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Last updated at Jan. 3, 2020 by Teachoo

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Ex 5.1, 16 Discuss the continuity of the function f, where f is defined by π(π₯)={β(β&2, ππ π₯β€β1@2&π₯, ππ β1<π₯β€1@2, ππ π₯>1 )β€ Since we need to find continuity at of the function We check continuity for different values of x When x < β1 When x = β1 When β1 < x < 1 When x = 1 When x > 1 Case 1 : When x < β1 For x < β1, f(x) = β2 Since this constant It is continuous β΄ f(x) is continuous for x < β1 Case 2 : When x = β1 f(x) is continuous at π₯ =β1 if L.H.L = R.H.L = π(β1) if limβ¬(xβγβ1γ^β ) π(π₯)=limβ¬(xβγβ1γ^+ ) " " π(π₯)= π(β1) Since there are two different functions on the left & right of β1, we take LHL & RHL . LHL at x β β1 limβ¬(xβγβ1γ^β ) f(x) = limβ¬(hβ0) f(β1 β h) = limβ¬(hβ0) 2(ββ) = 2 Γ 0 = 0 RHL at x β 0 limβ¬(xβγβ1γ^+ ) f(x) = limβ¬(hβ0) f(β1 + h) = limβ¬(hβ0) 0 = 0 & π(β1) = β2 Hence, L.H.L = R.H.L = π(β1) β΄ f is continuous at x=β1 Case 3 : When β1 < x < 1 For β1 < x < 1 f(x) = 2x Since this a polynomial It is continuous β΄ f(x) is continuous for β1 < x < 1 Case 4 : When x = 1 f(x) is continuous at π₯ =1 if L.H.L = R.H.L = π(1) if limβ¬(xβ1^β ) π(π₯)=limβ¬(xβ1^+ ) " " π(π₯)= π(1) Since there are two different functions on the left & right of 1, we take LHL & RHL . LHL at x β 1 limβ¬(xβ1^β ) f(x) = limβ¬(hβ0) f(1 β h) = limβ¬(hβ0) 0 = 0 RHL at x β 1 limβ¬(xβ1^+ ) f(x) = limβ¬(hβ0) f(1 + h) = limβ¬(hβ0) 4(1+β) = 4(1 + 0) = 4 & π(1) = 2(1) = 2 Hence, L.H.L = R.H.L = π(1) β΄ f is continuous at x=1 Case 5: When x > 1 For x > 1, f(x) = 2 Since this constant It is continuous β΄ f(x) is continuous for x > 1 Thus, f(x) is continuous for all real numbers, i.e. f is continuous for all x β R limβ¬(xβγβ1γ^β ) π(π₯) = limβ¬(xββ1) (β2) = β2 limβ¬(xβγβ1γ^+ ) π(π₯) = limβ¬(xββ1) 2π₯ Putting x = β 1 = 2(β1) = β 2 And π(β2) = β2 Thus L.H.L = R.H.L = π(β2) = β2 Hence π(π) is continuous at x = βπ Case 1 At x = β1 A function is continuous at x = β 1 if A function is continuous at x = 1 if if L.H.L = R.H.L = π(1) i.e. limβ¬(xβ1^β ) π(π₯) = limβ¬(xβ1^+ ) π(π₯) = π(1) & π(1) = 2(1) = 2 Thus L.H.L = R.H.L = π(β2) Hence π(π) is continuous at x = 1 limβ¬(xβ1^β ) π(π₯) = limβ¬(xβ1^β ) 2π₯ Putting x = 1 = 2(1) = 2 limβ¬(xβ1^+ ) π(π₯) = limβ¬(xβ1^+ ) 2 = 2 Case 3 For π₯\<β1 π(π₯) = β2 Thus, π(π₯) is a constant function . & Every constant function is continuous for all real number. Hence π(π) is continuous at π\<βπ Case 4 For π₯>1 π(π₯) = 2 Thus, π(π₯) is a constant function . & Every constant function is continuous for all real number. Hence π(π) is continuous at π>π Case 5 For β1β€π₯β€1 π(π₯) = 2π₯ So, f(x) is a polynomial & Every polynomial is continuous. β π(π) is continuous at βπ<πβ€π Thus, f(x) is continuous for all real numbers, i.e. f is continuous for all x β R

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.