Last updated at March 10, 2021 by Teachoo

Transcript

Ex 5.1, 3 Examine the following functions for continuity. (a) f(x) = x β 5 f(x) = x β 5 To check continuity of π(π₯), We check itβs if it is continuous at any point x = c Let c be any real number f is continuous at π₯ =π if (π₯π’π¦)β¬(π±βπ) π(π)=π(π) (πππ)β¬(π±βπ) π(π) = limβ¬(xβπ) π₯ β 5 = c β 5 π(π) = c β 5 Since, L.H.S = R.H.S β΄ Function is continuous at x = c Thus, we can write that f is continuous for x = c , where c βπ β΄ f is continuous for every real number. Ex 5.1, 3 Examine the following functions for continuity. (b) f (x) = 1/(π₯ β 5) , x β 5 f (x) = 1/(π₯ β 5) At x = 5 f (x) = 1/(5 β 5) = 1/0 = β Hence, f(x) is not defined at x = 5 So, we check for continuity at all points except 5 Let c be any real number except 5. f is continuous at π₯ = π if (π₯π’π¦)β¬(π±βπ) π(π) = π(π) LHS (π₯π’π¦)β¬(π±βπ) π(π) = limβ¬(π₯βΆπ)β‘γ1/(π₯ β 5)γ = 1/(π β 5) RHS π(π) = 1/(π β 5) Since, L.H.S = R.H.S β΄ Function is continuous at x = c (except 5) Thus, we can write that f is continuous for all real numbers except 5 β΄ f is continuous at each π± β R β {π} Ex 5.1, 3 Examine the following functions for continuity. (c) f (x) = (π₯^(2 )β 25 )/(π₯ + 5), x β β5 f (x) = (π₯^(2 )β 25 )/(π₯ + 5) Putting x = β5 f (β5) = (γ(β5)γ^(2 )β 25 )/(β5 + 5) = (25β 25 )/(β5 + 5) = 0/0 = Undefined Hence, f(x) is not defined at x = β5 So, we check for continuity at all points except β5 Let c be any real number except β5. f is continuous at π₯ = π if (π₯π’π¦)β¬(π±βπ) π(π) = π(π) LHS (π₯π’π¦)β¬(π±βπ) π(π) = limβ¬(xβπ) (π₯^2β 25)/(π₯ + 5) = limβ¬(xβπ) ((π₯ β 5) (π₯ + 5))/(π₯ + 5) = limβ¬(xβπ) π₯β5 Putting x = c = c β 5 RHS f (c) = (π^(2 )β 25 )/(π + 5) = ((π β 5)(π + 5))/((π + 5)) = c β 5 Let c be any real number except β5. f is continuous at π₯ = π if (π₯π’π¦)β¬(π±βπ) π(π) = π(π) Since, L.H.S = R.H.S β΄ Function is continuous at x = c (except β5) Thus, we can write that f is continuous for all real numbers except β5 β΄ f is continuous at each π± β R β {βπ} Ex 5.1, 3 Examine the following functions for continuity. (d) f (x) = |x β 5| f(x) = |π₯β5| = {β((π₯β5), π₯β5β₯0@β(π₯β5), π₯β5<0)β€ = {β((π₯β5), π₯β₯5@β(π₯β5), π₯<5)β€ Since we need to find continuity at of the function We check continuity for different values of x When x = 5 When x < 5 When x > 5 Case 1 : When x = 5 f(x) is continuous at π₯ = 5 if L.H.L = R.H.L = π(5) if limβ¬(xβ5^β ) π(π₯)=limβ¬(xβ5^+ ) " " π(π₯)= π(5) Since there are two different functions on the left & right of 5, we take LHL & RHL . LHL at x β 5 limβ¬(xβ5^β ) f(x) = limβ¬(hβ0) f(5 β h) = limβ¬(hβ0) |(5ββ)β5| = limβ¬(hβ0) |ββ| = limβ¬(hβ0) β = 0 RHL at x β 5 limβ¬(xβ5^+ ) f(x) = limβ¬(hβ0) f(5 + h) = limβ¬(hβ0) |(5+β)β5| = limβ¬(hβ0) |β| = limβ¬(hβ0) β = 0 & π(π) = |π₯β5| = |5β5| = 0 Hence, L.H.L = R.H.L = π(5) β΄ f is continuous at x = 5 Case 2 : When x < 5 For x < 5, f(x) = β (x β 5) Since this a polynomial It is continuous β΄ f(x) is continuous for x < 5 Case 3 : When x > 5 For x > 5, f(x) = (x β 5) Since this a polynomial It is continuous β΄ f(x) is continuous for x > 5 Hence, π(π₯)= |π₯β5| is continuous at all points. i.e. f is continuous at π β R.

Ex 5.1

Ex 5.1 ,1

Ex 5.1 ,2

Ex 5.1 ,3 Important You are here

Ex 5.1 ,4

Ex 5.1 ,5 Important

Ex 5.1 ,6

Ex 5.1 ,7 Important

Ex 5.1 ,8

Ex 5.1, 9 Important

Ex 5.1, 10

Ex 5.1, 11

Ex 5.1, 12

Ex 5.1, 13

Ex 5.1, 14

Ex 5.1, 15 Important

Ex 5.1, 16

Ex 5.1, 17 Important

Ex 5.1, 18 Important

Ex 5.1, 19 Important

Ex 5.1, 20

Ex 5.1, 21

Ex 5.1, 22 (i) Important

Ex 5.1, 22 (ii)

Ex 5.1, 22 (iii)

Ex 5.1, 22 (iv)

Ex 5.1, 23

Ex 5.1, 24 Important

Ex 5.1, 25

Ex 5.1, 26 Important

Ex 5.1, 27

Ex 5.1, 28 Important

Ex 5.1, 29

Ex 5.1, 30 Important

Ex 5.1, 31

Ex 5.1, 32

Ex 5.1, 33

Ex 5.1, 34 Important

Chapter 5 Class 12 Continuity and Differentiability

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.