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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
Ex 5.1, 34 Important You are here
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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 5.5, 11 Important
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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 May 29, 2023 by Teachoo
Ex 5.1, 34 Find all the points of discontinuity of f defined by π(π₯)= |π₯| β |π₯+1|.Given π(π₯)= |π₯| β |π₯+1|. Here, we have 2 critical points x = 0 and x + 1 = 0 i.e. x = 0, and x = β1 So, our intervals will be When πβ€βπ When βπ<π<π When πβ₯π When πβ€βπ π(π₯)= |π₯| β |π₯+1|. Here, both will be negative π(π₯)=(βπ₯) β(β(π₯+1)) π(π₯)=βπ₯+(π₯+1) " " π(π)=π When βπ<πβ€π π(π₯)= |π₯| β |π₯+1|. Here, x will be negative, but (x + 1) will be positive π(π₯)=(βπ₯) β(π₯+1) π(π₯)=βπ₯βπ₯β1 " " π(π)=βππβπ |π₯| = {β(π₯, π₯ β₯0@βπ₯, π₯<0)β€ |π₯+1| = {β((π₯+1) , π₯+1β₯0@β(π₯+1) π₯+1<0)β€ = {β((π₯+1) , π₯β₯β1@β(π₯+1) π₯<1)β€ When πβ₯π π(π₯)= |π₯| β |π₯+1|. Here, both will be positive π(π₯)=π₯ β(π₯+1) π(π₯)=π₯βπ₯β1 " " π(π)=βπ Thus, our function becomes π(π)={β(π ππ πβ€βπ@βππβπ ππ βπ<π<π@βπ ππ πβ₯π)β€ Since we need to find continuity at of the function We check continuity for different values of x When x < β1 When x = β1 When β1 < x < 0 When x = 0 When x > 0 Checking continuity Case 1 : When x < β1 For x < β1, f(x) = 1 Since this constant It is continuous β΄ f(x) is continuous for x < β1 Case 2 : When x = β1 f(x) is continuous at π₯ =β1 if L.H.L = R.H.L = π(β1) if limβ¬(xβγβ1γ^β ) π(π₯)=limβ¬(xβγβ1γ^+ ) " " π(π₯)= π(β1) Since there are two different functions on the left & right of β1, we take LHL & RHL . LHL at x β β1 limβ¬(xβγβ1γ^β ) f(x) = limβ¬(hβ0) f(β1 β h) = limβ¬(hβ0) 1 = 1 RHL at x β 0 limβ¬(xβγβ1γ^+ ) f(x) = limβ¬(hβ0) f(β1 + h) = limβ¬(hβ0) (β2(β1+β))β1 = limβ¬(hβ0) (2β2β)β1 = (2 β 2(0)) β 1 = 2 β 0 β 1 = 1 & π(βπ) = 1 Hence, L.H.L = R.H.L = π(β1) β΄ f is continuous at x = β1 Case 3 : When β1 < x < 0 For β1 < x < 0 f(x) = β2x β 1 Since this a polynomial It is continuous β΄ f(x) is continuous for β1 < x < 0 Case 4 : When x = 0 f(x) is continuous at π₯ =0 if L.H.L = R.H.L = π(0) if limβ¬(xβ0^β ) π(π₯)=limβ¬(xβ0^+ ) " " π(π₯)= π(0) Since there are two different functions on the left & right of 0, we take LHL & RHL . LHL at x β 0 limβ¬(xβ0^β ) f(x) = limβ¬(hβ0) f(0 β h) = limβ¬(hβ0) f(βh) = limβ¬(hβ0) β2(ββ)β1 = limβ¬(hβ0) 2ββ1 = 2(0) β 1 = β1 RHL at x β 0 limβ¬(xβ0^+ ) f(x) = limβ¬(hβ0) f(0 + h) = limβ¬(hβ0) f(h) = limβ¬(hβ0) β1 = β1 & π(0) = β1 Hence, L.H.L = R.H.L = π(0) β΄ f is continuous at x = 0 Case 5: When x > 0 For x > 0, f(x) = β1 Since this constant It is continuous β΄ f(x) is continuous for x > 0 Since there is no point of discontinuity Therefore, f is continuous for all x β R