JEE Main · 2006 · 30 Apr 2006 · Offline · Quadratic Equations and Inequalities

x

is real, the maximum value of

\frac{3 x ^{2} + 9 x + 17}{3 x ^{2} + 9 x + 7}

Answer: (B)

Solution

Set

y = \frac{3 x ^{2} + 9 x + 17}{3 x ^{2} + 9 x + 7}

and clear the denominator to get

3 x^{2} (y - 1) + 9 x (y - 1) + (7 y - 17) = 0

. For this to have a real solution

x

its discriminant must be non-negative:

81 (y - 1)^{2} - 12 (y - 1) (7 y - 17) \geq 0

. Factoring out

(y - 1)

gives

(y - 1) [81 (y - 1) - 12 (7 y - 17)] = (y - 1) (123 - 3 y) \geq 0

, i.e.

3 (y - 1) (41 - y) \geq 0

. This holds precisely for

1 \leq y \leq 41

, so the expression stays between

1

and

41

and its greatest value is

41

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option