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  • Help me please, - Limit , continuity and differentiability - JEE Main-10

Aprroximate value of (999.99)^{\frac{1}{3}} equals ?

  • Option 1)

    9.999967

  • Option 2)

    9.99955

  • Option 3)

    9.99952

  • Option 4)

    9.999951

 

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) = x^{1/3}  with a small change \delta x  in x  , there will be a small change \delta y in y , where \delta x,\delta y are very small .

now y= x^{1/3} \Rightarrow \frac{dy}{dx} = \frac{1}{3x^{2/3}}

\frac{dy}{dx}= 1/3x^{2/3}\Rightarrow \delta y = \frac{\delta x}{3x^{2/3}}

For given question , x=1000, dx= -0.01

\therefore dy= \frac{-0.01}{3*100}= -0.000033

\therefore y+\delta y= 10-0.000033= 9.999967

 

 

 

 


Option 1)

9.999967

Option 2)

9.99955

Option 3)

9.99952

Option 4)

9.999951

Posted by

Himanshu

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