Application of Derivatives
NCERT Class 12
CBSE
Grade 12
Question:
The absolute maximum value of $f(x) = x^3$ on the closed interval $[-2, 2]$ is:
(a) $8$
(b) $-8$
(c) $0$
(d) $4$
Step-by-Step Solution
$f(-2) = -8, f(0) = 0, f(2) = 8 \Rightarrow \text{Abs Max} = 8$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Evaluating absolute maximum $= 8$: 1.0 Mark
Correct Answer: $8$
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