# 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) $2$

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.

Exams
Articles
Questions