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

A function f:RtoR where R is the set of real numbers,is defined by f(x)=(alphax^2+6x-8)/(alpha+6x-8x^2) value of alpha for which f is onto.

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) :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]

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]

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]

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]

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

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