Last updated at Dec. 16, 2024 by Teachoo
Ex 5.1, 17 Find the relationship between a and b so that the function f defined by π(π₯)={β(ππ₯+1, ππ π₯β€3@&ππ₯+3, ππ π₯>3)β€ is continuous at x = 3.Given function is continuous at x = 3 f(x) is continuous at π₯ =3 if L.H.L = R.H.L = π(π) if limβ¬(xβ3^β ) π(π₯)=limβ¬(xβ3^+ ) " " π(π₯)= π(3) Since there are two different functions on the left & right of 3, we take LHL & RHL . LHL at x β 3 limβ¬(xβ3^β ) f(x) = limβ¬(hβ0) f(3 β h) = limβ¬(hβ0) π(3ββ)+1 = π(3β0)+1 = 3a + 1 RHL at x β 3 limβ¬(xβ3^+ ) f(x) = limβ¬(hβ0) f(3 + h) = limβ¬(hβ0) π(3+β)+3 = b(3 + 0) + 3 = b + 3 And π(3)=ππ₯+1 π(π)=ππ +π Now, limβ¬(xβ3^β ) π(π₯) = limβ¬(xβ3^+ ) π(π₯) = π(1) 3π + 1 = 3π + 3 = 3π + 1 Comparing values ππ + π = ππ + π 3πβ3b=3β1 3π β3π=2 3(πβπ)=2 πβπ=2/3 π=π+ π/π Thus , for any value of b, We can find value of a
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About the Author
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