Check sibling questions

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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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 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.