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Ex 5.1 ,5 Is f(x) = {x x <=1, 5 x > 1 continuous at x = 0, 1, 2

Ex 5.1 ,5 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.1 ,5 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

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Ex 5.1, 5 Is the function f defined by 𝑓(π‘₯)={β–ˆ(π‘₯, 𝑖𝑓 π‘₯≀[email protected]&5, 𝑖𝑓 π‘₯>1)─ continuous at π‘₯ = 0 ? At π‘₯ = 1 ? At π‘₯ = 2 ? Given 𝑓(π‘₯)={β–ˆ(π‘₯, 𝑖𝑓 π‘₯≀[email protected]&5, 𝑖𝑓 π‘₯>1)─ At x = 0 For x = 0, f(x) = x Since this a polynomial It is continuous ∴ f(x) is continuous for x = 0 At x = 1 f(x) is continuous at π‘₯ =1 if L.H.L = R.H.L = 𝑓(1) if lim┬(xβ†’1^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’1^+ ) " " 𝑓(π‘₯)= 𝑓(1) 𝑓(π‘₯)={β–ˆ(π‘₯, 𝑖𝑓 π‘₯≀[email protected]&5, 𝑖𝑓 π‘₯>1)─ Since there are two different functions on the left & right of 1, we take LHL & RHL . LHL at x β†’ 1 lim┬(xβ†’1^βˆ’ ) f(x) = lim┬(hβ†’0) f(1 βˆ’ h) = lim┬(hβ†’0) (1βˆ’β„Ž) = 1 βˆ’ 0 = 1 RHL at x β†’ 1 lim┬(xβ†’1^+ ) f(x) = lim┬(hβ†’0) f(1 + h) = lim┬(hβ†’0) 5 = 5 Since L.H.L β‰  R.H.L f(x) is not continuous at x = 1 At x = 2 For x = 2, f(x) = 5 Since this a constant function It is continuous ∴ f(x) is continuous for x = 2

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