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If $x$ is real, then the value of $\frac{{{x^2} + 34x - 71}}{{{x^2} + 2x - 7}}$ does not lie between
$-9$ and $-5$
$-5$ and $9$
$0$ and $9$
$5$ and $9$
Solution
(d) Let $y = \frac{{{x^2} + 34x – 71}}{{{x^2} + 2x – 7}}$
==> ${x^2}(y – 1) + 2(y – 17)x + (71 – 7y) = 0$
For real values of $x$, its discriminant $D \ge 0$
$ \Rightarrow 4{(y – 17)^2} – 4(y – 1)(71 – 7y) \ge 0$
$ \Rightarrow ({y^2} – 3 + y + 289) – (71y – 7{y^2} – 71 + 7y) \ge 0$
$ \Rightarrow {y^2} – 14y + 45 \ge 0 \Rightarrow (y – 5)(y – 9) \ge 0$
It is possible when both $y – 5$ and$y – 9$ are negative or both positive. Let $y – 5 \le 0 \Rightarrow y \le 5$ and $y – 9 \le 0 \Rightarrow y \le 9$.
Hence $y \le 5$…..(i)
If $y – 5 \ge 0 \Rightarrow $$y \ge 5$and $y – 9 \ge 0 \Rightarrow y \ge 9$
Hence $y \ge 9$……(ii)
Therefore $y$ does not lie between $5$ and $9.$
