Question
Let $$f:\left[ {2,\,7} \right] \to \left[ {0,\,\infty } \right)$$ be a continuous and differentiable function. Then, $$\left( {f\left( 7 \right) - f\left( 2 \right)} \right)\frac{{{{\left( {f\left( 7 \right)} \right)}^2} + {{\left( {f\left( 2 \right)} \right)}^2} + f\left( 2 \right)f\left( 7 \right)}}{3}$$ where $$c\, \in \left[ {2,\,7} \right].$$
A.
$$5{f^2}\left( c \right)f'\left( c \right)$$
B.
$$5f'\left( c \right)$$
C.
$$f\left( c \right)f'\left( c \right)$$
D.
none of these
Answer :
$$5{f^2}\left( c \right)f'\left( c \right)$$
Solution :
$$\eqalign{
& {\text{Let }}g\left( x \right) = {f^3}\left( x \right) \cr
& \Rightarrow g'\left( x \right) = 3{f^2}\left( x \right).f'\left( x \right) \cr
& \because \,f:\left[ {2,\,7} \right] \to \left[ {0,\,\infty } \right) \Rightarrow g:\left[ {2,\,7} \right] \to \left[ {0,\,\infty } \right) \cr} $$
Using Lagrange's mean value theorem on $$g\left( x \right),$$ we get $$g'\left( c \right) = \frac{{g\left( 7 \right) - g\left( 2 \right)}}{5},\,c\, \in \left[ {2,\,7} \right]$$
$$ \Rightarrow 2{f^2}\left( c \right)f'\left( c \right) = \left( {f\left( 7 \right) - f\left( 2 \right)} \right)\frac{{{{\left( {f\left( 7 \right)} \right)}^2} + {{\left( {f\left( 2 \right)} \right)}^2} + f\left( 2 \right)f\left( 7 \right)}}{3}$$