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  • Provide solution for RD Sharma math class class 12 chapter Mean value theorem exercise 14.1 question 2 sub question (vii)

Provide solution for RD Sharma math class class 12 chapter Mean value theorem exercise 14.1 question 2 sub question (vii)

Answers (1)

Answer:

c=\frac{-5}{2} \in(-3,-2) , hence Rolle’s Theorem is verified.

Hint:

f(x) is continuous for all x and hence continuous in [-3,-2]

Given:

               f(x)=x^{2}+5 x+6 on [-3,-2]

Explanation:

We have,

               f(x)=x^{2}+5 x+6

  1. Being polynomial f(x) is continuous for all x and hence continuous in [-3,-2] 
  2. f^{\prime}(x)=2 x+5, which exists in [-3,-2] 

               \therefore f(x) is derivable in (-3,-2)

       3. f(-3)=(-3)^{2}+5(-3)+6

                            =9-15+6=-6+6=0

            f(-2)=(-2)^{2}+5(-2)+6

                            =4-10+6=-6+6=0

          \therefore f(-3) =f(-2) 

Thus all the conditions of Rolle’s Theorem are satisfied.

There exists at least onec \in(-3,-2)such thatf^{\prime}(c)=0Now,

f^{\prime}(c)=0\\\\ 2 c+5=0\\\\ 2 c=-5\\\\ c=\frac{-5}{2}\\\\ c = \frac{-5}{2}\in (-3,-2)

Hence, Rolle’s Theorem is verified.

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