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If $p$ and $q$ are positive real numbers such that $p^2+q^2=1$, then the maximum value of $(p+q)$ is
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2964 Upvotes
Verified Answer
The correct answer is:
$\sqrt{2}$
$\sqrt{2}$
Using $A.M.$ $\geq$ $G.M.$
$\begin{aligned}
& \frac{p^2+q^2}{2} \geq p q \\
& \Rightarrow p q \leq \frac{1}{2} \\
& (p+q)^2=p^2+q^2+2 p q \\
& \Rightarrow p+q \leq \sqrt{2} .
\end{aligned}$
$\begin{aligned}
& \frac{p^2+q^2}{2} \geq p q \\
& \Rightarrow p q \leq \frac{1}{2} \\
& (p+q)^2=p^2+q^2+2 p q \\
& \Rightarrow p+q \leq \sqrt{2} .
\end{aligned}$
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