Ex 6.4

Ex 6.4, 1 (i)
Important
Deleted for CBSE Board 2022 Exams
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Ex 6.4, 1 (ii) Deleted for CBSE Board 2022 Exams

Ex 6.4, 1 (iii) Deleted for CBSE Board 2022 Exams

Ex 6.4, 1 (iv) Deleted for CBSE Board 2022 Exams

Ex 6.4, 1 (v) Important Deleted for CBSE Board 2022 Exams

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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

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Ex 6.4,2 Deleted for CBSE Board 2022 Exams

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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 Dec. 8, 2021 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