If p \: and \: q are positive real numbers such that p^{2} +q^{2}=1, then the maximum value of \left ( p +q \right ) is

  • Option 1)

    \frac{1}{2}

  • Option 2)

    \frac{1}{\sqrt{2}}

  • Option 3)

    \sqrt{2}

  • Option 4)

    2

 

Answers (1)
P Prateek Shrivastava

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=p+q

=p+\sqrt{1-p^{2}}

\frac{dy}{dp}=1-\frac{2p}{2\sqrt{1-p^{2}}}

0=1-\frac{p}{\sqrt{1-p^{2}}}

1=\frac{p}{\sqrt{1-p^{2}}}

1-p^{2}=p^{2}

p^{2}=\frac{1}{2}

\therefore p=\frac{1}{\sqrt{2}}\:\:\:\:q=\frac{1}{\sqrt{2}}

\therefore p+q=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2} 

 


Option 1)

\frac{1}{2}

This option is incorrect.

Option 2)

\frac{1}{\sqrt{2}}

This option is incorrect.

Option 3)

\sqrt{2}

This option is correct.

Option 4)

2

This option is incorrect.

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