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The value of k (k < 0) for which the function f defined as
f (x)={((1 -cos⁑kx)/(x sin⁑x ),xβ‰ 0Β  1/2,x=0)─ is continuous at x = 0 is :

(a) Β±Β  1Β  (b) βˆ’1Β  Β  Β  (c) Β± Β 1/2Β  (d) 1/2

This question is inspired from - Β  Question 21 - CBSE Class 12 Sample Paper for 2021 Boards


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Question 2 The value of k (k < 0) for which the function f defined as 𝑓 (π‘₯)={β–ˆ((1 βˆ’cosβ‘π‘˜π‘₯)/(π‘₯ sin⁑π‘₯ ),π‘₯β‰ [email protected]/2, π‘₯=0)─ is continuous at x = 0 is : (a) Β± 1 (b) βˆ’1 (c) Β± 1/2 (d) 1/2 Given that function is continuous at x = 0 𝑓(π‘₯) is continuous at x = 0 i.e. (π₯𝐒𝐦)┬(π±β†’πŸŽ) 𝒇(𝒙)=𝒇(𝟎) Limit at x β†’ 0 (π‘™π‘–π‘š)┬(π‘₯β†’0) f(x) = (π‘™π‘–π‘š)┬(β„Žβ†’0) f(h) = lim┬(hβ†’0) (1 βˆ’ cosβ‘π‘˜β„Ž)/(β„Ž (sinβ‘β„Ž) ) = lim┬(hβ†’0) (𝟐 〖𝐬𝐒𝐧〗^πŸβ‘γ€–π’Œπ’‰/πŸγ€—)/(β„Ž (sinβ‘β„Ž)) = lim┬(hβ†’0) (2 sin^2β‘γ€–π‘˜β„Ž/2γ€—)/1 Γ—1/(β„Ž (sinβ‘β„Ž)) = (π’π’Šπ’Ž)┬(π‘β†’πŸŽ) (𝟐 γ€–π’”π’Šπ’γ€—^πŸβ‘γ€–π’Œπ’‰/πŸγ€—)/(π’Œπ’‰/𝟐)^𝟐 Γ— (π’Œπ’‰/𝟐)^𝟐/(𝒉 (π’”π’Šπ’β‘π’‰)) = lim┬(hβ†’0) (2 sin^2β‘γ€–π‘˜β„Ž/2γ€—)/(π‘˜β„Ž/2)^2 Γ— (π‘˜^2 β„Ž^2)/(4β„Ž (sinβ‘β„Ž)) = lim┬(hβ†’0) sin^2β‘γ€–π‘˜β„Ž/2γ€—/(π‘˜β„Ž/2)^2 Γ— (π’Œ^𝟐 𝒉)/(𝟐(π’”π’Šπ’β‘π’‰)) = π’Œ^𝟐/𝟐 lim┬(hβ†’0) sin^2β‘γ€–π‘˜β„Ž/2γ€—/(π‘˜β„Ž/2)^2 Γ— β„Ž/sinβ‘β„Ž = π‘˜^2/2 lim┬(hβ†’0) sin^2β‘γ€–π‘˜β„Ž/2γ€—/(π‘˜β„Ž/2)^2 Γ—lim┬(hβ†’0) β„Ž/sinβ‘β„Ž = π‘˜^2/2 Γ— 1 Γ— 1 = π’Œ^𝟐/𝟐 Now, (π₯𝐒𝐦)┬(π±β†’πŸŽ) 𝒇(𝒙)=𝒇(𝟎) π‘˜^2/2 = 1/2 π‘˜^2 =1 π’Œ =±𝟏 But, given that k < 0 Thus, only value is k = βˆ’1 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.