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Solution for Using Differential, Find the Approximate Value of the √ 0 . 48 ? - CBSE (Commerce) Class 12 - Mathematics

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Question

Using differential, find the approximate value of the  \[\sqrt{0 . 48}\] ?

Solution

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

\[\text { Let }: \]

\[ x = 0 . 49 \]

\[x + ∆ x = 0 . 48\]

\[\text { Then }, \]

\[ ∆ x = - 0 . 01\]

\[\text { For }x = 0 . 49, \]

\[ y = \sqrt{0 . 49} = 0 . 7\]

\[\text { Let }: \]

\[ dx = ∆ x = 0 . 01\]

\[\text { Now,} y = \left( x \right)^\frac{1}{2} \]

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

\[ \Rightarrow \left( \frac{dy}{dx} \right)_{x = 0 . 49} = \frac{1}{1 . 4}\]

\[ \therefore ∆ y = dy = \frac{dy}{dx}dx = \frac{1}{1 . 4} \times \left( - 0 . 01 \right) = - 0 . 007143\]

\[ \Rightarrow ∆ y = - 0 . 007143\]

\[ \therefore \sqrt{0 . 48} = y + ∆ y = 0 . 693\]

  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: 9.2 | Page no. 9

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Solution for question: Using Differential, Find the Approximate Value of the √ 0 . 48 ? concept: Approximations. For the courses CBSE (Commerce), CBSE (Arts), PUC Karnataka Science, CBSE (Science)
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