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  • Solve it, - Limit , continuity and differentiability - JEE Main-8

Approximate value of of \sqrt{400.1} will be ?

  • Option 1)

    20.01

  • Option 2)

    20.0025

  • Option 3)

    20.001

  • Option 4)

    20.0035

 

Answers (1)

best_answer

As we have learned

Approximation -

It gives approximate value of any  f(x)  at   x=x_{\circ }.  We break  x_{\circ }  to  x + \delta x

Such\; that f(x+\delta x)=f(x)+f'(x)

ex:\:\:\:\sqrt{25.5}\:\:\:we \:take

f(x)=\sqrt{x},\:\:\:x=25\:\:and\:\:\delta x=0.5

- wherein

Where fx is negative value of  f(x). It may be positive and negative.

 

 Let y=f (x) =\sqrt x 

with a small change \delta (x) in x , there will be small change in y  i.e \delta (y) , where \delta (y) and 

\delta (x) are very small 

f(x+\delta x)= f(x)+\delta y

Now y= \sqrt x \Rightarrow \frac{dy}{dx}= 1/2 \sqrt x\Rightarrow dy=\frac{dx}{2\sqrt x}

\Rightarrow \delta y=\frac{\delta x}{2 \sqrt x}

For a given question , x= 400 , \delta (x) = 0.1

\therefore \delta y=\frac{0.1}{2 \sqrt {400}}=0.1/40=0.0025

\therefore y+ \delta y=20+0.0025=20.0025

 

 

 

 

 

 


Option 1)

20.01

Option 2)

20.0025

Option 3)

20.001

Option 4)

20.0035

Posted by

Himanshu

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