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# Solution for Using Differential, Find the Approximate Value of the √ 26 ? - CBSE (Science) Class 12 - Mathematics

#### Question

Using differential, find the approximate value of the $\sqrt{26}$ ?

#### Solution

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

$\text { Let }:$

$x = 25$

$x + ∆ x = 26$

$\text { Then },$

$∆ x = 1$

$\text { For } x = 25,$

$y = \sqrt{25} = 5$

$\text { Let }:$

$dx = ∆ x = 1$

$\text { Now }, y = \left( x \right)^{1/2}$

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

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

$\therefore ∆ y = dy = \frac{dy}{dx}dx = \frac{1}{10} \times 1 = 0 . 1$

$\Rightarrow ∆ y = 0 . 1$

$\therefore \sqrt{26} = y + ∆ y = 5 . 1$

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: 9.18 | Page no. 9

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Solution for question: Using Differential, Find the Approximate Value of the √ 26 ? concept: Approximations. For the courses CBSE (Science), CBSE (Commerce), PUC Karnataka Science, CBSE (Arts)
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