Question

If x is real, then the maximum and minimum values of expression $\frac{x^2+14x+9}{x^2+2x+3}$will be

• A. - 4, 5
• B. - 4, - 5
• C. 5, - 4
• D. 4, - 5

Answer 1

The correct option is D

4, - 5

Let $y=\frac{x^2+14x+9}{x^2+2x+3}$

⇒ $y(x^2+2x+3)-x^2-14x-9=0$

⇒ $(y-1)x^2+(2y-14)x+3y-9=0$

For real x, its discriminant ≥ 0

i.e. $4(y-7{)}^2-4(y-1\left)3\right(y-3)\ge 0$

⇒ $y^2+y-20\le 0or(y-4)(y+5)\le 0$

Now, the product of two factors is negative if these are of opposite signs. So following two cases arise:

Case 1: $y-4\ge 0ory\ge 4y+5\le 0ory\le -5$

This is not possible.

Case 2: $y-4\le 0ory\le 4y+5\ge 0ory\ge -5$ Both of these are satisfied if -5≤y≤4

hence maximum value of y is 4 and minimum value is -5