Let f(x)=x2−3x+4x2+3x+4 (x is real)

y=x2−3x+4x2+3x+4yx2+3xy+4y=x2−3x+4(y−1)x2+3x(y+1)+4(y−1)=0x∈RΔ≥09(y+1)2−4(y−1)4(y−1)≥09(y2+2y+1)−16(y2−2y+1)≥09y2+18y+9−16y2+32y−16≥0−7y2+50y−7≥07y2−50y+7≤07y2−49y−y+7≤07y(y−7)−1(y−7)≤0(y−7)(7y−1)≤017≤y≤7Minimum value =17Maximum value =7

Let f(x)=x2−3x+4x2+3x+4 (x is real)

y=x2−3x+4x2+3x+4yx2+3xy+4y=x2−3x+4(y−1)x2+3x(y+1)+4(y−1)=0x∈RΔ≥09(y+1)2−4(y−1)4(y−1)≥09(y2+2y+1)−16(y2−2y+1)≥09y2+18y+9−16y2+32y−16≥0−7y2+50y−7≥07y2−50y+7≤07y2−49y−y+7≤07y(y−7)−1(y−7)≤0(y−7)(7y−1)≤017≤y≤7Minimum value =17Maximum value =7

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Let $f (x) = \frac{x^{2} - 3 x + 4}{x^{2} + 3 x + 4}$ (x is real)

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Least value of f(x) is $\frac{1}{7}$

Least value of f(x) is 7

Maximum value of f(x) is $\frac{1}{7}$

Maximum value of f(x) is 7

answer is A, D.

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Detailed Solution

$\begin{array}{l} y = \frac{x^{2} - 3 x + 4}{x^{2} + 3 x + 4} \\ y x^{2} + 3 x y + 4 y = x^{2} - 3 x + 4 \\ (y - 1) x^{2} + 3 x (y + 1) + 4 (y - 1) = 0 \\ x \in R \\ Δ \geq 0 \\ 9 {(y + 1)}^{2} - 4 (y - 1) 4 (y - 1) \geq 0 \\ 9 (y^{2} + 2 y + 1) - 16 (y^{2} - 2 y + 1) \geq 0 \\ 9 y^{2} + 18 y + 9 - 16 y^{2} + 32 y - 16 \geq 0 \\ - 7 y^{2} + 50 y - 7 \geq 0 \\ 7 y^{2} - 50 y + 7 \leq 0 \\ 7 y^{2} - 49 y - y + 7 \leq 0 \\ 7 y (y - 7) - 1 (y - 7) \leq 0 \\ (y - 7) (7 y - 1) \leq 0 \\ \frac{1}{7} \leq y \leq 7 \\ M i n i m u m v a l u e = \frac{1}{7} \\ M a x i m u m v a l u e = 7 \end{array}$