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Question

# If the function f(x)=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩(1+|tanx|)p|tanx|,−π3<x<0qx=0esin3xsin2x,0<x<π3 is continuous at x=0, then

A
p=32
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B
p=23
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C
logeq=p
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D
q=2
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Solution

## The correct options are A logeq=p B p=32Since f(x) is continuous at x=0 limx→0−f(x)=f(0)=limx→0+f(x)...(1)Now , LHL=limx→0−f(x)=limx→0(1−tanx)−ptanxSince limx→0(1+x)1/x=e⇒LHL=ep...(2)Now, RHL=limx→0+f(x)=limx→0esin3xsin2x=limx→0e32(sin3x3x)(2xsin2x)RHL=e3/2....(3)From (1), (2) and (3)ep=e3/2=q⇒p=32=logeq

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