Question

Set of all real values of $a$ such that $f\left(x\right)=\frac{(2a-1)x^2+2(a+1)x+(2a-1)}{x^2-2x+40}$ is always negative is

• A. $(-\infty ,0)$
• B. $(0,\infty )$
• C. $(-\infty ,\frac{1}{2})$
• D. None of these.

Answer 1

The correct option is A $(-\infty ,0)$

Given,
$f\left(x\right)=\frac{(2a-1)x^2+2(a+1)x+2a-1}{x^2-2x+40}<0$
$\Rightarrow \frac{(2a-1)x^2+2(a+1)x+2a-1}{(x-1{)}^2+39}<0$
$\Rightarrow (2a-1)x^2+2(a+1)x+2a-1<0$ (denominator is always positive)
(i) Co-efficient of $x^2<0$
$\Rightarrow 2a-1<0$
$a<\frac{1}{2}$
(ii)$D<0$
$\Rightarrow 4(a+1{)}^2-4(2a-1{)}^2<0$
$\Rightarrow (a+1+2a-1)(a+1-2a+1)<0$
$a(a-2)>0$
$\Rightarrow a<0$ or $a>2$
Intersection of (i) and (ii), we get $a<0$
$\Rightarrow a\in (-\infty ,0)$
Hence, option A is correct.