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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

`0`

B

5

C

10

D

25

Text Solution

Verified by Experts

The correct Answer is:
B
Since, ` f''(x) =- f(x) `
` rArr" " d/(dx) {f'(x)} =- f(x) `
` rArr" " g'(x) =- f(x) " " [:' g(x) = f'(x)," given "]`…(i)
Also, `" "F(x) = {f(x/2)}^(2) +{g(x/2)}^(2) `
` rArr" " F'(x) = 2(f(x/2))*f'(x/2)*1/2+2(g(x/2))* g'(x/2)*1/2 = 0 ` [from Eq. (i) ]
` :." F (x) is conatant " rArr F (10) - F (5) - 5`
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