Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{x-a}{|x-a|},& -\infty <x\le 0\\ 2x^2+3bx+c,& 0<x<1\\ a^{2}x+c-2,& 1\le x<\infty \end{array}$
is a differentiable function, then

• A. $a+c=-1$
• B. $a+c=1$
• C. $a^2+b+c=3$
• D. $a^2+b+c=5$

Answer 1

Answer: (B), (C)

$a^2+b+c=3$

For $\frac{x-a}{|x-a|}$ to be continuous $\forall x\le 0,x-a$ should be either $>0$ or $<0$
$\because x\le 0$ (given)
If $a>0$ then $x-a<0$ (always)
Now $\frac{x-a}{|x-a|}=-1$ which is continuous and differentiable $\forall x\le 0$

For continuity at $x=0$
$f\left(0^{-}\right)=f\left(0^{+}\right)=f\left(0\right)\Rightarrow \frac{-a}{|a|}=c$
$\Rightarrow c=-1(\because a>0)\dots \left(1\right)$
For differentiability at $x=0,$
$0=4x+3b\Rightarrow b=0\dots \left(2\right)$

For continuity at $x=1,$
$f\left(1^{-}\right)=f\left(1^{+}\right)=f\left(1\right)\Rightarrow 2+3b+c=a^2+c-2\Rightarrow a^2-3b=4\dots \left(3\right)$
from equation $\left(2\right)$
$a^2=4\Rightarrow a=2(\because a>0)$
For differentiability at $x=1,$ we get
${(4x+3b)}_{x=1}={\left(a^2\right)}_{x=1}$
$4+3b=a^2\Rightarrow a^2-3b=4,$
which is same as equation $\left(3\right)$
$\therefore a=2,c=-1$ and $b=0$
$\Rightarrow a+c=2-1=1,a^2+b+c=4+0+(-1)=3$