Consider the function $\mathrm{f}(\mathrm{x})=\frac{27\left(\mathrm{x}^{2 / 3}-\mathrm{x}\right)}{4}$ How many solutions does the function $\mathrm{f}(\mathrm{x})=1$ have?

• A

  One
• Correct answer
  B

  Two
• C

  Three
• D

  Four

Solution

$\begin{aligned} f(x) & =\frac{27}{4}\left(x^{2 / 3}-x\right) \\ & =\frac{27}{4}\left(\left(x^2\right)^{1 / 3}-x\right)\end{aligned}$ For $f(x)=1$ $\begin{aligned} & \frac{27}{4}\left(\left(x^2\right)^{1 / 3}-x\right)=1 \\ & \left(x^{1 / 3}\right)^2-x=\frac{4}{27} \end{aligned}$ Put $x^{1 / 3}=y$ $x=y^3$ $\begin{aligned} & y^2-y^3=\frac{4}{27} \\ & \Rightarrow y^3-y^2+\frac{4}{27}=0 \\ & \Rightarrow 27 y^3-27 y^2+4=0 \end{aligned}$ This is a cubic equation. If we put $y=-\frac{1}{3}$ then $(3 y+1)=0$ is a factor of cubic equation.
$\begin{aligned} & (3 y+1)\left(9 y^2-12 y+4\right)=0 \\ & (3 y+1)(3 y-2)^2=0 \end{aligned}$ Hence $y=-\frac{1}{3}, \frac{2}{3}$ Thus $f(x)=1$ has two solutions.

Hence, the correct answer is option: 'B'.