FIND THE RANGE

Dear student,

the solution of your quesry is as follows -

Since (y-1) is there in the denominator, therefore x will tend to infinity when y=1. So, x will not be defined at y=1. => y ≠ 1.

Also, for x to be real, (9y²-11-6y) must be greater than or equal to 0.

therefore,

We note that (1+12^1/2)/3 > 1,

therefore the range of y is .