If x is real, then the maximum and minimum va | vedclass

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(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{

Vedclass · Accepted Answer

$7,\frac{1}{7}$

Vedclass · Answer

(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 $\frac{1}{7}$

If $x$ is real, then the maximum and minimum values of the expression $\frac{{{x^2} - 3x + 4}}{{{x^2} + 3x + 4}}$ will be

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