2021 H2 Mathematics Paper 2 Question 3

Functions

Answers

(ai)
gh(2)=14{gh(2)=\frac{1}{4}}
(aii)
x=0.4{x = 0.4}
(bi)
k=−b2{k=-\frac{b}{2}} since f(−b2){f\left(-\frac{b}{2}\right)} is undefined
(bii)
b=−1{b=-1}
a∈R,a≠−12{a \in \mathbb{R}, a \neq -\frac{1}{2}}
(biii)
f−1(−4)=4−a9{f^{-1}(-4) = \frac{4-a}{9}}

Full solutions

(ai)
gh(2)=g(12(2)2+3)=g(5)=5+15(5)−1=14  ■\begin{align*} gh(2) &= g\left( \frac{1}{2}(2)^2 + 3 \right) \\ &= g\left( 5 \right) \\ &= \frac{5+1}{5\left(5\right)-1} \\ &= \frac{1}{4} \; \blacksquare \end{align*}
(aii)
g(x)=1.4x+15x−1=1.4x+1=7x−1.46x=2.4x=0.4  ■\begin{align*} g(x) &= 1.4 \\ \frac{x + 1}{5 x - 1} &= 1.4 \\ x + 1 &= 7x - 1.4 \\ 6x &= 2.4 \\ x &= 0.4 \; \blacksquare \end{align*}
(bi)
k=−b2  ■k=-\frac{b}{2} \; \blacksquare
since f(−b2){f\left(-\frac{b}{2}\right)} is undefined ■{\blacksquare}
(bii)
y=x+a2x+b2xy+by=x+ax(2y−1)=a−byx=a−by2y−1f−1(x)=−bx+a2x−1\begin{gather*} y = \frac{x+a}{2x+b} \\ 2xy + by = x + a \\ x(2y-1) = a - by \\ x = \frac{a-by}{2y-1} \\ f^{-1}(x) = \frac{-bx+a}{2x-1} \end{gather*}
If f(x)=f−1(x),{f(x)=f^{-1}(x),}
−bx+a2x−1=x+a2x+b\frac{-bx+a}{2x-1} = \frac{x+a}{2x+b}
Comparing,
b=−1  ■b=-1 \; \blacksquare
If a=−12,{a=-\frac{1}{2}, } then
f(x)=x−122x−1=12\begin{align*} f(x) &= \frac{x-\frac{1}{2}}{2x-1} \\ &= \frac{1}{2} \end{align*}
which does not have an inverse as it is not one-one
Hence a{a} can take any real value except −12{-\frac{1}{2}}
a∈R,a≠−12  ■a \in \mathbb{R}, a \neq -\frac{1}{2} \; \blacksquare
(biii)
f−1(−4)=f(−4)=−4+a2(−4)−1=−4+a−9=4−a9  ■\begin{align*} f^{-1}(-4) &= f(-4) \\ &= \frac{-4+a}{2(-4)-1} \\ &= \frac{-4+a}{-9} \\ &= \frac{4-a}{9} \; \blacksquare \end{align*}