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Q.
If the function $f: R \rightarrow R$ defined by $f(x)=\frac{a x^{2}+6 x-8}{a+6 x-8 x^{2}}$ is onto for $a \in[m, n]$, then $\sqrt{m n-3}$ is equal to.
Relations and Functions - Part 2
Solution:
Let $y=\frac{a x^{2}+6 x-8}{a+6 x-8 x^{2}}$.
We want $y$ to be real for every real $x$
$\Rightarrow (8 y+a) x^{2}-6 y x+6 x-8-a y=0$ for $x, y \in R \ldots(1)$
If $8 y+a=0$, i.e., $y=-\frac{a}{8}$
$\Rightarrow -6\left(\frac{-a}{8}\right) x+6 x-8-a\left(\frac{-a}{8}\right)=0$
$\Rightarrow \frac{3 a}{4} x+6 x-8+\frac{a^{2}}{8}=0$
$ \Rightarrow x\left(\frac{3 a}{4}+6\right)=8-\frac{a^{2}}{8}$
$\Rightarrow x=\frac{64-a^{2}}{8} \times \frac{4}{3 a+24}$
$=\frac{64-a^{2}}{6 a+12} \in R$ for $a \neq-2$
$\therefore $ If $a \neq-2$ and $y=-\frac{a}{8}$, then $x \in R$.
$\Rightarrow $ For $a=-2$ and $y=\frac{1}{4}$ ;
$ x \in R \Rightarrow a \neq-2 \dots$(2)
If $(8 y+a) \neq 0$, then for real roots ' $x'$ of $(1)$; Disc.
$\geq 0 \forall y \in R$
$\Rightarrow (6-6 y)^{2}+4(8 y+a)(8+a y) \geq 0 \forall y \in R$
$\Rightarrow (9+8 a) y^{2}+\left(a^{2}+46\right) y+(9+8 a) > 0 \forall y \in R .$
If $9+8 a=0$, then $y \geq 0 \forall y \in R$,
which is false, so, $a \neq-\frac{9}{8} ....(3)$
Now for $9 a+8 a \neq 0 $
$\Rightarrow \left(a^{2}+46\right)^{2}-4(9+8 a)(9+8 a) \leq 0$
$\Rightarrow \left(a^{2}+46\right)^{2}-[2(9+8 a)]^{2} \leq 0$
$\Rightarrow -2(9+8 a) \leq\left(a^{2}+46\right) \leq 2(9+8 a)$
$\Rightarrow a^{2}+16 a+64 \leq 0$ and $a^{2}-16 a+28 \leq 0$
$\Rightarrow (a+8)^{2} \leq 0$ and $(a-2)(a-14) \leq 0$
$\Rightarrow a \in[2,14] .$
$\therefore $ From(ii), (iii) and (iv), we have
$a \in[2,14]=[m, n]$ (given)
$\Rightarrow m=2, n=14$
$ \Rightarrow \sqrt{m n-3}=5$