Let f:(-1,1) rightarrow R be a differentiable | vedclass

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(D) Given the differential equation $(f^{\prime}(x))^4 = 16(f(x))^2$,we have $(f^{\prime}(x))^2 = \pm 4f(x)$.Since $(f^{\prime}(x))^2 \geq 0$,we must

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more than $4$

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(D) Given the differential equation $(f^{\prime}(x))^4 = 16(f(x))^2$,we have $(f^{\prime}(x))^2 = \pm 4f(x)$.Since $(f^{\prime}(x))^2 \geq 0$,we must have $4f(x) \geq 0$,so $f(x) \geq 0$.Thus,$(f^{\prime}(x))^2 = 4f(x)$,which implies $f^{\prime}(x) = \pm 2\sqrt{f(x)}$.If $f(x) > 0$,then $\frac{f^{\prime}(x)}{\sqrt{f(x)}} = \pm 2$. Integrating both sides,we get $2\sqrt{f(x)} = \pm 2x + C$. Since $f(0)=0$,$C=0$,so $\sqrt{f(x)} = \pm x$,which gives $f(x) = x^2$.We can construct solutions by patching $f(x) = x^2$,$f(x) = -x^2$ (only if $f(x) \geq 0$ is not strictly required,but here $f(x) \geq 0$ is forced by the square),and $f(x) = 0$.Specifically,for any $a, b \in [0, 1)$,we can define $f(x)$ piecewise using $x^2$ and $0$ such that it remains differentiable at $x=0$. Since there are infinitely many such combinations,the number of such functions is more than $4$.

Column $I$	Column $II$
$(A)$ The number of solutions of the equation $x e^{\sin x}-\cos x=0$ in the interval $(0, \frac{\pi}{2})$	$(p)$ $1$
$(B)$ Value$(s)$ of $k$ for which the planes $k x+4 y+z=0, 4 x+k y+2 z=0$ and $2 x+2 y+z=0$ intersect in a straight line	$(q)$ $2$
$(C)$ Value$(s)$ of $k$ for which $\|x-1\|+\|x-2\|+\|x+1\|+\|x+2\|=4 k$ has integer solution$(s)$	$(r)$ $3$
$(D)$ If $y^{\prime}=y+1$ and $y(0)=1$,then value$(s)$ of $y(\ln 2)$	$(s)$ $4$
	$(t)$ $5$

Let $f:(-1,1) \rightarrow R$ be a differentiable function satisfying $(f^{\prime}(x))^4 = 16(f(x))^2$ for all $x \in (-1,1)$ and $f(0)=0$. The number of such functions is:

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