Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

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 You are here

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 May 29, 2023 by Teachoo

Ex 5.1, 9 Find all points of discontinuity of f, where f is defined by 𝑓(𝑥)={█(𝑥/|𝑥| , 𝑖𝑓 𝑥<0@&−1 , 𝑖𝑓 𝑥≥ 0)┤ 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 x < 0 Case 1 : When x = 0 f(x) is continuous at 𝑥 =0 if L.H.L = R.H.L = 𝑓(0) Since there are two different functions on the left & right of 0, we take LHL & RHL . if lim┬(x→0^− ) 𝑓(𝑥)=lim┬(x→0^+ ) " " 𝑓(𝑥)= 𝑓(0) And, f(0) = −1 LHL at x → 0 lim┬(x→0^− ) f(x) = lim┬(h→0) f(0 − h) = lim┬(h→0) f(−h) = lim┬(h→0) (−ℎ)/|−ℎ| = lim┬(h→0) (−ℎ)/ℎ = lim┬(h→0) −1 = −1 RHL at x → 0 lim┬(x→0^+ ) f(x) = lim┬(h→0) f(0 + h) = lim┬(h→0) f(h) = lim┬(h→0) −1 = −1 Hence, L.H.L = R.H.L = 𝑓(0) ∴ f is continuous at x = −3 Case 2 : When x < 0 For x < 0, f(x) = 𝑥/(|𝑥|) f(x) = 𝑥/((−𝑥)) f(x) = −1 Since this constant It is continuous ∴ f(x) is continuous for x < 0 (As x < 0, x is negative) Case 3 : When x > 0 For x > 0, f(x) = −1 Since this constant It is continuous ∴ f(x) is continuous for x > 0 ∴ f is continuous for all real numbers Thus, f is continuous for 𝒙∈ R