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

$\frac{1}{2}$

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

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\sqrt{2}

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Solution

The correct option is C $\sqrt{2}$
Since , p and q are positive real numbers $p^{2} + q^{2} = 1$ (given)
Using $A M \geq G M$
$∴ {(\frac{p + q}{2})}^{2} \geq \sqrt{(p q)^{2}} = \frac{p^{2} + q^{2} + 2 p q}{4} \geq p q \frac{1 + 2 p q}{4} \geq o r, 1 + 2 p q \geq 4 p q 1 \geq 2 p q o r p q \leq \frac{1}{2}$
Now $(p + q)^{2} = p^{2} + q^{2} + 2 p q \Rightarrow (p + q)^{2} \leq 1 + 2 \times \frac{1}{2} \Rightarrow p + q \leq \sqrt{2}$

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