2024 HSC Maths Extension 1 Exam Solutions

The Matrix Maths Team has put together solutions for the 2024 HSC Maths Extension 1 Exam - with full explanations included!

Written by:
Oak Ukrit
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In this article, we reveal 2024 HSC Maths Extension 1 Exam Solutions, complete with full explanations written by the Matrix Maths Team.

 

2024 HSC Maths Extension 1 Exam Solutions

Have you seen the 2024 HSC Mathematics Extension 1 exam paper yet? For reference, you can find it here.

Doing past papers? Find all the solutions to the 2017 – 2021 HSC Maths Ext 1 Exams here.

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2024 HSC Maths Extension 1 Exam Solutions – Multiple Choice Questions

QuestionAnswerSolution
1BWe use the formula \(\alpha \beta \gamma = -\frac{d}{a}\) for product of roots to obtain

\begin{align*}
(\alpha) (-1) (-3) &= -\frac{(-6)}{(1)} \\
3 \alpha &= 6 \\
\alpha &= 2.
\end{align*}

2DWe split the shaded area into the regions:

From \(x = -4\) to \(x = -3\), with area given by:
\(
\int_{-4}^{-3} g(x) – f(x) \, dx = – \int_{-4}^{-3} f(x) – g(x) \, dx \text{;}
\)

From \(x = -3\) to \(x = -1\), with area given by:
\(
\int_{-3}^{-1} f(x) – g(x) \, dx \text{;}
\)

From \(x = -1\) to \(x = 1\), with area given by:
\(
\int_{-1}^1 g(x) – f(x) \, dx = – \int_{-1}^1 f(x) – g(x) \, dx \text{; and}
\)

From \(x = 1\) to \(x = 4\), with area given by:
\(
\int_1^4 f(x) – g(x) \, dx \text{.}
\)

Adding these areas together gives answer (D).

3BWe can fill each of the \(4\) school choirs with at most \(19\) students each without any one choir having \(20\) students. Adding one more guarantees that there are at least \(20\) students in at least one of the choirs. The answer is thus

\(4 \times 19 + 1 = 77\).

4AWe’ll first note that for any real \(\theta\),

\begin{align*}
\cos^{-1} \theta &= \frac{\pi}{2} – \sin^{-1} \theta \\
\cos^{-1} \theta + \sin^{-1} \theta &= \frac{\pi}{2} \\
2 \cos^{-1} \theta + 2 \sin^{-1} \theta &= \pi,
\end{align*}

and the range of the function \(y = 2 \cos^{-1} (2 x) + 2 \sin^{-1} (2 x)\) is \(\{\pi\}\). Both \(\sin^{-1}\) and \(\cos^{-1}\) have domain \([-1, 1]\). So, \(x\) must satisfy

\begin{align*}
-1 \leq 2 x \leq 1 \\
-0.5 \leq x \leq 0.5,
\end{align*}

and the domain of our function is \([-0.5, 0.5]\).

5CThe coefficient of \(2\) outside of the function \(2 \sin^{-1}(3 x)\) is responsible for vertically stretching (dilating) the function by a factor of \(2\). The coefficient of \(3\) on the \(3 x\) is responsible for horizontally contracting by a factor of \(3\), or horizontally dilating by a factor of \(\frac{1}{3}\).
6DSimplifying the equation given, we can see that

\begin{align*}
|b| &= |b \sin(4 x)| \\
|b| |\sin(4 x)| &= |b| \\
|\sin(4 x)| &= 1 \\
\sin(4 x) &= \pm 1.
\end{align*}

A quick sketch of \(\sin(4 x)\) for \(0 \leq x \leq 2 \pi\), noting that \(4\) periods will elapse in this domain, gives the following points of intersection between \(y = \sin(4 x)\) and and the lines \(y = 1\) and \(y = -1\):

Graph of a sinusoidal wave with marked points on the peaks and troughs from the 2024 Maths Extensions 1 Exam Solutions

There are \(8\) such points of intersection, and the answer is (D).

