Functions Ques 10
- A function $f: I R \rightarrow I R$, where $I R$, is the set of real numbers, is defined by $f(x)=\frac{\alpha x^2+6 x-8}{\alpha+6 x-8 x^2}$.
Find the interval of values of $\alpha$ for which is onto. Is the functions one-to-one for $\alpha=3$ ? Justify your answer.
$(1996,5 M)$
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Answer:
Correct Answer: 10.$(2 \geq \alpha \geq 14 $ , No)
Solution: Let $y=\frac{\alpha x^2+6 x-8}{\alpha+6 x-8 x^2}$
$ \begin{array}{lc} \Rightarrow & \alpha y+6 x y-8 x^2 y=\alpha x^2+6 x-8 \\ \Rightarrow & -\alpha x^2-8 x^2 y+6 x y-6 x+\alpha y+8=0 \\ \Rightarrow & \alpha x^2+8 x^2 y-6 x y+6 x-\alpha y-8=0 \\ \Rightarrow & x^2(\alpha+8 y)+6 x(1-y)-(8+\alpha y)=0 \end{array} $
Since, $x$ is real.
$ \begin{aligned} & \Rightarrow \quad B^2-4 A C \geq 0 \\ & \Rightarrow \quad 36(1-y)^2+4(\alpha+8 y)(8+\alpha y) \geq 0 \\ & \Rightarrow \quad 9\left(1-2 y+y^2\right)+\left[8 \alpha+\left(64+\alpha^2\right) y+8 \alpha y^2\right] \geq 0 \\ & \Rightarrow \quad y^2(9+8 \alpha)+y\left(46+\alpha^2\right)+9+8 \alpha \geq 0 \\ & \Rightarrow \quad A>0, D \leq 0, \Rightarrow 9+8 \alpha>0 \\ & \left(46+\alpha^2\right)^2-4(9+8 \alpha)^2 \leq 0 \\ \end{aligned} $
$\Rightarrow \quad$ $ \alpha>-9 / 8 $
and $\left[46+\alpha^2-2(9+8 \alpha)\right]\left[46+\alpha^2+2(9+8 \alpha)\right] \leq 0$
$ \begin{aligned} & \Rightarrow \quad \alpha>-9 / 8 \\ & \text { and }\left(\alpha^2-16 \alpha+28\right)\left(\alpha^2+16 \alpha+64\right) \leq 0 \\ & \Rightarrow \quad \alpha>-9 / 8 \\ \end{aligned} $
$ \Rightarrow \quad \alpha>-9 / 8 $
and $[(\alpha-2)(\alpha-14)](\alpha + 8)^2 \leq 0$
$ \Rightarrow \quad \alpha>-9 / 8 $
and $(\alpha-2)(\alpha-14) \leq 0 \quad\left[\because(\alpha+8)^2 \geq 0\right]$
$ \Rightarrow \quad \alpha>-9 / 8 $
and $2 \leq \alpha \leq 14$
$\Rightarrow \quad 2 \leq \alpha \leq 14$
Thus, $f(x)=\frac{\alpha x^2+6 x-8}{\alpha+6 x-8 x^2}$ will be onto, if $2 \leq \alpha \leq 14$
Again, when $\alpha=3$
$ \begin{aligned} & f(x)=\frac{3 x^2+6 x-8}{3+6 x-8 x^2} \text {, in this case } f(x)=0 \\ & \Rightarrow \quad 3 x^2+6 x-8=0 \\ & \Rightarrow \quad x=\frac{-6 \pm \sqrt{36+96}}{6}=\frac{-6 \pm \sqrt{132}}{6}=\frac{1}{3}(-3 \pm \sqrt{33}) \\ \end{aligned} $
This shows that
$ f\left[\frac{1}{3}(-3+\sqrt{33)}]=f\left[\frac{1}{3}(-3-\sqrt{33)}]=0\right.\right. $
Therefore, $f$ is not one-to-one.
BETA
