Question

Let $f:[2,7]\to [0,\infty )$ be a continuous and differentiable function. Then,

$(f\left(7\right)-f\left(2\right))\frac{({\left(f\left(7\right)\right)}^2+{\left(f\left(2\right)\right)}^2+f\left(2\right)f\left(7\right))}{3}$ is equal to ( where $c\in (2,7)$)

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

Answer 1

Answer: (A)

$5f^2\left(c\right)f^{\prime }\left(c\right)$

Let $g\left(x\right)=f^3\left(x\right)\Rightarrow g^{\prime }\left(x\right)=3f^2\left(x\right).f^{\prime }\left(x\right)$

$\because f:[2,7]\to [0,\infty )\Rightarrow g:[2,7]\to [0,\infty )$

Using LMV theorem on $g\left(x\right)$, we get

$g^{\prime }\left(c\right)=\frac{g\left(7\right)-g\left(2\right)}{5};c\in (2,7)$

$\Rightarrow 5f^2\left(c\right)f^{\prime }\left(c\right)=(f\left(7\right)-f\left(2\right))\frac{({\left(f\left(7\right)\right)}^2+{\left(f\left(2\right)\right)}^2+f\left(2\right)f\left(7\right))}{3}$