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October 15, 2026 by

Free practice / JC1 H2 Maths Promo

Functions: VJC 2024 Promo Q11

Attempt the question on paper, then open the worked solution to check your method.

Victoria Junior College2024 PromoPromo Q11●●● Challenging11 marks

What this question tests

Functions, inverses and composition.

Question

The functions $f$ and $g$ are defined by $f:x\mapsto\frac{5x+13}{x^2-1}\quad\text{for }x\in\mathbb R,\ x\ne\pm1,\qquad g:x\mapsto-\ln(x-1)\quad\text{for }x\geq2.$ (a)Sketch the graph of $y=f(x)$ , stating the equations of any asymptotes, the coordinates of any turning points and points of intersections with the axes. [4]

(b)Explain why $gf$ does not exist. [1]

(c)Show that $g$ has an inverse and find $g^{-1}(x)$ . [4]

(d)Find the range of $fg^{-1}$ . [2]

Show full worked solution▾

(a)

$Victoria Junior College 2024 Promo Q11: solution diagram$

11(b) $0\in R_f$ but $0\notin D_g$ $R_f\nsubseteq D_g$ , $gf$ does not exist.

11(c)

$Victoria Junior College 2024 Promo Q11: solution diagram$

Every horizontal line $y=k$ cuts the graph of $y=g(x)$ at most once. Hence, $g$ is one-to-one. Inverse of $g$ exists.

$y=-\ln(x-1)\qquad e^{-y}=x-1\qquad x=e^{-y}+1$ $g^{-1}(x)=e^{-x}+1$

11(c) $R_{g^{-1}}=D_g=[2,\infty)$ $R_{g^{-1}}$ be the new domain of $f$ the graph in part (a), $R_{fg^{-1}}=\left(0,\dfrac{23}{3}\right]$

Alternatively

Sketch the graph $y=fg^{-1}(x)=\dfrac{5(e^{-x}+1)+13}{(e^{-x}+1)^2-1}$ for

$D_{g^{-1}}=R_g=(-\infty,0]$ and $R_{fg^{-1}}=\left(0,\dfrac{23}{3}\right]$

$Victoria Junior College 2024 Promo Q11: solution diagram$

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