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

(A) f is everywhere differentiable

(B) f is everywhere continuous but not differentiable at x = 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 18 Let f (x) = |sin x|. Then (A) f is everywhere differentiable (B) f is everywhere continuous but not differentiable at x = nšœ‹, nāˆˆ Z. (C) f is everywhere continuous but not differentiable at x = (2n + 1) šœ‹/2, nāˆˆ Z. (d) None of these f(š‘„) = |sin š‘„| We need to check continuity and differentiability of f(š‘„) Continuity of f(š’™) Let š‘”(š‘„)=|š‘„| & ā„Ž(š‘„)=sinā”š‘„ Then, š’ˆš’š’‰(š’™)=š‘”(ā„Ž(š‘„)) =š‘”(sinā”š‘„ ) =|sinā”š‘„ | =š’‡(š’™) āˆ“ š‘“(š‘„)=š‘”š‘œā„Ž(š‘„) We know that, š’‰(š’™)=š¬š¢š§ā”š’™ is continuous as sin is continuous š’ˆ(š’™)=|š’™| is continuous as it is a modulus function Hence, g(š‘„) & h(š‘„) both are continuous And If two functions g(š‘„) & h(š‘„) are continuous then their composition š‘”š‘œā„Ž(š‘„) is also continuous āˆ“ š’‡(š’™) is continuous Differentiability of š’‡(š’™) š‘“(š‘„)=|sinā”š‘„ | Since, it is a modulus function so we check differentiability when sinā”š‘„=0 i.e., š’™=š’š…, š’āˆˆš’ š‘“(š‘„) is differentiable at š‘„=š‘›šœ‹, if LHD = RHD (š’š’Šš’Ž)ā”¬(š”ā†’šŸŽ) (š’‡(š’™) āˆ’ š’‡(š’™ āˆ’ š’‰))/š’‰ = (š‘™š‘–š‘š)ā”¬(hā†’0) (š‘“(š‘›šœ‹) āˆ’ š‘“(š‘›šœ‹āˆ’ ā„Ž))/ā„Ž = (š‘™š‘–š‘š)ā”¬(hā†’0) (|sinā”š‘›šœ‹ | āˆ’|sinā”怖(š‘›šœ‹ āˆ’ ā„Ž)怗 |)/ā„Ž = (š‘™š‘–š‘š)ā”¬(hā†’0) (|0| āˆ’ |sinā”怖(š‘›šœ‹ āˆ’ ā„Ž)怗 |)/ā„Ž Using sin (A āˆ’ B) = sin Acos B āˆ’ cos Asin B = (š‘™š‘–š‘š)ā”¬(hā†’0) (0 āˆ’ |sinā”怖š‘›šœ‹ cosā”怖ā„Ž āˆ’ cosā”怖š‘›šœ‹ sinā”ā„Ž 怗 怗 怗 |)/ā„Ž = (š‘™š‘–š‘š)ā”¬(hā†’0) ( āˆ’ |0 āˆ’ cosā”怖š‘›šœ‹ sinā”ā„Ž 怗 |)/ā„Ž = (š‘™š‘–š‘š)ā”¬(hā†’0) ( āˆ’cosā”怖š‘›šœ‹ sinā”ā„Ž 怗)/ā„Ž = āˆ’cos š‘›šœ‹ Ɨ(š‘™š‘–š‘š)ā”¬(hā†’0) sinā”ā„Ž/ā„Ž Using (š‘™š‘–š‘š)ā”¬(š‘„ā†’0) š‘ š‘–š‘›ā”š‘„/š‘„=1 =āˆ’cos š‘›šœ‹ Ɨ1 =āˆ’šœšØš¬ š’š… (š’š’Šš’Ž)ā”¬(š”ā†’šŸŽ) (š’‡(š’™ + š’‰) āˆ’ š’‡(š’™ ))/š’‰ = (š‘™š‘–š‘š)ā”¬(hā†’0) (š‘“(š‘›šœ‹+ ā„Ž) āˆ’ š‘“(š‘›šœ‹))/ā„Ž = (š‘™š‘–š‘š)ā”¬(hā†’0) (|sinā”怖(š‘›šœ‹+ā„Ž)怗 |āˆ’|sinā”š‘›šœ‹ |)/ā„Ž = (š‘™š‘–š‘š)ā”¬(hā†’0) (|sinā”怖(š‘›šœ‹+ā„Ž)怗 |āˆ’|0|)/( ā„Ž) Using sin (A + B) = sin Acos B + cos Asin B = (š‘™š‘–š‘š)ā”¬(hā†’0) (|sinā”怖š‘›šœ‹ cosā”怖ā„Ž + cosā”怖š‘›šœ‹ sinā”ā„Ž 怗 怗 怗 | āˆ’ 0)/ā„Ž = (š‘™š‘–š‘š)ā”¬(hā†’0) |0 + cosā”怖š‘›šœ‹ sinā”ā„Ž 怗 |/ā„Ž = (š‘™š‘–š‘š)ā”¬(hā†’0) cosā”怖š‘›šœ‹ sinā”ā„Ž 怗/ā„Ž = cos š‘›šœ‹ Ɨ(š‘™š‘–š‘š)ā”¬(hā†’0) sinā”ā„Ž/ā„Ž Using (š‘™š‘–š‘š)ā”¬(š‘„ā†’0) š‘ š‘–š‘›ā”š‘„/š‘„=1 = cos š‘›šœ‹ Ɨ1 = šœšØš¬ š’š… Since, LHD ā‰  RHD āˆ“ š‘“(š‘„) is not differentiable at š’™=š’š… Hence, š‘“(š‘„) is continuous everywhere but not differentiable at š’™āˆˆš’š…, š‘›āˆˆš‘ 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.