# Ex 6.5,10 - 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 6.5,10 Find the maximum value of 2 3 24 + 107 in the interval [1, 3]. Find the maximum value of the same function in [ 3, 1]. Let f( )=2 ^3 24 +107 Step 1: Finding f ( ) f ( )= (2 ^3 24 +107)/ = 2 3 ^2 24 = 6 ^2 24 = 6 ( ^2 4" " ) Step 2: Putting f( )=0 6 ( ^2 4" " )=0 ^2 4" = 0 " ^2=4 = 4 = 2 Thus, =2 , 2 Step 3: Since x is in interval [1 , 3] = 2 is only Critical point Also, since given the interval = [1 , 3] We calculate f(x) at = 1 , 2 & 3 Hence maximum value of f( )= at = 3 in the interval [1 , 3] For the interval [ , ] = 2 is only Critical Point Also, since given the interval = [ 3, 1] We calculate f(x) at = 1 , 2 & 3 Hence maximum value of f( )= at = 2 in the interval [ 3, 1]

Ex 6.5

Ex 6.5,1
Important

Ex 6.5,2 Important

Ex 6.5,3

Ex 6.5,4

Ex 6.5,5 Important

Ex 6.5,6

Ex 6.5,7 Important

Ex 6.5,8

Ex 6.5,9 Important

Ex 6.5,10 You are here

Ex 6.5,11 Important

Ex 6.5,12 Important

Ex 6.5,13

Ex 6.5,14 Important

Ex 6.5,15 Important

Ex 6.5,16

Ex 6.5,17

Ex 6.5,18 Important

Ex 6.5,19 Important

Ex 6.5, 20 Important

Ex 6.5,21

Ex 6.5,22 Important

Ex 6.5,23 Important

Ex 6.5,24 Important

Ex 6.5,25 Important

Ex 6.5,26 Important

Ex 6.5, 27

Ex 6.5,28 Important

Ex 6.5,29

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 9 years. He provides courses for Maths and Science at Teachoo.