Question 5
Let f (x) = |cos x|. Then,
(A) f is everywhere differentiable.
(B) f is everywhere continuous but not differentiable at n = nš, n ā Z
(C) f is everywhere continuous but not differentiable at x = (2n + 1)š/2, n ā Z
(D) None of these
f(š„) = |cos š„|
We need to check continuity and differentiability of f(š„)
Continuity of f(š)
Let š(š„)=|š„| & ā(š„)=cosā”š„
Then,
ššš(š)=š(ā(š„))
=š(cosā”š„ )
=|cosā”š„ |
=š(š)
ā“ š(š„)=ššā(š„)
We know that,
š(š)=šššā”š is continuous as cos is continuous
š(š)=|š| is continuous as it is a modulus function
Hence, g(š„) & h(š„) both are continuous
And
If two functions g(š„) & h(š„) are continuous then their composition ššā(š„) is also continuous
ā“ š(š) is continuous
Differentiability of š(š)
š(š„)=|cosā”š„ |
Since, it is a modulus function so we check differentiability when cosā”š„=0
i.e., š=((šš +š)š )/š, šāš
š(š„) is differentiable at š„=((2š +1)š)/2, if
LHD = RHD
(ššš)ā¬(š”āš) (š(š) ā š(š ā š))/š
= (ššš)ā¬(hā0) (š(((šš +š)š )/š) ā š(((šš +š)š )/š ā ā))/ā
= (ššš)ā¬(hā0) (|cosā”ć((šš +š)š )/šć | ā|cosā”ć((šš +š)š /š ā ā)ć |)/ā
= (ššš)ā¬(hā0) (|0| ā |cosā”ć((šš +š)š /š ā ā)ć |)/ā
Using cos (A ā B) = cos Acos B + sin Asin B
= (ššš)ā¬(hā0) (0 ā |cosā”ć((šš +š)š )/š cosā”ćā+ sinā”ć((šš +š)š )/š sinā”ā ć ć ć |)/ā
= (ššš)ā¬(hā0) ( ā |0 ā ćsin ćā”ć((šš +š)š )/š sinā”ā ć |)/ā
= (ššš)ā¬(hā0) ( āsinā”ć((šš +š)š )/š sinā”ā ć)/ā
= āš šš ((2š +1)š)/2Ć(ššš)ā¬(hā0) sinā”ā/ā
Using (ššš)ā¬(š„ā0) š ššā”š„/š„=1
=āš šš ((2š +1)š)/2Ć1
=āššš ((šš +š)š )/š
(ššš)ā¬(š”āš) (š(š + š) ā š(š ))/š
= (ššš)ā¬(hā0) (š(((šš +š)š )/š + ā) ā š(((šš +š)š )/š))/ā
= (ššš)ā¬(hā0) (|cosā”ć((2š + 1)š/2 + ā )ć |ā|cosā”ć((2š +1)š)/2ć |)/ā
= (ššš)ā¬(hā0) (|ššš ā”(((2š + 1)š)/2 +ā) |ā|0|)/( ā)
Using cos (A + B) = cos Acos B -- sin Asin B
= (ššš)ā¬(hā0) (|ćcos ćā”ć((2š +1)š)/2 cosā”ćā ā sinā”ć((2š +1)š)/2 sinā”ā ć ć ć | ā 0)/ā
= (ššš)ā¬(hā0) |0 ā sinā”ć((2š + 1)š)/2 sinā”ā ć |/ā
= (ššš)ā¬(hā0) ćsin ćā”ć((2š +1)š)/2 sinā”ā ć/ā
=š in ((2š +1)š)/2Ć(ššš)ā¬(hā0) sinā”ā/ā
Using (ššš)ā¬(š„ā0) š ššā”š„/š„=1
= š in ((2š +1)š)/2Ć1
= sin ((šš +š)š )/š
Since,
LHD ā RHD
ā“ š(š„) is not differentiable at š=((šš +š)š )/š
Hence, š(š„) is continuous everywhere but not differentiable at š„ā((2š +1)š)/2, šāš
So, the correct answer is (C)

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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