2021 H2 Maths Paper 2 Q3 — Worked Solution

Part (a)(i)

2 marks

We first evaluate $h(2)$ .

h(2) = \frac{1}{2}(2)^{2} + 3

= 2 + 3

= 5

Since $5 \neq 0.2$ , this lies in the domain of $g$ , so we substitute into $g$ .

gh(2) = g(5)

= \frac{5 + 1}{5(5) - 1}

= \frac{6}{24}

= \frac{1}{4}

To check your answer:

h(2) = 5

is within the domain of

g

(since

5 \neq 0.2

), and

g(5) = \frac{6}{24} = 0.25

, a valid output. ✓

$\therefore$ $gh(2) = \frac{1}{4}$ .

Part (a)(ii)

1 mark

x = \frac{2}{5}

Explanation: setting $g(x) = 1.4 = \frac{7}{5}$ and solving,

\frac{x + 1}{5x - 1} = \frac{7}{5}

\Rightarrow 5(x + 1) = 7(5x - 1)

\Rightarrow 5x + 5 = 35x - 7

\Rightarrow 30x = 12

\Rightarrow x = \frac{2}{5}

Part (b)(i)

2 marks

k = - \frac{b}{2}

Explanation: $f$ is undefined when its denominator vanishes, i.e. when $2x + b = 0$ , giving $x = - \frac{b}{2}$ . Since division by zero is undefined, this value of $x$ must be excluded from the domain of $f$ .

Part (b)(ii)

3 marks

Approach

$f = f^{- 1}$ means applying $f$ twice returns $x$ for every $x$ in the domain, so we find $f^{- 1}$ explicitly and require $f^{- 1}(x) = f(x)$ identically. This produces an equation that must hold for every $x$ , not just specific values, so we collect the result by powers of $x$ and set each coefficient to zero.

Let $y = \frac{x + a}{2x + b}$ and solve for $x$ in terms of $y$ .

y(2x + b) = x + a

\Rightarrow 2xy + by = x + a

\Rightarrow 2xy - x = a - by

\Rightarrow x(2y - 1) = a - by

\Rightarrow x = \frac{a - by}{2y - 1}

Replacing $y$ by $x$ , $f^{- 1}(x) = \frac{a - bx}{2x - 1}$ . Setting $f(x) = f^{- 1}(x)$ for all $x$ in the domain,

\frac{x + a}{2x + b} = \frac{a - bx}{2x - 1}

Cross-multiplying and expanding,

(x + a)(2x - 1) = (a - bx)(2x + b)

\Rightarrow 2x^{2} - x + 2ax - a = 2ax + ab - 2bx^{2} - b^{2}x

The $2ax$ terms cancel from both sides. Since this must hold for every $x$ , we collect by powers of $x$ and equate each coefficient to zero.

(2 + 2b)x^{2} + \left( b^{2} - 1 \right)x - a(1 + b) = 0\ \text{for all}\ x

\Rightarrow 2 + 2b = 0,\quad b^{2} - 1 = 0,\quad a(1 + b) = 0

From the first equation, $b = - 1$ , which also satisfies $b^{2} - 1 = 0$ . With $b = - 1$ , the third equation becomes $a(0) = 0$ , true for every value of $a$ , so $a$ is not fixed by this condition.

Note: $f$ must still be a genuine function with an inverse, so its numerator cannot be a scalar multiple of its denominator (this would make $f$ constant). With $b = - 1$ , $x + a = \frac{1}{2}(2x - 1)$ when $a = - \frac{1}{2}$ , giving $f(x) = \frac{1}{2}$ identically, which has no inverse, so this value of $a$ is excluded.

To check your answer: with

b = - 1

and, say,

a = 5

:

f(x) = \frac{x + 5}{2x - 1}

; computing

f\left( f(x) \right)

symbolically gives exactly

x

, confirming

f

is self-inverse. ✓

$\therefore$ $b = - 1$ , and $a$ may be any real number except $a = - \frac{1}{2}$ .

Part (b)(iii)

1 mark

f^{- 1}( - 4) = \frac{4 - a}{9}

Explanation: since $f = f^{- 1}$ , $f^{- 1}( - 4) = f( - 4)$ .

f( - 4) = \frac{- 4 + a}{2( - 4) + ( - 1)}

= \frac{a - 4}{- 9}

= \frac{4 - a}{9}

2021 H2 Maths Paper 2, Question 3

Part (a)(i)

Part (a)(ii)

Part (b)(i)

Part (b)(ii)

Part (b)(iii)

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