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

(A) 3     

(B) 2

(C) 1     

(D) 1.5

 

This question is similar to Ex 5.1, 18 - 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 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 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.