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

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 𝑓(𝑥)={█(−&2, 𝑖𝑓 𝑥≤−1@2&𝑥, 𝑖𝑓 −1<𝑥≤1@2, 𝑖𝑓 𝑥>1 )┤ Since we need to find continuity at of the function We check continuity for different values of x When x < −1 When x = −1 When −1 < x < 1 When x = 1 When x > 1 Case 1 : When x < −1 For x < −1, f(x) = −2 Since this constant It is continuous ∴ f(x) is continuous for x < −1 Case 2 : When x = −1 f(x) is continuous at 𝑥 =−1 if L.H.L = R.H.L = 𝑓(−1) if lim┬(x→〖−1〗^− ) 𝑓(𝑥)=lim┬(x→〖−1〗^+ ) " " 𝑓(𝑥)= 𝑓(−1) Since there are two different functions on the left & right of −1, we take LHL & RHL . LHL at x → −1 lim┬(x→〖−1〗^− ) f(x) = lim┬(h→0) f(−1 − h) = lim┬(h→0) −2 = −2 RHL at x → 0 lim┬(x→〖−1〗^+ ) f(x) = lim┬(h→0) f(−1 + h) = lim┬(h→0) 2(−1+ℎ) = 2(−1+0) = −2 & 𝑓(−1) = −2 Hence, L.H.L = R.H.L = 𝑓(−1) ∴ f is continuous at x = −1 Case 3 : When −1 < x < 1 For −1 < x < 1 f(x) = 2x Since this a polynomial It is continuous ∴ f(x) is continuous for −1 < x < 1 Case 4 : When x = 1 f(x) is continuous at 𝑥 =1 if L.H.L = R.H.L = 𝑓(1) if lim┬(x→1^− ) 𝑓(𝑥)=lim┬(x→1^+ ) " " 𝑓(𝑥)= 𝑓(1) & 𝑓(1) = 2(1) = 2 Since there are two different functions on the left & right of 1, we take LHL & RHL . LHL at x → 1 lim┬(x→1^− ) f(x) = lim┬(h→0) f(1 − h) = lim┬(h→0) 2(1−ℎ) = 2(1−0) = 2 RHL at x → 1 lim┬(x→1^+ ) f(x) = lim┬(h→0) f(1 + h) = lim┬(h→0) 2 = 2 Hence, L.H.L = R.H.L = 𝑓(1) ∴ f is continuous at x=1 Case 5: When x > 1 For x > 1, f(x) = 2 Since this constant It is continuous ∴ f(x) is continuous for x > 1 Thus, f(x) is continuous for all real numbers, i.e. f is continuous for all x ∈ R