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Mathematics · 2026 · 1 mark
CBSE 2026 · Region 5 · Set 1 · Q9
Absolute minimum value of f ( x )=( x -2)^2+5 in the interval [-3,2] is:
Absolute minimum value of $\displaystyle \mathrm{f}(\mathrm{x})=(\mathrm{x}-2)^{2}+5$ in the interval $\displaystyle [-3,2]$ is :
Official answer
(C)
$\displaystyle 5$
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