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Ex 6.4

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

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 March 22, 2023 by Teachoo

Ex 6.4, 1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (i) √25.3Let y = √𝒙 Thus, √(𝟐𝟓.𝟑) = y + ∆𝒚 Here, ∆𝒚 = 𝒅𝒚/𝒅𝒙 △x where x = 25 & △x = 0.3 Since y = √𝑥 𝒅𝒚/𝒅𝒙 = (𝑑(√𝑥))/𝑑𝑥 = 𝟏/(𝟐√𝒙) Now, ∆𝒚 = 𝒅𝒚/𝒅𝒙 △x = 1/(2√𝑥) △x Putting x = 25 & △x = 0.3 = 1/(2√25) (0.3) = 1/(2 × 5) × 0.3 = 0.3/10 = 0.03 Therefore, √25.3 = y + ∆𝑦 Putting values √25.3 =√25+0.03 √(𝟐𝟓. 𝟑)=𝟓. 𝟎𝟑 Hence, approximate value of √25.3 is 5.03