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QJEEM22Ma01b2: JEE Main 2022 Mathematics - Class 12 JEEM22Ma01b2 Solution: If , then the value of (a b) is equal to ___________.

If lim⁡x→1sin⁡(3x2−4x+1)−x2+12x3−7x2+ax+b=−2\mathop {\lim }\limits_{x \to 1} {{\sin (3{x^2} - 4x + 1) - {x^2} + 1} \over {2{x^3} - 7{x^2} + ax + b}} = - 2 , then the value of (a −- b) is equal to ___________.

Step-by-Step Verified Solution

amp;lim⁡x→1(sin⁡(3x2−4x+1)3x2−4x+1)(3x2−4x+1)−x2+12x3−7x2+ax+b=−2⇒amp;lim⁡x→13x2−4x+1−x2+12x3−7x2+ax+b=−2⇒amp;lim⁡x→12(x−1)22x3−7x2+ax+b=−2\begin{aligned}& \lim _{x \rightarrow 1} \frac{\left(\frac{\sin \left(3 x^{2}-4 x+1\right)}{3 x^{2}-4 x+1}\right)\left(3 x^{2}-4 x+1\right)-x^{2}+1}{2 x^{3}-7 x^{2}+a x+b}=-2 \\\\\Rightarrow & \lim _{x \rightarrow 1} \frac{3 x^{2}-4 x+1-x^{2}+1}{2 x^{3}-7 x^{2}+a x+b}=-2 \\\\\Rightarrow & \lim _{x \rightarrow 1} \frac{2(x-1)^{2}}{2 x^{3}-7 x^{2}+a x+b}=-2\end{aligned}

So f(x)=2x3−7x2+ax+b=0f(x)=2 x^{3}-7 x^{2}+a x+b=0 has x=1x=1 as repeated root, therefore f(1)=0f(1)=0 and f′(1)=0f^{\prime}(1)=0 gives

a+b+5 and a=8a+b+5 \text { and } a=8

So, a−b=11a-b=11

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