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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
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Ex 5.3, 14
Example 29 Important
Example 30 Important
Ex 5.5,6 Important
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Ex 5.6, 7 Important
Ex 5.6, 11 Important
Example 38 Important
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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 May 29, 2023 by Teachoo
Ex 5.1, 30 Find the values of a and b such that the function defined by π(π₯)={β(5, ππ π₯β€2@ππ₯+π, ππ 2<π₯<[email protected], ππ π₯β₯10)β€ 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 = π(2) i.e. limβ¬(xβ2^β ) π(π₯)=limβ¬(xβ2^+ ) " " π(π₯)= π(2) LHL at x β 2 (πππ)β¬(π₯β2^β ) f(x) = (πππ)β¬(ββ0) f(2 β h) = limβ¬(hβ0) 5 = 5 RHL at x β 2 (πππ)β¬(π₯β2^+ ) f(x) = (πππ)β¬(ββ0) f(2 + h) = limβ¬(hβ0) a(2 + h) + b = a(2 + 0) + b = 2a + b Since, LHL = RHL 2a + b = 5 At x = 10 π is continuous at x = 10 if L.H.L = R.H.L = π(10) i.e. limβ¬(xβ10^β ) π(π₯)=limβ¬(xβ10^+ ) " " π(π₯)= π(10) LHL at x β 10 (πππ)β¬(π₯β10^β ) f(x) = (πππ)β¬(ββ0) f(10 β h) = limβ¬(hβ0) a(10 β h) + b = a(10 β 0) + b = 10a + b RHL at x β 10 (πππ)β¬(π₯β10^+ ) f(x) = (πππ)β¬(ββ0) f(10 + h) = limβ¬(hβ0) 21 = 21 Since, L.H.L = R.H.L 10a + b = 21 Now, our equations are 2a + b = 5 β¦(1) 10a + b = 21 β¦(2) From (1) 2a + b = 5 b = 5 β 2a Putting value of b in (2) 10π+(5β2π) = 21 10π+5β2π = 21 8π = 21β5 8π = 16 π = 16/8 π = π Putting value of a in (1) 2π+π=5 2(2)+π=5 4+π=5 π=5β4 π=π Hence, a = 2 & b = 1