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Ex 6.4

Ex 6.4, 1 (i)
Important
Deleted for CBSE Board 2023 Exams

Ex 6.4, 1 (ii) Deleted for CBSE Board 2023 Exams

Ex 6.4, 1 (iii) Deleted for CBSE Board 2023 Exams

Ex 6.4, 1 (iv) Deleted for CBSE Board 2023 Exams

Ex 6.4, 1 (v) Important Deleted for CBSE Board 2023 Exams

Ex 6.4, 1 (vi) Deleted for CBSE Board 2023 Exams

Ex 6.4, 1 (vii) Deleted for CBSE Board 2023 Exams

Ex 6.4, 1 (viii) Deleted for CBSE Board 2023 Exams

Ex 6.4, 1 (ix) Deleted for CBSE Board 2023 Exams

Ex 6.4, 1 (x) Deleted for CBSE Board 2023 Exams

Ex 6.4, 1 (xi) Important Deleted for CBSE Board 2023 Exams

Ex 6.4, 1 (xii) Deleted for CBSE Board 2023 Exams

Ex 6.4, 1 (xiii) Deleted for CBSE Board 2023 Exams

Ex 6.4, 1 (xiv) Important Deleted for CBSE Board 2023 Exams

Ex 6.4, 1 (xv) Deleted for CBSE Board 2023 Exams

Ex 6.4,2 Deleted for CBSE Board 2023 Exams You are here

Ex 6.4,3 Important Deleted for CBSE Board 2023 Exams

Ex 6.4,4 Deleted for CBSE Board 2023 Exams

Ex 6.4,5 Important Deleted for CBSE Board 2023 Exams

Ex 6.4,6 Deleted for CBSE Board 2023 Exams

Ex 6.4,7 Deleted for CBSE Board 2023 Exams

Ex 6.4,8 (MCQ) Important Deleted for CBSE Board 2023 Exams

Ex 6.4,9 (MCQ) Deleted for CBSE Board 2023 Exams

Last updated at April 15, 2021 by Teachoo

Ex 6.4, 2 Find the approximate value of f (2.01), where f (x) = 4x2 + 5x + 2.Let x = 2 and â x = 0.01 Given f(x) = 4x2 + 5x + 2 fâ(x) = 8x + 5 Now, âđŚ = fâ(x) â x = (8x + 5) 0.01 Also, ây = f (x + âx) â f(x) f(x + â x) = f (x) + âđŚ f (2.01) = 4x2 + 5x + 2 + (8x + 5)(0.01) Putting value of x = 2 đ (2.01)=4 (2)^2+ 5(2)+2+(0.01)[8(2)+5] = (16+10+2)+(21) (0.01) = 28+0.21 =28.21 Hence, the approximate value of f (2.01) is 28.21