2016 PJC P2 Q3

2016 PJC P2 Q3

11 marks
Prelims

The function \(\mathrm{f}\) is defined by

\(\mathrm{f}(x)=\left\{ \begin{matrix} \frac{x}{2} \qquad \text{ if}\hspace{0.5em} x\le 0 \\ 2\sin x \qquad \text{ if}\hspace{0.5em} 0<x\le 4 \\ \end{matrix} \right.\)

  1. Sketch the graph of \(y=\mathrm{f}(x)\).[2]
  2. If the domain of \(\mathrm{f}\) is restricted to \(x\le k\), state the largest value of \(k\), in exact form, for which the function \({{\mathrm{f}}^{-1}}\) exist.[1]
  3. Using the domain from part (ii), define \({{\mathrm{f}}^{-1}}\) in a similar form.[4]
  4. Solve \({{\mathrm{f}}^{-1}}(x)=\mathrm{f}(x)\).[2]

In the rest of the question, the domain of f is as originally defined.

The function \(\mathrm{g}\) is defined by \(\mathrm{g}:x\mapsto -{{x}^{3}},\mathrm{ }x\in \mathbb{R},\mathrm{ }x>0\)

  1. Find an expression for \(\mathrm{fg}\left( x \right)\).[2]

Video Solution:

Video Solution

Video solution locked

A free account saves your progress; this video needs tuition enrolment.

Solution:

Solution locked

680+ questions here come with free worked solutions. A free account saves your questions and progress; this one needs tuition enrolment.

Create a free account to practise similar questions.

Answer:(ii) \(k=\frac\pi2\); (iii) \(f^{-1}(x)=2x\) for \(x\le0\), \(f^{-1}(x)=\sin^{-1}(x/2)\) for \(0<x\le2\); (iv) \(x=0\); (v) \(fg(x)=-\frac{x^3}{2}\)

Need help? Join our JC Math tuition classes.

Learn more