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

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Answer:  \frac{1}{2}

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

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


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

f(x)  is a polynomial function.

It is continuous in [0, 1] 

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

(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} \\ &2-2 c=\frac{\left(2(1)-(1)^{2}\right)-\left(2(0)-(0)^{2}\right)}{1} \end{aligned}

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

\begin{aligned} &2 c=1 \\ &c=\frac{1}{2} \end{aligned}


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