Mathematics · 2024
JEE Main · 31 January 2024, Shift 1 · Q6
lim_x → 0 (e^2| sin x|-2| sin x|-1)/x^2
$$\lim _{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{x^2}
$$
Official answer
From NTA’s final answer key for this paper.
(1)
is equal to $\displaystyle 2$
More from Limits
- The value of lim_x → 0 (log_e( sec (e x) · sec (e^2 x) · … · sec (e^10 x)))/(e^2-e^2 cos x) is equal to2026
- If lim_x → ∞((e/(1-e))(1/e-x/(1+x)))^x=α, then the value of (log_e α)/(1+ log_e α) equals:2025
- Let [t] be the greatest integer less than or equal to t. Then the least value of p ∈ N for which lim_x →…2025
- If lim_x → 0 (e^(a-1) x+2 cos b x+(c-2) e^-x)/(x cos x- log_e(1+x))=2, then a^2+b^2+c^2 is equal to:2026
- Let f:[-(π)/2, (π)/2] → R be a differentiable function such that f(0)=1/2. If the lim_x → 0 (x ∫_0^x f(t) d t)/(e^x^2-1)=α, then 8 α^2 is equal to:2024
- For α, β, γ ∈ R, if lim_x → 0 (x^2 sin α x+(γ-1) e^x^2)/(sin 2 x-β x)=3, then β+γ-α is equal to:2025
- The product of all possible values of α, for which lim_x → 0((1- cos (α x) cos ((α+1) x) cos ((α+2) x))/(sin^2((α+1) x)))=2, is:2026
- Let f:(-∞, ∞)-{0} → R be a differentiable function such that f^′(1)= lim_a → ∞ a^2 f(1/a). Then lim_a → ∞ (a(a+1))/2 tan^-1(1/a)+a^2-2 log_e a is…2024
JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.