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If f^(x)=-f(x)a n dg(x)=f^(prime)(x)a n ...

If `f^(x)=-f(x)a n dg(x)=f^(prime)(x)a n d` `F(x)=(f(x/2))^2+(g(x/2))^2` and given that `F(5)=5,` then `F(10)` is 5 (b) 10 (c) 0 (d) 15

A

5

B

10

C

0

D

15

Text Solution

Verified by Experts

`F'(x)=[f((x)/(2))f'((x)/(2))+g((x)/(2))g'((x)/(2))]`
`"Here, "g(x) =f'(x)`
`"and "g'(x)=f'''(x)=-f(x)`
`"So, F'(x)=f((x)/(2))g((x)/(2))-f((x)/(2))g((x)/(2))=0`
Hence, F(x) is a constant function.
Therefore, F(10) =5.
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