Question

Let $f:[2,7]\to [0,\infty ]$ be a continuous and differentiable function. Then, the value of $\left(f\right(7)-f(2\left)\right)\frac{{\left(f\right(7\left)\right)}^2+{\left(f\right(2\left)\right)}^2+f\left(2\right).f\left(7\right)}{3}$, is (where $cϵ(2,7)$)

• A. $3f^2\left(c\right)f^{\prime }\left(c\right)$
• B. $5f^2\left(c\right).f\left(c\right)$
• C. $5f^2\left(c\right).f^{\prime }\left(c\right)$
• D. none of these

Answer 1

The correct option is C $5f^2\left(c\right).f^{\prime }\left(c\right)$

Let $h\left(x\right)=\left(f\right(x){)}^3$
Since, $f:[2,7]\to [0,\infty )$
$\Rightarrow$$h:[2,7]$ $\to [0,\infty )$
Also, $h\left(x\right)$ is continuous on $[2,7]$ and differentiable on $(2,7)$. So, by Lagrange's mean value theorem on $h\left(x\right)$, there exists $c$$\in$ $(2,7)$ such that
$h^{\prime }\left(c\right)=\frac{h\left(7\right)-h\left(2\right)}{5}$
$\Rightarrow 3f^2\left(c\right)f^{\prime }\left(c\right)=\frac{\left(f\right(7){)}^3-(f\left(2\right){)}^3}{5}$
$\Rightarrow 3f^2\left(c\right)f^{\prime }\left(c\right)=\frac{\left(f\right(7)-f(2\left)\right)\left[\right(\left(f\right(7){)}^2-f(7\left)f\right(2)+(f\left(2\right){)}^2]}{5}$
$\Rightarrow \frac{\left(f\right(7)-f(2\left)\right)\left[\right(\left(f\right(7){)}^2-f(7\left)f\right(2)+(f\left(2\right){)}^2]}{3}=5f^2\left(c\right)f^{\prime }\left(c\right)$