Ex 5.1

Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

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Ex 5.1, 3 Examine the following functions for continuity. (d) f (x) = |x β 5| f(x) = |π₯β5| = {β((π₯β5), π₯β5β₯[email protected]β(π₯β5), π₯β5<0)β€ = {β((π₯β5), π₯β₯[email protected]β(π₯β5), 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¬(xβ5^β ) π(π₯)=limβ¬(xβ5^+ ) " " 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¬(xβ5^β ) f(x) = limβ¬(hβ0) f(5 β h) = limβ¬(hβ0) |(5ββ)β5| = limβ¬(hβ0) |ββ| = limβ¬(hβ0) β = 0 RHL at x β 5 limβ¬(xβ5^+ ) f(x) = limβ¬(hβ0) f(5 + h) = limβ¬(hβ0) |(5+β)β5| = limβ¬(hβ0) |β| = limβ¬(hβ0) β = 0 & π(π) = |π₯β5| = |5β5| = 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₯β5| is continuous at all points. i.e. f is continuous at π β R.

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.