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Solution for Using Differential, Find the Approximate Value of the Following: √ 25 . 02 - CBSE (Science) Class 12 - Mathematics

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Question

1 Using differential, find the approximate value of the following:

\[\sqrt{25 . 02}\]

Solution

\[\text { Consider the function y } = f\left( x \right) = \sqrt{x} . \]

\[\text { Let }: \]

\[ x = 25 \]

\[ x + ∆ x = 25 . 02\]

\[\text { Then, } \]

\[ ∆ x = 0 . 02\]

\[\text { For} x = 25, \]

\[ y = \sqrt{25} = 5\]

\[\text { Let }: \]

\[ dx = ∆ x = 0 . 02\]

\[\text { Now,} y = \sqrt{x}\]

\[ \Rightarrow \frac{dy}{dx} = \frac{1}{2\sqrt{x}}\]

\[ \Rightarrow \left( \frac{dy}{dx} \right)_{x = 25} = \frac{1}{10}\]

\[ \therefore ∆ y = dy = \frac{dy}{dx}dx = \frac{1}{10} \times 0 . 02 = 0 . 002\]

\[ \Rightarrow ∆ y = 0 . 002\]

\[ \therefore \sqrt{25 . 02} = y + ∆ y = 5 . 002\]

  Is there an error in this question or solution?

APPEARS IN

 RD Sharma Mathematics for Class 12 by R D Sharma (Set of 2 Volume) (2018-19 Session) (with solutions)
Chapter 14: Differentials, Errors and Approximations
Q: 9.01 | Page no. 9

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Solution for question: Using Differential, Find the Approximate Value of the Following: √ 25 . 02 concept: Approximations. For the courses CBSE (Science), CBSE (Commerce), PUC Karnataka Science, CBSE (Arts)
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