Question

If $f:[-3,2]\to \left[0,{}^3\sqrt{x}\right]$ is an onto function defined by
$f(n)=\{\begin{array}{ll}2+\sqrt[3]{n},& -3\le n\le -1\\ n^{2/3},& -1\le n\le 2\end{array}$
then $x=$

• A. 1
• B. 2
• C. 4
• D. 6

Answer 1

Answer: (C)

Given function $f:[-3,2]\to \left[0,{}^3\sqrt{x}\right]$, such that
$f(n)=\{\begin{array}{ll}2+{}^3\sqrt{n}& -3\le n\le -1\\ n^{2/3}& -1\le n\le 2\end{array}$
So
$f^{\prime }(n)=\{\begin{array}{ll}\frac{1}{3}{n}^{-2/3}& -3<n<-1\\ \frac{2}{3}{n}^{-1/3}& -1<n<2\end{array}$
$\because f(n)$ is strictly increasing in $(-3,-1)$ and $(0,2)$ because $f^{\prime }(n)$ is positive,
for $n\in (-3,-1)\cup (0,2)$ and $f$ is strictly decreasing in $(-1,0)$.
$\because f(-1)$ or $f(2)$ is the maximum value of the function and $f(-1)=1$ and $f(2)=2^{2/3}$
$\therefore \sqrt[3]{x}=2^{2/3}$
$\Rightarrow x=4$