Question

If the equation $2^{2x}+a\cdot 2^{x+1}+a+1=0$ has roots of opposite sign, then ${}^{\prime }{a}^{\prime }$ lies in the interval

• A. $(-\infty ,-\frac{2}{3})$
• B. $(-1,-\frac{2}{3})$
• C. $(-\infty ,-1)$
• D. $(-1,0)$

Answer 1

The correct option is B $(-1,-\frac{2}{3})$

$2^{2x}+a\cdot 2^{x+1}+a+1=0\dots \dots \left(1\right)$
Let the roots be $\alpha ,\beta$
Assuming $2^x=t$,
$t^2+2at+(a+1)=0\dots \dots \left(2\right)$
Let the roots of equation be $t_1,t_2$
As equation $\left(1\right)$ have roots of opposite signs, so
$-\infty <\alpha <0<\beta <\infty \Rightarrow 0<2^{\alpha }<1<2^{\beta }<\infty$
$\Rightarrow 0<t_1<1<t_2<\infty$
$f\left(0\right)>0\Rightarrow a+1>0\Rightarrow a>-1\cdots \left(3\right)$
$1$ lies in between the roots for equation $\left(2\right)$,
$f\left(1\right)<0$
$\Rightarrow (1+2a+a+1)<0$
$\Rightarrow a<-\frac{2}{3}\cdots \left(4\right)$

From equation $\left(1\right)$ and $\left(2\right)$,
$a\in (-1,-\frac{2}{3})$