For α, β, γ, ∈ R , if _ x arrow 0 x^2 α x+(γ-1) e^ x^2 2 x-β x =3 , then β+γ-α is equal to:

Correct answer: A. 7 aligned & _ x arrow 10 (x^2(α x)+(γ-1)(1+(x^2)/(1)))/(2 x-(8 x^3)/(6)-β x)=3 & _ x arrow 0 ((γ-1)+(γ-1) x^2+α x^3)/((2-β) x-(4)/(3) x^3)=3 & γ-1, β=2, (-3 α)/(4)=+3 α=-4 & β+γ-α=7 aligned

JEE Main 20252 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 2 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Limits

JEE Main 2025 · 2 April · Morning Shift · Mathematics · Limits

For α, β, γ, ∈ R , if _ x arrow 0 x^2 α x+(γ-1) e^ x^2 2 x-β x =3 ,...

For

\alpha, \beta, \gamma, \in \mathbf{R}

, if

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

, then

\beta+\gamma-\alpha

is equal to:

Options

A7
B4
C6
D $-1$

Correct answer

A. 7

Step-by-step solution

\begin{aligned} & \lim _{x \rightarrow 10} \frac{x^2(\alpha x)+(\gamma-1)\left(1+\frac{x^2}{1}\right)}{2 x-\frac{8 x^3}{6}-\beta x}=3 \\ & \lim _{x \rightarrow 0} \frac{(\gamma-1)+(\gamma-1) x^2+\alpha x^3}{(2-\beta) x-\frac{4}{3} x^3}=3 \\ & \gamma-1, \beta=2, \frac{-3 \alpha}{4}=+3 \Rightarrow \alpha=-4 \\ & \beta+\gamma-\alpha=7\end{aligned}

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