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  • Need solution for RD Sharma math class 12 chapter Mean value theorem exercise 14.1 question 7 sub question (i)

Need solution for RD Sharma math class 12 chapter Mean value theorem exercise 14.1 question 7 sub question (i)

Answers (1)

Answer:

Geometrical interpretation of Rolle’s Theorem, (0,16) is the point on curvey = 16 - x ^2 , where tangent is parallel to x-axis.

Hint:

All conditions of Rolle’s Theorem are satisfied.

Given:

               y=16-x^{2}, x \in[-1,1]

Explanation:

              y=16-x^{2}, x \in[-1,1]

  1. Being polynomial f(x) is continuous for all x and hence continuous in[-1,1].
  2. f(x) is derivable in (-1,1)
  3. f(-1)=15=f(1)

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

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

Let’s find the derivative of f(x) 

\Rightarrow \quad f^{\prime}(x)=-2 x\) \ \ \ \ \left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right] \\\\ 

Now,

\(f^{\prime}(c)=0\) \\\\ \Rightarrow \quad-2 c=0 \\\\ \Rightarrow \quad c=0

 

Also,      f(0) = 16 -0 = 16

Hence by geometrical interpretation of Rolle’s Theorem (0,0)is the point on curvey = 16 - x^2, where tangent is parallel to x-axis.

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