Q165 / 270 · NDA 2016 (Phase 1) · Mathematics

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

  • Correct answer
    A
    One
  • B
    Two
  • C
    Three
  • D
    Four

Solution

$\begin{aligned} f(x) & =\frac{27}{4}\left(x^{2 / 3}-x\right) \\ & =\frac{27}{4}\left(\left(x^2\right)^{1 / 3}-x\right)\end{aligned}$ Similarly for $f(x)=-1$ we will get $27 y^3-27 y^2-4=0$ and after solving it we will find that it has one real solution. $\begin{aligned} & y_0=1.1184, \\ & x=\left(\mathrm{y}_0\right)^3=(1.1184)^3=1.4 \end{aligned}$

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

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