Application of Derivatives
NCERT Class 12
CBSE
Grade 12
Question:
The maximum slope of the curve $y = -x^3 + 3x^2 + 9x - 27$ is:
(a) $12$
(b) $0$
(c) $3$
(d) $16$
Step-by-Step Solution
$m = -3x^2 + 6x + 9$. Critical point $x = 1 \Rightarrow m_{\max} = 12$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Evaluating maximum slope $= 12$: 1.0 Mark
Correct Answer: $12$
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