Ex 5.1

Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

### Transcript

Ex 5.1, 28 Find the values of k so that the function f is continuous at the indicated point π(π₯)={β(ππ₯+1 , ππ π₯β€π@cosβ‘γπ₯, γ ππ π₯>π)β€ at x = π Given that function is continuous at π₯ =π π is continuous at π₯ =π If L.H.L = R.H.L = π(π) i.e. limβ¬(xβπ^β ) π(π₯)=limβ¬(xβπ^+ ) " " π(π₯)= π(π) LHL at x β Ο (πππ)β¬(π₯βπ^β ) f(x) = (πππ)β¬(ββ0) f(Ο β h) = limβ¬(hβ0) k (Ο β h) + 1 = k(Ο β 0) + 1 = kΟ + 1 RHL at x β Ο (πππ)β¬(π₯βπ^+ ) f(x) = (πππ)β¬(ββ0) f(Ο + h) = limβ¬(hβ0) cos (Ο + h) = cos (Ο + 0) = cos (Ο) = β1 Since L.H.L = R.H.L ππ+1=β1 ππ=β2 π= (βπ)/π

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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, Social Science, Physics, Chemistry, Computer Science at Teachoo.