The set of points where the function f given by f (x) = |2x−1| sin x is differentiable is

(A)R     

(B) R − {1/2} 

(C) (0, ∞)    

(D) none of these

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

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

Transcript

Question 15 The set of points where the function f given by f (x) = |2xβˆ’1| sin x is differentiable is R (B) R βˆ’ {1/2} (C) (0, ∞) (D) none of these f(x) = |2π‘₯βˆ’1| sin⁑π‘₯ = {β–ˆ((2π‘₯βˆ’1) sin⁑π‘₯, 2π‘₯βˆ’1β‰₯0@βˆ’(2π‘₯βˆ’1) sin⁑π‘₯, 2π‘₯βˆ’1<0)─ = {β–ˆ((2π‘₯βˆ’1) sin⁑π‘₯, π‘₯β‰₯1/2@βˆ’(2π‘₯βˆ’1) sin⁑〖π‘₯ ,γ€— π‘₯<1/2)─ Now, f(x) is a differentiable at x = 1/2 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(1/2) βˆ’ 𝑓(1/2 βˆ’ β„Ž))/β„Ž =(π‘™π‘–π‘š)┬(hβ†’0) (|2(1/2)βˆ’1| sin⁑(1/2)βˆ’|2(1/2 βˆ’ β„Ž)βˆ’ 1| sin⁑(1/2 βˆ’β„Ž))/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (0 βˆ’|1 βˆ’ β„Ž βˆ’ 1| γ€–sin 〗⁑(1/2 βˆ’ β„Ž))/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (0 βˆ’|βˆ’ β„Ž| sin⁑(1/2 βˆ’ β„Ž))/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (βˆ’β„Ž γ€–sin 〗⁑(1/2 βˆ’ β„Ž))/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) βˆ’sin⁑(1/2 βˆ’ β„Ž) = βˆ’sin⁑(1/2βˆ’0) = βˆ’π‘ π‘–π‘› 1/2 = βˆ’ 𝝅/πŸ” (π’π’Šπ’Ž)┬(π‘β†’πŸŽ) (𝒇(𝒙+𝒉) βˆ’ 𝒇(𝒙 ))/𝒉 = (π‘™π‘–π‘š)┬(hβ†’0) (𝑓(1/2+β„Ž) βˆ’ 𝑓(1/2))/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (|(2(1/2+β„Ž)βˆ’1| 𝑠𝑖𝑛⁑(1/2+β„Ž)βˆ’|2(1/2)βˆ’1| 𝑠𝑖𝑛⁑(1/2))/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (|1+ β„Ž βˆ’1| γ€–sin 〗⁑〖(1/2 + β„Ž) γ€—βˆ’ 0 )/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (|β„Ž| γ€–sin 〗⁑(1/2+β„Ž))/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (β„Ž γ€–sin 〗⁑(1/2+β„Ž))/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) sin⁑(1/2+β„Ž) = γ€–sin 〗⁑(1/2+0) = sin 1/2 = 𝝅/πŸ” Since LHD β‰  RHD ∴ f(x) is not differentiable at x = 1/2 Hence, we can say that f(x) is differentiable on R βˆ’ {𝟏/𝟐} So, the correct answer is (B)

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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.