If F(x) = left(fleft(frac{x}{2}right)right)^2 | vedclass

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(A) Given $F(x) = \left(f\left(\frac{x}{2}\right)\right)^2 + \left(g\left(\frac{x}{2}\right)\right)^2$.Differentiating with respect to $x$,we get:$F^{

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$5$

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(A) Given $F(x) = \left(f\left(\frac{x}{2}\right)\right)^2 + \left(g\left(\frac{x}{2}\right)\right)^2$.Differentiating with respect to $x$,we get:$F^{\prime}(x) = 2 f\left(\frac{x}{2}\right) \cdot f^{\prime}\left(\frac{x}{2}\right) \cdot \frac{1}{2} + 2 g\left(\frac{x}{2}\right) \cdot g^{\prime}\left(\frac{x}{2}\right) \cdot \frac{1}{2}$$F^{\prime}(x) = f\left(\frac{x}{2}\right) \cdot f^{\prime}\left(\frac{x}{2}\right) + g\left(\frac{x}{2}\right) \cdot g^{\prime}\left(\frac{x}{2}\right)$Since $g(x) = f^{\prime}(x)$,we have $g^{\prime}(x) = f^{\prime \prime}(x) = -f(x)$.Substituting these into the derivative expression:$F^{\prime}(x) = f\left(\frac{x}{2}\right) \cdot g\left(\frac{x}{2}\right) + g\left(\frac{x}{2}\right) \cdot (-f\left(\frac{x}{2}\right))$$F^{\prime}(x) = f\left(\frac{x}{2}\right) \cdot g\left(\frac{x}{2}\right) - g\left(\frac{x}{2}\right) \cdot f\left(\frac{x}{2}\right) = 0$Since $F^{\prime}(x) = 0$,$F(x)$ is a constant function.Given $F(5) = 5$,it follows that $F(x) = 5$ for all $x$.Therefore,$F(10) = 5$.

Column $I$	Column $II$
$(A)$ $f(x) = x\|x\|$	$(p)$ continuous in $(-1, 1)$
$(B)$ $f(x) = \sqrt{\|x\|}$	$(q)$ differentiable in $(-1, 1)$
$(C)$ $f(x) = x + [x]$	$(r)$ strictly increasing in $(-1, 1)$
$(D)$ $f(x) = \|x - 1\| + \|x + 1\|$	$(s)$ not differentiable at least at one point in $(-1, 1)$

If $F(x) = \left(f\left(\frac{x}{2}\right)\right)^2 + \left(g\left(\frac{x}{2}\right)\right)^2$,where $f^{\prime \prime}(x) = -f(x)$ and $g(x) = f^{\prime}(x)$,and given $F(5) = 5$,then $F(10)$ is equal to

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