Check sibling questions

Let f (x) = |sin x|. Then

(A) f is everywhere differentiable

(B) f is everywhere continuous but not differentiable at x = nΟ€, n∈ Z .

(C) f is everywhere continuous but not differentiable at x = (2n + 1) Ο€/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 20 Let f (x) = |sin x|. Then (A) f is everywhere differentiable (B) f is everywhere continuous but not differentiable at x = 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‘₯) = |sin π‘₯| We need to check continuity and differentiability of f(π‘₯) Continuity of f(𝒙) Let 𝑔(π‘₯)=|π‘₯| & β„Ž(π‘₯)=sin⁑π‘₯ Then, π’ˆπ’π’‰(𝒙)=𝑔(β„Ž(π‘₯)) =𝑔(sin⁑π‘₯ ) =|sin⁑π‘₯ | =𝒇(𝒙) ∴ 𝑓(π‘₯)=π‘”π‘œβ„Ž(π‘₯) We know that, 𝒉(𝒙)=𝐬𝐒𝐧⁑𝒙 is continuous as sin 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‘₯)=|sin⁑π‘₯ | Since, it is a modulus function so we check differentiability when sin⁑π‘₯=0 i.e., 𝒙=𝒏𝝅, π’βˆˆπ’ 𝑓(π‘₯) is differentiable at π‘₯=π‘›πœ‹, 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‘›πœ‹) βˆ’ 𝑓(π‘›πœ‹βˆ’ β„Ž))/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (|sinβ‘π‘›πœ‹ | βˆ’|sin⁑〖(π‘›πœ‹ βˆ’ β„Ž)γ€— |)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (|0| βˆ’ |sin⁑〖(π‘›πœ‹ βˆ’ β„Ž)γ€— |)/β„Ž Using sin (A βˆ’ B) = sin Acos B βˆ’ cos Asin B = (π‘™π‘–π‘š)┬(hβ†’0) (0 βˆ’ |sinβ‘γ€–π‘›πœ‹ cosβ‘γ€–β„Ž βˆ’ cosβ‘γ€–π‘›πœ‹ sinβ‘β„Ž γ€— γ€— γ€— |)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) ( βˆ’ |0 βˆ’ cosβ‘γ€–π‘›πœ‹ sinβ‘β„Ž γ€— |)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) ( βˆ’cosβ‘γ€–π‘›πœ‹ sinβ‘β„Ž γ€—)/β„Ž = βˆ’cos π‘›πœ‹ Γ—(π‘™π‘–π‘š)┬(hβ†’0) sinβ‘β„Ž/β„Ž Using (π‘™π‘–π‘š)┬(π‘₯β†’0) 𝑠𝑖𝑛⁑π‘₯/π‘₯=1 =βˆ’cos π‘›πœ‹ Γ—1 =βˆ’πœπ¨π¬ 𝒏𝝅 (π’π’Šπ’Ž)┬(π‘β†’πŸŽ) (𝒇(𝒙 + 𝒉) βˆ’ 𝒇(𝒙 ))/𝒉 = (π‘™π‘–π‘š)┬(hβ†’0) (𝑓(π‘›πœ‹+ β„Ž) βˆ’ 𝑓(π‘›πœ‹))/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (|sin⁑〖(π‘›πœ‹+β„Ž)γ€— |βˆ’|sinβ‘π‘›πœ‹ |)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (|sin⁑〖(π‘›πœ‹+β„Ž)γ€— |βˆ’|0|)/( β„Ž) Using sin (A + B) = sin Acos B + cos Asin B = (π‘™π‘–π‘š)┬(hβ†’0) (|sinβ‘γ€–π‘›πœ‹ cosβ‘γ€–β„Ž + cosβ‘γ€–π‘›πœ‹ sinβ‘β„Ž γ€— γ€— γ€— | βˆ’ 0)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) |0 + cosβ‘γ€–π‘›πœ‹ sinβ‘β„Ž γ€— |/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) cosβ‘γ€–π‘›πœ‹ sinβ‘β„Ž γ€—/β„Ž = cos π‘›πœ‹ Γ—(π‘™π‘–π‘š)┬(hβ†’0) sinβ‘β„Ž/β„Ž Using (π‘™π‘–π‘š)┬(π‘₯β†’0) 𝑠𝑖𝑛⁑π‘₯/π‘₯=1 = cos π‘›πœ‹ Γ—1 = 𝐜𝐨𝐬 𝒏𝝅 Since, LHD β‰  RHD ∴ 𝑓(π‘₯) is not differentiable at 𝒙=𝒏𝝅 Hence, 𝑓(π‘₯) is continuous everywhere but not differentiable at π’™βˆˆπ’π…, π‘›βˆˆπ‘ 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.