Question

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

• A. 1
• B. 2
• C. 4
• D. 6

Answer 1

Answer: (C)

Given function $f:[-3,2] \rightarrow\left[0,{ }^{3} \sqrt{x}\right]$, such that
$f(n)=\begin{cases}2+{ }^{3} \sqrt{n} & -3 \leq n \leq-1 \\ n^{2 / 3} & -1 \leq n \leq 2\end{cases}$
So
$f'(n)=\begin{cases}\frac{1}{3} n^{-2 / 3} & -3< n< -1 \\ \frac{2}{3} n^{-1 / 3} & -1< n< 2\end{cases}$
$\because f(n)$ is strictly increasing in $(-3,-1)$ and $(0,2)$ because $f'(n)$ 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$