N2022 P1 Q7

N2022 P1 Q7

Junior College 2
9 marks
Free

A curve \(C\) has equation \(y=x^{-3}\ln x\).

  1. Show that \(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1-3\ln x}{x^4}\) and hence find the coordinates of the turning point of \(C\).[4]
  2. Find the exact area enclosed by \(C\), the \(x\)-axis and the line \(x=3\).[5]

Default solution

  1. \(\frac{\mathrm{d}y}{\mathrm{d}x}=x^{-3}\frac1x-3x^{-4}\ln x=\frac{1-3\ln x}{x^4}\) (shown)
    \(\frac{\mathrm{d}y}{\mathrm{d}x}=0\Rightarrow \ln x=\frac13\Rightarrow x=\mathrm{e}^{1/3}\)
    \(y=(\mathrm{e}^{1/3})^{-3}\ln(\mathrm{e}^{1/3})=\frac{1}{3\mathrm{e}}\)
    \(\text{Turning point}=\left(\mathrm{e}^{1/3},\frac1{3\mathrm{e}}\right)\)
  2. \(y=0\Rightarrow \ln x=0\Rightarrow x=1\)
    \(A=\int_1^3x^{-3}\ln x\,\mathrm{d}x\)
    \(=\left[-\frac{\ln x}{2x^2}-\frac1{4x^2}\right]_1^3\) (integration by parts)
    \(=\frac29-\frac{\ln3}{18}\)
Answer:(a) \(\left(\mathrm{e}^{1/3},\frac1{3\mathrm{e}}\right)\) (b) \(\frac29-\frac{\ln3}{18}\)

Need help? Join our JC Math tuition classes.

Learn more