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

Last updated at April 16, 2024 by

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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo