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The value of ‘k’ for which the function f(x)={(1 -cos⁡4x/8x 2 ,if x≠0 k,if x=0 is continuous at x = 0 is

(a) 0              (b) -1                 (c) 1                   (d) 2

 

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Question 4 The value of β€˜k’ for which the function 𝑓(π‘₯)={β–ˆ((1 βˆ’cos⁑4π‘₯)/γ€–8π‘₯γ€—^2 , 𝑖𝑓 π‘₯β‰ 0@&π‘˜, 𝑖𝑓 π‘₯=0)─ is continuous at x = 0 is (a) 0 (b) -1 (c) 1 (d) 2At 𝒙=𝟎 f(x) is continuous at x = 0 if (π₯𝐒𝐦)┬(π±β†’πŸŽ) 𝒇(𝒙) = 𝒇(𝟎) L.H.S (π₯𝐒𝐦)┬(π±β†’πŸŽ) 𝒇(𝒙) "= " lim┬(xβ†’0) (1 βˆ’ π’„π’π’”β‘πŸ’π’™)/γ€–8π‘₯γ€—^2 Using cos 2ΞΈ = 1 βˆ’ 2sin2 ΞΈ Putting ΞΈ = 2x "= " lim┬(xβ†’0) (1 βˆ’ (πŸβˆ’ 𝟐 〖𝐬𝐒𝐧〗^πŸβ‘πŸπ’™ ))/γ€–8π‘₯γ€—^2 "= " lim┬(xβ†’0) (1 βˆ’ 1 + 2 sin^2⁑2π‘₯)/γ€–8π‘₯γ€—^2 "= " lim┬(xβ†’0) ( 2 sin^2⁑2π‘₯)/γ€–8π‘₯γ€—^2 "= " lim┬(xβ†’0) ( sin^2⁑2π‘₯)/γ€–4π‘₯γ€—^2 "= " (π₯𝐒𝐦)┬(π±β†’πŸŽ) (π’”π’Šπ’β‘πŸπ’™/πŸπ’™)^𝟐 = 1 R.H.S 𝒇(𝟎) = k Since f(x) is continuous at x = 0. 1 = k k = 1 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.