Question

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

• A. $2$
• B. $3$
• C. $4$
• D. more than $4$

Answer 1

The correct option is D more than $4$

$\text{Given that,}\left(f^{\prime }\right(x){)}^4=16(f\left(x\right){)}^2\Rightarrow \left(f^{\prime }\right(x){)}^2=\pm 4(f\left(x\right))$
$Case-If\left(x\right)\ge 0\left(f^{\prime }\right(x){)}^2=4(f\left(x\right))\Rightarrow f^{\prime }(x)=\pm 2\sqrt{f\left(x\right)}$
$f^{\prime }\left(x\right)=2\sqrt{f\left(x\right)}\text{or}{f}^{\prime }\left(x\right)=-2\sqrt{f\left(x\right)}$
$f\left(x\right)=x^2,1>x\ge 0\text{or}f\left(x\right)=x^2,-1\le x<0$

$Case-IIf\left(x\right)<0\left(f^{\prime }\right(x){)}^2=-4(f\left(x\right))\Rightarrow f^{\prime }(x)=\pm 2\sqrt{-f\left(x\right)}$
$f^{\prime }\left(x\right)=2\sqrt{-f\left(x\right)}\text{or}{f}^{\prime }\left(x\right)=-2\sqrt{-f\left(x\right)}$
$f\left(x\right)=-x^2,1>x\ge 0\text{or}f\left(x\right)=-x^2,-1\le x<0$

Hence function can be
$f\left(x\right)=x^2,-1<x<1f\left(x\right)=-x^2,-1<x<1$

$f_2\left(x\right)=\{\begin{array}{cc}{x}^2,-1\le x<0& \\ -x^2,0\le x<1& \end{array}$

$f\left(x\right)=\{\begin{array}{cc}-x^2,-1<x<0& \\ x^2,0\le x<1& \end{array}$

$f\left(x\right)=0\forall x\in (-1,1)$

$f\left(x\right)=\{\begin{array}{cc}{x}^2,-1<x<0& \\ 0,0\le x<1& \end{array}$
Therefore more functions are also possible.
Hence,number of such functions are more than $4.$



Alternative solution
$\text{Given that,}\left(f^{\prime }\right(x){)}^4=16(f\left(x\right){)}^2\Rightarrow f^{\prime }\left(x\right)=\pm 2\sqrt{(\pm f(x\left)\right)}$
$\Rightarrow \sqrt{\pm f\left(x\right)}=\pm x\Rightarrow f\left(x\right)=\pm x^2$

So, there will be four cases.

Case $1$: $x>0,f\left(x\right)>0$ $\Rightarrow f\left(x\right)=x^2$
Case $2$: $x>0,f\left(x\right)<0$ $\Rightarrow f\left(x\right)=-x^2$
Case $3$: $x<0,f\left(x\right)>0$ $\Rightarrow f\left(x\right)=x^2$
Case $4$: $x<0,f\left(x\right)<0$ $\Rightarrow f\left(x\right)=-x^2$

For these cases, four different function possible

$f_1\left(x\right)=x^2,\forall -1\le x\le 1$

$f_2\left(x\right)=\{\begin{array}{cc}{x}^2,0\le x\le 1& \\ -x^2,-1\le x<0& \end{array}$

$f_3\left(x\right)=\{\begin{array}{cc}-x^2,0\le x\le 1& \\ x^2,-1\le x<0& \end{array}$

$f_4\left(x\right)=-x^2,\forall -1\le x\le 1$

Apart from this, there is one more function satisfying the conditions

$f_5\left(x\right)=0,\forall -1\le x\le 1$

So, there are total five functions