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Ex 6.4
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