Let $R$ be the set of all real numbers and let $f$ be a function from $R$ to $R$ such that $f(x)+\left(x+\frac{1}{2}\right) f(1-x)=1$, for all $x \in R$. Then $2 f(0)+3 f(1)$ is equal to

Which of the following is correct

Let $A = \left\{ {{x_1},{x_2},{x_3},.....,{x_7}} \right\}$ and $B = \left\{ {{y_1},{y_2},{y_3}} \right\}$ be two sets containing seven and three distinct elements respectively. Then the total number of functions $f:A \to B$ which are onto, if there exist exactly three elements $x$ in $A$ such that $f(x) = {y_2}$ , is equal to

Let $f(x) = {\cos ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right) + {\sin ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)$ then the value of $f(1) + f(2)$, is -

Which of the following is function

The domain of the function

$f(x)=\frac{\cos ^{-1}\left(\frac{x^{2}-5 x+6}{x^{2}-9}\right)}{\log _{e}\left(x^{2}-3 x+2\right)} \text { is }$