7CThe applicant has \(5\) questions to go and needs to get at least \(4\) correct. Let \(X \sim \operatorname{Bin} \left( 5, \frac{1}{4} \right)\) be the number of questions they correctly answer out of those \(5\) questions, with a probability of \(\frac{1}{4}\) of correctly answering any individual question. We can see that

\(
\begin{align*}
\mathrm{P}(X \geq 4)
&= \mathrm{P}(X = 4) + \mathrm{P}(X = 5) \\
&= \binom{5}{4} \left( \frac{1}{4} \right)^4 \left( \frac{3}{4} \right)^1 + \binom{5}{5} \left( \frac{1}{4} \right)^5 \left( \frac{3}{4} \right)^0 \\
&= \frac{1}{64}.
\end{align*}
\)

8BWe will use the formula

\(\sigma = \sqrt \frac{p (1 – p)}{n}\)

for standard deviation of the proportion \(\hat p\), where \(n\) is the number of households and the probability \(p\) that any one household will have a dog is \(\frac{7}{12}\). Solving \(\sigma < 0.06\) gives

\begin{align*}
\sqrt \frac{\frac{7}{12} (1 – \frac{7}{12})}{n} &< 0.06 \\
\frac{\frac{35}{144}}{n} &< 0.0036 \\
n &> \frac{\frac{35}{144}}{0.0036} \approx 67.5.
\end{align*}

The smallest value \(n\) can thus take such that \(\sigma < 0.06\) is \(n = 68\).

9AThe probability that both coins drawn are made of the same metal is the same as the probability that both coins are silver or both coins are bronze. There are \(\binom{k}{2}\) ways to pick two silver coins, \(\binom{n – k}{2}\) ways to pick two bronze coins, and \(\binom{n}{2}\) ways to pick any two coins. This gives that our probability can be expressed as:

\(
\begin{aligned}
\frac{\binom{k}{2} + \binom{n – k}{2}}{\binom{n}{2}}
&= \frac{\frac{k!}{2! (k – 2)!} + \frac{(n – k)!}{2! (n – k – 2)!}}{\frac{n!}{2! (n – 2)!}} \\
&= \frac{\frac{1}{2} k (k – 1) + \frac{1}{2} (n – k) (n – k – 1)}{\frac{1}{2} n (n – 1)} \\
&= \frac{k (k – 1) + (n – k) (n – k – 1)}{n (n – 1)},
\end{aligned}
\)

which is answer \(A\).

10DSince all four forms:

\(
\begin{align}
R \sin(x + \alpha) \text{,} \tag{1} \\
R \sin(x – \beta) \text{,} \tag{2} \\
R \cos(x + \gamma) \text{,} \tag{3} \\
R \cos(x – \delta) \tag{4}
\end{align}
\)

are equal, we’ll try to find connections between pairs using the fact that \(\sin \theta = \sin \phi\) if and only if \(\theta – \phi = 2 k \pi\) for some integer \(k\), and an analogous fact for \(\cos\). We’ll equate forms (1) with (2) and (3) with (4) to see that:

\(
\begin{align*}
(x + \alpha) – (x – \beta) &= 2 m \pi & (x + \gamma) – (x – \delta) &= 2 n \pi \\
\alpha + \beta &= 2 m \pi & \gamma + \delta &= 2 n \pi
\end{align*}
\)

for some integers \(m\) and \(n\). Since \(0 < \alpha, \beta, \gamma, \delta < 2 \pi\), then both:

\(
\begin{align*}
0 < \alpha &+ \beta < 4 \pi \text{, and} \\
0 < \gamma &+ \delta < 4 \pi.
\end{align*}
\)

This gives that both \(\alpha + \beta\) and \(\gamma + \delta\) must equal \(2 \pi\), as these sums are multiples of \(2 \pi\) in the range \((0, 4 \pi)\). Hence:

\(
\alpha + \beta + \gamma + \delta = 2 \pi + 2 \pi = 4 \pi.
\)

2024 HSC Maths Extension 1 Exam Solutions – Long Response Questions

Question 11

(a).

(i).

\(
\begin{align*}
2 \mathbf{a} – \mathbf{b}
&= 2 (3 \mathbf{i} + 2 \mathbf{j}) – (-\mathbf{i} + 4 \mathbf{j}) \\
&= 6 \mathbf{i} + 4 \mathbf{j} + \mathbf{i} – 4 \mathbf{j} \\
&= 7 \mathbf{i}.
\end{align*}
\)

(ii).

\(
\begin{align*}
\vec{a} \cdot \vec{b}
&= \begin{pmatrix} 3 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 4 \end{pmatrix} \\
&= (3)(-1) + (2)(4) \\
&= 5.
\end{align*}
\)

(b).

Factorising we can see that \(x^2 – 8 x – 9 = (x + 1) (x – 9)\), which corresponds to the following concave up quadratic with \(x\)-intercepts at \(x = -1\) and \(x = 9\):

Parabola with roots at -1 and 9, showing the quadratic equation (x+1)(x-9)

We’re looking for where this curve is intersecting or below the \(x\)-axis, which occurs for \(-1 \leq x \leq 9\) from the diagram.

