Question

If x is real and $\frac{(x-1)(x+3)}{(x-2)(x+4)}$ cannot lie between a and b, then

• A. $a=\frac{1}{3},b=-3$
• B. $a=1,b=\frac{4}{9}$
• C. $a=\frac{1}{3},b=\frac{2}{3}$
• D. None of these

Answer 1

The correct option is B $a=1,b=\frac{4}{9}$

Let $y=\frac{(x-1)(x+3)}{(x-2)(x+4)}$

$\Rightarrow y=\frac{x^2+2x-3}{x^2+2x-8}$

$\Rightarrow y(x^2+2x-8)=x^2+2x-3$

$\Rightarrow (y-1)x^2+2(y-1)x+(-8y+3)=0$ is of the form $a{x}^2+bx+c=0$

Since $x$ is real, discriminant$>0$

$\Rightarrow b^2-4ac>0$

$\Rightarrow {\left(2(y-1)\right)}^2-4\times (y-1)\times (-8y+3)>0$

$\Rightarrow 4{(y-1)}^2-4(y-1)(3-8y)>0$

$\Rightarrow 4(y-1)[y-1-3+8y]>0$

$\Rightarrow y-1<0,9y-4>0$

$\Rightarrow y<1,9y>4$

$\Rightarrow y<1,y>\frac{4}{9}$

$\Rightarrow 1<y<\frac{4}{9}$

Given that $\frac{(x-1)(x+3)}{(x-2)(x+4)}$ lies between $a$ and $b$

$\therefore a=1$ and $b=\frac{4}{9}$