Question 5 - NCERT Exemplar - MCQs - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Last updated at Nov. 17, 2021 by Teachoo

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Let
f
(x) = |cos x|. Then,

(A)
f
is everywhere differentiable.

(B)
f
is everywhere continuous but not differentiable at n = nΟ, n β
Z

(C)
f
is everywhere continuous but not differentiable at x = (2n + 1)Ο/2, nΒ β
Z

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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)

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