(A) Given $y = \frac{x^2 + 14x + 9}{x^2 + 2x + 3}$.$y(x^2 + 2x + 3) = x^2 + 14x + 9$.$(y - 1)x^2 + 2(y - 7)x + 3y - 9 = 0$.Since $x \in R$,the discrim

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(A) Given $y = \frac{x^2 + 14x + 9}{x^2 + 2x + 3}$.$y(x^2 + 2x + 3) = x^2 + 14x + 9$.$(y - 1)x^2 + 2(y - 7)x + 3y - 9 = 0$.Since $x \in R$,the discriminant $D \geq 0$.$4(y - 7)^2 - 4(y - 1)(3y - 9) \geq 0$.$(y^2 - 14y + 49) - (3y^2 - 12y + 9) \geq 0$.$-2y^2 - 2y + 40 \geq 0$.$y^2 + y - 20 \leq 0$.$(y + 5)(y - 4) \leq 0$.Thus,$y \in [-5, 4]$.

Class 11 Mathematics 4-2.Quadratic Equations and Inequations Mix Examples-Quadratic Equations and Inequations

Easy

If $y = \frac{x^2 + 14x + 9}{x^2 + 2x + 3}$ for all $x \in R$,then the interval of maximum length in which $y$ lies is

TS EAMCET 2021 All TS EAMCET

A
$[-5, 4]$
B
$[-4, 5]$
C
$[\frac{1}{3}, 3]$
D
$[-\frac{1}{3}, 3]$

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If $y = \frac{x^2 + 14x + 9}{x^2 + 2x + 3}$ for all $x \in R$,then the interval of maximum length in which $y$ lies is

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