Mathematics · 2025
JEE Main · 8 April 2025, Shift 2 · Q16
Given below are two statements: Statement I: lim_x → 0((tan^-1 x+ log_e √((1+x)/(1-x))-2 x)/x^5)=2/5 Statement II: lim_x → 1(x^(2/(1-x)))=1/e^2 In…
Given below are two statements:
Statement I : $\displaystyle \lim _{x \rightarrow 0}\left(\frac{\tan ^{-1} x+\log _e \sqrt{\dfrac{1+x}{1-x}}-2 x}{x^5}\right)=\frac{2}{5}$
Statement II : $\displaystyle \lim _{x \rightarrow 1}\left(x^{\frac{2}{1-x}}\right)=\frac{1}{e^2}$
In the light of the above statements, choose the correct answer from the options given below
Official answer
From NTA’s final answer key for this paper.
(1)
Both Statement I and Statement II are true
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JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.