Question

If the function $f:R\to R$ defined by $f(x)=\frac{a{x}^2+6x-8}{a+6x-8x^2}$ is onto for $a\in [m,n]$, then $\sqrt{mn-3}$ is equal to.

• A. 4
• B. 19
• C. 5
• D. None of these

Answer 1

Answer: (C)

Let $y=\frac{a{x}^2+6x-8}{a+6x-8x^2}$.
We want $y$ to be real for every real $x$
$\Rightarrow (8y+a)x^2-6yx+6x-8-ay=0$ for $x,y\in R\dots (1)$
If $8y+a=0$, i.e., $y=-\frac{a}{8}$
$\Rightarrow -6\left(\frac{-a}{8}\right)x+6x-8-a\left(\frac{-a}{8}\right)=0$
$\Rightarrow \frac{3a}{4}x+6x-8+\frac{a^2}{8}=0$
$\Rightarrow x\left(\frac{3a}{4}+6\right)=8-\frac{a^2}{8}$
$\Rightarrow x=\frac{64-a^2}{8}\times \frac{4}{3a+24}$
$=\frac{64-a^2}{6a+12}\in R$ for $a\ne -2$
$\therefore$ If $a\ne -2$ and $y=-\frac{a}{8}$, then $x\in R$.
$\Rightarrow$ For $a=-2$ and $y=\frac{1}{4}$ ;
$x\in R\Rightarrow a\ne -2\dots$(2)
If $(8y+a)\ne 0$, then for real roots ' $x^{\prime }$ of $(1)$; Disc.
$\ge 0\forall y\in R$
$\Rightarrow (6-6y{)}^2+4(8y+a)(8+ay)\ge 0\forall y\in R$
$\Rightarrow (9+8a)y^2+\left(a^2+46\right)y+(9+8a)>0\forall y\in R.$
If $9+8a=0$, then $y\ge 0\forall y\in R$,
which is false, so, $a\ne -\frac{9}{8}....(3)$
Now for $9a+8a\ne 0$
$\Rightarrow {\left(a^2+46\right)}^2-4(9+8a)(9+8a)\le 0$
$\Rightarrow {\left(a^2+46\right)}^2-[2(9+8a){]}^2\le 0$
$\Rightarrow -2(9+8a)\le \left(a^2+46\right)\le 2(9+8a)$
$\Rightarrow a^2+16a+64\le 0$ and $a^2-16a+28\le 0$
$\Rightarrow (a+8{)}^2\le 0$ and $(a-2)(a-14)\le 0$
$\Rightarrow a\in [2,14].$
$\therefore$ From(ii), (iii) and (iv), we have
$a\in [2,14]=[m,n]$ (given)
$\Rightarrow m=2,n=14$
$\Rightarrow \sqrt{mn-3}=5$