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

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Answer:  2

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

Given: f(x)=x^{2}+x-1 \text { on }[0,4]

Solution:

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

f(x)  is a polynomial function.

It is continuous in [0, 4] 

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

(Which is defined in [0, 4])

\therefore f(x) is differentiable in [0, 4]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

\begin{aligned} &c \in[0,4] \\ &f^{\prime}(c)=\frac{f(4)-f(2)}{4-2} \end{aligned}

\begin{aligned} &2 c+1=\frac{\left((4)^{2}+4-1\right)-\left((0)^{2}+0-1\right)}{4} \\ &2 c+1=\frac{(16+4-1)-(0+0-1)}{4} \end{aligned}

\begin{aligned} &2 c+1=\frac{19+1}{4} \\ &2 c+1=\frac{20}{4} \end{aligned}

\begin{aligned} &2 c+1=5 \\ &2 c=5-1 \\ &2 c=4 \\ &c=\frac{4}{2} \\ &c=2 \end{aligned}

 

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