2012 H2 Mathematics Paper 1 Question 7

Functions

Answers

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

Full solutions

(i)

y=x+kx1xyy=x+kxyx=y+kx(y1)=y+kx=y+ky1\begin{gather*} y = \frac{x+k}{x-1} \\ xy-y = x + k \\ xy-x = y+k \\ x(y-1) = y + k \\ x = \frac{y+k}{y-1} \end{gather*}
g1(x)=x+kx1g^{-1}(x) = \frac{x+k}{x-1}
Hence g(x)=g1(x){g(x)=g^{-1}(x)} for all x{x} in the domain of g{g} (since Dg1=Rg=Dg=R{1}D_{g^{-1}} = R_g = D_g = \mathbb{R}\setminus\{1\}).
Hence g{g} is self-inverse {\blacksquare}

(ii)

(iii)

Line of symmetry of the curve:
y=x  y=x \; \blacksquare
By long division,
g(x)=1+k+1x1g(x)=1+\frac{k+1}{x-1}
Sequence of transformations required:
  • Translate the curve y=1x{y=\frac{1}{x}} by 1{1} unit in the positive x-{x\textrm{-}}axis direction.
  • Scale the resulting curve with scale factor k+1{k+1} unit parallel to the y-{y\textrm{-}}axis.
  • Translate the resulting curve by 1{1} unit in the positive y-{y\textrm{-}}axis direction.