For x∈R, the expression x2+2x+cx2+4x+4c can take all real values if c∈

Let y=x2+2x+cx2+4x+4c⇒(x2+4x+3c)y=x2+2x+c⇒(y−1)x2+2(2y−1)x+(4cy−c)=0Δ≥0⇒(2(2y−1))2−4(y−1)(4cy−c)≥0⇒4(4y2+1−4y)−16cy2+4cy+16cy−4c≥0⇒16y2+4−16y−16cy2+16cy−4c≥0⇒16y2(1−c)+16(c−1)y+4(1−c)≥0⇒(c−1)(−16y2−16y+4)≥0∴(c−1)<0⇒c<1 (∵(−16y2−16y+4)<0)c∈(0,1)

For x∈R, the expression x2+2x+cx2+4x+4c can take all real values if c∈

Let y=x2+2x+cx2+4x+4c⇒(x2+4x+3c)y=x2+2x+c⇒(y−1)x2+2(2y−1)x+(4cy−c)=0Δ≥0⇒(2(2y−1))2−4(y−1)(4cy−c)≥0⇒4(4y2+1−4y)−16cy2+4cy+16cy−4c≥0⇒16y2+4−16y−16cy2+16cy−4c≥0⇒16y2(1−c)+16(c−1)y+4(1−c)≥0⇒(c−1)(−16y2−16y+4)≥0∴(c−1)<0⇒c<1 (∵(−16y2−16y+4)<0)c∈(0,1)

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For $x \in R$ , the expression $\frac{x^{2} + 2 x + c}{x^{2} + 4 x + 4 c}$ can take all real values if $c \in$

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$(- 1, 0)$

$[0, 1]$

$(1, 2)$

$(0, 1)$

answer is C.

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Detailed Solution

Let $y = \frac{x^{2} + 2 x + c}{x^{2} + 4 x + 4 c}$

$\begin{array}{l} \Rightarrow (x^{2} + 4 x + 3 c) y = x^{2} + 2 x + c \\ \Rightarrow (y - 1) x^{2} + 2 (2 y - 1) x + (4 c y - c) = 0 \\ Δ \geq 0 \end{array}$

$\begin{array}{l} \Rightarrow {(2 (2 y - 1))}^{2} - 4 (y - 1) (4 c y - c) \geq 0 \\ \Rightarrow 4 (4 y^{2} + 1 - 4 y) - 16 c y^{2} + 4 c y + 16 c y - 4 c \geq 0 \\ \Rightarrow 16 y^{2} + 4 - 16 y - 16 c y^{2} + 16 c y - 4 c \geq 0 \\ \Rightarrow 16 y^{2} (1 - c) + 16 (c - 1) y + 4 (1 - c) \geq 0 \\ \Rightarrow (c - 1) (- 16 y^{2} - 16 y + 4) \geq 0 \end{array}$