(c).

Letting \(u = x – 1\), we can see that \(x = u + 1\) and \(du = dx\). So,

\begin{align*}
\int x \sqrt{x – 1} \, dx
&= \int (u + 1) \sqrt u \, du \\
&= \int u^\frac{3}{2} + u^\frac{1}{2} \, du \\
&= \frac{2}{5} u^\frac{5}{2} + \frac{2}{3} u^\frac{3}{2} + C \tag{for some real \(C\)} \\
&= \frac{2}{5} (x – 1)^\frac{5}{2} + \frac{2}{3} (x – 1)^\frac{3}{2} + C.
\end{align*}

(d).

\begin{align*}
\frac{dy}{dx} &= x y \\
\frac{1}{y} \, dy &= x \, dx \\
\int \frac{1}{y} \, dy &= \int x \, dx \\
\ln |y| &= \frac{1}{2} x^2 + C \tag{for some real \(C\)} \\
|y| &= e^{\frac{1}{2} x^2 + C}.
\end{align*}

But since \(y > 0\), \(|y| = y\), and we can write \(y = e^{f(x)}\) where \(f(x) = \frac{1}{2} x^2 + C\).

(e).

We will use chain rule

\(\frac{d}{dx} \left[ f(g(x)) \right] = g'(x) f'(g(x))\)

where
\begin{align*}
f(x) &= \arcsin(x) & g(x) &= x^5 \\
f'(x) &= \frac{1}{\sqrt{1 – x^2}} & g'(x) &= 5 x^4
\end{align*}

This gives
\begin{align*}
\frac{d}{dx} \left[ \arcsin(x^5) \right]
&= 5 x^4 \times \frac{1}{\sqrt{1 – (x^5)^2}} \\
&= \frac{5 x^4}{\sqrt{1 – x^{10}}}
\end{align*}

(f).

We are given that \(\frac{dV}{dt} = 10\) and want to find \(\frac{dr}{dt}\). Using chain rule, we can set up the equation

\begin{align*}
\frac{dr}{dt}
&= \frac{dr}{dV} \times \frac{dV}{dt} \\
&= \left( \frac{dV}{dr} \right)^{-1} \times 10.
\end{align*}

Starting with \(V = \frac{4}{3} \pi r^3\), we can see that

\(\frac{dV}{dr} = 4 \pi r^2\),

and that
\begin{align*}
\frac{dr}{dt}
&= 10 \left( \frac{dV}{dr} \right)^{-1} \\
&= 10 \left( 4 \pi r^2 \right)^{-1} \\
&= \text{\(\frac{5}{2 \pi r^2}\) cm s\(^{-1}\),}
\end{align*}

as required.

(g).

Using the formula \(\mathrm{Area} = \int_a^b \operatorname{top}(x) – \operatorname{bottom}(x) \, dx\) for the area between two curves, we can see that our area is

\begin{align*}
\int_0^\frac{\pi}{2} x – \sin(x) \, dx
&= \left[ \frac{1}{2} x^2 + \cos(x) \right]_0^\frac{\pi}{2} \\
&= \frac{1}{2} \left( \left( \frac{\pi}{2} \right)^2 – 0^2 \right) + \left( \cos \left( \frac{\pi}{2} \right)^2 – \cos(0)^2 \right) \\
&= \text{\(\left( \frac{\pi^2}{8} – 1 \right)\) units\(^2\)}.
\end{align*}

Question 12

(a).

Since these vectors are perpendicular, their dot product must be zero. So:

\(
\begin{aligned}
\begin{bmatrix} a^2 \\ 2 \end{bmatrix} \cdot \begin{bmatrix} a + 5 \\ a – 4 \end{bmatrix} &= 0 \\
a^2 (a + 5) + 2 (a – 4) &= 0 \\
a^3 + 5a^2 + 2a – 8 &= 0.
\end{aligned}
\)

We can now use the remainder theorem on the factors \(\{1, -1, 2, -2, 3, -3, 4, -4, 8, -8\}\) of the constant term \(-8\) to potentially find some of the roots of the polynomial \(f(a) = a^3 + 5a^2 + 2a – 8\).

Testing these values yields \(f(1) = 0\), \(f(-2) = 0\), and \(f(-4) = 0\).

