Let f (x) = |cos x|. Then,

(A) f is everywhere differentiable.

(B) f is everywhere continuous but not differentiable at n = 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

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

Transcript

Question 5 Let f (x) = |cos x|. Then, (A) f is everywhere differentiable. (B) f is everywhere continuous but not differentiable at n = 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‘₯) = |cos π‘₯| We need to check continuity and differentiability of f(π‘₯) Continuity of f(𝒙) Let 𝑔(π‘₯)=|π‘₯| & β„Ž(π‘₯)=cos⁑π‘₯ Then, π’ˆπ’π’‰(𝒙)=𝑔(β„Ž(π‘₯)) =𝑔(cos⁑π‘₯ ) =|cos⁑π‘₯ | =𝒇(𝒙) ∴ 𝑓(π‘₯)=π‘”π‘œβ„Ž(π‘₯) We know that, 𝒉(𝒙)=𝒄𝒐𝒔⁑𝒙 is continuous as cos 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‘₯)=|cos⁑π‘₯ | Since, it is a modulus function so we check differentiability when cos⁑π‘₯=0 i.e., 𝒙=((πŸπ’ +𝟏)𝝅)/𝟐, π’βˆˆπ’ 𝑓(π‘₯) is differentiable at π‘₯=((2𝑛 +1)πœ‹)/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(((πŸπ’ +𝟏)𝝅)/𝟐) βˆ’ 𝑓(((πŸπ’ +𝟏)𝝅)/𝟐 βˆ’ β„Ž))/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (|cos⁑〖((πŸπ’ +𝟏)𝝅)/πŸγ€— | βˆ’|cos⁑〖((πŸπ’ +𝟏)𝝅/𝟐 βˆ’ β„Ž)γ€— |)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (|0| βˆ’ |cos⁑〖((πŸπ’ +𝟏)𝝅/𝟐 βˆ’ β„Ž)γ€— |)/β„Ž Using cos (A βˆ’ B) = cos Acos B + sin Asin B = (π‘™π‘–π‘š)┬(hβ†’0) (0 βˆ’ |cos⁑〖((πŸπ’ +𝟏)𝝅)/𝟐 cosβ‘γ€–β„Ž+ sin⁑〖((πŸπ’ +𝟏)𝝅)/𝟐 sinβ‘β„Ž γ€— γ€— γ€— |)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) ( βˆ’ |0 βˆ’ γ€–sin 〗⁑〖((πŸπ’ +𝟏)𝝅)/𝟐 sinβ‘β„Ž γ€— |)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) ( βˆ’sin⁑〖((πŸπ’ +𝟏)𝝅)/𝟐 sinβ‘β„Ž γ€—)/β„Ž = βˆ’π‘ π‘–π‘› ((2𝑛 +1)πœ‹)/2Γ—(π‘™π‘–π‘š)┬(hβ†’0) sinβ‘β„Ž/β„Ž Using (π‘™π‘–π‘š)┬(π‘₯β†’0) 𝑠𝑖𝑛⁑π‘₯/π‘₯=1 =βˆ’π‘ π‘–π‘› ((2𝑛 +1)πœ‹)/2Γ—1 =βˆ’π’”π’Šπ’ ((πŸπ’ +𝟏)𝝅)/𝟐 (π’π’Šπ’Ž)┬(π‘β†’πŸŽ) (𝒇(𝒙 + 𝒉) βˆ’ 𝒇(𝒙 ))/𝒉 = (π‘™π‘–π‘š)┬(hβ†’0) (𝑓(((πŸπ’ +𝟏)𝝅)/𝟐 + β„Ž) βˆ’ 𝑓(((πŸπ’ +𝟏)𝝅)/𝟐))/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (|cos⁑〖((2𝑛 + 1)πœ‹/2 + β„Ž )γ€— |βˆ’|cos⁑〖((2𝑛 +1)πœ‹)/2γ€— |)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (|π‘π‘œπ‘ β‘(((2𝑛 + 1)πœ‹)/2 +β„Ž) |βˆ’|0|)/( β„Ž) Using cos (A + B) = cos Acos B -- sin Asin B = (π‘™π‘–π‘š)┬(hβ†’0) (|γ€–cos 〗⁑〖((2𝑛 +1)πœ‹)/2 cosβ‘γ€–β„Ž βˆ’ sin⁑〖((2𝑛 +1)πœ‹)/2 sinβ‘β„Ž γ€— γ€— γ€— | βˆ’ 0)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) |0 βˆ’ sin⁑〖((2𝑛 + 1)πœ‹)/2 sinβ‘β„Ž γ€— |/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) γ€–sin 〗⁑〖((2𝑛 +1)πœ‹)/2 sinβ‘β„Ž γ€—/β„Ž =𝑠in ((2𝑛 +1)πœ‹)/2Γ—(π‘™π‘–π‘š)┬(hβ†’0) sinβ‘β„Ž/β„Ž Using (π‘™π‘–π‘š)┬(π‘₯β†’0) 𝑠𝑖𝑛⁑π‘₯/π‘₯=1 = 𝑠in ((2𝑛 +1)πœ‹)/2Γ—1 = sin ((πŸπ’ +𝟏)𝝅)/𝟐 Since, LHD β‰  RHD ∴ 𝑓(π‘₯) is not differentiable at 𝒙=((πŸπ’ +𝟏)𝝅)/𝟐 Hence, 𝑓(π‘₯) is continuous everywhere but not differentiable at π‘₯∈((2𝑛 +1)πœ‹)/2, π‘›βˆˆπ‘ So, the correct answer is (C)

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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.