1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise
  3. Ex 5.1
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Ex 5.1, 26 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at Dec. 16, 2024 by Teachoo

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

    3. Ex 5.1

    Ex 5.2 →
Ex 5.2 →
Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo