Question

Consider the function $f\left(x\right)=x^2+mx+n,$ where $D=m^2-4n>0.$
$\begin{array}{cc}Column1& Column2\\ \text{Condition on}m\text{and}n& \text{Number of points of non-}\\ & \text{differentiability of}g\left(x\right)=|f\left(|x|\right)|\\ \text{a.}m<0,n>0& \text{p.}1\\ \text{b.}n=0,m<0& \text{q.}2\\ \text{c.}n=0,m>0& \text{r.}3\\ \text{d.}m=0,n<0& \text{s.}5\end{array}$
Then which of the following is correct ?

• A. $a\to s,b\to r,c\to p,d\to q$
• B. $a\to r,b\to s,c\to p,d\to q$
• C. $a\to s,b\to r,c\to q,d\to p$
• D. $a\to s,b\to p,c\to r,d\to q$

Answer 1

The correct option is A $a\to s,b\to r,c\to p,d\to q$

$f\left(x\right)=x^2+mx+n$
$g\left(x\right)=|f\left(|x|\right)|=|x^2+m|x|+n|$
We know that, derivative at sharp point does not exist.

$\text{a.}m<0,n>0$
From the graph number of points of non-differentiability of $g\left(x\right)=|f\left(|x|\right)|$ is $5$.
$a\to s$

$\text{b.}n=0,m<0$
From the graph number of points of non-differentiability of $g\left(x\right)=|f\left(|x|\right)|$ is $3$.
$b\to r$

$\text{c.}n=0,m>0$
From the graph number of points of non-differentiability of $g\left(x\right)=|f\left(|x|\right)|$ is $1$.
$c\to p$

$\text{d.}m=0,n<0$
From the graph number of points of non-differentiability of $g\left(x\right)=|f\left(|x|\right)|$ is $2$.
$d\to q$