2012 Paper 1 Question 7

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Sketch.

Line of symmetry: $y=x.$

Translate the curve $y=\frac{1}{x}$ by $1$ unit in the positive $x\text{-axis}$ direction.
Scale the resulting curve by a factor of $1+k$ parallel to the $y\text{-axis}.$
Translate the resulting curve by $1$ unit in the positive $y\text{-axis}$ direction.

$$\begin{array}{c}\text{Let }y=\frac{x+k}{x-1}\\ y(x-1)=x+k\\ yx-y=x+k\\ yx-x=y+k\\ x(y-1)=y+k\\ x=\frac{y+k}{y-1}\\ g^{-1}(x)=\frac{x+k}{x-1}\end{array}$$

Hence $g$ is self-inverse as $g(x)=g^{-1}(x)$ and $D_{g^{-1}}=R_g=D_g∎$

Line of symmetry: $y=x∎$

$$\frac{x+k}{x-1}=1+\frac{1+k}{x-1}$$

• Translate the curve $y=\frac{1}{x}$ by $1$ unit in the positive $x\text{-axis}$ direction
• Scale the resulting curve by factor $1+k$ parallel to the $y\text{-axis}$
• Translate the resulting curve by $1$ unit in the positive $y\text{-axis}$ direction