JEE MAIN - Mathematics (2007 - No. 14)
If $$p$$ and $$q$$ are positive real numbers such that $${p^2} + {q^2} = 1$$, then the maximum value of $$(p+q)$$ is
$${1 \over 2}$$
$${1 \over {\sqrt 2 }}$$
$${\sqrt 2 }$$
$$2$$
Explanation
Given that $${p^2} + {q^2} = 1$$
$$\therefore$$ $$p = \cos \theta $$ and $$q = \sin \theta $$
Then $$p+q$$ $$ = \cos \theta + \sin \theta $$
We know that
$$ - \sqrt {{a^2} + {b^2}} \le a\cos \theta + b\sin \theta \le \sqrt {{a^2} + {b^2}} $$
$$\therefore$$ $$ - \sqrt 2 \le \cos \theta + \sin \theta \le \sqrt 2 $$
Hence max. value of $$p + q$$ is $$\sqrt 2 $$
$$\therefore$$ $$p = \cos \theta $$ and $$q = \sin \theta $$
Then $$p+q$$ $$ = \cos \theta + \sin \theta $$
We know that
$$ - \sqrt {{a^2} + {b^2}} \le a\cos \theta + b\sin \theta \le \sqrt {{a^2} + {b^2}} $$
$$\therefore$$ $$ - \sqrt 2 \le \cos \theta + \sin \theta \le \sqrt 2 $$
Hence max. value of $$p + q$$ is $$\sqrt 2 $$
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