Ex 5.1

Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

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Ex 5.1, 17 Find the relationship between a and b so that the function f defined by π(π₯)={β(ππ₯+1, ππ π₯β€[email protected]&ππ₯+3, ππ 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¬(xβ3^β ) π(π₯)=limβ¬(xβ3^+ ) " " 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¬(xβ3^β ) f(x) = limβ¬(hβ0) f(3 β h) = limβ¬(hβ0) π(3ββ)+1 = π(3β0)+1 = 3a + 1 RHL at x β 3 limβ¬(xβ3^+ ) f(x) = limβ¬(hβ0) f(3 + h) = limβ¬(hβ0) π(3+β)+3 = b(3 + 0) + 3 = b + 3 And π(3)=ππ₯+1 π(π)=ππ +π Now, limβ¬(xβ3^β ) π(π₯) = limβ¬(xβ3^+ ) π(π₯) = π(1) 3π + 1 = 3π + 3 = 3π + 1 Comparing values ππ + π = ππ + π 3πβ3b=3β1 3π β3π=2 3(πβπ)=2 πβπ=2/3 π=π+ π/π Thus , for any value of b, We can find value of a