Ex 5.1

Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

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Ex 5.1, 27 Find the values of k so that the function f is continuous at the indicated point π(π₯)={β(ππ₯2 , ππ π₯β€[email protected], ππ π₯>2)β€ at x = 2Given that function is continuous at π₯ = 2 π is continuous at π₯ = 2 if L.H.L = R.H.L = π(2) i.e. limβ¬(xβ2^β ) π(π₯)=limβ¬(xβ2^+ ) " " π(π₯)= π(2) LHL at x β 2 limβ¬(xβ2^β ) f(x) = limβ¬(hβ0) f(2 β h) = limβ¬(hβ0) γπ(2ββ)γ^2 = γπ(2β0)γ^2 = γπ(2)γ^2 = ππ RHL at x β 2 limβ¬(xβ2^+ ) f(x) = limβ¬(hβ0) f(2 + h) = limβ¬(hβ0) 3 = 3 RHL at x β 2 limβ¬(xβ2^+ ) f(x) = limβ¬(hβ0) f(2 + h) = limβ¬(hβ0) 3 = 3 RHL at x β 2 limβ¬(xβ2^+ ) f(x) = limβ¬(hβ0) f(2 + h) = limβ¬(hβ0) 3 = 3 RHL at x β 2 limβ¬(xβ2^+ ) f(x) = limβ¬(hβ0) f(2 + h) = limβ¬(hβ0) 3 = 3 Since LHL = RHL 4k = 3 k = π/π