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Question

# If f(x)=⎛⎝1−cos xx sin x, x≠012, x=0 then at x = 0, f(x) is

A
continuous
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B
not continuous
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C
differentiable
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D
not differentiable
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Solution

## The correct options are A continuous C differentiable We have, f(x)=⎛⎝1−cos xx sin x, x≠012, x=0⇒f(x)=⎛⎜ ⎜⎝2 sin2 (x/2)x (2 sin x/2 cos x/2), x≠012, x=0⇒f(x)=⎛⎜⎝tan x/2x , x≠012, x=0(LHL at x=0)=limx→0−f(x)=limh→0f(0−h)=limh→0tan −h/2−h =12limh→0 tan −h/2−h/2 =12(RHL at x=0)=limx→0+f(x)=limh→0f(0+h)=limh→0tan h/2h =12limh→0 tan h/2h/2 =12(LHL at x=0)=(RHL at x=0)=f(0)∴f(x) is continuous at x=0.(LHD at x=0)=limx→0−f(x)−f(0)x−0=limh→0f(0−h)−f(0)0−h−0=limh→0tan (−h2)−h−12−h =limh→0tan (−h2)+h2h2 =limh→0⎛⎜ ⎜ ⎜⎝−h2+(−h2)33+2(−h2)515....⎞⎟ ⎟ ⎟⎠+h2h2 =limh→0(−h2)33+2(−h2)515....h2 =limh→0−h83+2−h33215....= 0(RHD at x=0)=limx→0+f(x)−f(0)x−0=limh→0f(0+h)−f(0)0+h−0=limh→0tan (h2)h−12h =limh→0tan (h2)−h2h2 = limh→0⎛⎜ ⎜ ⎜⎝h2+(h2)33+2(h2)515....⎞⎟ ⎟ ⎟⎠−h2h2= limh→0(h2)33+2(h2)515....h2= limh→0h83+2h33215....= 0(LHD at x=0)=(RHD at x=0)∴f(x) is differentiable at x=0

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