Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12

Last updated at March 10, 2021 by Teachoo

Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12

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.

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.