(D) Given $p^2 + q^2 = 1$. We know that $(p + q)^2 = p^2 + q^2 + 2pq = 1 + 2pq$. By the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality,$\frac{p

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(D) Given $p^2 + q^2 = 1$. We know that $(p + q)^2 = p^2 + q^2 + 2pq = 1 + 2pq$. By the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality,$\frac{p^2 + q^2}{2} \geq \sqrt{p^2 q^2} = pq$. Substituting the given value,$\frac{1}{2} \geq pq$. Therefore,$(p + q)^2 = 1 + 2pq \leq 1 + 2(\frac{1}{2}) = 1 + 1 = 2$. Taking the square root on both sides,$p + q \leq \sqrt{2}$. Thus,the maximum value of $(p + q)$ is $\sqrt{2}$.

Class 11 Mathematics 4-2.Quadratic Equations and Inequations Mix Examples-Quadratic Equations and Inequations

Easy

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

A
$2$
B
$1/2$
C
$\frac{1}{\sqrt{2}}$
D
$\sqrt{2}$

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4-2.Quadratic Equations and Inequations

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Mix Examples-Quadratic Equations and Inequations

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

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