The function `f(x) = x^2+ lambdax^2 + 5x + sin 2x` will be bijective function if `lambda` belongs to

The function f(x)=x^(2)+(lambda)/(x) has a

The function f(x) = (x^(2)-4)/(sin x-2) is

The function f(x) = (x^(2)-4)/(sin x-2) is

For the function f (x) = (x + 3)/( x - 5) to be a bijective function, the domain and range, respectively of f (x) must be :

Show that the function f(x) = {:{(x+lambda, x < 1),(lambdax^2 + 1, x ge1):} is a continous function. Regardless of the choice of lambda in R .

The period of the function f(x) = sin 2x is-

The period of the function f(x) = lambda| sin x| + lambda^(2) | cos x| + g (lambda) is (lambdai)/(2) , if lambda is

Let f:ArarrB is a function defined by f(x)=(2x)/(1+x^(2)) . If the function f(x) is a bijective function, than the correct statement can be