

Ex 6.4
Ex 6.4, 1 (ii) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (iii) Deleted for CBSE Board 2022 Exams You are here
Ex 6.4, 1 (iv) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (v) Important Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (vi) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (vii) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (viii) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (ix) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (x) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (xi) Important Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (xii) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (xiii) Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (xiv) Important Deleted for CBSE Board 2022 Exams
Ex 6.4, 1 (xv) Deleted for CBSE Board 2022 Exams
Ex 6.4,2 Deleted for CBSE Board 2022 Exams
Ex 6.4,3 Important Deleted for CBSE Board 2022 Exams
Ex 6.4,4 Deleted for CBSE Board 2022 Exams
Ex 6.4,5 Important Deleted for CBSE Board 2022 Exams
Ex 6.4,6 Deleted for CBSE Board 2022 Exams
Ex 6.4,7 Deleted for CBSE Board 2022 Exams
Ex 6.4,8 (MCQ) Important Deleted for CBSE Board 2022 Exams
Ex 6.4,9 (MCQ) Deleted for CBSE Board 2022 Exams
Last updated at April 15, 2021 by Teachoo
Ex 6.4, 1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (iii) √0.6 Let y = √𝑥 where x = 0.64 & △x = –0.04 Since y = √𝑥 𝑑𝑦/𝑑𝑥 = (𝑑(√𝑥))/𝑑𝑥 = 1/(2√𝑥) Now, ∆𝑦 = 𝑑𝑦/𝑑𝑥 △x = 1/(2√0.64) (–0.04) = 1/(2√(64/100)) (–0.04) = 1/(2 ×8/10) × (–0.04) = (−10 × 0.04)/16 = (−10 × 4)/(16 × 100) = –0.025 Also, ∆𝑦=𝑓(𝑥+∆𝑥)−𝑓(𝑥) Putting values ∆𝑦=√(𝑥+∆𝑥)−√𝑥 −0. 025=√(0.64−0.04)−√0.64 −0.025=√0.60−0.8 0.8 – 0. 025=√0.60 √0.60=0.775 Hence, approximate value of √0.60 is 0.775