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If $x$ is real, then the maximum and minimum values of expression $\frac{{{x^2} + 14x + 9}}{{{x^2} + 2x + 3}}$ will be
$4, -5$
$5, -4$
$-4, 5$
$-4, -5$
Solution
(a) Let $y = \frac{{{x^2} + 14x + 9}}{{{x^2} + 2x + 3}}$
==> $y({x^2} + 2x + 3) – {x^2} – 14x – 9 = 0$
$ \Rightarrow $ $(y – 1){x^2} + (2y – 14)x + 3y – 9 = 0$
For real $x$, its discriminant $ \ge 0$
i.e. $4{(y – 7)^2} – 4(y – 1)3(y – 3) \ge 0$
==> ${y^2} + y – 20 \le 0$ or $(y – 4)(y + 5) \le 0$
Now, the product of two factors is negative if these are of opposite signs. So following two cases arise:
Case I : $y – 4 \ge 0$or $y \ge 4$ and $y + 5 \le 0$or $y \le – 5$
This is not possible.
Case II : $y – 4 \le 0$ or $y \le 4$and $y + 5 \ge 0$or $y \ge – 5$ Both of these are satisfied if $ – 5 \le y \le 4$
Hence maximum value of $y$ is $4$ and minimum value is $-5.$
