2023 H2 Maths Paper 1, Question 7

Writing $2x + 4 = - 2(3 - x) + 10$,

As $x \rightarrow \pm \infty$, $\frac{10}{3 - x} \rightarrow 0$, so $g(x) \rightarrow - 2$: $g$ has horizontal asymptote $y = - 2$, so $f = |g|$ has horizontal asymptote $y = 2$. There is a vertical asymptote $x = 3$, where the denominator vanishes.

so $g$ is strictly increasing on each branch. Since $g( - 2) = 0$, $g(x) < 0$ for $x < - 2$ and $g(x) > 0$ for $- 2 < x < 3$; and $g(x) < 0$ throughout $x > 3$, since $g \rightarrow - \infty$ as $x \rightarrow 3^{\text{+}}$ and $g \rightarrow - 2$ as $x \rightarrow \infty$.

Taking $f = |g|$: $f$ decreases from $2$ (as $x \rightarrow - \infty$) to $f( - 2) = 0$, a cusp where $g$ changes sign, then increases from $0$ to $\text{+}\infty$ as $x \rightarrow 3^{-}$; for $x > 3$, $f$ decreases from $\text{+}\infty$ towards $2$.

$x$-intercept: $g(x) = 0 \Rightarrow 2x + 4 = 0 \Rightarrow x = - 2$, giving $( - 2,0)$. $y$-intercept: $f(0) = \left| \frac{4}{3} \right| = \frac{4}{3}$, giving $\left( 0,\frac{4}{3} \right)$.

From the sketch, the branch $- 2 \leq x < 3$ alone rises continuously from $f( - 2) = 0$ to $\text{+}\infty$, covering every value in $\lbrack 0,\infty)$.

$f^{2}$ does not exist.

Explanation: for $f \circ f$ to exist, the range of $f$ must be a subset of the domain of $f$. The domain of $f$ is all real $x$ except $x = 3$, but the range $\lbrack 0,\infty)$ includes $3$ — e.g. $f(1) = \left| \frac{6}{2} \right| = 3$ — so $f\left( f(1) \right) = f(3)$ is undefined.

$a = - 2$

Explanation: from the sketch, $f$ decreases on $( - \infty, - 2\rbrack$ and increases on $\lbrack - 2,3)$, with a cusp (minimum) at $x = - 2$. Restricting to $x \leq a$ keeps $f$ one-one, so that $f^{- 1}$ exists, only while the domain avoids containing both a decreasing and an increasing portion; the greatest such $a$ is where the turning point occurs.

With domain restricted to $x \leq - 2$, $g(x) \leq 0$ throughout, so $f(x) = - g(x) = 2 - \frac{10}{3 - x}$.

Let $y = 2 - \frac{10}{3 - x}$ and solve for $x$ in terms of $y$.

Replacing $y$ by $x$: $f^{- 1}(x) = 3 + \frac{10}{x - 2}$.

The domain of $f^{- 1}$ equals the range of the restricted $f$: on $x \leq - 2$, $f( - 2) = 0$ and $f \rightarrow 2$ (not attained) as $x \rightarrow - \infty$, so the range of the restricted $f$ is $\lbrack 0,2)$.