Maharashtra State BoardHSC Arts 12th Board Exam
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Find the approximate value of √8.95 - Mathematics and Statistics

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Find the approximate value of ` sqrt8.95 `

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Solution

`Let f(x)=sqrtx`

`f(a+h)=sqrt(a+h)`

we choose a=9 and h=-0.05

Then `sqrt8.95=f(a+h) `

But `, f(a+h)~~f(a)+hf'(a)`

`sqrt8.95~~f(a)+hf'(a)...........(1)`

Now ` f(a)=sqrta=sqrt9 and h=-0.05`

we have `f(X)=sqrtx `

`f'(x)=1/(2sqrtx)`

`f'(a)=f'(9)=1/(2sqrt9)=1/(2xx3)=1/6`

from(1) ` sqrt(8.95)~~3+(-0.05)(1/6)`

=3-0.0083=2.9917

`=> sqrt8.95 ~~2.9917`

Concept: Approximations
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