Ex 5.1, 28 - Chapter 5 Class 12 Continuity and Differentiability (Important Question)

Last updated at Dec. 16, 2024 by

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

Go Ad-free

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

Davneet Singh's photo - Co-founder, Teachoo

Made by

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