NIMCET Differentibility Previous Year Questions (PYQs)

Given,

$f(x)=|x+1|e^{-x^2}$

For $x<-1$,

$f(x)=-(x+1)e^{-x^2}$

Differentiate:

$f'(x)=e^{-x^2}(2x^2+2x-1)$

For critical points,

$2x^2+2x-1=0$

Using quadratic formula,

$x=\frac{-2\pm\sqrt{4+8}}{4}$

$x=\frac{-2\pm 2\sqrt{3}}{4}$

$x=\frac{-1\pm\sqrt{3}}{2}$

For $x<-1$, the valid critical point is

$x=\frac{-1-\sqrt{3}}{2}$

This lies in the interval $(-2,-1)$.

At this point, $f$ has a point of maxima.

Using expansion near $x=0$,

$(1+x)^{\frac{1}{3}}=1+\frac{x}{3}-\frac{x^2}{9}+O(x^3)$

Also,

$(1+2x)^{\frac{1}{4}}=1+\frac{x}{2}-\frac{3x^2}{8}+O(x^3)$

Now,

$(1+x)^{\frac{1}{3}}-(1+2x)^{\frac{1}{4}}$

$=\left(1+\frac{x}{3}-\frac{x^2}{9}\right)-\left(1+\frac{x}{2}-\frac{3x^2}{8}\right)+O(x^3)$

$=-\frac{x}{6}+\left(-\frac{1}{9}+\frac{3}{8}\right)x^2+O(x^3)$

$=-\frac{x}{6}+\frac{19x^2}{72}+O(x^3)$

Therefore,

$f(x)=\frac{-\frac{x}{6}+\frac{19x^2}{72}+O(x^3)}{x}$

$f(x)=-\frac{1}{6}+\frac{19x}{72}+O(x^2)$

For differentiability at $x=0$, function must be continuous at $x=0$.

So,

$a=\lim_{x\to 0}f(x)=-\frac{1}{6}$

Also,

$f'(0)=\frac{19}{72}$

Hence,

$a+f'(0)=-\frac{1}{6}+\frac{19}{72}$

$=-\frac{12}{72}+\frac{19}{72}$

$=\frac{7}{72}$