Mathematics · 2026
JEE Main · 5 April 2026, Shift 2 · Q19
Let f:[1, ∞) → R be a differentiable function defined as f(x)=∫_1^x f( t ) dt +(1-x)( log_e x-1)+ e. Then the value of f(f(1)) is:
Let $\displaystyle f:[1, \infty) \rightarrow \mathbf{R}$ be a differentiable function defined as $\displaystyle f(x)=\int_1^x f(\mathrm{t}) \mathrm{dt}+(1-x)\left(\log _{\mathrm{e}} x-1\right)+\mathrm{e}$. Then the value of $\displaystyle f(f(1))$ is :
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle \left(1+\mathrm{e}^{\mathrm{e}}\right)$
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.