#### provide solution for RD Sharma maths class 12 chapter Mean Value Theoram exercise 14.2 question 1 sub question (xvi)

Answer:  $\frac{7}{3}$

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

Given: $f(x)=x^{3}-5 x^{2}-3 x \text { on }[1,3]$

Solution:

$f(x)=x^{3}-5 x^{2}-3 x$

$f(x)$  is a polynomial function.

It is continuous in [1, 3]

$f^{\prime}(x)=3 x^{2}-10 x-3$

(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} \end{aligned}

\begin{aligned} &3 c^{2}-10 c-3=\frac{\left((3)^{3}-5(3)^{2}-3(3)\right)-\left((1)^{3}-5(1)^{2}-3(1)\right)}{2} \\ &3 c^{2}-10 c-3=\frac{(27-45-9)-(1-5-3)}{2} \end{aligned}

\begin{aligned} &3 c^{2}-10 c-3=\frac{(27-54)-(1-8)}{2} \\ &3 c^{2}-10 c-3=\frac{(-27)-(-7)}{2} \end{aligned}

\begin{aligned} &3 c^{2}-10 c-3=\frac{-27+7}{2} \\ &3 c^{2}-10 c-3=\frac{-20}{2} \end{aligned}

\begin{aligned} &3 c^{2}-10 c-3=-10 \\ &3 c^{2}-10 c-3+10=0 \\ &3 c^{2}-10 c+7=0 \\ &3 c^{2}-3 c-7 c+7=0 \end{aligned}

\begin{aligned} &3 c(c-1)-7(c-1)=0 \\ &(3 c-7)(c-1)=0 \end{aligned}

$c=\frac{7}{3} \text { And } c=1$

$c=\frac{7}{3}$  Value exists.

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