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

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Answer:   \frac{3}{2}

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

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



f(x) is a polynomial function.

It is continuous in [1, 2] 

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

(Which is defined in [1, 2])

\therefore f(x)  is differentiable in [1, 2]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

\begin{aligned} &c \in[1,2] \\ &f^{\prime}(c)=\frac{f(2)-f(1)}{2-1} \\ &2 c-1=\frac{\left(2^{2}-2\right)-\left(1^{2}-1\right)}{1} \end{aligned}

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


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