Ex 1.3, 2 - Chapter 1 Class 12 Relation and Functions
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Ex 1.3, 2 (Introduction) Let f, g and h be functions from R to R. Show that (f + g)oh = foh + goh (f.g)oh = (foh).(goh) Let f(x) = x, g(x) = sin x , h(x) = log x We will show (f + g) oh = foh + goh Let f(x) = x, g(x) = sin x , h(x) = log x To prove: We will show (f .g) oh = foh . goh (f .g) oh = foh . goh Ex 1.3, 2 Let f, g and h be functions from R to R. Show that (f + g)oh = foh + goh & (f.g) oh = (foh).(goh) Proving (f + g) oh = foh + goh L.H.S (f + g)oh = (f + g) (h(x)) = f (h(x)) + g (h(x)) = foh + goh = R.H.S Hence, (f + g) oh = foh + goh Proving (f .g) oh = foh . goh L.H.S (f . g)oh = (f . g) (h(x)) = f (h(x)) . g (h(x)) = foh . goh = R.H.S Hence, (f . g) oh = (foh) . (goh)
Inverse of a function
Ex 1.3, 2 You are here
Ex 1.3, 3 (i) Important
Ex 1.3, 3 (ii)
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Ex 1.3, 5 (i)
Ex 1.3, 5 (ii) Important
Ex 1.3, 5 (iii) Important
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Ex 1.3 , 7
Ex 1.3 , 8 Important
Ex 1.3 , 9 Important
Ex 1.3, 10 Important
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Ex 1.3, 12
Ex 1.3, 13 (MCQ) Important
Ex 1.3, 14 (MCQ) Important
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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