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Question

If fx=xsinxcosxx2tanxx32xsin2x5x∣ ∣ ∣,then limx0f'(x)x equals


A

1

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B

4

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C

3

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D

-4

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Solution

The correct option is B

4


limx0f'(x)x=limx0f''(x)1=f''0,Now, fx=xsinxcosxx2tanxx32xsin2x5x∣ ∣ ∣Upon differentiating f w.r.t x twiceand substituting x=0,f''0=4limx0f'(x)x=f''0=4


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November 2016
MATHEMATICS
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