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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

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The correct Answer is:
5
`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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