Ex 5.1, 28 - Find k so that f(x) = { kx + 1, cos x at x = pi

Ex 5.1, 28 - Chapter 5 Class 12 Continuity and Differentiability - Part 2

  1. Chapter 5 Class 12 Continuity and Differentiability (Term 1)
  2. Concept wise

Transcript

Ex 5.1, 28 Find the values of k so that the function f is continuous at the indicated point 𝑓(π‘₯)={β–ˆ(π‘˜π‘₯+1 , 𝑖𝑓 π‘₯β‰€πœ‹@cos⁑〖π‘₯, γ€— 𝑖𝑓 π‘₯>πœ‹)─ at x = πœ‹ Given that function is continuous at π‘₯ =πœ‹ 𝑓 is continuous at π‘₯ =πœ‹ If L.H.L = R.H.L = 𝑓(πœ‹) i.e. lim┬(xβ†’πœ‹^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’πœ‹^+ ) " " 𝑓(π‘₯)= 𝑓(πœ‹) LHL at x β†’ Ο€ (π‘™π‘–π‘š)┬(π‘₯β†’πœ‹^βˆ’ ) f(x) = (π‘™π‘–π‘š)┬(β„Žβ†’0) f(Ο€ βˆ’ h) = lim┬(hβ†’0) k (Ο€ – h) + 1 = k(Ο€ βˆ’ 0) + 1 = kΟ€ + 1 RHL at x β†’ Ο€ (π‘™π‘–π‘š)┬(π‘₯β†’πœ‹^+ ) f(x) = (π‘™π‘–π‘š)┬(β„Žβ†’0) f(Ο€ + h) = lim┬(hβ†’0) cos (Ο€ + h) = cos (Ο€ + 0) = cos (Ο€) = βˆ’1 Since L.H.L = R.H.L π‘˜πœ‹+1=βˆ’1 π‘˜πœ‹=βˆ’2 π’Œ= (βˆ’πŸ)/𝝅

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.