2015 H2 Maths Paper 1, Question 2

Approach: Write $f(x) = \frac{x + 1}{1 - x}$ so that the required curve is $y = \left| f(x) \right|$. It is easiest to sketch $y = f(x)$ first and then reflect the part below the $x$-axis.

Since $f$ is undefined at $x = 1$, there is a vertical asymptote $x = 1$.

As $x \rightarrow \pm \infty$, $\frac{2}{x - 1} \rightarrow 0$, so $f(x) \rightarrow - 1$: the horizontal asymptote of $f$ is $y = - 1$, so for $y = \left| f(x) \right|$ the horizontal asymptote is $y = 1$.

$f(x) = 0$ at $x = - 1$, so $y = \left| f(x) \right|$ meets the $x$-axis at $( - 1,\ 0)$, and $y = \left| f(0) \right| = 1$ gives the $y$-intercept $(0,\ 1)$.

Note: $\left| f(x) \right|$ has a sharp corner (not a smooth turning point) at $( - 1,\ 0)$, since this is where the expression inside the modulus changes sign.

Approach: $\left| \frac{x + 1}{1 - x} \right| < x + 2$ is exactly the inequality $\left| f(x) \right|\text{<}$ (line), so its solution is read off the sketch in (i) as the set of $x$ for which the curve lies strictly below the line — provided we also solve algebraically for the exact intersection points, since a sketch alone does not give exact values.

Since the left side is non-negative, we need $x + 2 > 0$, i.e. $x > - 2$, for the inequality to have any chance of holding.

We find where the curve meets the line, splitting by the sign of $f(x)$.

For $- 1 \leq x < 1$, $f(x) \geq 0$, so $\left| f(x) \right| = f(x)$:

Only $x = \sqrt{2} - 1 \approx 0.414$ lies in $- 1 \leq x < 1$; $x = - 1 - \sqrt{2}$ is rejected as outside this range.

For $x < - 1$ or $x > 1$, $f(x) < 0$, so $\left| f(x) \right| = - f(x)$:

Both $x = - \sqrt{3} \approx - 1.732$ and $x = \sqrt{3} \approx 1.732$ lie in the valid range ($x < - 1$ or $x > 1$ respectively), so both are kept.

From the sketch, the curve $y = \left| f(x) \right|$ lies below the line $y = x + 2$ on $\left( - \sqrt{3},\ \sqrt{2} - 1 \right)$ (passing continuously through the corner at $x = - 1$, where both graphs are defined), and again for $x > \sqrt{3}$, where the line has overtaken the curve beyond the vertical asymptote.