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The set of points where the functions f given by f (x) = |x – 3| cos x is differentiable is

(A) R Β 

(B) R βˆ’ {3}

(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 8 The set of points where the functions f given by f (x) = |x – 3| cos x is differentiable is (A) R (B) R βˆ’ {3} (C) (0, ∞) (D) None of these f(x) = |π‘₯βˆ’3| cos⁑π‘₯ = {β–ˆ((π‘₯βˆ’3) cos⁑π‘₯, π‘₯βˆ’3β‰₯[email protected]βˆ’(π‘₯βˆ’3) cos⁑π‘₯, π‘₯βˆ’3<0)─ = {β–ˆ((π‘₯βˆ’3) cos⁑π‘₯, π‘₯β‰₯[email protected]βˆ’(π‘₯βˆ’3) cos⁑π‘₯, π‘₯<3)─ Now, f(x) is a differentiable at x = 3 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(3) βˆ’ 𝑓(3 βˆ’ β„Ž))/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (|3 βˆ’ 3| cos⁑3βˆ’|(3 βˆ’ β„Ž)βˆ’3| cos⁑〖(3 βˆ’ β„Ž)γ€—)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (0 βˆ’|3 βˆ’ β„Ž βˆ’3| cos⁑〖(3 βˆ’ β„Ž)γ€—)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (0 βˆ’|βˆ’β„Ž| cos⁑〖(3 βˆ’ β„Ž)γ€—)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (βˆ’β„Ž cos⁑〖(3 βˆ’ β„Ž)γ€—)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) βˆ’cos⁑〖(3 βˆ’β„Ž)γ€— = βˆ’cos⁑〖(3 βˆ’0)γ€— = βˆ’π’„π’π’”β‘πŸ‘ (π’π’Šπ’Ž)┬(π‘β†’πŸŽ) (𝒇(𝒙+𝒉) βˆ’ 𝒇(𝒙 ))/𝒉 = (π‘™π‘–π‘š)┬(hβ†’0) (𝑓(3+β„Ž) βˆ’ 𝑓(3))/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (|(3+β„Ž) βˆ’ 3| cos⁑〖(3+β„Ž)γ€—βˆ’|3 βˆ’ 3| cos⁑〖(3)γ€—)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (|3 + β„Ž βˆ’3| cos⁑(3 + β„Ž)βˆ’0 )/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (| β„Ž| cos⁑〖(3+β„Ž)γ€—)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (β„Ž cos⁑〖(3 + β„Ž)γ€—)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) cos⁑〖(3+β„Ž)γ€— = cos⁑〖(3+0)γ€— = π’„π’π’”β‘πŸ‘ Since LHD β‰  RHD ∴ f(x) is not differentiable at x = 3 Hence, we can say that f(x) is differentiable on R βˆ’ {πŸ‘} 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.