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

(A)RΒ  Β  Β 

(B) R βˆ’ {1/2}Β 

(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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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‰₯[email protected]βˆ’(2π‘₯βˆ’1) sin⁑π‘₯, 2π‘₯βˆ’1<0)─ = {β–ˆ((2π‘₯βˆ’1) sin⁑π‘₯, π‘₯β‰₯1/[email protected]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 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.