Mathematics · 2026
JEE Main · 21 January 2026, Shift 2 · Q17
Let f(x)=x^3+x^2 f^′(1)+2 x f^′ ′(2)+f^′ ′ ′(3), x ∈ R. Then the value of f^′(5) is:
Let $\displaystyle f(x)=x^3+x^2 f^{\prime}(1)+2 x f^{\prime \prime}(2)+f^{\prime \prime \prime}(3), x \in \mathbf{R}$. Then the value of $\displaystyle f^{\prime}(5)$ is :
Official answer
From NTA’s final answer key for this paper.
(3)
$\displaystyle \frac{117}{5}$
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.