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If $x$ is real, then the maximum and minimum values of the expression $\frac{{{x^2} - 3x + 4}}{{{x^2} + 3x + 4}}$ will be
$2, 1$
$5,\frac{1}{5}$
$7,\frac{1}{7}$
None of these
Solution
(c) Let $y = \frac{{{x^2} – 3x + 4}}{{{x^2} + 3x + 4}}$
==> $(y – 1){x^2} + 3(y + 1)x + 4(y – 1) = 0$
For $x$ is real $D \ge 0$
==> $9{(y + 1)^2} – 16{(y – 1)^2} \ge 0$
==> $ – 7{y^2} + 50y – 7 \ge 0$==>$7{y^2} – 50y + 7 \le 0$
==> $(y – 7)(7y – 1) \le 0$
Now, the product of two factors is negative if one in $ – ve$and one in $ + ve$.
Case I : $(y – 7) \ge 0$ and $(7y – 1) \le 0$
==> $y \ge 7$and $y \le \frac{1}{7}$. But it is impossible
Case II : $(y – 7) \le 0$and $(7y – 1) \ge 0$
==> $y \le 7$and $y \ge \frac{1}{7} \Rightarrow \frac{1}{7} \le y \le 7$
Hence maximum value is $7$ and minimum value is $\frac{1}{7}$
