Verified Solution • 2025–26 Session moderate

QJEEM21Ma8932: JEE Main 2021 Mathematics - Class 12 JEEM21Ma8932 Solution: If , then the value of + + is _____________.

If lim⁡x→0αxex−βlog⁡e(1+x)+γx2e−xxsin⁡2x=10,α,β,γ∈R\mathop {\lim }\limits_{x \to 0} {{\alpha x{e^x} - \beta {{\log }_e}(1 + x) + \gamma {x^2}{e^{ - x}}} \over {x{{\sin }^2}x}} = 10,\alpha ,\beta ,\gamma \in R , then the value of α\alpha + β\beta + γ\gamma is _____________.

Step-by-Step Verified Solution

lim⁡x→0αx(1+x+x2x)−β(x−x22+x33)+γx2(1−x)x3\mathop {\lim }\limits_{x \to 0} {{\alpha x\left( {1 + x + {{{x^2}} \over x}} \right) - \beta \left( {x - {{{x^2}} \over 2} + {{{x^3}} \over 3}} \right) + \gamma {x^2}(1 - x)} \over {{x^3}}}

lim⁡x→0x(α−β)+x2(α+β2+γ)+x3(α2−β3−γ)x3=10\mathop {\lim }\limits_{x \to 0} {{x(\alpha - \beta ) + {x^2}\left( {\alpha + {\beta \over 2} + \gamma } \right) + {x^3}\left( {{\alpha \over 2} - {\beta \over 3} - \gamma } \right)} \over {{x^3}}} = 10

For limit to exist

α−β=0,α+β2+γ=0\alpha - \beta = 0,\alpha + {\beta \over 2} + \gamma = 0 α2−β3−γ=10{\alpha \over 2} - {\beta \over 3} - \gamma = 10 ..... (i)

β=α,γ=−3α2\beta = \alpha ,\gamma = - 3{\alpha \over 2}

Put in (i)

α2−α3+3α2=10{\alpha \over 2} - {\alpha \over 3} + {{3\alpha } \over 2} = 10

α6+3α2=10⇒α+9α6=10{\alpha \over 6} + {{3\alpha } \over 2} = 10 \Rightarrow {{\alpha + 9\alpha } \over 6} = 10

⇒α=6\Rightarrow \alpha = 6

α\alpha = 6, β\beta = 6, γ\gamma = −- 9

α\alpha + β\beta + γ\gamma = 3

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