JEE 2025 — Mathematics PYQ

${\lim }_{x\to 0}\frac{x^2\sin \alpha x+(\gamma -1)e^{x^2}}{\sin 2x-\beta x}=3$

Applying following Frenyes expansion

$\begin{array}{rl}& amp;\sin t=t-\frac{t^3}{3!}+\dots \\ & amp;e^{x^2}=1+\frac{x^2}{1}+\frac{x^4}{2!}\dots \end{array}$

Neglecting higher degree terms,

$\begin{array}{rl}& amp;\underset{x\to 10}{\lim }\frac{x^2(\alpha x)+(\gamma -1)\left(1+\frac{x^2}{1}\right)}{2x-\frac{8{\alpha }^3}{6}-\beta x}=3\\ & amp;\underset{x\to 0}{\lim }\frac{(\gamma -1)+(\gamma -1)x^2+\alpha x^3}{(2-\beta )x-\frac{4}{3}{x}^3}\end{array}$

Limit will exist if coefficients of $x^0,x^1,x^2$ are zero.

$\gamma -1=0,2-\beta =0$

and

$\frac{\alpha }{\left(-\frac{4}{3}\right)}=3$

$\gamma =1,\beta =2,\frac{-3\alpha }{4}=3<br>$
$\Rightarrow \alpha =-4<br>$
$\beta +\gamma -\alpha =7$