If x is real, then the value of x2−3x+4x2+3x+4 lies in the interval

y=x2−3x+4x2+3x+4⇒yx2+3yx+4y=x2−3x+4⇒(y−1)x2+(3y+3)x+(4y−4)=0x is real ⇒(3y+3)2−4(y−1)(4y−4)≥0⇒9y2+9+18y−16y2+32y−16≥0⇒−7y2+50y+7≥0⇒−(7y−1)(y−7)≥0⇒1/7≤y≤7

If x is real, then the value of x2−3x+4x2+3x+4 lies in the interval

y=x2−3x+4x2+3x+4⇒yx2+3yx+4y=x2−3x+4⇒(y−1)x2+(3y+3)x+(4y−4)=0x is real ⇒(3y+3)2−4(y−1)(4y−4)≥0⇒9y2+9+18y−16y2+32y−16≥0⇒−7y2+50y+7≥0⇒−(7y−1)(y−7)≥0⇒1/7≤y≤7

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If x is real, then the value of $\frac{x^{2} - 3 x + 4}{x^{2} + 3 x + 4}$ lies in the interval

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$[\frac{1}{3}, 3]$

$[\frac{1}{5}, 5]$

$[\frac{1}{6}, 6]$

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

answer is D.

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Detailed Solution

$\begin{array}{l} y = \frac{x^{2} - 3 x + 4}{x^{2} + 3 x + 4} \Rightarrow {yx}^{2} + 3 yx + 4 y = x^{2} - 3 x + 4 \\ \Rightarrow (y - 1) x^{2} + (3 y + 3) x + (4 y - 4) = 0 \\ x is real \Rightarrow (3 y + 3)^{2} - 4 (y - 1) (4 y - 4) \geq 0 \Rightarrow 9 y^{2} + 9 + 18 y - 16 y^{2} + 32 y - 16 \geq 0 \\ \Rightarrow - 7 y^{2} + 50 y + 7 \geq 0 \Rightarrow - (7 y - 1) (y - 7) \geq 0 \Rightarrow 1 / 7 \leq y \leq 7 \end{array}$