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The function $\text { f: R->R, }$ $f(x)=\frac{x^2+2 x-15}{x^2-4 x+9}, x \in \mathbb{R}$ is
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The correct answer is:
neither one-one nor onto.
$f(x)=\frac{(x+5)(x-3)}{x^2-4 x+9}$
$\begin{aligned} & \text { Let } g(x)=x^2-4 x+9 \\ & D < 0 \\ & g(x)>0 \text { for } x \in R\end{aligned}$
$\therefore\left[\begin{array}{l}
\mathrm{f}(-5)=0 \\
\mathrm{f}(3)=0
\end{array}\right.$
So, $\mathrm{f}(\mathrm{x})$ is many-one.
again,
$\begin{aligned}
& y x^2-4 x y+9 y=x^2+2 x-15 \\
& x^2(y-1)-2 x(2 y+1)+(9 y+15)=0 \\
& \text { for } \forall x \in R \Rightarrow D \geq 0 \\
& D=4(2 y+1)^2-4(y-1)(9 y+15) \geq 0 \\
& 5 y^2+2 y+16 \leq 0 \\
& (5 y-8)(y+2) \leq 0
\end{aligned}$
$\begin{aligned} & \text { Let } g(x)=x^2-4 x+9 \\ & D < 0 \\ & g(x)>0 \text { for } x \in R\end{aligned}$
$\therefore\left[\begin{array}{l}
\mathrm{f}(-5)=0 \\
\mathrm{f}(3)=0
\end{array}\right.$
So, $\mathrm{f}(\mathrm{x})$ is many-one.
again,
$\begin{aligned}
& y x^2-4 x y+9 y=x^2+2 x-15 \\
& x^2(y-1)-2 x(2 y+1)+(9 y+15)=0 \\
& \text { for } \forall x \in R \Rightarrow D \geq 0 \\
& D=4(2 y+1)^2-4(y-1)(9 y+15) \geq 0 \\
& 5 y^2+2 y+16 \leq 0 \\
& (5 y-8)(y+2) \leq 0
\end{aligned}$
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