JEE Advanced 2024 Paper 2 Q03 Mathematics Differentiation & Applications Limits & Continuity Medium

JEE Advanced 2024 Paper 2 · Q03 · Limits & Continuity

Let $k \in \mathbb{R}$. If $\displaystyle \lim_{x \to 0^{+}} \bigl(\sin(\sin kx) + \cos x + x\bigr)^{\frac{2}{x}} = e^{6}$, then the value of $k$ is

  1. A. $1$
  2. B. $2$
  3. C. $3$
  4. D. $4$
Reveal answer + step-by-step solution

Correct answer:B

Solution

This is a $1^{\infty}$ form. Limit equals $\exp\!\left(\displaystyle\lim_{x\to 0^{+}}\dfrac{2\bigl(\sin(\sin kx) + \cos x + x - 1\bigr)}{x}\right)$. Expand near $0$: $\sin(\sin kx) \approx kx$, $\cos x \approx 1 - x^{2}/2$, so the bracket $\approx kx + x - x^{2}/2$. Dividing by $x$ and taking the limit gives $k + 1$. Therefore $2(k+1) = 6$, so $k = 2$. Answer: (B).

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