Ex 5.1

Ex 5.1 ,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 You are here

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

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, 6 Find all points of discontinuity of f, where f is defined by π(π₯)={β(2π₯+3, ππ π₯β€2@&2π₯β3, ππ π₯>2)β€ π(π₯)={β(2π₯+3, ππ π₯β€2@&2π₯β3, ππ π₯>2)β€ Since we need to find continuity at of the function We check continuity for different values of x When x = 2 When x < 2 When x > 2 Case 1 : When x = 2 f(x) is continuous at π₯ =2 if L.H.L = R.H.L = π(2) if limβ¬(xβ2^β ) π(π₯)=limβ¬(xβ2^+ ) " " π(π₯)= π(2) Since there are two different functions on the left & right of 2, we take LHL & RHL . LHL at x β 2 limβ¬(xβ2^β ) f(x) = limβ¬(hβ0) f(2 β h) = limβ¬(hβ0) (2(2ββ)+3) = limβ¬(hβ0) (4β2β+3) = limβ¬(hβ0) (7β2β) = 7 β 0 = 7 RHL at x β 2 limβ¬(xβ2^+ ) f(x) = limβ¬(hβ0) f(2 + h) = limβ¬(hβ0) (2(2+β)β3) = limβ¬(hβ0) (4β2ββ3) = limβ¬(hβ0) (1β2β) = 1 β 0 = 1 Since L.H.L β R.H.L f(x) is not continuous at x = 2 Case 2 : When x < 2 For x < 2, f(x) = 2x + 3 Since this a polynomial It is continuous β΄ f(x) is continuous for x < 2 Case 3 : When x > 2 For x > 2, f(x) = 2x β 3 Since this a polynomial It is continuous β΄ f(x) is continuous for x > 2 Hence, only x = 2 is point is discontinuity. f is continuous at all real numbers except 2. Thus, f is continuous for π± β R β {2}