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Please solve RD Sharma class 12 chapter Mean Value Theoram exercise 14.2 question 1 sub question (v) maths textbook solution

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Answer:  2

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

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


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

f(x)  is a polynomial function.

It is continuous in [1, 3] 

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

(Which is defined in [1, 3])

\therefore f(x) is differentiable in [1, 3]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

\begin{aligned} &c \in[1,3] \\ &f^{\prime}(c)=\frac{f(3)-f(1)}{3-1} \\ &4 c-3=\frac{\left(2(3)^{2}-3(3)+1\right)-\left(2(1)^{2}-3(1)+1\right)}{2} \end{aligned}

\begin{aligned} &4 c-3=\frac{(2(9)-9+1)-(2-3+1)}{2} \\ &4 c-3=\frac{(18-9+1)-(2-3+1)}{2} \end{aligned}

\begin{aligned} &4 c-3=\frac{10-0}{2} \\ &4 c-3=5 \\ &4 c=5+3 \\ &c=\frac{8}{4} \\ &c=2 \end{aligned}


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