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  • Need solution for RD Sharma math class 12 chapter Mean value theorem exercise 14 point 1 question 8 sub question (iii)

Need solution for RD Sharma math class 12 chapter Mean value theorem exercise 14 point 1 question 8 sub question (iii)

Answers (1)

Answer:

               \left ( \frac{1}{2}, -27 \right )

 

Hint:

f' (x) = 124x -12 exist(-1,2)

Given:

               12(x+1)(x-2), x \in[-1,2]

Explanation:

 f(x) = 12(x+1)(x-2), x \in[-1,2]   

  1. Being polynomial f(x) is continuous for all x and hence continuous in[-1,2].

        2. 

            \\ f^{\prime}(x)=12[(x+1)(1)+(x-2)(1)] \\\\ f^{\prime}(x)=12[x+1+x-2] \\\\ f^{\prime}(x)=12[2 x-1]=24 x-12

          \therefore f(x) is derivable in 

         3.

          \\ f(-1)=12(-1+1)(-1-2)=0\\\\ f(2)=12(2+1)(2-2)=0\\\\ \therefore f(-1)=f(2)

Thus all conditions of Rolle’s Theorem are satisfied.

Therefore, there exists at least onec \in(-1,2) such thatf'(c) = 0

Now f'(c) = 0

\Rightarrow 24 c-12=0 \\\\ \Rightarrow 24 c=12 \\\\ \Rightarrow \quad c=\frac{12}{24}=\frac{1}{2} \\\\ \Rightarrow f\left(\frac{1}{2}\right)=12\left(\frac{1}{2}+1\right)\left(\frac{1}{2}-2\right) \\\\ \Rightarrow f\left(\frac{1}{2}\right)=12\left(\frac{3}{2}\right)\left(\frac{-3}{2}\right) \\\\ f\left (\frac{1}{2} \right ) = -27

Therefore, the tangent is parallel to x-axis and the point will be\left ( \frac{1}{2}, -27 \right )

Posted by

infoexpert22

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