Practicing Success
Practicing Success
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If the equation $e^{\left||x| – 2\right| + b} = 2$ has four solution then b lies in |
(ln 2 – 2, ln 2) (–2, ln 2) (0, ln 2) None of these |
(ln 2 – 2, ln 2) |
$e^{\left||x| – 2\right| + b} = 2$ so $\left||x| – 2\right| + b=\log 2$ $\left||x| – 2\right|=\log 2-b$
so for 4 solutions $0<\log 2-b<2$ so $-2<b-\log 2<0$ $\log|2|-2<b<\log|2|$ |
