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Ex 1.3
Ex 1.3, 2 Deleted for CBSE Board 2023 Exams You are here
Ex 1.3, 3 (i) Important Deleted for CBSE Board 2023 Exams
Ex 1.3, 3 (ii) Deleted for CBSE Board 2023 Exams
Ex 1.3 , 4 Deleted for CBSE Board 2023 Exams
Ex 1.3, 5 (i) Deleted for CBSE Board 2023 Exams
Ex 1.3, 5 (ii) Important Deleted for CBSE Board 2023 Exams
Ex 1.3, 5 (iii) Important Deleted for CBSE Board 2023 Exams
Ex 1.3 , 6 Deleted for CBSE Board 2023 Exams
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Ex 1.3 , 8 Important Deleted for CBSE Board 2023 Exams
Ex 1.3 , 9 Important Deleted for CBSE Board 2023 Exams
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Ex 1.3, 12 Deleted for CBSE Board 2023 Exams
Ex 1.3, 13 (MCQ) Important Deleted for CBSE Board 2023 Exams
Ex 1.3, 14 (MCQ) Important Deleted for CBSE Board 2023 Exams
Last updated at March 16, 2023 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)