Get live Maths 1-on-1 Classs - Class 6 to 12

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 You are here

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 March 30, 2023 by Teachoo

Ex 5.1, 15 Discuss the continuity of the function f, where f is defined by 𝑓(𝑥)={ █(2𝑥, 𝑖𝑓 𝑥<[email protected] 0, 𝑖𝑓 0≤𝑥≤[email protected] 4𝑥, 𝑖𝑓 𝑥>1 )┤ Since we need to find continuity at of the function We check continuity for different values of x When x < 0 When x = 0 When 0 < x < 1 When x = 1 When x > 1 Case 1 : When x < 0 For x < 0, f(x) = 2x Since this a polynomial It is continuous ∴ f(x) is continuous for x < 0 Case 2 : When x = 0 f(x) is continuous at 𝑥 =0 if L.H.L = R.H.L = 𝑓(0) if lim┬(x→0^− ) 𝑓(𝑥)=lim┬(x→0^+ ) " " 𝑓(𝑥)= 𝑓(0) Since there are two different functions on the left & right of 0, we take LHL & RHL . LHL at x → 0 lim┬(x→0^− ) f(x) = lim┬(h→0) f(0 − h) = lim┬(h→0) f(−h) = lim┬(h→0) 2(−ℎ) = 2 × 0 = 0 RHL at x → 0 lim┬(x→1^+ ) f(x) = lim┬(h→0) f(0 + h) = lim┬(h→0) f(h) = lim┬(h→0) 0 = 0 & 𝑓(0) = 0 Hence, L.H.L = R.H.L = 𝑓(0) ∴ f is continuous at x = 0 Case 3 : When 0 < x < 1 For 0 < x < 1 f(x) = 0 Since this constant It is continuous ∴ f(x) is continuous for 0 < 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) Since L.H.L ≠ R.H.L f(x) is not continuous at x = 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) 0 = 0 RHL at x → 1 lim┬(x→1^+ ) f(x) = lim┬(h→0) f(1 + h) = lim┬(h→0) 4(1+ℎ) = 4(1 + 0) = 4 Case 5: When x > 1 For x > 1, f(x) = 4x Since this a polynomial It is continuous ∴ f(x) is continuous for x > 1 Hence points of discontinuity are x = 1 Thus, f is continuous for all x ∈ R − {𝟏}