If `x` is real, then `x//(x^2-5x+9)` lies between `-1a n d-1//11` b. `1a n d-1//11` c. `1a n d1//11` d. none of these

Options

A. `-1 and -1//11`
B. `1 and -1//11`
C. `1 and 1//11`
D. none of these

Answer

2

Explanation

Let `(x)/(x^(2) - 5x + 9) = y`
or `yx^(2) - 5yx + 9y = x`
or `yx^(2) - (5y + 1)x + 9y = 0`
Now, x is real, so
`D ge 0`
`rArr (-(5y + 1))^(2) - 4 * y *(9y) ge 0`
or `-11y^(2) + 10y + 1 ge 0`
or `11y^(2) - 10y - 1 le 0`
or `(11y + 1) (y - 1) le 0`
or `-1/11 le y le 1`