2024 H2 Maths Paper 2 Q5 — Worked Solution

Part (a)(i)

1 mark

Since $y = f(x)$ meets the axes at $(a,0)$ and $(0,b)$ , and $y = f^{- 1}(x)$ is the reflection of $y = f(x)$ in the line $y = x$ , each labelled point swaps its coordinates: $(a,0)$ maps to $(0,a)$ , and $(0,b)$ maps to $(b,0)$ .

$\therefore$ $y = f^{- 1}(x)$ is the reflection of $y = f(x)$ in $y = x$ , cutting the axes at $(0,a)$ and $(b,0)$ .

Part (a)(ii)

2 marks

Approach

$y = 2f^{- 1}(x - 1)$ is built from $y = f^{- 1}(x)$ by two transformations: replace $x$ with $x - 1$ (translation $\text{+}1$ in the $x$ -direction), then double the resulting $y$ -values (stretch, scale factor $2$ , parallel to the $y$ -axis). Only the $x$ -intercept is required, and a $y$ -stretch does not move a point already on the $x$ -axis, so we only need to track how the translation moves the $x$ -intercept of $y = f^{- 1}(x)$ .

From part (i), $y = f^{- 1}(x)$ has $x$ -intercept $(b,0)$ . Setting $y = 0$ in the transformed curve,

2f^{- 1}(x - 1) = 0

\Rightarrow f^{- 1}(x - 1) = 0

\Rightarrow x - 1 = b

\Rightarrow x = b + 1

$\therefore$ $y = 2f^{- 1}(x - 1)$ has $x$ -intercept $(b + 1,0)$ .

Part (b)(i)

2 marks

$y = \left| g(x) \right|$ agrees with $y = g(x)$ wherever $g(x) \geq 0$ , and reflects upward (in the $x$ -axis) the part where $g(x) < 0$ . The dip between $s$ and $t$ , where $g(x) < 0$ , is reflected up so the curve touches the $x$ -axis at $s$ and $t$ instead of crossing it; the arch from $t$ to $u$ , already non-negative, is unchanged; beyond $u$ , where $g(x)$ falls below the axis, that branch is reflected to rise instead.

$\therefore$ $y = \left| g(x) \right|$ touches the $x$ -axis at $(s,0)$ and $(t,0)$ , meets it at $(u,0)$ , and cuts the $y$ -axis at $(0,v)$ .

Part (b)(ii)

2 marks

Approach

$y = g\left( |x| \right)$ is an even function, so the sketch has two parts: for $x \geq 0$ it is identical to $y = g(x)$ , and for $x < 0$ it is the mirror image, in the $y$ -axis, of that same $x \geq 0$ branch — the original branch of $g$ for $x < 0$ is discarded entirely.

The right-hand branch of $g$ passes through $(0,v)$ , rises to a local maximum, then falls through $(u,0)$ . Reflecting this branch in the $y$ -axis gives a symmetric curve with a local minimum at $(0,v)$ , rising to two symmetric maxima, and crossing the $x$ -axis at $(u,0)$ and $( - u,0)$ .

$\therefore$ $y = g\left( |x| \right)$ is the even function agreeing with $g$ for $x \geq 0$ and mirrored for $x < 0$ , cutting the axes at $( - u,0)$ , $(u,0)$ and $(0,v)$ .

Part (c)(i) and (ii)

3 marks

Approach

to sketch a parametric curve we eliminate $t$ to find its Cartesian shape, then use the endpoints of the given range of $t$ and the axis-crossings to fix where the sketch starts, stops, and meets the axes.

Since $y = 2t + 3$ ,

t = \frac{y - 3}{2}

Substituting into $x = t^{2}$ ,

x = \left( \frac{y - 3}{2} \right)^{2}

= \frac{(y - 3)^{2}}{4}

This is a sideways parabola, opening in the positive $x$ -direction, with vertex at $(0,3)$ (at $t = 0$ ).

Endpoints: $t = - 4$ gives $(16, - 5)$ ; $t = 4$ gives $(16,11)$ .

$y$ -axis: $x = 0 \Rightarrow t^{2} = 0 \Rightarrow t = 0 \Rightarrow y = 3$ , giving $(0,3)$ — this is also the vertex, since $t = 0$ is the only value of $t$ with $x = 0$ .

$x$ -axis: $y = 0 \Rightarrow 2t + 3 = 0 \Rightarrow t = - 1.5$ , which lies in $\lbrack - 4,4\rbrack$ , giving $x = ( - 1.5)^{2} = 2.25$ , so $(2.25,0)$ .

$\therefore$ $C$ runs from $(16, - 5)$ through $(2.25,0)$ and the vertex $(0,3)$ to $(16,11)$ .

Part (c)(iii)

1 mark

The curve is symmetric about the horizontal line through its vertex $(0,3)$ .

$\therefore$ the line of symmetry is $y = 3$ .

2024 H2 Maths Paper 2, Question 5

Part (a)(i)

Part (a)(ii)

Part (b)(i)

Part (b)(ii)

Part (c)(i) and (ii)

Part (c)(iii)

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