If x is real, the maximum value of `(3x^(2)+9x+17)/(3x^(2)+9x+7)` is
(a) `(17)/(7)`
(b) `(1)/(4)`
(c) 41
(d) 1

Options

A. `(1)/(4)`
B. 41
C. 1
D. `(17)/(7)`

Answer

B

Explanation

Let `y=(3x^(2)+9x+17)/(3x^(2)+9x+7)`. Then,
`3x^(2)(y-1) + 9x(y-1)+7y - 17 = 0`
Since x is ral. Therefore, Disc `ge` 0 i.e.
`81 (y-1)^(2) -12(y-1)(7y-17= 0`
`rArr" "-3(y-1)(y-41) ge 0 rarr (y-1)(y-41) le 0 rArr 1 le y le 41`
Hence, the maximum value of y is 41.