JEE MAIN - Mathematics (2022 - 28th June Evening Shift - No. 23)
If $$\mathop {\lim }\limits_{x \to 1} {{\sin (3{x^2} - 4x + 1) - {x^2} + 1} \over {2{x^3} - 7{x^2} + ax + b}} = - 2$$, then the value of (a $$-$$ b) is equal to ___________.
Answer
11
Explanation
$$
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{\left(\frac{\sin \left(3 x^{2}-4 x+1\right)}{3 x^{2}-4 x+1}\right)\left(3 x^{2}-4 x+1\right)-x^{2}+1}{2 x^{3}-7 x^{2}+a x+b}=-2 \\\\
\Rightarrow & \lim _{x \rightarrow 1} \frac{3 x^{2}-4 x+1-x^{2}+1}{2 x^{3}-7 x^{2}+a x+b}=-2 \\\\
\Rightarrow & \lim _{x \rightarrow 1} \frac{2(x-1)^{2}}{2 x^{3}-7 x^{2}+a x+b}=-2
\end{aligned}
$$
So $f(x)=2 x^{3}-7 x^{2}+a x+b=0$ has $x=1$ as repeated root, therefore $f(1)=0$ and $f^{\prime}(1)=0$ gives
$$ a+b+5 \text { and } a=8 $$
So, $a-b=11$
So $f(x)=2 x^{3}-7 x^{2}+a x+b=0$ has $x=1$ as repeated root, therefore $f(1)=0$ and $f^{\prime}(1)=0$ gives
$$ a+b+5 \text { and } a=8 $$
So, $a-b=11$
Comments (0)
