Question

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

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

Answer 1

The correct option is D

$\sqrt{2}$

Explanation for the correct option

Step 1: Relationship between Arithmetic Mean (AM) and Geometric Mean (GM)

As per the relationship, $\mathbf{\text{AM}}\mathbf{\ge }\mathbf{\text{GM}}$. Here arithmetic mean for the two numbers $a\text{and}b$ is given by

$\mathit{A}\mathit{M}\mathbf{=}\frac{\mathbf{a}\mathbf{+}\mathbf{b}}{\mathbf{2}}$

The geometric mean for the same numbers is given by

$\mathit{G}\mathit{M}\mathbf{=}\sqrt{\mathbf{a}\mathbf{b}}$

Step 2: Calculation to find the value of $\mathit{p}\mathbf{+}\mathit{q}$.

Here, let the numbers $a\text{and}b$ as $p^2\text{and}{q}^2$ respectively, then by the above relationship we have,

$\frac{p^2+q^2}{2}\ge \sqrt{p^2{q}^2}$

From this, we have

$pq\le \frac{1}{2}\left[\because p^2+q^2=1\right]$

Now, apply the algebraic identity ${\left(a+b\right)}^{\mathbf{2}}\mathbf{=}{\mathit{a}}^{\mathbf{2}}\mathbf{+}{\mathit{b}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{a}\mathit{b}$, we get

$\begin{array}{rcl}{\left(p+q\right)}^2& =& p^2+q^2+2pq\\ & =& 1+2\left(\frac{1}{2}\right)\\ & =& 2\end{array}$

Therefore, the value of $p+q$ is $\sqrt{2}$.

Hence, the correct option is (D).