If $f^{''}(x) = - f(x),\ where\ f(x)$is a continuous double differentiable function and $g(x) = f^{'}(x).If\ F(x) = \left( f\left( \frac{x}{2} \right) \right)^{2} + \left( g\left( \frac{x}{2} \right) \right)^{2}and\ F(5) = 5\ then\ F(10)$ is

Question

Dr. Abubacker · Accepted Answer

(b)
Given, $f^{''}(x) = - f(x)$
$\Rightarrow g^{'}(x) = - f(x)$ and $f^{'}(x) = g(x)$ …(i)
Now, $F(x) = \left( f\left( \frac{x}{2} ight) ight)^{2} + \left( g\left( \frac{x}{2} ight) ight)^{2}$
$	herefore F^{'}(x) = 2\left( f\left( \frac{x}{2} ight) ight).f^{'}\left( \frac{x}{2} ight).\frac{1}{2}$
$+ 2\left( g\left( \frac{x}{2} ight) ight).g^{'}\left( \frac{x}{2} ight).\frac{1}{2} = 0$
[using Eq.(i)]
$	herefore\ F(x)$ is a constant $\Rightarrow$ $F(10) = F(5) = 5$

Q. If $f^{''}(x) = - f(x),\ where\ f(x)$is a continuous double differentiable function and $g(x) = f^{'}(x).If\ F(x) = \left( f\left( \frac{x}{2} \right) \right)^{2} + \left( g\left( \frac{x}{2} \right) \right)^{2}and\ F(5) = 5\ then\ F(10)$ is

Explanation

Reviewed by Dr. Abubacker

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