Question 20
Let f (x) = |sin x|. Then
(A) f is everywhere differentiable
(B) f is everywhere continuous but not differentiable at x = n𝜋, n∈ Z.
(C) f is everywhere continuous but not differentiable at x = (2n + 1) 𝜋/2, n∈ Z.
(d) None of these
f(𝑥) = |sin 𝑥|
We need to check continuity and differentiability of f(𝑥)
Continuity of f(𝒙)
Let 𝑔(𝑥)=|𝑥| & ℎ(𝑥)=sin𝑥
Then,
𝒈𝒐𝒉(𝒙)=𝑔(ℎ(𝑥))
=𝑔(sin𝑥 )
=|sin𝑥 |
=𝒇(𝒙)
∴ 𝑓(𝑥)=𝑔𝑜ℎ(𝑥)
We know that,
𝒉(𝒙)=𝐬𝐢𝐧𝒙 is continuous as sin is continuous
𝒈(𝒙)=|𝒙| is continuous as it is a modulus function
Hence, g(𝑥) & h(𝑥) both are continuous
And
If two functions g(𝑥) & h(𝑥) are continuous then their composition 𝑔𝑜ℎ(𝑥) is also continuous
∴ 𝒇(𝒙) is continuous
Differentiability of 𝒇(𝒙)
𝑓(𝑥)=|sin𝑥 |
Since, it is a modulus function so we check differentiability when sin𝑥=0
i.e., 𝒙=𝒏𝝅, 𝒏∈𝒁
𝑓(𝑥) is differentiable at 𝑥=𝑛𝜋, if
LHD = RHD
(𝒍𝒊𝒎)┬(𝐡→𝟎) (𝒇(𝒙) − 𝒇(𝒙 − 𝒉))/𝒉
= (𝑙𝑖𝑚)┬(h→0) (𝑓(𝑛𝜋) − 𝑓(𝑛𝜋− ℎ))/ℎ
= (𝑙𝑖𝑚)┬(h→0) (|sin𝑛𝜋 | −|sin〖(𝑛𝜋 − ℎ)〗 |)/ℎ
= (𝑙𝑖𝑚)┬(h→0) (|0| − |sin〖(𝑛𝜋 − ℎ)〗 |)/ℎ
Using sin (A − B) = sin Acos B − cos Asin B
= (𝑙𝑖𝑚)┬(h→0) (0 − |sin〖𝑛𝜋 cos〖ℎ − cos〖𝑛𝜋 sinℎ 〗 〗 〗 |)/ℎ
= (𝑙𝑖𝑚)┬(h→0) ( − |0 − cos〖𝑛𝜋 sinℎ 〗 |)/ℎ
= (𝑙𝑖𝑚)┬(h→0) ( −cos〖𝑛𝜋 sinℎ 〗)/ℎ
= −cos 𝑛𝜋 ×(𝑙𝑖𝑚)┬(h→0) sinℎ/ℎ
Using (𝑙𝑖𝑚)┬(𝑥→0) 𝑠𝑖𝑛𝑥/𝑥=1
=−cos 𝑛𝜋 ×1
=−𝐜𝐨𝐬 𝒏𝝅
(𝒍𝒊𝒎)┬(𝐡→𝟎) (𝒇(𝒙 + 𝒉) − 𝒇(𝒙 ))/𝒉
= (𝑙𝑖𝑚)┬(h→0) (𝑓(𝑛𝜋+ ℎ) − 𝑓(𝑛𝜋))/ℎ
= (𝑙𝑖𝑚)┬(h→0) (|sin〖(𝑛𝜋+ℎ)〗 |−|sin𝑛𝜋 |)/ℎ
= (𝑙𝑖𝑚)┬(h→0) (|sin〖(𝑛𝜋+ℎ)〗 |−|0|)/( ℎ)
Using sin (A + B) = sin Acos B + cos Asin B
= (𝑙𝑖𝑚)┬(h→0) (|sin〖𝑛𝜋 cos〖ℎ + cos〖𝑛𝜋 sinℎ 〗 〗 〗 | − 0)/ℎ
= (𝑙𝑖𝑚)┬(h→0) |0 + cos〖𝑛𝜋 sinℎ 〗 |/ℎ
= (𝑙𝑖𝑚)┬(h→0) cos〖𝑛𝜋 sinℎ 〗/ℎ
= cos 𝑛𝜋 ×(𝑙𝑖𝑚)┬(h→0) sinℎ/ℎ
Using (𝑙𝑖𝑚)┬(𝑥→0) 𝑠𝑖𝑛𝑥/𝑥=1
= cos 𝑛𝜋 ×1
= 𝐜𝐨𝐬 𝒏𝝅
Since,
LHD ≠ RHD
∴ 𝑓(𝑥) is not differentiable at 𝒙=𝒏𝝅
Hence, 𝑓(𝑥) is continuous everywhere but not differentiable at 𝒙∈𝒏𝝅, 𝑛∈𝑍
So, the correct answer is (B)

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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