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

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

Answers (1)

Answer:

               c=\left(4, \frac{4}{3}\right)\in[0,4], hence Rolle’s Theorem is verified.

Hint:

Using discrimination method,

\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

Given:

               f(x)=x(x-4)^{2}on [0,4]

Explanation:

We have,

               f(x)=x(x-4)^{2}

  1. Being polynomial f(x) is continuous for all x and hence continuous in [0,4]
  2.  

            \\f^{\prime}(x)=x \frac{d}{d x}(x-4)^{2}+(x-4)^{2} \frac{d}{d x} x \\\\ x \cdot 2 (x-4)+ ( x-4)^2 (1)

              \\ = 2 x^{2}-8 x+x^{2}+16-8 x \\ \quad=3 x^{2}-16 x+16 \\

               \therefore f(x) is derivable in (0,4) 

  1.  

                \\(0)=0(0-4)^{2}=0\\\\ f(4)=4(4-4)^{2}=0\\\\ \therefore f(0)=f(4)

 

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

There exists at least onec \in (0,4)such that f' (c ) = 0

Now, f' (c ) = 0

               3 c^{2}-16 c+16

Using discrimination method,

\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\\\\\\ \begin{array}{l} c=\frac{16 \pm \sqrt{256-192}}{6}=\frac{16 \pm \sqrt{64}}{6} \\\\ c=\frac{16 \pm 8}{6} \\\\ c=\frac{24}{6}, \frac{8}{6} \\\\ c=4, \frac{4}{3} \\\\ c=\left(4, \frac{4}{3}\right) \in[0,4] \end{array}

Hence, Rolle’s Theorem is verified.

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