Ex 1.3, 2 - Chapter 1 Class 12 Relation and Functions

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Ex 1.3, 2 - Show that (f + g)oh = foh + goh (f.g)oh = (foh).(goh) - Composite funcions

Ex 1.3, 2 - Chapter 1 Class 12 Relation and Functions - Part 2
Ex 1.3, 2 - Chapter 1 Class 12 Relation and Functions - Part 3
Ex 1.3, 2 - Chapter 1 Class 12 Relation and Functions - Part 4

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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)

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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, Social Science, Physics, Chemistry, Computer Science at Teachoo.