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

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

Answers (1)

Answer:

               c=\left(1, \frac{1}{3}\right) \in(0,1), 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-1)^{2}on  [0,1]

Explanation:

We have,

               f(x)=x(x-1)^{2}

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

       2.

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

             \\=2 x(x-1)+(x-1)^{2} \\ =2 x^{2}-2 x+x^{2}+1-2 x \\ =3 x^{2}-4 x+1 \\ =3 x^{2}-3 x-x+1 \\

       3.

          \\f(0)=0(0-1)^{2}=0 \\\\ \quad f(1)=1(1-1)^{2}=0 \\\\ \therefore f(0) =f(1)

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

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

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

               3 c^{2}-4 c+1

Using discrimination method,

\\ \frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\\\\\\ \begin{aligned} c &=\frac{-(-4) \pm \sqrt{16-12}}{6} \\\\ c &=\frac{4 \pm \sqrt{4}}{6} \\\\ c &=\frac{4 \pm 2}{6} \\\\ &=\frac{6}{6}, \frac{2}{6} \\\\ c &=1, \frac{1}{3} \\\\ c &=\left(1, \frac{1}{3}\right) \in(0,1) \end{aligned}

Hence, Rolle’s Theorem is verified.

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