JEE Advanced 2016 · Paper 1 · Mathematics Q53

MathematicsInteger+3Question 53 of 54

Let $\alpha,\beta\in\mathbb R$ be such that $\displaystyle\lim_{x\to0}\frac{x^2\sin(\beta x)}{\alpha x-\sin x}=1$ . Then $6(\alpha+\beta)$ equals

Your answer

Answer and worked solution

Official answer: 7

Expand. Near 0, the numerator is $x^2\sin\beta x\approx\beta x^3$ . The denominator is
$\alpha x-\sin x=(\alpha-1)x+\frac{x^3}6-\cdots$

Finite non-zero limit. We need $\alpha=1$ . Then the limit is $6\beta=1$ , so $\beta=\frac16$ .

Answer. $6(\alpha+\beta)=6+1=7$ .

Answer: 7

Topics:Limits

JEE Advanced 2016 · Paper 1, Mathematics question 53