#### Explain Solution RD Sharma Class 12 Chapter 14 Inverse Trigonometric Function Exercise Very Short Answer Question 4 Maths.

$n=3$

Hint:

Find differentiate of $f\left ( x \right )$ and then apply Rolle’s Theorem.

Given:

$f\left ( x \right )=2x\left ( x-3 \right )^{n},x\: \epsilon \left [ 0,2\sqrt{3} \right ]$and value of $c$ is $\frac{3}{4}$.

Solution:

According to Rolle’s Theorem

$\begin{array}{ll} & f(a)=f(b) \\\\ \Rightarrow & f(0)=f(2 \sqrt{3}) \\\\ \Rightarrow & 0=2 \cdot 2 \sqrt{3}(2 \sqrt{3}-3)^{n} \end{array}$

$\Rightarrow \left ( 2\sqrt{3}-3 \right )^{n}=0$

Now,     $f\left ( x \right )=2x\left ( x-3 \right )^{n}$

Differentiate with respect to $x$,

${f}'\left ( x \right )=2\left ( x-3 \right )^{n}+2x\cdot n\left ( x-3 \right )^{n-1}$                                           $\left [ \because \frac{d\left ( uv \right )}{dx}=u\frac{dv}{dx} +v\frac{du}{dx}\right ]$

Applying Rolle’s Theorem,

${f}'\left ( c \right )=2\left ( c-3 \right )^{n}+2c\cdot n\left ( c-3 \right )^{n-1}$

${f}'\left ( c \right )=0$

$2\left ( c-3 \right )^{n}+2c\cdot n\left ( c-3 \right )^{n-1}=0$

$\Rightarrow 2\left ( \frac{3}{4}-3 \right )^{n}+2\cdot \frac{3}{4}\cdot n\left ( \frac{3}{4}-3 \right )^{n-1}=0$                                                                                                   $\left [ \because c=\frac{3}{4} \right ]$

$\Rightarrow \left ( \frac{3}{4}-3 \right )^{n}\left [ 2\left ( \frac{3}{4}-3 \right )+\frac{3n}{2}\right ]=0$

$\Rightarrow \left ( \frac{3}{4}-3 \right )^{n-1}=0$          or  $2\left ( \frac{3-12}{4} \right )+\frac{3n}{2}=0$

$\Rightarrow$                                                $\frac{3-12+3n}{2}=0$

$3-12+3n=0$

$3n=9$

$n=3$