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explain solution RD Sharma class 12 chapter Mean Value Theoram exercise 14.2 question 1 sub question (iv) maths

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Answer:  \frac{1}{2}

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

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


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

f(x)  is a polynomial function.

It is continuous in [-1, 2] 

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

(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)} \end{aligned}

\begin{aligned} &2 c-3=\frac{\left(2^{2}-3(2)+2\right)-\left((-1)^{2}-3(-1)+2\right)}{2+1} \\ &2 c-3=\frac{(4-6+2)-(1+3+2)}{3} \end{aligned}

\begin{aligned} &2 c-3=\frac{0-6}{3} \\ &2 c-3=\frac{-6}{3} \\ &2 c-3=-2 \\ &2 c=-2+3 \\ &2 c=1 \\ &c=\frac{1}{2} \end{aligned}



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