Solve : $f(x)=\int \limits_0^2 \large\frac{\sqrt {2-x}}{\sqrt {2-x}+\sqrt {x}}dx$

Question

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

Answer 1

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.