Question

If $x$ is real, then $\frac{x^2+2x+c}{x^2+4x+3c}$ can take all real values if

• A. $0<c<2$
• B. $0<c<1$
• C. $-1<c<1$
• D. None of the above

Answer 1

The correct option is B $0<c<1$

Let us assume that

$\frac{x^2+2x+c}{x^2+4x+3c}=y$

$\Rightarrow y{x}^2+4xy+3yc=x^2+2x+c$

$\Rightarrow x^2(y-1)+2x(2y-1)+3yc-c=0$

For $x$ to be real, Discriminant of this equation should be $\ge 0$

$\therefore 4(2y-1{)}^2-4\times (y-1)\times (3yc-c)\ge 0$

$4(4y^2-4y+1)-4(3c{y}^2-4cy+c)\ge 0$

Dividing both sides by $4$, we get

$(4-3c)y^2+4y(c-1)+1-c\ge 0$

For this equation to hold true, coefficient of $y^2$ should be greater than $0$ and $D<0$

$\Rightarrow 4-3c>0$ i.e. $c<\frac{4}{3}.....\left(1\right)$

Also,

$16(c^2-2c+1)-4\times (1-c)\times (4-3c)<0$

$\Rightarrow 16c^2-32c+16-4(4-7a+3a^2)<0$

$\Rightarrow 16c^2-32c+16-12c^2+28c-16<0$

$\Rightarrow 4c(c-1)<0$

$\therefore 0<c<1.....\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, we can say that $0<c<1$ is the required condition.