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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Transcript

Ex 5.1, 26 Find the values of k so that the function f is continuous at the indicated point 𝑓(𝑥)={█((𝑘 cos⁡𝑥)/(𝜋 − 2𝑥 ) , 𝑖𝑓 𝑥≠𝜋/2@& 3, 𝑖𝑓 𝑥=𝜋/2)┤ at 𝑥 = 𝜋/2 Given that function is continuous at 𝑥 =𝜋/2 𝑓 is continuous at =𝜋/2 if L.H.L = R.H.L = 𝑓(𝜋/2) i.e. lim┬(x→〖𝜋/2〗^− ) 𝑓(𝑥)=lim┬(x→〖𝜋/2〗^+ ) " " 𝑓(𝑥)= 𝑓(𝜋/2) LHL at x → 𝝅/𝟐 (𝑙𝑖𝑚)┬(𝑥→〖𝜋/2〗^− ) 𝑓(𝑥) = (𝑙𝑖𝑚)┬(ℎ→0) 𝑓(𝜋/2−ℎ) = lim┬(h→0) (𝑘 cos⁡(𝜋/2 − ℎ))/(𝜋 − 2(𝜋/2 − ℎ) ) = lim┬(h→0) (𝑘 sin⁡ℎ)/(𝜋 − 𝜋 + 2ℎ ) = lim┬(h→0) (𝑘 sin⁡ℎ)/(2ℎ ) = k/2 (𝒍𝒊𝒎)┬(𝐡→𝟎) 𝐬𝐢𝐧⁡𝒉/(𝒉 ) = 𝑘/2 × 1 = 𝒌/𝟐 RHL at x → 𝝅/𝟐 (𝑙𝑖𝑚)┬(𝑥→〖𝜋/2〗^+ ) 𝑓(𝑥) = (𝑙𝑖𝑚)┬(ℎ→0) 𝑓(𝜋/2+ℎ) = lim┬(h→0) (𝑘 cos⁡(𝜋/2 + ℎ))/(𝜋 − 2(𝜋/2 + ℎ) ) = lim┬(h→0) (𝑘 〖(−sin〗⁡ℎ))/(𝜋 − 𝜋 − 2ℎ ) = lim┬(h→0) (−𝑘 sin⁡ℎ)/(−2ℎ ) = k/2 (𝒍𝒊𝒎)┬(𝐡→𝟎) 𝒔𝒊𝒏⁡𝒉/(𝒉 ) = 𝑘/2 × 1 = 𝑘/2 And 𝑓(𝜋/2) = 3 Now, L.H.L = R.H.L = 𝑓(𝜋/2) 𝑘/2 = 𝑘/2 = 3 Hence, 𝑘/2 = 3 k = 3 × 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.