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Q.If x∈Rx \in R, then determine the range of the expression x+22x2+3x+6\dfrac{x+2}{2x^2 + 3x + 6}.

Telangana TsbieTelangana Board of Intermediate Education 2019Subjective· 4mImportance★★★★★
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Set yy equal to the expression, clear denominators to get a quadratic in xx, and require its discriminant to be ≥0\ge 0 (since xx must be real).

Let y=x+22x2+3x+6y = \dfrac{x+2}{2x^2+3x+6} (note 2x2+3x+62x^2+3x+6 has discriminant 9−48<09-48<0, so it is never zero — the domain is all reals).

Cross-multiplying: y(2x2+3x+6)=x+2y(2x^2+3x+6) = x+2

⇒2y x2+(3y−1)x+(6y−2)=0\Rightarrow 2y\,x^2 + (3y-1)x + (6y-2) = 0.

For xx to be real, the discriminant of this quadratic in xx must be ≥0\ge 0:

(3y−1)2−4(2y)(6y−2)≥0(3y-1)^2 - 4(2y)(6y-2) \ge 0

9y2−6y+1−48y2+16y≥09y^2 - 6y + 1 - 48y^2 + 16y \ge 0

−39y2+10y+1≥0  ⇒  39y2−10y−1≤0-39y^2 + 10y + 1 \ge 0 \;\Rightarrow\; 39y^2 - 10y - 1 \le 0.

Solving 39y2−10y−1=039y^2-10y-1=0: y=10±100+15678=10±1678y = \dfrac{10\pm\sqrt{100+156}}{78} = \dfrac{10\pm16}{78}, giving y=13y=\dfrac{1}{3} or y=−113y=-\dfrac{1}{13}. …

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