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Confused! kindly explain, - Limit , continuity and differentiability - JEE Main-3

Suppose the cubic x^{3}-px+q has three distinct real roots

where p> 0\: and \: q> 0. Then which one of the following holds?

  • Option 1)

    The\: cubic\: has \: maxima \: at \: both\sqrt{\frac{p}{3}}\: and -\sqrt{\frac{p}{3}}

  • Option 2)

    The\: cubic\: has \:minima \: at \: \sqrt{\frac{p}{3}}\: and \: maxima\: at-\sqrt{\frac{p}{3}}

  • Option 3)

    The\: cubic\: has \:minima \: at \: -\sqrt{\frac{p}{3}}\: and \: maxima\: at\sqrt{\frac{p}{3}}

  • Option 4)

    The\: cubic\: has \: minima \: at \: both\sqrt{\frac{p}{3}}\: and -\sqrt{\frac{p}{3}}

 
Answers (1)
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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^{3}-px+q

\frac{dy}{dx}=3x^{2}-p=0

x^{2}=\frac{p}{\sqrt{3}}

x=\pm \frac{p}{\sqrt{3}}

\frac{d^{2}y}{dx^{2}}=6x

\therefore  at x=\frac{p}{\sqrt{3}} min

x=-\frac{p}{\sqrt{3}}  max


Option 1)

The\: cubic\: has \: maxima \: at \: both\sqrt{\frac{p}{3}}\: and -\sqrt{\frac{p}{3}}

Incorrect option

Option 2)

The\: cubic\: has \:minima \: at \: \sqrt{\frac{p}{3}}\: and \: maxima\: at-\sqrt{\frac{p}{3}}

Correct option

Option 3)

The\: cubic\: has \:minima \: at \: -\sqrt{\frac{p}{3}}\: and \: maxima\: at\sqrt{\frac{p}{3}}

Incorrect option

Option 4)

The\: cubic\: has \: minima \: at \: both\sqrt{\frac{p}{3}}\: and -\sqrt{\frac{p}{3}}

Incorrect option

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