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# Solution for Using Differential, Find the Approximate Value of the ( 1 . 999 ) 5 ? - CBSE (Science) Class 12 - Mathematics

#### Question

Using differential, find the approximate value of the $\left( 1 . 999 \right)^5$ ?

#### Solution

$\text { Consider the function } y = f\left( x \right) = x^5 .$

$\text { Let }:$

$x = 2$

$x + ∆ x = 1 . 999$

$\text { Then },$

$∆ x = - 0 . 001$

$\text { For } x = 2,$

$y = 2^5 = 32$

$\text { Let }:$

$dx = ∆ x = - 0 . 001$

$\text { Now }, y = x^5$

$\Rightarrow \frac{dy}{dx} = 5 x^4$

$\Rightarrow \left( \frac{dy}{dx} \right)_{x = 2} = 80$

$\therefore ∆ y = dy = \frac{dy}{dx}dx = 80 \times \left( - 0 . 001 \right) = - 0 . 08$

$\Rightarrow ∆ y = - 0 . 08$

$\therefore 1 . {999}^5 = y + ∆ y = 31 . 92$

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#### 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: 9.28 | Page no. 9

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Solution for question: Using Differential, Find the Approximate Value of the ( 1 . 999 ) 5 ? concept: Approximations. For the courses CBSE (Science), PUC Karnataka Science, CBSE (Arts), CBSE (Commerce)
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