Since \(f(a)\) is a polynomial of degree 3, the three values \(1\), \(-2\), and \(-4\) will be the only roots of \(f(a)\). Importantly, none of these roots make either of the vectors above equal to the zero vector, so:

\(
\begin{bmatrix} a^2 \\ 2 \end{bmatrix} \text{ and } \begin{bmatrix} a + 5 \\ a – 4 \end{bmatrix}
\)

are perpendicular only for \(a = 1\), \(a = -2\), and \(a = -4\).

(b).

We can use the formula \(\mathrm{Volume} = \pi \int_a^b y^2 \, dx\) for finding the volume of a solid of revolution rotated about the \(x\)-axis to find that our desired volume is

\begin{align*}
\pi \int_1^2 (x^3)^2 \, dx
&= \pi \int_1^2 x^6 \, dx \\
&= \pi \left[ \frac{1}{7} x^7 \right]_1^2 \\
&= \frac{\pi}{7} (2^7 – 1^7) \\
&= \text{\(\frac{127 \pi}{7}\) units\(^3\)}
\end{align*}

(c).

We first need to check that we can approximate this distribution with a normal distribution. Letting the number of people the charity talked to be \(n = 100\) and the probability that this person makes a donation be \(p = 0.31\), we can see that:

\(
\begin{align*}
n p &= 100 \times 0.31 \\
&= 31 > 5
\end{align*}
\)

and

\(
\begin{align*}
n (1 – p) &= 100 \times 0.69 \\
&= 69 > 5,
\end{align*}
\)

Now calculate the standard deviation:

\(
\begin{align*}
\sigma
&= \sqrt{\frac{p (1 – p)}{n}} \\
&= \sqrt{\frac{0.31 \times 0.69}{100}} \\
&= \sqrt{0.002139}.
\end{align*}
\)

 

To find the probability:

\(
\begin{align*}
P(\hat{p} \geq 0.35)
&= P\left(Z \geq \frac{0.35 – 0.31}{\sqrt{0.002139}}\right) \\
&= P(Z \geq 0.87) \\
&= 1 – P(Z < 0.87) \\
&= 1 – 0.8078 \tag{Using the table of values} \\
&= 0.1922.
\end{align*}
\)

(d).

Let’s take the base case where \(n = 1\). We can see that in this case:

\(
2^{3n} + 13 = 21,
\)

which is divisible by \(7\). The expression \(2^{3n} + 13\) is thus divisible by \(7\) for \(n = 1\). Let us now assume that \(2^{3n} + 13\) is divisible by \(7\) for \(n = k\), where \(k\) is some positive integer. That is:

\(
2^{3k} + 13 = 7M,
\)

where \(M\) is an integer. Take \(n = k + 1\). We can see in this case that:

\(
\begin{align*}
2^{3n} + 13
&= 2^{3(k + 1)} + 13 \\
&= 2^{3k + 3} + 13 \\
&= 8(2^{3k}) + 13 \\
&= 8(7M – 13) + 13 \tag{by the inductive assumption} \\
&= 56M – 91 \\
&= 7(8M – 13),
\end{align*}
\)

which, since \(M\) is an integer and \(8M – 13\) is thus an integer:

\(
2^{3n} + 13 = 7N,
\)

for some integer \(N\). The expression \(2^{3n} + 13\) is thus divisible by \(7\) for \(n = k + 1\), assuming that is also the case for \(n = k\). This result in addition with the base case concludes our inductive proof.

(e).

We first observe that \(x \neq 5\), before we manipulate the inequality to obtain

\begin{align*}
\frac{x}{6} &\geq \frac{1}{|x – 5|} \\
x |x – 5| &\geq 6 \\
x |x – 5| – 6 &\geq 0.
\end{align*}

We proceed then in two cases. In our first case, take \(|x – 5| = x – 5\). This means that \(x \geq 5\). Under our restriction, though, that \(x \neq 5\), this becomes \(x > 5\). Our main inequality then becomes

\begin{align*}
x |x – 5| – 6 &\geq 0 \\
x (x – 5) – 6 &\geq 0 \\
x^2 – 5 x – 6 &\geq 0 \\
(x + 1) (x – 6) &\geq 0.
\end{align*}

A quick sketch of \((x + 1) (x – 6)\) gives the diagram below:

Parabola with roots at -1 and 6, labeled with the quadratic equation (x+1)(x-6)

