Let f(x)=1−|cosx|∀x∈R, then which among the following options is/are correct?
A
f(x) is not differentiable everywhere
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B
f(x) is continuous everywhere
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C
f′(π2) does not exist
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D
limx→π2+f(x)=1
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Solution
The correct option is Dlimx→π2+f(x)=1 f(x)=1−|cosx| is continuous as cosx is continuous for all x which gives |cosx| is also continuous.
Hence, 1−|cosx| is continuous.
Also, limx→π2+f(x)=limx→π2+1−|cosx|=1
Also, since f(x) has sharp corner at x=π2. Hence, f′(π2) does not exist.
Similar sharp corners occur at all the points where cosx=0 i.e. at x=(2n+1)π2,n∈Z.
Hence, f(x) is not differentiable everywhere.