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

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

Answers (1)

Answer:

               c=\left(\frac{2-\sqrt{7}}{3}, \frac{2+\sqrt{7}}{3}\right) \in[-1,2], Hence Rolle’s Theorem is verified.

Hint:

Using discrimination method,\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

Given:

               f(x)=\left(x^{2}-1\right)(x-2) on [1,-2] 

Explanation:

We have,

               f(x)=\left(x^{2}-1\right)(x-2)

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

          \\f^{\prime}(x) =\left(x^{2}-1\right) \frac{d}{d x}(x-2)+(x-2) \frac{d}{d x}\left(x^{2}-1\right) \\\\ =\left(x^{2}-1\right)(1)+(x-2)(2 x) \\\\ =x^{2}-1+2 x^{2}-4 x \\\\ =3 x^{2}-4 x-1 \\\\

 

3.

         \\ f(-1)=\left((-1)^{2}-1\right)(-1-2)=(1-1)(-1-2)=0\\\\ f(2)=\left(2^{2}-1\right)(2-2)=(4-1)(0)=0 \\\\ \therefore f(-1)=f(2)

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

There exists at least onec \in(-1,2)such thatf^{\prime}(c)=0

Now, f^{\prime}(c)=0

               3 c^{2}-4 c-1

 

Using discrimination method,

 \begin{array}{l} \frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\\\ c=\frac{4 \pm \sqrt{16+12}}{6}=\frac{4 \pm \sqrt{28}}{6} \\\\ c=\frac{4 \pm 2 \sqrt{7}}{6} \\\\ c=\frac{2(2 \pm \sqrt{7})}{6} \\\\ c=\left(\frac{2-\sqrt{7}}{3}, \frac{2+\sqrt{7}}{3}\right) \in[-1,2] \end{array}

Hence, Rolle’s Theorem is verified.

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