Last updated at Dec. 16, 2024 by Teachoo
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
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Ex 6.2, 6 (a)
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Ex 6.2, 12 (A)
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Ex 6.2, 13 (MCQ) Important
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Ex 6.2,19 (MCQ) Important
About the Author
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