If `f: R->R ,f(x)=(alphax^2+6x-8)/(alpha+6x-8x^2)` is onto then `alpha in`

If f:R rarr R,f(x)=(alpha x^(2)+6x-8)/(alpha+6x-8x^(2)) is onto then alpha in

Let f(x) :R->R,f(x)={(2x+alpha^2,xge2),((alpha x)/2+10,x lt 2)) If f(x) is onto function then alpha belongs to (A) [1,4] (B) [-2,3] (C) [0,3] (D) [2,5]

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

Define a function f:R-{9/4}rarr R by f(x)=(3+alpha-alpha^(2)x)/(9-4x) The number of values of alpha in R such that f(f(x))=x

If f:Rto(1,10) , where f(x)=(x^(2)+1-alpha)/(x^(2)+2) is an onto function, then the value alpha is

a in R and (alpha-2)x^(2)+8x+alpha+4<0,AA x in R, then alpha cannot be

consider the function f:R rarr R,f(x)=(x^(2)-6x+4)/(x^(2)+2x+4) which of the following is not true about f(x)?

Let f:R rarr R be defined by f(x)=2+alpha x,alpha!=0 , If f(x)=f^(-1)(x) , for all x in R , then alpha equals

Define a function f : R-{9/4}rarr R by f(x)=(3+alpha-1)/(9-4x) The number of values of alpha in R such that f(f(x))=x is