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

This question is similar to Ex 5.1, 32 - Chapter 5 Class 12 and Ex 5.2, 9 - Chapter 5 Class 12 - Continuity and Differentiability

Slide90.JPG

Slide91.JPG
Slide92.JPG
Slide93.JPG
Slide94.JPG

  1. Chapter 5 Class 12 Continuity and Differentiability (Term 1)
  2. Serial order wise

Transcript

Question 20 Let f (x) = |sin x|. Then (A) f is everywhere differentiable (B) f is everywhere continuous but not differentiable at x = nπœ‹, n∈ Z. (C) f is everywhere continuous but not differentiable at x = (2n + 1) πœ‹/2, n∈ Z. (d) None of these f(π‘₯) = |sin π‘₯| We need to check continuity and differentiability of f(π‘₯) Continuity of f(𝒙) Let 𝑔(π‘₯)=|π‘₯| & β„Ž(π‘₯)=sin⁑π‘₯ Then, π’ˆπ’π’‰(𝒙)=𝑔(β„Ž(π‘₯)) =𝑔(sin⁑π‘₯ ) =|sin⁑π‘₯ | =𝒇(𝒙) ∴ 𝑓(π‘₯)=π‘”π‘œβ„Ž(π‘₯) We know that, 𝒉(𝒙)=𝐬𝐒𝐧⁑𝒙 is continuous as sin is continuous π’ˆ(𝒙)=|𝒙| 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 π‘”π‘œβ„Ž(π‘₯) is also continuous ∴ 𝒇(𝒙) is continuous Differentiability of 𝒇(𝒙) 𝑓(π‘₯)=|sin⁑π‘₯ | Since, it is a modulus function so we check differentiability when sin⁑π‘₯=0 i.e., 𝒙=𝒏𝝅, π’βˆˆπ’ 𝑓(π‘₯) is differentiable at π‘₯=π‘›πœ‹, if LHD = RHD (π’π’Šπ’Ž)┬(π‘β†’πŸŽ) (𝒇(𝒙) βˆ’ 𝒇(𝒙 βˆ’ 𝒉))/𝒉 = (π‘™π‘–π‘š)┬(hβ†’0) (𝑓(π‘›πœ‹) βˆ’ 𝑓(π‘›πœ‹βˆ’ β„Ž))/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (|sinβ‘π‘›πœ‹ | βˆ’|sin⁑〖(π‘›πœ‹ βˆ’ β„Ž)γ€— |)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (|0| βˆ’ |sin⁑〖(π‘›πœ‹ βˆ’ β„Ž)γ€— |)/β„Ž Using sin (A βˆ’ B) = sin Acos B βˆ’ cos Asin B = (π‘™π‘–π‘š)┬(hβ†’0) (0 βˆ’ |sinβ‘γ€–π‘›πœ‹ cosβ‘γ€–β„Ž βˆ’ cosβ‘γ€–π‘›πœ‹ sinβ‘β„Ž γ€— γ€— γ€— |)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) ( βˆ’ |0 βˆ’ cosβ‘γ€–π‘›πœ‹ sinβ‘β„Ž γ€— |)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) ( βˆ’cosβ‘γ€–π‘›πœ‹ sinβ‘β„Ž γ€—)/β„Ž = βˆ’cos π‘›πœ‹ Γ—(π‘™π‘–π‘š)┬(hβ†’0) sinβ‘β„Ž/β„Ž Using (π‘™π‘–π‘š)┬(π‘₯β†’0) 𝑠𝑖𝑛⁑π‘₯/π‘₯=1 =βˆ’cos π‘›πœ‹ Γ—1 =βˆ’πœπ¨π¬ 𝒏𝝅 (π’π’Šπ’Ž)┬(π‘β†’πŸŽ) (𝒇(𝒙 + 𝒉) βˆ’ 𝒇(𝒙 ))/𝒉 = (π‘™π‘–π‘š)┬(hβ†’0) (𝑓(π‘›πœ‹+ β„Ž) βˆ’ 𝑓(π‘›πœ‹))/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (|sin⁑〖(π‘›πœ‹+β„Ž)γ€— |βˆ’|sinβ‘π‘›πœ‹ |)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (|sin⁑〖(π‘›πœ‹+β„Ž)γ€— |βˆ’|0|)/( β„Ž) Using sin (A + B) = sin Acos B + cos Asin B = (π‘™π‘–π‘š)┬(hβ†’0) (|sinβ‘γ€–π‘›πœ‹ cosβ‘γ€–β„Ž + cosβ‘γ€–π‘›πœ‹ sinβ‘β„Ž γ€— γ€— γ€— | βˆ’ 0)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) |0 + cosβ‘γ€–π‘›πœ‹ sinβ‘β„Ž γ€— |/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) cosβ‘γ€–π‘›πœ‹ sinβ‘β„Ž γ€—/β„Ž = cos π‘›πœ‹ Γ—(π‘™π‘–π‘š)┬(hβ†’0) sinβ‘β„Ž/β„Ž Using (π‘™π‘–π‘š)┬(π‘₯β†’0) 𝑠𝑖𝑛⁑π‘₯/π‘₯=1 = cos π‘›πœ‹ Γ—1 = 𝐜𝐨𝐬 𝒏𝝅 Since, LHD β‰  RHD ∴ 𝑓(π‘₯) is not differentiable at 𝒙=𝒏𝝅 Hence, 𝑓(π‘₯) is continuous everywhere but not differentiable at π’™βˆˆπ’π…, π‘›βˆˆπ‘ So, the correct answer is (B)

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.