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  • How to solve this problem- - Limit , continuity and differentiability - JEE Main-2

The equation of the tangent to the curve y=x+\frac{4}{x^{2}}, that is parallel to the x-axis, is

  • Option 1)

    y=0

  • Option 2)

    y=1

  • Option 3)

    y=2

  • Option 4)

    y=3

 

Answers (1)

As we learnt in

Rate Measurement -

Rate of any of variable with respect to time is rate of measurement. Means according to small change in time how much other factors change is rate measurement:

\Rightarrow \frac{dx}{dt},\:\frac{dy}{dt},\:\frac{dR}{dt},(linear),\:\frac{da}{dt}


\Rightarrow \frac{dS}{dt},\:\frac{dA}{dt}(Area)


\Rightarrow \frac{dV}{dt}(Volume)


\Rightarrow \frac{dV}{V}\times 100(percentage\:change\:in\:volume)

- wherein

Where dR / dt  means Rate of change of radius.

 

 y = x +\frac{4}{x^{2}}

\frac{dy}{dx} = 1 +\frac{4.(-2)}{x^{3}}

0 = 1 -\frac{8}{x^{3}}

\therefore x^{3} = +8 \ \ \ x=+2

y = +2 +\frac{4}{4} = +2 +1 = 3

 


Option 1)

y=0

This option is incorrect.

Option 2)

y=1

This option is incorrect.

Option 3)

y=2

This option is incorrect.

Option 4)

y=3

This option is correct.

Posted by

Vakul

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