MCQ
Let $f:(-1,1) \rightarrow R$ be a differentiable function satisfying $\left(f^{\prime}(x)\right)^4=16(f(x))^2$ for all $x \in(-1,1)$ $f(0)=0$ The number of such functions is
- A$2$
- B$3$
- C$4$
- ✓more than $4$
It is given that,
$\left(f^{\prime}(x)\right)^4 =16(f(x))^2, \forall x \in(-1,1)$
$\Rightarrow \quad \left(f^{\prime}(x)\right)^2 =\pm 4 f(x)$
Case $I$
If $f(x) \geq 0 \forall x \in(-1,1)$
Then, $f^{\prime}(x)=\pm 2 \sqrt{f(x)}$
$\Rightarrow \quad \frac{f^{\prime}(x)}{\sqrt{f(x)}}=\pm 2$
$\Rightarrow 2 \sqrt{f(x)}=\pm 2 x, \quad[\because f(0)=0]$
$\Rightarrow \quad f(x)=x^2 \forall x \in(-1,1)$
Case $II$
If $f(x)<0 \forall x \in(-1,1)$
Then, $f^{\prime}(x)=\pm 2 \sqrt{-f(x)}$
$\therefore \quad f(x)=-x^2 \forall x \in(-1,1)$
$\therefore \quad f_1(x)=\left[\begin{array}{l}x^2, 0 \leq x < 1 \\ x^2,-1 < x < 0\end{array}\right.$
$f_2(x)=\left[\begin{array}{cc}x^2 & , 0 \leq x < 1 \\ -x^2 & -1 < x < 0\end{array}\right.$
$f_3(x)=\left[\begin{array}{cc}-x^2 & , 0 \leq x < 1 \\ x^2 & ,-1 < x < 0\end{array}\right.$
$f_4(x)=\left[\begin{array}{ll}-x^2 & , 0 \leq x < 1 \\ -x^2 & ,-1 < x < 0\end{array}\right.$
$f_5(x)=0$
$f_6(x)=\left[\begin{array}{cc}0 & , 0 \leq x < 1 \\ x^2 & ,-1 < x < 0\end{array}, \ldots\right.$ and so on
Hence, there are more than $4$ functions possible.
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