# Ex 6.2,2 - Chapter 6 Class 12 Application of Derivatives

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

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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.