Checking continuity using LHL and RHL
Example 10
Example 13 Important
Ex 5.1, 10
Ex 5.1, 11
Ex 5.1 ,6
Ex 5.1, 13
Ex 5.1, 12 Important
Example 11 Important
Example 7
Ex 5.1, 3 (a)
Ex 5.1, 14
Ex 5.1, 16
Ex 5.1, 15 Important
Ex 5.1 ,7 Important
Ex 5.1, 25
Ex 5.1, 23
Ex 5.1, 24 Important
Ex 5.1 ,8
Ex 5.1, 9 Important
Ex 5.1, 29
Ex 5.1, 27
Ex 5.1, 28 Important
Ex 5.1, 17 Important
Ex 5.1, 18 Important
Ex 5.1, 26 Important
Ex 5.1, 30 Important You are here
Example 15 Important
Checking continuity using LHL and RHL
Last updated at April 16, 2024 by Teachoo
Ex 5.1, 30 Find the values of a and b such that the function defined by 𝑓(𝑥)={█(5, 𝑖𝑓 𝑥≤2@𝑎𝑥+𝑏, 𝑖𝑓 2<𝑥<10@21, 𝑖𝑓 𝑥≥10)┤ is a continuous function Since f(x) is a continuous function, It will be continuous for all values of x At x = 2 A function is continuous at x = 2 if L.H.L = R.H.L = 𝑓(2) i.e. lim┬(x→2^− ) 𝑓(𝑥)=lim┬(x→2^+ ) " " 𝑓(𝑥)= 𝑓(2) LHL at x → 2 (𝑙𝑖𝑚)┬(𝑥→2^− ) f(x) = (𝑙𝑖𝑚)┬(ℎ→0) f(2 − h) = lim┬(h→0) 5 = 5 RHL at x → 2 (𝑙𝑖𝑚)┬(𝑥→2^+ ) f(x) = (𝑙𝑖𝑚)┬(ℎ→0) f(2 + h) = lim┬(h→0) a(2 + h) + b = a(2 + 0) + b = 2a + b Since, LHL = RHL 2a + b = 5 At x = 10 𝑓 is continuous at x = 10 if L.H.L = R.H.L = 𝑓(10) i.e. lim┬(x→10^− ) 𝑓(𝑥)=lim┬(x→10^+ ) " " 𝑓(𝑥)= 𝑓(10) LHL at x → 10 (𝑙𝑖𝑚)┬(𝑥→10^− ) f(x) = (𝑙𝑖𝑚)┬(ℎ→0) f(10 − h) = lim┬(h→0) a(10 − h) + b = a(10 − 0) + b = 10a + b RHL at x → 10 (𝑙𝑖𝑚)┬(𝑥→10^+ ) f(x) = (𝑙𝑖𝑚)┬(ℎ→0) f(10 + h) = lim┬(h→0) 21 = 21 Since, L.H.L = R.H.L 10a + b = 21 Now, our equations are 2a + b = 5 …(1) 10a + b = 21 …(2) From (1) 2a + b = 5 b = 5 − 2a Putting value of b in (2) 10𝑎+(5−2𝑎) = 21 10𝑎+5−2𝑎 = 21 8𝑎 = 21−5 8𝑎 = 16 𝑎 = 16/8 𝒂 = 𝟐 Putting value of a in (1) 2𝑎+𝑏=5 2(2)+𝑏=5 4+𝑏=5 𝑏=5−4 𝒃=𝟏 Hence, a = 2 & b = 1