JEE Main 2022 · 26 Jul · Shift 2 · Mathematics Q7

← JEE Main 2022 · 26 Jul · Shift 2

MathematicsSingle correct+4 / -1Question 7 of 90

Let $\beta=\displaystyle\lim_{x\to0}\frac{\alpha x-\left(e^{3x}-1\right)}{\alpha x\left(e^{3x}-1\right)}$ for some $\alpha\in\mathbf R$ . Then the value of $\alpha+\beta$ is :

\dfrac{14}5

\dfrac32

\dfrac52

\dfrac72

Answer and worked solution

Official answer: C

Expand. $e^{3x}-1=3x+\frac92x^2+\dots$ , so the numerator is $(\alpha-3)x-\frac92x^2+\dots$ and the denominator $3\alpha x^2+\dots$

Finite limit needs $\alpha=3$ ; then $\beta=\frac{-9/2}{9}=-\frac12$ .

$\alpha+\beta=\frac52$ . (C)

Topics:Limits

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JEE Main 2022 · 26 Jul · Shift 2, Mathematics question 7