Mathematics · 2022
JEE Main · 26 June 2022, Shift 1 · Q8
The sum of the absolute minimum and the absolute maximum values of the function f(x)=|3 x-x^2+2|-x in the interval [-1,2] is:
The sum of the absolute minimum and the absolute maximum values of the function $\displaystyle f(x)=\left|3 x-x^2+2\right|-x$ in the interval $\displaystyle [-1,2]$ is :
Official answer
(1)
$\displaystyle \frac{\sqrt{17}+3}{2}$
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.