2026 EJC P1 Q6

2026 EJC P1 Q6

Junior College 2
11 marks

The function \(\mathrm{f}\) is defined by \(\mathrm{f}(x)=\sqrt{|2-x|+1}\), \(x\in\mathbb{R}\), \(x\geq-1\).

  1. Show that \(\mathrm{f}^{-1}\) does not exist.[2]

The function \(\mathrm{f}\) has an inverse if its domain is restricted to \(x\geq2\).

  1. Find an expression for \(\mathrm{f}^{-1}\) and state its domain.[4]
  2. Sketch the graphs of \(y=\mathrm{f}(x)\) and \(y=\mathrm{f}^{-1}(x)\) on the same diagram, showing clearly the relationship between the two graphs.[3]

The functions \(\mathrm{g}\) and \(\mathrm{h}\) are defined by \(\mathrm{g}(x)=\frac{4}{4x^2-1}\), \(x\in\mathbb{R}\), \(x\neq-\frac12\), \(x\neq\frac12\), and \(\mathrm{h}(x)=1+e^{-x}\), \(x\in\mathbb{R}\).

  1. Find the exact range of \(\mathrm{hg}\).[2]

Solution:

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Answer:(a) No inverse (b) \(\mathrm{f}^{-1}(x)=x^2+1\), \(x\geq1\) (d) \((1,2)\cup[1+e^4,\infty)\)

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