It can be seen from this diagram that \((x + 1) (x – 6) \geq 0\) for \(x \leq -1\) or \(x \geq 6\). But, since \(x > 5\), we get \(x \geq 6\) for our first case. In our second case, take \(|x – 5| = 5 – x\). This means that \(x \leq 5\). Under our restriction, though, that \(x \neq 5\), this becomes \(x < 5\). Our main inequality then becomes

\begin{align*}
x |x – 5| – 6 &\geq 0 \\
x (5 – x) – 6 &\geq 0 \\
-x^2 + 5 x – 6 &\geq 0 \\
x^2 – 5 x + 6 &\leq 0 \\
(x – 2) (x – 3) &\leq 0.
\end{align*}

A quick sketch of \((x – 2) (x – 3)\) gives the diagram below:

Parabola with roots at 2 and 3, showing the quadratic equation (x-2)(x-3)

It can be seen from this diagram that \((x – 2) (x – 3) \leq 0\) for \(2 \leq x \leq 3\). The restriction \(x < 5\) does not affect this inequality, so \(2 \leq x \leq 3\) in our second case. Taking both cases together gives \(2 \leq x \leq 3\) or \(x \geq 6\).

Question 13

(a).

(i).

We can see that there is a horizontal asymptote at \(P = 2000\) which prevents a curve satisfying this differential equation passing through both the point \(S\) and the point \(T\).

(ii).

Slope field with a curve showing the solution trajectory and equilibrium point

(iii).

The rate of population growth is given by the quadratic \(\frac{dP}{dt} = P (2000 – P)\) in \(P\), graphed below:

Graph of a derivative function peaking at 1000 and returning to zero at 2000

The maximum occurs at the axis of symmetry, which is \(P = 1000\).

(b).

(i).

Working on the right-hand-side, we can see that

\(
\begin{align*}
\mathrm{RHS}
&= \frac{1}{2} \left( 1 + \cos^2(2x) \right) \\
&= \frac{1}{2} \left( 1 + \left( \cos^2(x) – \sin^2(x) \right)^2 \right) \\
&= \frac{1}{2} \left({1} + \cos^4(x) + \sin^4(x) – 2 \sin^2(x) \cos^2(x) \right) \\
&= \frac{1}{2} \left({\left(\sin^2(x) + \cos^2(x)\right)^2} + \cos^4(x) + \sin^4(x) – 2 \sin^2(x) \cos^2(x) \right) \\
&= \frac{1}{2} \left( 2 \cos^4(x) + 2 \sin^4(x) \right) \\
&= \cos^4(x) + \sin^4(x) \\
&= \mathrm{LHS}
\end{align*}
\)

(ii).

\begin{align*}
{\int_0}^\frac{\pi}{4} \cos^4(x) + \sin^4(x) \, dx &= \frac{1}{2}{\int_0}^\frac{\pi}{4} 1+\cos^2(2x) \, dx \\
&= \frac{1}{2} {\int_0}^\frac{\pi}{4} 1\, dx +\frac{1}{2}{\int_0}^\frac{\pi}{4} \cos^2(2x) \, dx \\
&=\frac{1}{2}\biggl[x\biggr]^\frac{\pi}{4}_0+\frac{1}{2}\times \frac{1}{2}{\int_0}^\frac{\pi}{4} 1+\cos(4x) \, dx \\
&=\frac{1}{2} \left\{ \frac{\pi}{4}-0 \right\}+\frac{1}{4}\biggl[x+\frac{\sin (4x)}{4}\biggr]^\frac{\pi}{4}_0 \\
&=\frac{\pi}{8}+\frac{1}{4}\left\{\frac{\pi}{4}+0-0\right\} \\
&=\frac{\pi}{8} +\frac{\pi}{16} \\
&=\frac{3\pi}{16}
\end{align*}

(c).

Let \(\vec{x}\) be the vector \(m \vec{i} + n \vec{j}\), and let’s turn these two pieces of information about the projections of \(\vec{x}\) into simultaneous linear equations of the form:

\(
\begin{align*}
\frac{\vec{x} \cdot \vec{a}}{\vec{a} \cdot \vec{a}} \vec{a} &= k \vec{a}
&
\frac{\vec{x} \cdot \vec{b}}{\vec{b} \cdot \vec{b}} \vec{b} &= p \vec{b}
\\
\left( \frac{\vec{x} \cdot \vec{a}}{\vec{a} \cdot \vec{a}} \right) \vec{a} &= k \vec{a}
&
\left( \frac{\vec{x} \cdot \vec{b}}{\vec{b} \cdot \vec{b}} \right) \vec{b} &= p \vec{b}
\\
k &= \frac{\vec{x} \cdot \vec{a}}{\vec{a} \cdot \vec{a}} \quad \text{(since $\vec{a} \neq \vec{0}$)}
&
p &= \frac{\vec{x} \cdot \vec{b}}{\vec{b} \cdot \vec{b}} \quad \text{(since $\vec{b} \neq \vec{0}$)}
\\
&= \frac{m + 3 n}{10}
&
&= \frac{2 m – n}{5}
\\
m + 3 n &= 10 k \quad \text{(1)}
&
2 m – n &= 5 p \quad \text{(2)}
\end{align*}
\)

