If F(x)=(f(x2))2+(g(x2))2 where fā²(x)=āf(x) and g(x)=fā²(x) and given that F(5)=5, then F(10) is equal to
5
F(x)=(f(x2)2)+(g(x2))2
⇒F′(x)=2f(x2).f′(x2).12+2g(x2).g′(x2).12
=f(x2).f′(x2)+f′(x2).f′(x2)
[∵g(x)=f′(x)⇒g(x)=f"(x)]
=f(x2).f′(x2)−f′(x2)f(x2)
=0 [∵f′(x)=−f(x)]
⇒F(x) is a constant fucntion.
∴F(x)=F(5)=5 ∀x∈R⇒F(10)=5