Example 13 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at December 16, 2024 by

Ā  Example 13 - Discuss continuity of f(x) = {x, x >= 0 and x^2, x < 0 - Examples

part 2 - Example 13 - Examples - Serial order wise - Chapter 5 Class 12 Continuity and Differentiability
part 3 - Example 13 - Examples - Serial order wise - Chapter 5 Class 12 Continuity and Differentiability
part 4 - Example 13 - Examples - Serial order wise - Chapter 5 Class 12 Continuity and Differentiability

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Example 13 Discuss the continuity of the function f given by š‘“(š‘„)={ā–ˆ(& š‘„, š‘–š‘“ š‘„ā‰„0@& š‘„2 , š‘–š‘“ š‘„<0)ā”¤ š‘“(š‘„)={ā–ˆ(& š‘„, š‘–š‘“ š‘„ā‰„0@& š‘„2 , š‘–š‘“ š‘„<0)ā”¤ Since we need to find continuity at of the function We check continuity for different values of x When x = 0 When x < 0 When x > 0 Case 1 : When x = 0 f(x) is continuous at š‘„ =0 if L.H.L = R.H.L = š‘“(0) if limā”¬(xā†’0^āˆ’ ) š‘“(š‘„)=limā”¬(xā†’0^+ ) " " š‘“(š‘„)= š‘“(0) Since there are two different functions on the left & right of 0, we take LHL & RHL . LHL at x ā†’ 0 limā”¬(xā†’0^āˆ’ ) f(x) = limā”¬(hā†’0) f(0 āˆ’ h) = limā”¬(hā†’0) f(āˆ’h) = limā”¬(hā†’0) (āˆ’h)2 = (āˆ’0)2 = 0 RHL at x ā†’ 0 limā”¬(xā†’0^+ ) f(x) = limā”¬(hā†’0) f(0 + h) = limā”¬(hā†’0) f(h) = limā”¬(hā†’0) h = 0 & š‘“(0) = š‘„ = 0 Hence, L.H.L = R.H.L = š‘“(0) āˆ“ f is continuous at x = 0 Case 2 : When x < 0 For x < 0, f(x) = š‘„^2 Since this a polynomial It is continuous āˆ“ f(x) is continuous for x < 0Case 3 : When x > 0 For x > 0, f(x) = x Since this a polynomial It is continuous āˆ“ f(x) is continuous for x > 0 Hence, there is no point of discontinuity Thus, f is continuous for all š’™āˆˆš‘

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo