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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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Solve : $f(x)=\int \limits_0^2 \large\frac{\sqrt {2-x}}{\sqrt {2-x}+\sqrt {x}}dx$

\[\begin {array} {1 1} (a)\;1 \\ (b)\;\frac{1}{2} \\ (c)\;2 \\ (d)\;\frac{3}{2} \end {array}\]
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1 Answer

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By using property $f(x)= f(a-x)$
$f(2-x) =\int \limits_0^2 \large\frac{\sqrt {2-2+x}}{\sqrt {2-2+x}+\sqrt {2-x}}$$dx$---(1)
$\qquad =\int \limits_0^2 \large\frac{\sqrt x}{\sqrt x+\sqrt {2-x}}$$dx$----(2)
adding (1) and (2)
$2f(x)=\int \limits_0^2 dx$
$f(x)=\bigg[\large\frac{x}{2}\bigg]_0^2$
$f(x) => 1$
Hence a is the correct answer.
answered Dec 14, 2013 by meena.p
 
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