Chapter 5 Class 12 Continuity and Differentiability
Ex 5.1, 13
Ex 5.1, 16
Ex 5.1, 18 Important
Ex 5.1, 28 Important
Ex 5.1, 30 Important You are here
Ex 5.1, 34 Important
Ex 5.2, 5
Ex 5.2, 9 Important
Ex 5.2, 10 Important
Ex 5.3, 10 Important
Ex 5.3, 14
Example 29 Important
Example 30 Important
Ex 5.5,6 Important
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Ex 5.5, 11 Important
Ex 5.5, 16 Important
Ex 5.6, 7 Important
Ex 5.6, 11 Important
Example 38 Important
Ex 5.7, 14 Important
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 5 Important Deleted for CBSE Board 2024 Exams
Example 39 (i)
Example 40 (i)
Example 42 Important
Misc 6 Important
Misc 15 Important
Misc 16 Important
Misc 22 Important
Chapter 5 Class 12 Continuity and Differentiability
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