 

Then solve for \(m\) and \(n\) as:

\(
\begin{align*}
(1) + 3 \times (2) :
m + 6 m &= 10 k + 15 p
&
2 \times (1) – (2) :
6 n + n &= 20 k – 5 p
\\
7 m &= 5 (2 k + 3 p)
&
7 n &= 5 (4 k – p)
\\
m &= \frac{5}{7} (2 k + 3 p)
&
n &= \frac{5}{7} (4 k – p)
\end{align*}
\)

We may thus write \(\vec{x}\) as:

\(
\vec{x} = \begin{bmatrix} m \\ n \end{bmatrix} = \frac{5}{7} \begin{bmatrix} 2 k + 3 p \\ 4 k – p \end{bmatrix}.
\)

 

(d).

With the substitution \(u = e^x + 2 e^{-x}\), we can see that
$$
du = \left( e^x – 2 e^{-x} \right) dx
\qquad \text{and} \qquad
u^2 = e^{2 x} + 4 e^{-2 x} + 4,
$$
so that our integral becomes
\begin{align*}
\int \frac{e^{3 x} – 2 e^x}{4 + 8 e^{2 x} + e^{4 x}} \, dx
&= \int \frac{e^x – 2 e^{-x}}{4 e^{2 x} + 8 + e^{2 x}} \, dx \\
&= \int \frac{1}{4 + u^2} \, du \\
&= \frac{1}{2} \tan^{-1} \left( \frac{u}{2} \right) + C \tag{for some real $C$} \\
&= \frac{1}{2} \tan^{-1} \left( \frac{1}{2} e^x + e^{-x} \right) + C.
\end{align*}

Question 14

(a).

We can rewrite \(\frac{dy}{dx}\) using index laws:

\(
\begin{align*}
\frac{dy}{dx} &= e^{x+y} \\
&= e^x \cdot e^y
\end{align*}
\)

 

Integrating to find \(y\):

\begin{align*}
\int dy &= \int e^x \cdot e^y \, dy \\
\int e^{-y} \, dy &= \int e^x \, dy \\
-e^{-y} &= e^x + C \tag{for some real \(C\)}
\end{align*}

Substitute \((0, 0)\):

\(
\begin{align*}
-e^{0} &= e^0 + C \\
-1 &= 1 + C \\
C &= -2 \\
\unicode{x2234} -e^{-y} &= e^x – 2 \\
e^{-y} &= 2 – e^x \\
e^y &= \left(2 – e^x \right)^{-1} \\
y &= \ln \left[\left(2 – e^x \right)^{-1}\right] \\
&= -\ln \left(2 – e^x \right)
\end{align*}
\)

 

\(
\begin{align*}
0 &< e^x < 2 \\
0 &< e^x < e^{\ln 2} \\
\unicode{x2234} x &< \ln 2
\end{align*}
\)

 

Since \(e^x > 0\):
\(
\begin{align*}
2 – e^x &< 2 \\
\ln(2 – e^x) &< \ln(2) \\
-\ln(2 – e^x) &> -\ln(2) \\
y &> -\ln 2
\end{align*}
\)

\(\unicode{x2234}\) The domain is the interval \((-\infty, \ln 2)\), and the range is the interval \((-\ln 2, \infty)\).

(b).

For the function to have an inverse, \(f(x)\) must be a one-to-one function. So either \(f'(x)\geq0\) for all \(x\) or \(f'(x)\leq0\) for all \(x\).

If \(f(x)=\frac{kx}{1+x^2}+\arctan (x)\) then:

\begin{align*}
f'(x) &= \frac{k(1+x^2)-2x(kx)}{(1+x^2)^2}+\frac{1}{1+x^2}\\
&=\frac{k+kx^2-2kx^2+1+x^2}{(1+x^2)^2} \\
&=\frac{x^2-kx^2+1+k}{(1+x^2)^2} \\
&=\frac{x^2(1-k)+1+k}{(1+x^2)^2} \\
\end{align*}

Since \((1+x^2)^2\geq0\) for all \(x\), we need to investigate the sign of the numerator of \(f'(x)\).

Let \(g(x)=x^2(1-k)+1+k\).

