If x is real, then the value of frac{{{x^2} + | vedclass

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(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 discrim

Vedclass · Accepted Answer

$5$ and $9$

Vedclass · Answer

(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 $9.$

If $x$ is real, then the value of $\frac{{{x^2} + 34x - 71}}{{{x^2} + 2x - 7}}$ does not lie between

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