(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 $9.$