JEE Main 2022 · 25 Jul · Shift 1 · Mathematics Q5

MathematicsSingle correct+4 / -1Question 5 of 90

If $\displaystyle\lim_{n\to\infty}\left(\sqrt{n^2-n-1}+n\alpha+\beta\right)=0$ , then $8(\alpha+\beta)$ is equal to :

-8

-4

Answer and worked solution

Official answer: C

Expansion. $\sqrt{n^2-n-1}=n-\frac12+O\!\left(\frac1n\right)$ , so the limit is 0 only if $\alpha=-1$ and $\beta=\frac12$ .

$8(\alpha+\beta)=-4$ .

(C)

Topics:Limits

JEE Main 2022 · 25 Jul · Shift 1, Mathematics question 5