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
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
Ex 5.5, 7 Important
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, 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