# Ex 6.2,2 - Class 12

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 6.2,2 (Method 1) Show that the function given by f (x) = e2x is strictly increasing on R. We need to show that f (š„) is strictly increasing on R f(š„) = e2š„ fā (š„) = (e2š„)ā fā(š„) = 2e2š„ Case 1 When š = 0 fā (0) = 2e2(0) fā (0) = 2e0 fā (0) = 2 . 1 fā (0) = 2 > 0 ā“ fā (š„) > 0 when š„ = 0 Case 2 When š > 0 fā (š„) = 2e2š„ fā (š„) = 2ļ·1+ļ·2š„ļ·Æ+ ļ·ļ·2š„ļ·Æ2ļ·®2!ļ·Æ+ ļ·ļ·2š„ļ·Æ3ļ·®3!ļ·Æ+ ā¦ļ·Æ Since š„ > 0 So, far every value of š„ greater than 0 fā (š„) = 2 ļ·1+ļ·2š„ļ·Æ+ ļ·ļ·ļ·2š„ļ·Æļ·®2ļ·Æļ·®2!ļ·Æ+ ā¦ļ·Æ > 0 Hence fā (š„) > 0 when š„>0 ā Thus, f (x) is strictly increasing for x > 0 Case 3 When š < 0 ā š„ = some negative value Let š„ =ā š¦ fā(š„) = 2ļ·eļ·®ā2yļ·Æ fā(š„) = ļ·2ļ·®š2š¦ļ·Æ fā(š„) = ļ·2ļ·®ļ·1+ļ·2š¦ļ·Æ+ ļ·ļ·2š¦ļ·Æ2ļ·®2ļ·Æļ·Æļ·Æ ā fā(š„) = ļ·2ļ·®š ššš ļ·+ļ·Æ šš¢ššš”šš”š¦ļ·Æ >0 ā“ f (š„) > 0 for š„ < 0 ā Thus fā (š„) > 0 for all š„ ā R f is strictly increasing on R Ex 6.2,2 (Method 2) Show that the function given by f (x) = e2x is strictly increasing on R. Let š„1 & š„2 be any two real number such that š„1 < š„2 š„1 < š„2 ļ·šļ·®2š„1ļ·Æ<ļ·šļ·®2š„2ļ·Æ f (š„1) < f (š„2) Thus, if š„1 < š„2 then, f (š„1) < f (š„2) Hence f is strictly increasing on R

Chapter 6 Class 12 Application of Derivatives

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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.