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

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Question

(a) Let $y = \frac{{{x^2} + 14x + 9}}{{{x^2} + 2x + 3}}$

==> $y({x^2} + 2x + 3) - {x^2} - 14x - 9 = 0$

$ \Rightarrow $ $(y - 1){x^2} + (2y -

Vedclass · Accepted Answer

$4, -5$

Vedclass · Answer

(a) Let $y = \frac{{{x^2} + 14x + 9}}{{{x^2} + 2x + 3}}$

==> $y({x^2} + 2x + 3) - {x^2} - 14x - 9 = 0$

$ \Rightarrow $ $(y - 1){x^2} + (2y - 14)x + 3y - 9 = 0$

For real $x$, its discriminant $ \ge 0$

i.e. $4{(y - 7)^2} - 4(y - 1)3(y - 3) \ge 0$

==> ${y^2} + y - 20 \le 0$ or $(y - 4)(y + 5) \le 0$

Now, the product of two factors is negative if these are of opposite signs. So following two cases arise:

Case I : $y - 4 \ge 0$or $y \ge 4$ and $y + 5 \le 0$or $y \le - 5$
This is not possible.

Case II : $y - 4 \le 0$ or $y \le 4$and $y + 5 \ge 0$or $y \ge - 5$ Both of these are satisfied if $ - 5 \le y \le 4$

Hence maximum value of $y$ is $4$ and minimum value is $-5.$

If $x$ is real, then the maximum and minimum values of expression $\frac{{{x^2} + 14x + 9}}{{{x^2} + 2x + 3}}$ will be

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