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Ex 5.1, 26 - Find values of k so that f(x) = k cos x / pi - 2x

Ex 5.1, 26 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.1, 26 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

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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(π‘₯) = (π‘™π‘–π‘š)┬(β„Žβ†’0) 𝑓(πœ‹/2βˆ’β„Ž) = 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 (π’π’Šπ’Ž)┬(π‘β†’πŸŽ) 𝐬𝐒𝐧⁑𝒉/(𝒉 ) = π‘˜/2 Γ— 1 = π’Œ/𝟐 RHL at x β†’ 𝝅/𝟐 (π‘™π‘–π‘š)┬(π‘₯β†’γ€–πœ‹/2γ€—^+ ) 𝑓(π‘₯) = (π‘™π‘–π‘š)┬(β„Žβ†’0) 𝑓(πœ‹/2+β„Ž) = 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 (π’π’Šπ’Ž)┬(π‘β†’πŸŽ) π’”π’Šπ’β‘π’‰/(𝒉 ) = π‘˜/2 Γ— 1 = π‘˜/2 And 𝑓(πœ‹/2) = 3 Now, L.H.L = R.H.L = 𝑓(πœ‹/2) π‘˜/2 = π‘˜/2 = 3 Hence, π‘˜/2 = 3 k = 3 Γ— 2 k = 6 Hence, k = 6

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.