Question

Let $f:(-1,1)\to R$ be a differentiable function satisfying
$(f^{\prime }(x){)}^4=16(f(x){)}^2$ 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,
${\left(f^{\prime }\left(x\right)\right)}^4=16{\left(f\left(x\right)\right)}^2,\forall x\in \left(-1,1\right)$
$\Rightarrow {\left(f^{\prime }\left(x\right)\right)}^2=\pm 4f\left(x\right)$
Case I
If $f\left(x\right)\ge 0\forall x\in \left(-1,1\right)$
Then, $f^{\prime }\left(x\right)=\pm 2\sqrt{f\left(x\right)}$
$\Rightarrow \frac{f^{\prime }\left(x\right)}{\sqrt{f\left(x\right)}}=\pm 2$
$\Rightarrow 2\sqrt{f\left(x\right)}=\pm 2x$,
$\left[\because f\left(0\right)=0\right]$
$\Rightarrow f\left(x\right)=x^2\forall x\in \left(-1,1\right)$
Case II
If $f\left(x\right)<0\forall x\in \left(-1,1\right)$
Then, $f^{\prime }\left(x\right)=\pm 2\sqrt{-f\left(x\right)}$
$\therefore f\left(x\right)=-x^2\forall x\in \left(-1,1\right)$
$\therefore$ $f_1(x)=\{\begin{array}{ll}{x}^2,& \text{ 0≤x<1 }\\ x^2,& \text{ −1<x<0 }\end{array}$
$f_2(x)=\{\begin{array}{ll}{x}^2,& \text{ 0≤x<1 }\\ -x^2,& \text{ −1<x<0 }\end{array}$
$f_3(x)=\{\begin{array}{ll}-x^2,& \text{ 0≤x<1 }\\ x^2,& \text{ −1<x<0 }\end{array}$
$f_4(x)=\{\begin{array}{ll}-x^2,& \text{ 0≤x<1 }\\ -x^2,& \text{ −1<x<0 }\end{array}$
$f_5(x)=0$
$f_6(x)=\{\begin{array}{ll}0,& \text{ 0≤x<1 }\\ x^2,& \text{ −1<x<0 }\end{array},\dots$ and so on
Hence, there are more than $4$ functions possible