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Question

If f(x)=(3x2+12x−1, −1≤x≤237−x, 2<x≤3, then

A
f(x) is increasing in [-1, 2]
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B
f(x) is continuous in [-1, 3]
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C
f '(2) does not exist
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D
f(x) has a maximum value at x = 2
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Solution

The correct options are A f(x) is increasing in [-1, 2] B f(x) is continuous in [-1, 3] C f '(2) does not exist D f(x) has a maximum value at x = 2We have, f(x)=(3x2+12x−1, −1≤x≤237−x, 2<x≤3 Then, in [-1, 2], f′(x)=6x+12.f′(x)=0⇒x=−2Thus, f(x) decreases in (−∞,−2) and increases in (−2,∞) Also, f(2−)=3(2)2+12(2)−1=35and f(2+)=37−2=35 Hence, f(x) is continuous. f'(x)=(6x+12, −1≤x≤2−1, 2<x≤3∴f'(2−)=24 and f'(2+)=−1Hence, f(x) is non−differentiable at x=2.Also, f'(2+)<f(2) and f'(2−)<f(2).Hence, x=2 is the point of maxima.

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