Question 20 - NCERT Exemplar - MCQs - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Last updated at Nov. 18, 2021 by Teachoo

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Let f (x) = |sin x|. Then

(A) f is everywhere differentiable

(B) f is everywhere continuous but not differentiable at x = nΟ, nβ
Z
.

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

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Question 20
Let f (x) = |sin x|. Then
(A) f is everywhere differentiable
(B) f is everywhere continuous but not differentiable at x = nπ, nβ Z.
(C) f is everywhere continuous but not differentiable at x = (2n + 1) π/2, nβ Z.
(d) None of these
f(π₯) = |sin π₯|
We need to check continuity and differentiability of f(π₯)
Continuity of f(π)
Let π(π₯)=|π₯| & β(π₯)=sinβ‘π₯
Then,
πππ(π)=π(β(π₯))
=π(sinβ‘π₯ )
=|sinβ‘π₯ |
=π(π)
β΄ π(π₯)=ππβ(π₯)
We know that,
π(π)=π¬π’π§β‘π is continuous as sin 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₯)=|sinβ‘π₯ |
Since, it is a modulus function so we check differentiability when sinβ‘π₯=0
i.e., π=ππ , πβπ
π(π₯) is differentiable at π₯=ππ, if
LHD = RHD
(πππ)β¬(π‘βπ) (π(π) β π(π β π))/π
= (πππ)β¬(hβ0) (π(ππ) β π(ππβ β))/β
= (πππ)β¬(hβ0) (|sinβ‘ππ | β|sinβ‘γ(ππ β β)γ |)/β
= (πππ)β¬(hβ0) (|0| β |sinβ‘γ(ππ β β)γ |)/β
Using sin (A β B) = sin Acos B β cos Asin B
= (πππ)β¬(hβ0) (0 β |sinβ‘γππ cosβ‘γβ β cosβ‘γππ sinβ‘β γ γ γ |)/β
= (πππ)β¬(hβ0) ( β |0 β cosβ‘γππ sinβ‘β γ |)/β
= (πππ)β¬(hβ0) ( βcosβ‘γππ sinβ‘β γ)/β
= βcos ππ Γ(πππ)β¬(hβ0) sinβ‘β/β
Using (πππ)β¬(π₯β0) π ππβ‘π₯/π₯=1
=βcos ππ Γ1
=βππ¨π¬ ππ
(πππ)β¬(π‘βπ) (π(π + π) β π(π ))/π
= (πππ)β¬(hβ0) (π(ππ+ β) β π(ππ))/β
= (πππ)β¬(hβ0) (|sinβ‘γ(ππ+β)γ |β|sinβ‘ππ |)/β
= (πππ)β¬(hβ0) (|sinβ‘γ(ππ+β)γ |β|0|)/( β)
Using sin (A + B) = sin Acos B + cos Asin B
= (πππ)β¬(hβ0) (|sinβ‘γππ cosβ‘γβ + cosβ‘γππ sinβ‘β γ γ γ | β 0)/β
= (πππ)β¬(hβ0) |0 + cosβ‘γππ sinβ‘β γ |/β
= (πππ)β¬(hβ0) cosβ‘γππ sinβ‘β γ/β
= cos ππ Γ(πππ)β¬(hβ0) sinβ‘β/β
Using (πππ)β¬(π₯β0) π ππβ‘π₯/π₯=1
= cos ππ Γ1
= ππ¨π¬ ππ
Since,
LHD β RHD
β΄ π(π₯) is not differentiable at π=ππ
Hence, π(π₯) is continuous everywhere but not differentiable at πβππ , πβπ
So, the correct answer is (B)

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