Question 5
Let f (x) = |cos x|. Then,
(A) f is everywhere differentiable.
(B) f is everywhere continuous but not differentiable at n = nπ, n β Z
(C) f is everywhere continuous but not differentiable at x = (2n + 1)π/2, n β Z
(D) None of these
f(π₯) = |cos π₯|
We need to check continuity and differentiability of f(π₯)
Continuity of f(π)
Let π(π₯)=|π₯| & β(π₯)=cosβ‘π₯
Then,
πππ(π)=π(β(π₯))
=π(cosβ‘π₯ )
=|cosβ‘π₯ |
=π(π)
β΄ π(π₯)=ππβ(π₯)
We know that,
π(π)=πππβ‘π is continuous as cos is continuous
π(π)=|π| is continuous as it is a modulus function
Hence, g(π₯) & h(π₯) both are continuous
And
If two functions g(π₯) & h(π₯) are continuous then their composition ππβ(π₯) is also continuous
β΄ π(π) is continuous
Differentiability of π(π)
π(π₯)=|cosβ‘π₯ |
Since, it is a modulus function so we check differentiability when cosβ‘π₯=0
i.e., π=((ππ +π)π )/π, πβπ
π(π₯) is differentiable at π₯=((2π +1)π)/2, if
LHD = RHD
(πππ)β¬(π‘βπ) (π(π) β π(π β π))/π
= (πππ)β¬(hβ0) (π(((ππ +π)π )/π) β π(((ππ +π)π )/π β β))/β
= (πππ)β¬(hβ0) (|cosβ‘γ((ππ +π)π )/πγ | β|cosβ‘γ((ππ +π)π /π β β)γ |)/β
= (πππ)β¬(hβ0) (|0| β |cosβ‘γ((ππ +π)π /π β β)γ |)/β
Using cos (A β B) = cos Acos B + sin Asin B
= (πππ)β¬(hβ0) (0 β |cosβ‘γ((ππ +π)π )/π cosβ‘γβ+ sinβ‘γ((ππ +π)π )/π sinβ‘β γ γ γ |)/β
= (πππ)β¬(hβ0) ( β |0 β γsin γβ‘γ((ππ +π)π )/π sinβ‘β γ |)/β
= (πππ)β¬(hβ0) ( βsinβ‘γ((ππ +π)π )/π sinβ‘β γ)/β
= βπ ππ ((2π +1)π)/2Γ(πππ)β¬(hβ0) sinβ‘β/β
Using (πππ)β¬(π₯β0) π ππβ‘π₯/π₯=1
=βπ ππ ((2π +1)π)/2Γ1
=βπππ ((ππ +π)π )/π
(πππ)β¬(π‘βπ) (π(π + π) β π(π ))/π
= (πππ)β¬(hβ0) (π(((ππ +π)π )/π + β) β π(((ππ +π)π )/π))/β
= (πππ)β¬(hβ0) (|cosβ‘γ((2π + 1)π/2 + β )γ |β|cosβ‘γ((2π +1)π)/2γ |)/β
= (πππ)β¬(hβ0) (|πππ β‘(((2π + 1)π)/2 +β) |β|0|)/( β)
Using cos (A + B) = cos Acos B -- sin Asin B
= (πππ)β¬(hβ0) (|γcos γβ‘γ((2π +1)π)/2 cosβ‘γβ β sinβ‘γ((2π +1)π)/2 sinβ‘β γ γ γ | β 0)/β
= (πππ)β¬(hβ0) |0 β sinβ‘γ((2π + 1)π)/2 sinβ‘β γ |/β
= (πππ)β¬(hβ0) γsin γβ‘γ((2π +1)π)/2 sinβ‘β γ/β
=π in ((2π +1)π)/2Γ(πππ)β¬(hβ0) sinβ‘β/β
Using (πππ)β¬(π₯β0) π ππβ‘π₯/π₯=1
= π in ((2π +1)π)/2Γ1
= sin ((ππ +π)π )/π
Since,
LHD β RHD
β΄ π(π₯) is not differentiable at π=((ππ +π)π )/π
Hence, π(π₯) is continuous everywhere but not differentiable at π₯β((2π +1)π)/2, πβπ
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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