =========================



Question 12(a) iii iv: =============================



Question 12b: =============================



Question 12c: =============================





Question 13: =============================


Question 14: =============================

Harder Inequalities question from Cambridge Book Show $x^2+xy+y^2>0$ for $x, y\ne 0$.

Harder Inequalities question from Cambridge Book Show $$x^2+xy+y^2>0$$ for $x, y\ne 0$.

3 Methods:

Method 1: Consider two cases. (Keep in mind $x,y$ are both non zero)
Case 1: $xy>0$. In this case the result is trivial as each summand is positive. That is since $x^2>0, y^2>0$ and $xy>0$ then their sum is also positive!//

Case 2: $xy<0$. Then $-xy>0$.
So $x^2+xy+y^2=(x+y)^2-xy$. But $(x+y)^2>0$ and so the LHS is the sum of two positive numbers and so is itself positive.//

Method 2: Use completing the square

$\begin{array}{l} x^2+xy+y^2&=x^2+xy+y^2/4+3y^2/4\\ &=\left( x+y/2 \right)^2 +3y^2/4\end{array}$

Each summand is clearly positive and hence the sum is also.//

Method 3: Use the discriminant by considering $x^2+xy+y^2$ as a quadratic in $x$.
$\Delta = y^2-4y^2=-3y^2<0$.
Since the coefficient of $x$ is $1>0$ and $\Delta<0$ then the expression must be positive.

[Remember that if you have a quadratic $Ax^2+Bx+C$ then $\Delta=B^2-4AC$]

FRACTIONS - Basic Operations

${2\over 7}+{3\over 7}={2+3\over 7}={5\over 7}$

 ${7\over 10}-{4\over 10}={7-4\over 10}={3\over 10}$

 ${4\over 5}\times{5\over 6}={4\times 5\over 5\times 6}={20\over 30}={2\over 3}$

 ${4\over 5}\div{5\over 6}= {4\over 5}\times{6\over 5}={4\times 6\over 5\times 5}={24\over 25}$

${{2\over 5}\over 6} ={2\over 5}\div 6={2\over 5} \times {1\over 6}={2\over 30}={1\over 15}$

${2\over {5\over 6}} =2\div {5\over 6}=2\times {6\over 5}={12\over 5}=2 {2\over 5}$

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

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

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

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

Question 12(a): ===================================

Question 12(b):  ===================================

Question 12(c):  ===================================

Question 12(d):  ===================================

Question 13(a): ==================================




Question 13(b(c): ================================

Question 13(d): ==================================

Question 14(a): ==================================



Question 14(b): ==================================





Question 14(c): ==================================





Question 14(d,e): ==================================





Question 15(a):  =================================


Question 15(b):  =================================

Question 15(c):  =================================

(from BOS notes)

Question 16(a): ================================




Question 16( b): ================================