Question

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

The correct option is A $0\le a\le 1$

We have, $y=\frac{x^2+2x+a}{x^2+4x+3a}$

$\Rightarrow {yx}^2+4xy+3ay=x^2+2x+a$

$\Rightarrow {yx}^2+4xy+3ay-x^2-2x-a=0$

$\Rightarrow x^2(y-1)+2x(2y-1)+3ay-a=0$

Give $x$ is real

$\therefore$ Discriminant $\ge 0$

$\Rightarrow {\left\{2(2y-1)\right\}}^2-4(3ay-a)(y-1)\ge 0$

$\Rightarrow 4{(2y-1)}^2-4(3ay-a)(y-1)\ge 0$

$\Rightarrow {(2y-1)}^2-(3ay-a)(y-1)\ge 0$ [Divide by $4$]

$\Rightarrow {4y}^2+1-4y-({3ay}^2-3ay-ay+a)\ge 0$

$\Rightarrow {4y}^2+1-4y-({3ay}^2-4ay+a)\ge 0$

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

For the above equation to be always true,

$(4-3a)>0$ and Discriminant $\le 0$

$\Rightarrow (4-3a)>0$

$\Rightarrow a<\frac{4}{3}$

For, Discriminant $\le 0$

$\Rightarrow {\left\{4(a-1)\right\}}^2-4(4-3a)(1-a)\le 0$

$\Rightarrow 4\times 4{(a-1)}^2-4(4-3a)(1-a)\le 0$

$\Rightarrow 4(a^2+1-2a)-(4-3a)(1-a)\le 0$

$\Rightarrow {4a}^2+4-8a-(4-7a+{3a}^2)\le 0$

$\Rightarrow {4a}^2+4-8a-4+7a-{3a}^2\le 0$

$\Rightarrow a^2-a\le 0$

$\Rightarrow a(a-1)\le 0$

$aϵ[0,1]$

Taking intersection of $a<\frac{4}{3}$ and $aϵ[0,1]$, we get $aϵ[0,1]$