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provide solution for RD Sharma maths class 12 chapter Mean Value Theoram exercise 14.2 question 1 sub question (ii)

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Answer:    \frac{1}{3}

Hint:You must know the formula of Lagrange’s Mean Value Theorem.

Given:  f(x)=x^{3}-2 x^{2}-x+3 \text { on }[0,1]


f(x)=x^{3}-2 x^{2}-x+3

f(x)  is a polynomial function.

It is continuous in [0, 1] 

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

(Which is defined in [0, 1])

\therefore f(x) is differentiable in [0, 1]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

\begin{aligned} &c \in[0,1] \\ &f^{\prime}(c)=\frac{f(1)-f(0)}{1-0} \end{aligned}

3 c^{2}-4 c-1=\frac{\left((1)^{3}-2(1)^{2}-1+3\right)-\left((0)^{3}-2(0)^{2}-0+3\right)}{1}

\begin{aligned} &3 c^{2}-4 c-1=(1-2-1+3)-(3) \\ &3 c^{2}-4 c-1=1-3 \\ &3 c^{2}-4 c-1=-2 \\ &3 c^{2}-4 c-1+2=0 \\ &3 c^{2}-4 c+1=0 \end{aligned}

\begin{aligned} &3 c^{2}-3 c-c+1=0 \\ &3 c(c-1)-(c-1)=0 \\ &(3 c-1)(c-1)=0 \end{aligned}

Therefore, c=\frac{1}{3} \text { and } \mathrm{c}=1


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