Question

Let $f: D \rightarrow R$ be defined as $f(x)=\frac{x^2+2 x+a}{x^2+4 x+3 a}$ 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. $0 \leq$ a $\leq 1$
• B. $0< $ a $\leq 1$
• C. $0 \leq$ a $< 1$
• D. $0< a< 1$

Answer 1

Answer: (D)

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