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  • Please Solve RD Sharma Class 12 Chapter 14 Inverse Trigonometric Function Exercise Fill in the Blanks Question 3 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 14 Inverse Trigonometric Function Exercise Fill in the Blanks Question 3 Maths Textbook Solution.

Answers (1)

Answer:

                a=11,b\: \epsilon \: R

Hint:

Use Rolle’s Theorem formula,

Given:

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

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 0=-11+a        

\Rightarrow a=11        

Now f\left ( 1 \right )=f\left ( 3 \right )=0

                1-6+a+b=27-54+3a+b

\Rightarrow -5+a+b=27-54+3a+b        

\Rightarrow a+b=-22+3a+b                                                                                                                               \left [ \because a=11 \right ]

\Rightarrow 11+b=-22+33+b            

\Rightarrow 11+b=11+b          

This shows that b\: \epsilon \: R.

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