If x is real, the maximum value of (3x2 + 9x + 17/3x2 + 9x + 7) is

Question

If x is real, the maximum value of (3x2 + 9x + 17/3x2 + 9x + 7) is (A) (1/4) (B) 41 (C) 1 (D) (17/7)

Tardigrade Admin · Accepted Answer

y = (3x2 + 9x + 17/3x2 + 9x +7) 3x2 (y -1)+9x(y-1)+7y -17 = 0 D ge 0 ∵ x is real 81(y-1)2 - 4 × 3(y - 1)(7y - 17) ge 0 ⇒ ( y -1)( y - 41) le 0 ⇒ 1 le y le 41 ∴ Max value of y is 41

Q. If x is real, the maximum value of $\frac{3 x ^{2} + 9 x + 17}{3 x ^{2} + 9 x + 7}$ is

$\frac{1}{4}$

41

1

$\frac{17}{7}$

$y = \frac{3 x ^{2} + 9 x + 17}{3 x ^{2} + 9 x + 7}$
$3 x^{2} (y - 1) + 9 x (y - 1) + 7 y - 17 = 0$
$D \geq 0 ∵ x$ is real
$81 (y - 1)^{2} - 4 \times 3 (y - 1) (7 y - 17) \geq 0$
$\Rightarrow (y - 1) (y - 41) \leq 0 \Rightarrow 1 \leq y \leq 41$
$∴$ Max value of $y$ is $41$

Q. If x is real, the maximum value of 3x2+9x+73x2+9x+17​ is

41​

41

1

717​

y=3x2+9x+73x2+9x+17​ 3x2(y−1)+9x(y−1)+7y−17=0 D≥0∵x is real 81(y−1)2−4×3(y−1)(7y−17)≥0 ⇒(y−1)(y−41)≤0⇒1≤y≤41 ∴ Max value of y is 41

Q. If x is real, the maximum value of $\frac{3 x ^{2} + 9 x + 17}{3 x ^{2} + 9 x + 7}$ is

$\frac{1}{4}$

$\frac{17}{7}$

$y = \frac{3 x ^{2} + 9 x + 17}{3 x ^{2} + 9 x + 7}$
$3 x^{2} (y - 1) + 9 x (y - 1) + 7 y - 17 = 0$
$D \geq 0 ∵ x$ is real
$81 (y - 1)^{2} - 4 \times 3 (y - 1) (7 y - 17) \geq 0$
$\Rightarrow (y - 1) (y - 41) \leq 0 \Rightarrow 1 \leq y \leq 41$
$∴$ Max value of $y$ is $41$