2017 H2 Maths Paper 2, Question 3

Since $y = f(x)$ cuts the axes at $(a,0)$ and $(0,b)$, we have $f(a) = 0$ and $f(0) = b$. Only these two values of $f$ are known, so a point can be stated only when the transformed curve’s axis-crossing depends on $f$ at $x = a$ or $x = 0$ alone.

$y = f(2x)$

$x$-intercept: $f(2x) = 0 \Rightarrow 2x = a \Rightarrow x = \frac{a}{2}$, giving $\left( \frac{a}{2},0 \right)$.

$y$-intercept: at $x = 0$, $y = f(0) = b$, giving $(0,b)$.

$y = f(x - 1)$

$x$-intercept: $f(x - 1) = 0 \Rightarrow x - 1 = a \Rightarrow x = a + 1$, giving $(a + 1,0)$.

$y$-intercept: at $x = 0$, $y = f( - 1)$, which is not one of the given values of $f$, so it is not possible to state this point.

$y = f(2x - 1)$

$x$-intercept: $f(2x - 1) = 0 \Rightarrow 2x - 1 = a \Rightarrow x = \frac{a + 1}{2}$, giving $\left( \frac{a + 1}{2},0 \right)$.

$y$-intercept: at $x = 0$, $y = f( - 1)$, again not determinable, so it is not possible to state this point.

$y = f^{- 1}(x)$

Since $f^{- 1}$ exists, its graph is the reflection of $y = f(x)$ in the line $y = x$, so every point $(u,v)$ on $y = f(x)$ gives a point $(v,u)$ on $y = f^{- 1}(x)$. Since $(a,0)$ and $(0,b)$ lie on $y = f(x)$, the points $(0,a)$ and $(b,0)$ lie on $y = f^{- 1}(x)$.

Explanation: $g$ is defined by $1 - \frac{1}{1 - x}$, and at $x = 1$ the denominator $1 - x = 0$, so $\frac{1}{1 - x}$ is undefined there. Hence $x = 1$ must be excluded from the domain of $g$.

Using this simplified form,

So $g^{2}(x) = x$ for $x \neq 1$: $g$ is its own inverse.

To find $g^{- 1}(x)$, let $y = g(x) = \frac{x}{x - 1}$ and solve for $x$ in terms of $y$.

Replacing $y$ by $x$:

Using $g^{2}(x) = x$ and $g^{- 1}(x) = \frac{x}{x - 1}$ from part (ii), the equation $g^{2}(b) = g^{- 1}(b)$ becomes

Both values satisfy $b \neq 1$, so both lie in the domain.