Mathematics · 2026
JEE Main · 22 January 2026, Shift 1 · Q16
Let f(x)=x^2025-x^2000, x ∈[0,1] and the minimum value of the function f(x) in the interval [0,1] be (80)^80(n)^-81. Then n is equal to
Let $\displaystyle f(x)=x^{2025}-x^{2000}, x \in[0,1]$ and the minimum value of the function $\displaystyle f(x)$ in the interval $\displaystyle [0,1]$ be $\displaystyle (80)^{80}(n)^{-81}$. Then $\displaystyle n$ is equal to
Official answer
From NTA’s final answer key for this paper.
(3)
$\displaystyle - 81$
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.