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

(a) ±  1  (b) −1      (c) ±  1/2  (d) 1/2

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

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  1. Class 12
  2. Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Transcript

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‰ 0@1/2, 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 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.