Question

Let $f:D\to R$ be defined as $f(x)=\frac{x^2+2x+a}{x^2+4x+3a}$ where $D$ and $R$ denote the domain of $f$ and the set of all real numbers respectively. If $f$ is surjective mapping then the range of $a$ is

• A. $0\le$ a $\le 1$
• B. $0<$ a $\le 1$
• C. $0\le$ a $<1$
• D. $0<a<1$

Answer 1

Answer: (D)

Let $y=\frac{x^2+2x+a}{x^2+4x+3a}$....(i)
$\Rightarrow (y-1)x^2+(4y-2)x+(3ay-a)=0$....(ii)
If range of eq. (i) is $R$, then roots of eq. (ii) must be real for all $y\in R$.
$\Rightarrow (4y-2{)}^2-4(y-1)(3ay-a)\ge 0\forall y\in R$
$\Rightarrow (4-3a)y^2-4(1-a)y+(1-a)\ge 0\forall y\in R$.....(iii)
Now, eq. (iii) is true $\forall y\in R$, if $4-3a>0\Rightarrow a<\frac{4}{3}$
[If $x^2+2x+a=0$ and $x^2+4x+3a=0$ have a common root then $a=0,1$. So range of $f$ is not equal to $R$].
$\text{ and }16(1-a{)}^2-4(4-3a)(1-a)\le 0$
$\Rightarrow a(a-1)\le 0$
Hence, we get $0<a<1$.