  • Case 1: \(g(x)\) is a parabola. Then:
    \begin{align*}
    1 – k &\neq 0 \\
    k &\neq 1
    \end{align*}

If \(g(x)\) is positive definite, both the coefficient of \(x^2\) and the constant term should be greater than zero:

\begin{align*}
1-k &\geq 0 \\
k &\leq 1 \tag{but \(k\neq 1\)} \\
\unicode{x2234} k &<1
\end{align*}
\begin{align*}
1 + k &\geq 0 \\
k &\geq -1 \\
\unicode{x2234} -1 &\leq k < 1 \tag{Combining both inequalities}
\end{align*}

If  \(g(x)\) is negative definite:
\begin{align*}
1 – k &\leq 0 \\
k &\geq 1 \tag{but \(k \neq 1\)} \\
\unicode{x2234} k &> 1
\end{align*}
\begin{align*}
1 + k &\leq 0 \\
k &\leq -1
\end{align*}

But these two inequalities cannot both be true at the same time. Therefore, \(g(x)\) is not negative definite.

  • Case 2: \(g(x)\) is not a parabola. Then:
    \begin{align*}
    1 – k &= 0 \\
    k &= 1
    \end{align*}

Then: \(g(x) = 0x^2 + 1 + 1 = 2\), which is always positive.

\(\unicode{x2234}\) The values of  \(k\) for which \(f(x)\) has an inverse are: \(-1 \leq k \leq 1\).

(c).

(i).

\begin{align*}
\text{Let }y &= \tan^{-1}(3x)+\tan^{-1}(10x) \\
\frac{dy}{dx} &= \frac{1}{1+(3x)^2}+\frac{1}{1+(10x)^2} > 0 \tag{for all real \(x\)}
\end{align*}

\(\unicode{x2234} \, y = \tan^{-1}(3x) + \tan^{-1}(10x)\) is increasing for all \(x\) in its domain.

So for any constant \(\theta\), \(y = \theta\) and \(y = \tan^{-1}(3x) + \tan^{-1}(10x)\) will only have one point of intersection.

\(\unicode{x2234}\) There is only 1 solution for \(\theta = \tan^{-1}(3x) + \tan^{-1}(10x)\).

(ii).

\begin{align*}
\text{Let } a&= \tan^{-1}(3x) \to \tan(a) =3x\\
\text{Let } b&= \tan^{-1}(10x) \to \tan(b) =10x\\
\text{Where } a+b&=\frac{3\pi}{4} \in (-\pi,\pi) \\
\unicode{x2234} \tan(a+b) &= \tan\frac{3\pi}{4} \\
\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)} &= -1 \\
\frac{3x+10x}{1-3x\times 10x}&= -1 \\
\frac{13x}{1-30x^2}&= -1 \\
13x &= 30x^2-1 \tag{where \(30x^2\neq 1\). i.e. \(x\neq \pm \frac{1}{\sqrt{30}}\)} \\
30x^2-13x-1&=0 \\
x&= \frac{13\pm \sqrt{13^2+4\times 30 \times 1}}{60} \\
&= \frac{13\pm 17}{60} \\
x&= \frac{1}{2}, \, -\frac{1}{15}
\end{align*}

Check solution \(x = \frac{1}{2}\):

\(
\begin{align*}
\tan^{-1}(3x) + \tan^{-1}(10x) &= \tan^{-1}\left(\frac{3}{2}\right) + \tan^{-1}(5), \\
&\approx 0.9828 + 1.3734, \\
&\approx \frac{3\pi}{4}.
\end{align*}
\)

\(\unicode{x2234}\) the solution \(x = \frac{1}{2}\) is valid.

Check solution \(x = -\frac{1}{15}\):

\(
\begin{align*}
\tan^{-1}(3x) + \tan^{-1}(10x) &= \tan^{-1}\left(-\frac{1}{5}\right) + \tan^{-1}\left(-\frac{2}{3}\right), \\
&\approx -0.1974 + (-0.5880), \\
&\not\approx \frac{3\pi}{4}.
\end{align*}
\)

\(\unicode{x2234}\) the solution \(x = -\frac{1}{15}\) is not valid.

\(\unicode{x2234}\) the only solution is \(x = \frac{1}{2}\).

(d).

