2015 H2 Maths Paper 2 Q3 — Worked Solution

Part (a)(i)

2 marks

Approach

to show f has an inverse, we show f is one-one on its domain $x > 1$ , by showing f is strictly monotonic there.

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

Differentiating,

f'(x) = \frac{2x}{\left( 1 - x^{2} \right)^{2}}

For $x > 1$ , $2x > 0$ and $\left( 1 - x^{2} \right)^{2} > 0$ , so $f'(x) > 0$ for all $x$ in the domain. Thus f is strictly increasing on $x > 1$ , so f is one-one, and therefore f has an inverse. (shown)

Part (a)(ii)

3 marks

Let $y = f(x) = \frac{1}{1 - x^{2}}$ .

1 - x^{2} = \frac{1}{y}

x^{2} = 1 - \frac{1}{y}

x = \sqrt{1 - \frac{1}{y}}\ \ \ \left( \text{taking the positive root, since }x > 1 \right)

$\therefore$ $f^{- 1}(x) = \sqrt{1 - \frac{1}{x}}$ .

The domain of $f^{- 1}$ is the range of f. Since f is continuous and strictly increasing on $x > 1$ , its range runs between the limiting values

\lim_{x \rightarrow 1^{\text{+}}}f(x) = - \infty,\ \ \ \lim_{x \rightarrow \infty}f(x) = 0^{-}

so the range of f is $( - \infty,0)$ .

$\therefore$ the domain of $f^{- 1}$ is $\{ x \in \mathbb{R}:x < 0\}$ .

To check your answer:

f(2) = \frac{1}{1 - 4} = - \frac{1}{3}

, and

f^{- 1}\left( - \frac{1}{3} \right) = \sqrt{1 - \frac{1}{- 1/3}} = \sqrt{1 + 3} = 2

, recovering the original input. ✓

Part (b)

5 marks

Approach

to find the range of g algebraically, we set $y = g(x)$ , rearrange into a quadratic in $x$ , and use the condition that $x$ is real (discriminant $\text{≥}0$ ) to find the possible values of $y$ .

Let $y = g(x) = \frac{2 + x}{1 - x^{2}}$ .

y\left( 1 - x^{2} \right) = 2 + x

yx^{2} + x + (2 - y) = 0

Since $x$ is real, the discriminant of this quadratic in $x$ must be non-negative. (This also covers $y = 0$ , where the equation reduces to the linear equation $x = - 2$ , which does give a real $x$ .)

1 - 4y(2 - y) \geq 0

4y^{2} - 8y + 1 \geq 0

Solving $4y^{2} - 8y + 1 = 0$ ,

y = \frac{8 \pm \sqrt{64 - 16}}{8} = \frac{8 \pm 4\sqrt{3}}{8} = \frac{2 \pm \sqrt{3}}{2}

Since the coefficient of $y^{2}$ is positive, $4y^{2} - 8y + 1 \geq 0$ outside the roots:

y \leq \frac{2 - \sqrt{3}}{2}\ \ \text{or}\ \ y \geq \frac{2 + \sqrt{3}}{2}

Note: the excluded values $x = 1$ and $x = - 1$ never solve $yx^{2} + x + (2 - y) = 0$ for any $y$ — substituting $x = 1$ gives $0 = 3$ , and $x = - 1$ gives $0 = 1$ , both false regardless of $y$ — so no further values of $y$ need to be excluded from the range.

$\therefore$ the range of g is $\left\{ y \in \mathbb{R}:y \leq \frac{2 - \sqrt{3}}{2}\ \text{or}\ y \geq \frac{2 + \sqrt{3}}{2} \right\}$ .

2015 H2 Maths Paper 2, Question 3

Part (a)(i)

Part (a)(ii)

Part (b)

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