JEE Main 2022 — Limits Question with Solution

Given $\underset{x\to 1}{\lim }\left(\frac{\sin \left(3x^2-4x+1\right)-x^2+1}{2x^3-7x^2+ax+b}\right)=-2 ...\left(\mathrm{i}\right)$

Putting $x=1$ in limit we get

$\Rightarrow \underset{x\to 1}{\lim }\left(\frac{\sin 0-1+1}{2-7+a+b}\right)=-2$ $\Rightarrow \underset{x\to 1}{\lim }\left(\frac{0}{a+b-5}\right)=-2$

For limit to exist $a+b-5$ should be zero

so $a+b-5=0$ $\Rightarrow a+b=5 ...\left(\mathrm{ii}\right)$

Now Using L-Hospital in equation $\left(\mathrm{i}\right)$ we get

$\underset{x\to 1}{\lim }\frac{\cos \left(3x^2-4x+1\right)\left(6x-4\right)-2x}{6x^2-14x+a}=-2$

Again putting $x=1$ we get

$\Rightarrow \underset{x\to 1}{\lim }\frac{\left(\cos 0\right)\times \left(6-4\right)-2}{6-14+a}=-2$

$\Rightarrow \underset{x\to 1}{\lim }\frac{0}{a-8}=-2$

Again for limit to exist $a-8=0\Rightarrow a=8$

So from equation $\left(\mathrm{ii}\right)$ we get

$8+b=5\Rightarrow b=-3$

So $a-b=8-\left(-3\right)=11$