MCQ
The function $f(x)=\frac{x^2+2 x-15}{x^2-4 x+9}, x \in R$ is
- Aboth one $-$ one and onto.
- Bonto but not one $-$ one.
- ✓neither one $-$ one nor onto.
- Done $-$ one but not onto.
✓
Answer
Correct option: C.
neither one $-$ one nor onto.
$f(x)=\frac{(x+5)(x-3)}{x^2-4 x+9}$
Let $g(x)=x^2-4 x+9$
$ D < 0 $
$ g(x) > 0$ for $x \in R$
$\therefore\left[\begin{array}{l}\mathrm{f}(-5)=0 \\ \mathrm{f}(3)=0\end{array}\right.$
So, $\mathrm{f}(\mathrm{x})$ is many $-$ one.
again,
$ 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$
Image
$\mathrm{y} \in\left[-2, \frac{8}{5}\right]$ range
Note : If function is defined from $f: R \rightarrow R$ then only correct answer is option ($3$)
$\Rightarrow$ Bonus
Let $g(x)=x^2-4 x+9$
$ D < 0 $
$ g(x) > 0$ for $x \in R$
$\therefore\left[\begin{array}{l}\mathrm{f}(-5)=0 \\ \mathrm{f}(3)=0\end{array}\right.$
So, $\mathrm{f}(\mathrm{x})$ is many $-$ one.
again,
$ 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$
Image
$\mathrm{y} \in\left[-2, \frac{8}{5}\right]$ range
Note : If function is defined from $f: R \rightarrow R$ then only correct answer is option ($3$)
$\Rightarrow$ Bonus
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