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If $x$ is real,then maximum value of $y=2(a-x)(x+\sqrt{x^2+b^2})$ is

$\begin{array}{1 1}(A)\;a^2&(B)\;a^2+b^2\\(C)\;\sqrt{a^2+b^2}&(D)\;a\sqrt{a^2+b^2}\end{array} $

1 Answer

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Let $t=x+\sqrt{x^2+b^2}$
$\Rightarrow \large\frac{1}{t}=\frac{1}{x+\sqrt{x^2+b^2}}=\frac{\sqrt{x^2+b^2}-x}{b^2}$
$\Rightarrow t-\large\frac{b^2}{t}$$=2x$
$\Rightarrow 2at-t^2+b^2$
$\Rightarrow a^2+b^2-(a^2-2at+t^2)$
$\Rightarrow a^2+b^2-(a-t)^2$
Therefore $y=2(a-x)(x+\sqrt{x^2+b^2})\leq a^2+b^2$
Hence (B) is the correct answer.
answered Apr 15, 2014 by sreemathi.v