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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=32
Since f(x) is continuous at x=0
limx0f(x)=f(0)=limx0+f(x)...(1)
Now ,
LHL=limx0f(x)=limx0(1tanx)ptanx
Since limx0(1+x)1/x=e
LHL=ep...(2)
Now,
RHL=limx0+f(x)=limx0esin3xsin2x
=limx0e32(sin3x3x)(2xsin2x)
RHL=e3/2....(3)
From (1), (2) and (3)
ep=e3/2=q
p=32=logeq

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