Limits Continuity And Differentiability Question 52
Question: If $ f’’(x)=-f(x) $ and $ g(x)=f’(x) $ and $ F(x)={{( f( \frac{x}{2} ) )}^{2}}+{{( g( \frac{x}{2} ) )}^{2}} $ and given that $ F(5)=5 $ , then F (10) is equal to-
Options:
A) 5
B) 10
C) 0
D) 15
Show Answer
Answer:
Correct Answer: A
Solution:
$ F’(x)=[ f( \frac{x}{2} ).f’( \frac{x}{2} )+g( \frac{x}{2} )g’( \frac{x}{2} ) ] $
Here, $ g(x)=f’(x) $ and $ g’(x)=f’’(x)=-f(x) $
so $ F’(x)=f( \frac{x}{2} )g( \frac{x}{2} )-f( \frac{x}{2} )g( \frac{x}{2} )=0 $
$ \Rightarrow F(x) $ is constant function so $ F(10)=5 $
BETA
