#### Please Solve RD Sharma Class 12 Chapter 14 Inverse Trigonometric Function Exercise Fill in the Blanks Question 4 Maths Textbook Solution.

$a=11,b=-6$

Hint:

Find differentiation to find $a$ and use $f\left ( 1 \right )=f\left ( 3 \right )=0$ to find $b$.

Given:

$f\left ( x \right )=x^{3}-6x^{2}+ax+b,x\: \epsilon \left [ 1,3 \right ],c=2+\frac{1}{\sqrt{3}}$ and $f\left ( 1 \right )=f\left ( 3 \right )=0$

Solution:

Rolle’s Theorem states that,

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

Now,     $f\left ( x \right )=x^{3}-6x^{2}+ax+b$

On differentiating we get,

$\Rightarrow {f}'\left ( x \right )=3x^{2}-12x+a$

$\Rightarrow {f}'\left ( c \right )=3c^{2}-12c+a$

$\Rightarrow 0=3c^{2}-12c+a$

$\Rightarrow 0=3\left ( 2+\frac{1}{\sqrt{3}} \right )^{2}-12\left ( 2+\frac{1}{\sqrt{3}} \right )+a$                                                                                       $\left [ \because c=2+\frac{1}{\sqrt{3}} \right ]$

$\Rightarrow 0=3\left ( 4+\frac{1}{3}+\frac{4}{\sqrt{3}} \right )-24-\frac{12}{\sqrt{3}} +a$

$\Rightarrow 0=12+1+\frac{12}{\sqrt{3}} -24-\frac{12}{\sqrt{3}} +a$

$\Rightarrow a=24-13$

$\Rightarrow a=11$

Now $f\left ( 1 \right )=0$

$1-6\left ( 1 \right )+a+b=0$

$-5+11+b=0$                                                                                                                                $\left [ \because a=11 \right ]$

$b=-6$