Question

The equation $2^{2x}+(a-1)2^{x+1}+a=0$ has roots of opposite signs, then exhaustive set of values of $a$ is

• A. $a\in (-1,0)$
• B. $a<0$
• C. $a\in (-\infty ,\frac{1}{3})$
• D. $a\in (0,\frac{1}{3})$

Answer 1

The correct option is C $a\in (-\infty ,\frac{1}{3})$

The given equation is
$2^{2x}+(a-1{)}^{2^{x+1}}+a=0$ ...(1)
or $t^2+2(a-1)t+a=0$, where $2^x=t$ ...(2)
Now, $t=1$ should lie between the roots of this equation.
By using the above result $f\left(1\right)<0$ as coefficient of $x^2>0$ and both roots are of opposite signs (for $t$, one root is $>1$ and other is $<1$)
$\therefore 1+2(a-1)+a<0$ or $a<\frac{1}{3}$

Hence, option C.