Question

The number of positive integral value(s) of $p$ for which $p\cdot 2^{e^x+e^{-x}}-8\cdot 2^{\frac{e^x+e^{-x}}{2}}+p=0$ has atleast one solution is

Answer 1

Given: $p\cdot 2^{e^x+e^{-x}}-8\cdot 2^{\frac{e^x+e^{-x}}{2}}+p=0$
$2^{\frac{e^x+e^{-x}}{2}}=t$
As $\frac{e^x+e^{-x}}{2}\ge 1$, so
$2^{\frac{e^x+e^{-x}}{2}}\ge 2\Rightarrow t\ge 2$

Now,
$p{t}^2-8t+p=0\Rightarrow t+\frac{1}{t}=\frac{8}{p}$
As $t\ge 2$, so
$t+\frac{1}{t}\ge \frac{5}{2}\Rightarrow \frac{8}{p}\ge \frac{5}{2}\Rightarrow p\le \frac{16}{5}\therefore p=1,2,3$

Hence, there are $3$ positive integral values of $p$.