If f''x=-fx and gx=f'x and Fx=fx22+gx22 and given that f(5) = 5, then F(10) is

F'x=fx2f'x2+gx2g'x2Here, gx=f'xand g'x=f''x=-fxSo, F'x=fx2gx2-fx2gx2=0Hence, Fx is a constant functionTherefore, F10=5.

If f''x=-fx and gx=f'x and Fx=fx22+gx22 and given that f(5) = 5, then F(10) is

F'x=fx2f'x2+gx2g'x2Here, gx=f'xand g'x=f''x=-fxSo, F'x=fx2gx2-fx2gx2=0Hence, Fx is a constant functionTherefore, F10=5.

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$If f'' (x) = - f (x) and g (x) = f' (x) and F (x) = {(f (\frac{x}{2}))}^{2} + {(g (\frac{x}{2}))}^{2}$ and given that f(5) = 5, then F(10) is

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Detailed Solution

$\begin{array}{l} F' (x) = [f (\frac{x}{2}) f' (\frac{x}{2}) + g (\frac{x}{2}) g' (\frac{x}{2})] \\ Here, g (x) = f' (x) \\ and g' (x) = f'' (x) = - f (x) \\ So, F' (x) = f (\frac{x}{2}) g (\frac{x}{2}) - f (\frac{x}{2}) g (\frac{x}{2}) = 0 \\ Hence, F (x) is a constant function \\ Therefore, F (10) = 5 . \end{array}$