Question

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

• A. 2
• B. 3
• C. 4
• D. more than 4

Answer 1

Answer: (D)

It is given that,
$\left(f' \left(x\right)\right)^{4} =16 \left(f \left(x\right)\right)^{2}, \forall x \,\in\left(-1, 1\right)$
$\Rightarrow \left(f'\left(x\right)\right)^{2}=\pm4f \left(x\right)$
Case I
If $f \left(x\right) \ge\,0\forall x \,\in\left(-1,1\right)$
Then, $f' \left(x\right)=\pm 2\sqrt{f \left(x\right)}$
$\Rightarrow \frac{f'\left(x\right)}{\sqrt{f\left(x\right)}}=\pm2$
$\Rightarrow 2\sqrt{f\left(x\right)}=\pm2x$,
$\left[\because f \left(0\right)=0\right]$
$\Rightarrow f \left(x\right)=x^{2}\forall\,x \,\in\left(-1,1\right)$
Case II
If $f \left(x\right)<\,0 \forall x \in\left(-1, 1\right)$
Then, $f'\left(x\right)=\pm2\sqrt{-f \left(x\right)}$
$\therefore f \left(x\right)=-x^{2}\,\forall x\,\in\left(-1,1\right)$
$\therefore $ $f_{1}(x) = \begin{cases} x^{2}, & \text{ $0 \le\,x <\,1$ } \\[2ex] x^{2}, & \text{ $-1<\,x<\,0$ } \end{cases}$
$f_{2}(x) = \begin{cases} x^{2}, & \text{ $0 \le\,x <\,1$ } \\[2ex] -x^{2}, & \text{ $-1<\,x<\,0$ } \end{cases}$
$f_{3}(x) = \begin{cases} - x^{2}, & \text{ $0 \le\,x <\,1$ } \\[2ex] x^{2}, & \text{ $-1<\,x<\,0$ } \end{cases}$
$f_{4}(x) = \begin{cases} - x^{2}, & \text{ $0 \le\,x <\,1$ } \\[2ex] -x^{2}, & \text{ $-1<\,x<\,0$ } \end{cases}$
$f_{5} (x)=0$
$f_{6}(x) = \begin{cases} 0, & \text{ $0 \le\,x <\,1$ } \\[2ex] x^{2}, & \text{ $-1<\,x<\,0$ } \end{cases},\ldots$ and so on
Hence, there are more than $4$ functions possible