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

$(a)\;2\qquad(b)\;\large\frac{1}{2}\qquad$$(c)\;\large\frac{1}{\sqrt 2}\qquad$$(d)\;\sqrt 2$

1 Answer

Using AM $\geq$ GM
$\large\frac{p^2+q^2}{2}$$\geq pq$
$\Rightarrow pq\leq \large\frac{1}{2}$
$p^2+q^2=1$
We know that
$(p+q)^2=p^2+q^2+2pq$
we have $p^2+q^2=1$
$(p+q)^2=1+2pq$
$(p+q)^2\leq 1+1$
$p+q\leq \sqrt 2$
Hence (d) is the correct answer.
answered Oct 16, 2013 by sreemathi.v
 

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