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

Answer:  $\frac{5}{2}$

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

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

Solution:

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

$f(x)$  is a polynomial function.

It is continuous in [2, 3]

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

(Which is defined in [2, 3])

$f(x)$ is differentiable in [2, 3]. Hence it satisfies the condition of Lagrange’s Mean Value Theorem.

Now, there exists at least one value,

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

\begin{aligned} &2 c=\frac{\left(3^{2}-1\right)-\left(2^{2}-1\right)}{1} \\ &2 c=(9-1)-(4-1) \end{aligned}

\begin{aligned} &2 c=8-3 \\ &2 c=5 \\ &c=\frac{5}{2} \end{aligned}