Verified Solution • 2025–26 Session moderate QJEEM22Ma12e6: JEE Main 2022 Mathematics - Class 12 JEEM22Ma12e6 Solution: Let for some . Then the value of is : Let β=limx→0αx−(e3x−1)αx(e3x−1)\beta=\mathop {\lim }\limits_{x \to 0} \frac{\alpha x-\left(e^{3 x}-1\right)}{\alpha x\left(e^{3 x}-1\right)}β=x→0limαx(e3x−1)αx−(e3x−1) for some α∈R\alpha \in \mathbb{R}α∈R . Then the value of α+β\alpha+\betaα+β is : Multiple Choice Options A 145\frac{14}{5}514 B 32\frac{3}{2}23 C 52\frac{5}{2}25 D 72\frac{7}{2}27 Step-by-Step Verified Solution β=limx→0αx−(e3x−1)αx(e3x−1), α∈R\beta = \mathop {\lim }\limits_{x \to 0} {{\alpha x - ({e^{3x}} - 1)} \over {\alpha x({e^{3x}} - 1)}},\,\alpha \in Rβ=x→0limαx(e3x−1)αx−(e3x−1),α∈R =limx→0α3−(e3x−13x)αx(e3x−13x)= \mathop {\lim }\limits_{x \to 0} {{{\alpha \over 3} - \left( {{{{e^{3x}} - 1} \over {3x}}} \right)} \over {\alpha x\left( {{{{e^{3x - 1}}} \over {3x}}} \right)}}=x→0limαx(3xe3x−1)3α−(3xe3x−1) So, α=3\alpha = 3α=3 (to make independent form) β=limx→01−(e3x−13x)3x=1−(3x+9x22 + ......)3x3x\beta = \mathop {\lim }\limits_{x \to 0} {{1 - \left( {{{{e^{3x}} - 1} \over {3x}}} \right)} \over {3x}} = {{1 - {{\left( {3x + {{9{x^2}} \over 2}\, + \,......} \right)} \over {3x}}} \over {3x}}β=x→0lim3x1−(3xe3x−1)=3x1−3x(3x+29x2+......) =−(92x2+(3x)331 + ....)9x2=−12= {{ - \left( {{9 \over 2}{x^2} + {{{{(3x)}^3}} \over {31}}\, + \,....} \right)} \over {9{x^2}}} = {{ - 1} \over 2}=9x2−(29x2+31(3x)3+....)=2−1 ∴\therefore∴ α+β=3−12=52\alpha + \beta = 3 - {1 \over 2} = {5 \over 2}α+β=3−21=25 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring