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The function f (x) = {8(sin⁑x/x " + cos x, if x " β‰ " 0" k", if x " =" 0" )─ is continuous at x = 0, then the value of k is

(A) 3Β  Β Β Β 

(B) 2

(C) 1Β  Β  Β 

(D) 1.5

Β 

This question is similar to Ex 5.1, 18 - Chapter 5 Class 12 - Continuity and Differentiability

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Transcript

Question 1 The function f (x) = {β– 8(sin⁑π‘₯/π‘₯ " + cos x, if x " β‰ " 0" @π‘˜ ", if x " =" 0" )─ is continuous at x = 0, then the value of k is (A) 3 (B) 2 (C) 1 (D) 1.5 At 𝒙 = 0 f(x) is continuous at π‘₯=0 if L.H.L = R.H.L = 𝑓(0) if if lim┬(xβ†’0^βˆ’ ) 𝑓(π‘₯) = lim┬(xβ†’0^+ ) 𝑓(π‘₯) = 𝑓(0) LHL at x β†’ 0 (π’π’Šπ’Ž)┬(π±β†’πŸŽ^βˆ’ ) f(x) = lim┬(hβ†’0) f(0 βˆ’ h) = (π’π’Šπ’Ž)┬(π‘β†’πŸŽ) f(βˆ’h) = lim┬(hβ†’0) sin⁑〖(βˆ’β„Ž)γ€—/((βˆ’β„Ž)) " + cos (βˆ’ h)" = (π’π’Šπ’Ž)┬(π‘β†’πŸŽ) γ€–βˆ’π’”π’Šπ’γ€—β‘π’‰/(βˆ’π’‰) " +" (π’π’Šπ’Ž)┬(π‘β†’πŸŽ) " cos h" = lim┬(hβ†’0) sinβ‘β„Ž/β„Ž " + " (π‘™π‘–π‘š)┬(β„Žβ†’0) " cos h" Using lim┬(xβ†’0) sin⁑π‘₯/π‘₯=1 = 1 + "cos 0" = 1 + 1 = 2 RHL at x β†’ 0 (π’π’Šπ’Ž)┬(π’™β†’πŸŽ^+ ) f(x) = lim┬(hβ†’0) f(0 + h) = (π’π’Šπ’Ž)┬(π‘β†’πŸŽ) f(h) = lim┬(hβ†’0) sinβ‘β„Ž/β„Ž " + cos h" = (π’π’Šπ’Ž)┬(π‘β†’πŸŽ) π’”π’Šπ’β‘π’‰/𝒉 " + " (π’π’Šπ’Ž)┬(π‘β†’πŸŽ) " cos h" Using lim┬(xβ†’0) sin⁑π‘₯/π‘₯=1 = 1 + "cos 0" = 1 + 1 = 2 At 𝒙=𝟎 𝑓(π‘₯)=π‘˜ 𝒇(𝟎)=π’Œ Since f(x) is continuous at x = 0. L.H.L = R.H.L = 𝑓(0) 𝑓(0)=2 π’Œ=𝟐 So, the correct answer is (B)

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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.