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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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Correct option is A

F′(x)=fx2f′x2+gx2g′x2 Here, g(x)=f′(x) and g′(x)=f′′(x)=−f(x) So, F′(x)=fx2gx2−fx2gx2=0 Hence, F(x) is a constant function. Therefore, F(10)=5 since F(5=5 )

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If f′′(x)=−f(x) and g(x)=f′(x) and F(x)=fx22+gx22 and given that F(5)=5 , then F(10) is

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