Mathematics · 2026
JEE Main · 21 January 2026, Shift 2 · Q18
Let f: R → R be a twice differentiable function such that f^′ ′(x)>0 for all x ∈ R and f^′( a -1)=0, where a is a real number. Let g (x)=f( tan^2 x-2…
Let $\displaystyle f: \mathbf{R} \rightarrow \mathbf{R}$ be a twice differentiable function such that $\displaystyle f^{\prime \prime}(x)>0$ for all $\displaystyle x \in \mathbf{R}$ and $\displaystyle f^{\prime}(\mathrm{a}-1)=0$, where a is a real number. Let $\displaystyle \mathrm{g}(x)=f\left(\tan ^2 x-2 \tan x+\mathrm{a}\right), 0<x<\frac{\pi}{2}$.
Consider the following two statements :
(I)
g is increasing in $\displaystyle \left(0, \frac{\pi}{4}\right)$
(II)
g is deceasing in $\displaystyle \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$
Then,
Official answer
From NTA’s final answer key for this paper.
(1)
Neither (I) nor (II) is True
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.