JEE Main 2022 — Mathematics Calculus

Given $\underset{x\rightarrow 1}{\mathrm{lim}}(\frac{\mathrm{sin}(3{x}^{2}-4x+1)-{x}^{2}+1}{2{x}^{3}-7{x}^{2}+ax+b})=-2...(i)$x→1lim​(2x3−7x2+ax+bsin(3x2−4x+1)−x2+1​)=−2...(i)

Putting $x=1$x=1 in limit we get

$\Rightarrow \underset{x\rightarrow 1}{\mathrm{lim}}(\frac{\mathrm{sin}0-1+1}{2-7+a+b})=-2$⇒x→1lim​(2−7+a+bsin0−1+1​)=−2 $\Rightarrow \underset{x\rightarrow 1}{\mathrm{lim}}(\frac{0}{a+b-5})=-2$⇒x→1lim​(a+b−50​)=−2

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

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

Now Using L-Hospital in equation $(i)$(i) we get

$$\underset{x\rightarrow 1}{\mathrm{lim}}\frac{\mathrm{cos}(3{x}^{2}-4x+1)(6x-4)-2x}{6{x}^{2}-14x+a}=-2$$x→1lim​6x2−14x+acos(3x2−4x+1)(6x−4)−2x​=−2

Again putting $x=1$x=1 we get

$$\Rightarrow \underset{x\rightarrow 1}{\mathrm{lim}}\frac{(\mathrm{cos}0)\times (6-4)-2}{6-14+a}=-2$$⇒x→1lim​6−14+a(cos0)×(6−4)−2​=−2

$$\Rightarrow \underset{x\rightarrow 1}{\mathrm{lim}}\frac{0}{a-8}=-2$$⇒x→1lim​a−80​=−2

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

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

$$8+b=5\Rightarrow b=-3$$8+b=5⇒b=−3

So $a-b=8-(-3)=11$a−b=8−(−3)=11