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# Solution for Using Differential, Find the Approximate Value of the 1 ( 2 . 002 ) 2 ? - CBSE (Science) Class 12 - Mathematics

#### Question

Using differential, find the approximate value of the $\frac{1}{(2 . 002 )^2}$ ?

#### Solution

$\text { Consider the function y } = f\left( x \right) = \frac{1}{x^2} .$

$\text { Let }:$

$x = 2$

$x + ∆ x = 2 . 002$

$\text { Then },$

$∆ x = - 0 . 002$

$\text { For } x = 2 ,$

$y = \frac{1}{2^2} = \frac{1}{4}$

$\text { Let }:$

$dx = ∆ x = 0 . 002$

$\text { Now,} y = \frac{1}{x^2}$

$\Rightarrow \frac{dy}{dx} = \frac{2}{x^3}$

$\Rightarrow \left( \frac{dy}{dx} \right)_{x = 2} = \frac{1}{4}$

$\therefore ∆ y = dy = \frac{dy}{dx}dx = \frac{1}{4} \times - 0 . 002 = - 0 . 0005$

$\Rightarrow ∆ y = - 0 . 0005$

$\therefore \frac{1}{\left( 2 . 002 \right)^2} = y + ∆ y = 0 . 2495$

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: 9.07 | Page no. 9

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Solution for question: Using Differential, Find the Approximate Value of the 1 ( 2 . 002 ) 2 ? concept: Approximations. For the courses CBSE (Science), CBSE (Commerce), PUC Karnataka Science, CBSE (Arts)
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