Question:

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

Updated On: Jul 27, 2022
  • $\frac{1}{4}$
  • 41
  • 1
  • $\frac{17}{7}$
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The Correct Option is B

Solution and Explanation

$y = \frac{3x^{2} + 9x + 17}{3x^{2} + 9x +7}$ $3x^{2} \left(y -1\right)+9x\left(y-1\right)+7y -17 = 0$ $D \ge 0 \,\,\,\because\, x$ is real $81\left(y-1\right)^{2} - 4 \times 3\left(y - 1\right)\left(7y - 17\right) \ge 0$ $\Rightarrow \left( y -1\right)\left( y - 41\right) \le 0 \Rightarrow 1\le y \le 41$ $\therefore$ Max value of $y$ is $41$
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Concepts Used:

Complex Numbers and Quadratic Equations

Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.

Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.