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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise

Transcript

Ex 5.1, 26 Find the values of k so that the function f is continuous at the indicated point 𝑓(π‘₯)={β–ˆ((π‘˜ cos⁑π‘₯)/(πœ‹ βˆ’ 2π‘₯ ) , 𝑖𝑓 π‘₯β‰ πœ‹/2@& 3, 𝑖𝑓 π‘₯=πœ‹/2)─ at π‘₯ = πœ‹/2 Given that function is continuous at π‘₯ =πœ‹/2 𝑓 is continuous at =πœ‹/2 if L.H.L = R.H.L = 𝑓(πœ‹/2) i.e. lim┬(xβ†’γ€–πœ‹/2γ€—^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’γ€–πœ‹/2γ€—^+ ) " " 𝑓(π‘₯)= 𝑓(πœ‹/2) LHL at x β†’ 𝝅/𝟐 (π‘™π‘–π‘š)┬(π‘₯β†’γ€–πœ‹/2γ€—^βˆ’ ) f(x) = (π‘™π‘–π‘š)┬(β„Žβ†’0) f(πœ‹/2 βˆ’ h) = lim┬(hβ†’0) (π‘˜ cos⁑(πœ‹/2 βˆ’ β„Ž))/(πœ‹ βˆ’ 2(πœ‹/2 βˆ’ β„Ž) ) = lim┬(hβ†’0) (π‘˜ sinβ‘β„Ž)/(πœ‹ βˆ’ πœ‹ + 2β„Ž ) = lim┬(hβ†’0) (π‘˜ sinβ‘β„Ž)/(2β„Ž ) = k/2 lim┬(hβ†’0) sinβ‘β„Ž/(β„Ž ) = π‘˜/2 Γ— 1 = π‘˜/2 RHL at x β†’ 𝝅/𝟐 (π‘™π‘–π‘š)┬(π‘₯β†’γ€–πœ‹/2γ€—^+ ) f(x) = (π‘™π‘–π‘š)┬(β„Žβ†’0) f(πœ‹/2 + h) = lim┬(hβ†’0) (π‘˜ cos⁑(πœ‹/2 + β„Ž))/(πœ‹ βˆ’ 2(πœ‹/2 + β„Ž) ) = lim┬(hβ†’0) (π‘˜ γ€–(βˆ’sinγ€—β‘β„Ž))/(πœ‹ βˆ’ πœ‹ βˆ’ 2β„Ž ) = lim┬(hβ†’0) (βˆ’π‘˜ sinβ‘β„Ž)/(βˆ’2β„Ž ) = k/2 lim┬(hβ†’0) sinβ‘β„Ž/(β„Ž ) = π‘˜/2 Γ— 1 = π‘˜/2 And 𝑓(πœ‹/2) = 3 Now, L.H.L = R.H.L = 𝑓(πœ‹/2) Hence, π‘˜/2 = 3 k = 3 Γ— 2 k = 6 Hence, k = 6

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.