If the equation `2^(2x)+a*2^(x+1)+a+1=0` has roots of opposite sign, then the exhaustive set of real values of `a` is (a)(−∞,0) (b)(−1,−2/3) (c)(−∞,−2/3) (d)(−1,∞)

Options

A. `(-oo,0)`
B. `(-1,(-2)/(3))`
C. `(-oo,(-2)/(3))`
D. `(-1,oo)`

Answer

B

Explanation

`(b)` Put `2^(x)=t`
`implies t^(2)+2a*t+(a+1)=0`
Now, `t=1` should lie between the roots of above equation.
Let `f(t)=t^(2)+2a*t+(a+1)`
`:.f(1) lt 0` and `f(0) gt 0`
`implies a lt (-2)/(3)` and `a gt -1`
`:.a in (-1,(-2)/(3))`