Let `f:R -> ( 0,(2pi)/2]` defined as `f(x) = cot^-1 (x^2-4x + alpha)` Then the smallest integral value of `alpha` such that, `f(x)` is into function is

Let f:R -> ( 0,(2pi)/3] defined as f(x) = cot^-1 (x^2-4x + alpha) Then the smallest integral value of alpha such that, f(x) is into function is

Let f:R -> ( 0,(2pi)/3] defined as f(x) = cot^-1 (x^2-4x + alpha) Then the smallest integral value of alpha such that, f(x) is into function is

Let f:R rarr R be defined as f(x)={2kx+3,x =0 If f(x) is injective then find the smallest integral value of

Let f:R rarr(0,(pi)/(2)) be a function defined by f(x)=cot^(-1)(x^(2)+4x+alpha^(2)-alpha), complete set of valuesof alpha for which f(x) is onto,is (A) [(1-sqrt(17))/(2),(1+sqrt(17))/(2)] (B) (-oo,(1-sqrt(17))/(2)]uu[(1+sqrt(17))/(2),oo)(C)((1-sqrt(17))/(2),(1+sqrt(17))/(2))( D) none of these

Let f : R to [0, pi/2) be defined by f ( x) = tan^(-1) ( x^(2) + x + a. Then set of value of a for which f(x) is onto is

A function f : R rarr R, where R is the set of real numbers, is defined by : f(x)= (alpha x^2+6x-8)/(alpha+6x-8x^2) . Find the interval of values of alpha for which f is onto. Is the function one-one for alpha = 3 ? Justify your answer.

A function f : R rarr R is defined as f(x) = (alpha x^2 + 6x - 8 )/(alpha + 6x - 8x^2) . Find the set of values of alpha for which is onto.

Let f(x)=(alpha x)/(x+1) Then the value of alpha for which f(f(x)=x is

Let f:R->R be a function defined by f(x)=x^2-(x^2)/(1+x^2) . Then:

Let f:R to R be defined as f (x) =( 9x ^(2) -x +4)/( x ^(2) _ x+4). Then the range of the function f (x) is