Limits Continuity And Differentiability Question 9
Question: Let $ f:[2,7]\to [0,\infty ) $ be a continuous and differentiable function. Then, $ (f(7)-f(2))\frac{{{(f(7))}^{2}}+{{(f(2))}^{2}}+f(2)f(7)}{3} $ is, where $ c\in [2,7] $ [2, 7].
Options:
A) $ 5f^{2}(c)f’(c) $
B) $ 5f’(c) $
C) $ f(c)f’(c) $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ g(x)=f^{3}(x) $
$ \Rightarrow g’(x)=3f^{2}(x)\cdot f’(x) $
$ \because f:[2,7]\to [0,\infty )\Rightarrow g:[2,7]\to [0,\infty ) $
Using Lagrange’s mean value theorem on g(x), We get $ g’(c)=\frac{g(7)-g(2)}{5},c\in [2,7] $
$ \Rightarrow 2f^{2}(c)f’(c)=(f(7)-f(2)) $
$ \frac{{{(f(7))}^{2}}+{{(f(2))}^{2}}+f(2)f(7)}{3} $
BETA
