For , if , then is equal to: [2025]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
For $α, β, γ \in R$ , if $\lim_{x \to 0} \frac{x^{2} \sin α x + (γ - 1) e^{x^{2}}}{\sin 2 x - β x} = 3$ , then $β + γ - α$ is equal to: [2025]

1 7

2 4

3 6

4 –1

Ans.
(1)

As, $x \to 0 \Rightarrow \sin 2 x - β x \to 0$

To make the given limit in $\frac{0}{0}$ form; $(γ - 1) e^{0} + 0 \sin (α 0) = 0$

$\Rightarrow (γ - 1) = 0 \Rightarrow γ = 1$

So, $\lim_{x \to 0} \frac{x^{2} \sin (α x)}{(\sin 2 x - β x)} = 3$

$\Rightarrow \lim_{x \to 0} \frac{x^{2} [α x - \frac{{(α x)}^{3}}{3!} + \frac{{(α x)}^{5}}{5!} - . . . . .]}{[2 x - \frac{{(2 x)}^{3}}{3!} + \frac{{(2 x)}^{5}}{5!} - . . . . .] - β x} = 3$

$\Rightarrow \lim_{x \to 0} \frac{α x^{3} - \frac{α^{3} x^{5}}{3!} + \frac{α^{5} x^{7}}{5!} - . . . . .}{x (2 - β) - \frac{8 x^{3}}{6} + \frac{2^{5} \cdot x^{5}}{5!} - . . . . .} = 3$

$\Rightarrow 2 - β = 0 and \frac{α}{\frac{- 8}{6}} = 3$

$\Rightarrow β = 2 and α = 3 (- \frac{8}{6}) = - 4$

$∴ β + γ - α = 2 + 1 - (- 4) = 7$ .

Want Us to Email you About Special Offers & Updates?