Application of Derivatives
NCERT Class 12
CBSE
Grade 12
Question:
Find the maximum and minimum values of $f(x) = 3x^4 - 8x^3 + 12x^2 - 48x + 25$ on $[0, 3]$.
Step-by-Step Solution
Critical point $x = 2$. Evaluate $f(0)=25, f(2)=-39, f(3)=16$. [1.0 Mark]
Maximum $= 25$, Minimum $= -39$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Finding critical point $x=2$: 1.0 Mark
Evaluating maximum $= 25$ and minimum $= -39$: 1.0 Mark
Correct Answer:
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