Q

# 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)

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

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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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