Last updated at April 19, 2021 by Teachoo

Transcript

Ex 6.2, 1 (Method 1) Show that the function given by f (๐ฅ) = 3๐ฅ + 17 is strictly increasing on R. f(๐ฅ) = 3๐ฅ + 17 Finding fโ(๐) fโ(๐ฅ) = 3 Since fโ(๐) > 0 Hence, f is strictly increasing on R Ex 6.2, 1 (Method 2) Show that the function given by f (x) = 3x + 17 is strictly increasing on R. Let ๐ฅ1 and ๐ฅ2 be real numbers Such that ๐๐ < ๐2 Multiplying both sides by 3 3๐ฅ1 < 3 ๐ฅ2 Adding both sides by 17 3๐ฅ1 + 17 < 3๐ฅ2 + 17 f (๐๐) < f ( ๐2) Hence, when x1 < x2 , f(x1) < f(x2) Thus, f(x) is strictly increasing on R.

Ex 6.2

Ex 6.2, 1
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Ex 6.2,2

Ex 6.2,3

Ex 6.2,4

Ex 6.2, 5

Ex 6.2, 6 (a)

Ex 6.2, 6 (b)

Ex 6.2, 6 (c)

Ex 6.2, 6 (d)

Ex 6.2, 6 (e) Important

Ex 6.2, 7

Ex 6.2,8 Important

Ex 6.2,9 Important

Ex 6.2,10

Ex 6.2,11

Ex 6.2, 12 (A)

Ex 6.2, 12 (B) Important

Ex 6.2, 12 (C) Important

Ex 6.2, 12 (D)

Ex 6.2, 13 (MCQ) Important

Ex 6.2,14 Important

Ex 6.2,15

Ex 6.2, 16

Ex 6.2,17 Important

Ex 6.2,18

Ex 6.2,19 (MCQ) Important

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.