Ex 5.1
Ex 5.1 ,2
Ex 5.1, 3 (a)
Ex 5.1, 3 (b)
Ex 5.1, 3 (c) Important
Ex 5.1, 3 (d) Important
Ex 5.1 ,4
Ex 5.1 ,5 Important
Ex 5.1 ,6
Ex 5.1 ,7 Important
Ex 5.1 ,8
Ex 5.1, 9 Important
Ex 5.1, 10
Ex 5.1, 11
Ex 5.1, 12 Important
Ex 5.1, 13
Ex 5.1, 14
Ex 5.1, 15 Important
Ex 5.1, 16 You are here
Ex 5.1, 17 Important
Ex 5.1, 18 Important
Ex 5.1, 19 Important
Ex 5.1, 20
Ex 5.1, 21
Ex 5.1, 22 (i) Important
Ex 5.1, 22 (ii)
Ex 5.1, 22 (iii)
Ex 5.1, 22 (iv) Important
Ex 5.1, 23
Ex 5.1, 24 Important
Ex 5.1, 25
Ex 5.1, 26 Important
Ex 5.1, 27
Ex 5.1, 28 Important
Ex 5.1, 29
Ex 5.1, 30 Important
Ex 5.1, 31
Ex 5.1, 32
Ex 5.1, 33
Ex 5.1, 34 Important
Last updated at April 16, 2024 by Teachoo
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) β2 = β2 RHL at x β 0 limβ¬(xβγβ1γ^+ ) f(x) = limβ¬(hβ0) f(β1 + h) = limβ¬(hβ0) 2(β1+β) = 2(β1+0) = β2 & 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) & π(1) = 2(1) = 2 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) 2(1ββ) = 2(1β0) = 2 RHL at x β 1 limβ¬(xβ1^+ ) f(x) = limβ¬(hβ0) f(1 + h) = limβ¬(hβ0) 2 = 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