Question

If the function $f\left(x\right)=|x^2+a|x|+b|$ has exactly three points of non-differentiability, then which of the following may hold

• A. $b=0,a<0$
• B. $b<0,aϵR$
• C. $b>0,aϵR$
• D. $b<0,aϵR^{-}$

Answer 1

The correct option is A $b=0,a<0$

$f\left(x\right)=|x^2+a\left|x\right|+b|$

$f^{\prime }\left(x\right)=\frac{|x^2+a\left|x\right|+b|}{x^2+a\left|x\right|+b}\times (2x+\frac{a\left|x\right|}{x})$

for $f\left(x\right)$ to be non differentiable

$f^{\prime }\left(x\right)$ must not exist

$\Rightarrow (x^2+a\left|x\right|+b)\left(x\right)=0$

$x=0$ (or) $x^2+a\left|x\right|+b=0$

if $b\ne 0$

$x^2+a\left|x\right|+b$ will have $4$ roots

$\alpha ,\beta ,-\alpha ,-\beta$

if $b=0$

$x^2+a\left|x\right|$ will have $2$ roots (other than $0$)

i.e., $x^2+a\left|x\right|=0$

${\left|x\right|}^2=-a\left|x\right|$

$\left|x\right|=-a$

Value of mod function can only be positive.

$\Rightarrow -a>0\Rightarrow a<0$

$\left|x\right|=-a$

$x=\pm a$