\(
\vec{r}(t) = \begin{bmatrix} Vt\cos \theta \\ Vt\sin \theta – \frac{gt^2}{2} \end{bmatrix}
\)

 

\(
\begin{align*}
D(t) &= |\vec{r}(t)|, \\
&= \sqrt{\left(Vt\cos \theta\right)^2 + \left(Vt\sin \theta – \frac{gt^2}{2}\right)^2}, \\
&= \sqrt{V^2t^2\cos^2 \theta + V^2t^2\sin^2 \theta + \frac{g^2t^4}{4} – Vgt^3\sin \theta}, \\
&= \sqrt{V^2t^2 + \frac{g^2t^4}{4} – Vgt^3\sin \theta}, \\
&= \left(V^2t^2 + \frac{g^2t^4}{4} – Vgt^3\sin \theta \right)^{\frac{1}{2}}.
\end{align*}
\)

Using the chain rule:

\(
\begin{align*}
\frac{dD}{dt} &= \frac{2tV^2 + g^2t^3 – 3Vgt^2 \sin \theta}{2\sqrt{V^2t^2 + \frac{g^2t^4}{4} – Vgt^3\sin \theta}} \\
&= \frac{2tV^2 + g^2t^3 – 3Vgt^2 \sin \theta}{2t\sqrt{V^2 + \frac{g^2t^2}{4} – Vgt\sin \theta}} \\
&= \frac{2V^2 + g^2t^2 – 3Vgt \sin \theta}{2\sqrt{V^2 + \frac{g^2t^2}{4} – Vgt\sin \theta}}
\end{align*}
\)

 

If \(D(t)\) is increasing for all \(t > 0\), then \(\frac{dD}{dt} > 0\) for all \(t > 0\):

\(
\begin{align*}
\frac{2V^2 + g^2t^2 – 3Vgt \sin \theta}{2\sqrt{V^2 + \frac{g^2t^2}{4} – Vgt\sin \theta}} &> 0
\end{align*}
\)

 

So we need to solve for:

\(
\begin{align*}
2V^2 + g^2t^2 – 3Vgt\sin \theta &> 0 \tag{Since the denominator is always greater than 0}
\end{align*}
\)

 

Let \(h(t) = g^2t^2 – 3Vgt + 2V^2\).
\(h(t)\) is a concave-up parabola since \(g \neq 0\), \(g^2 > 0\).
\(\unicode{x2234}\) If the minimum turning point of \(h(t)\) has a height above zero, then \(h(t) > 0\) for all \(t > 0\).

To find the turning point, we can rewrite \(h(t)\) in vertex form.
\(
\begin{align*}
h(t) &= g^2\left(t^2 – \frac{3Vt}{g}\sin \theta\right) + 2V^2 \\
&= g^2\left[\left(t – \frac{3V}{2g}\sin \theta\right)^2 – \frac{9V^2}{4g^2}\sin^2 \theta\right] + 2V^2 \tag{By completing the square} \\
&= g^2\left(t – \frac{3V}{2g}\sin \theta\right)^2 – \frac{9V^2}{4}\sin^2 \theta + 2V^2.
\end{align*}
\)

So the minimum turning point of \(h(t)\) is at:

\(
\left(\frac{3V}{2g}\sin \theta, 2V^2 – \frac{9V^2}{4}\sin^2\theta \right).
\)

 

For \(h(t)\) to be positive definite:

\(
\begin{align*}
2V^2 – \frac{9V^2}{4}\sin^2 \theta &> 0, \\
2 – \frac{9}{4}\sin^2 \theta &> 0, \\
\frac{9}{4}\sin^2 \theta &< 2, \\
\sin^2 \theta &< \frac{8}{9}, \\
\unicode{x2234} -\sqrt{\frac{8}{9}} < \sin \theta &< \sqrt{\frac{8}{9}}.
\end{align*}
\)

But \(0 < \theta < \frac{\pi}{2}\):

\(
\begin{align*}
\unicode{x2234} 0 < \sin \theta &< \sqrt{\frac{8}{9}}, \\
\theta &< \sin^{-1} \left(\frac{2 \sqrt{2}}{3}\right).
\end{align*}
\)

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Written by Oak Ukrit

Oak is the Head of Mathematics at Matrix Education and has been teaching for over 12 years and has been helping students at Matrix since 2016. He has 1st class honours in Aeronautical Engineering from UNSW where he taught for over 4 years while he was undertaking a PhD. When not plane spotting he enjoys landscape photography.

© Matrix Education and www.matrix.edu.au, 2023. Unauthorised use and/or duplication of this material without express and written permission from this site’s author and/or owner is strictly prohibited. Excerpts and links may be used, provided that full and clear credit is given to Matrix Education and www.matrix.edu.au with appropriate and specific direction to the original content.

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