Q41 / 160 · TS EAMCET 2020 (11 Sep Shift 1) · Mathematics
If $f:[-3,2] \rightarrow[0, \sqrt[3]{x}]$ is an onto function defined by $f(n)=\left\{\begin{array}{cc}2+\sqrt[3]{n}, & -3 \leq n \leq-1 \\ n^{2 / 3}, & -1 \leq n \leq 2\end{array}\right.$, then $x=$
- A1
- B2
- Correct answerC4
- D6
Solution
Given function $f:[-3,2] \rightarrow[0, \sqrt[3]{x}]$, such that
$f(n)=\left\{\begin{array}{cc}2+\sqrt[3]{n}, & -3 \leq n \leq-1 \\ n^{2 / 3}, & -1 \leq n \leq 2\end{array}\right.$
So, $f^{\prime}(n)= \begin{cases}\frac{1}{3} n^{-2 / 3} & -3 < n < -1 \\ \frac{2}{3} n^{-1 / 3}, & -1 < n < 2\end{cases}$
$\because f(n)$ is strictly increasing in $(-3,-1)$ and $(0,2)$ because $f^{\prime}(n)$ 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 $f(2)=2^{2 / 3}$
$\therefore \quad \sqrt[3]{x}=2^{2 / 3} \Rightarrow x=4$
Hence, the correct answer is option: 'C'.