Mathematics · 2019
JEE Main · 11 January 2019, Shift 1 · Q73
The maximum value of the function f(x)=3 x^3-18 x^2+27 x-40 on the set S ={x ∈ R: x^2+30 ≤ 11 x} is:
The maximum value of the function $\displaystyle f(x)=3 x^3-18 x^2+27 x-40$ on the set $\displaystyle \mathrm{S}=\left\{x \in \mathbf{R}: x^2+30 \leq 11 x\right\}$ is :
Official answer
(1)
$\displaystyle 122$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.