Question

Let $a$ denote the number of non-negative values of $p$ for which the equation $p.2^x+2^{2-x}=5$ passes a unique solution.

If $\frac{1}{a^{}},\frac{1}{{\alpha }_1},\frac{1}{{\alpha }_2},...,\frac{1}{{\alpha }_{20}},\frac{1}{6}$ are in AP and $1,{\beta }_1,{\beta }_2,....,{\beta }_{20},6$ are in AP. Then, the number of digits in value of ${\alpha }_{18}{\beta }_3$ is

Answer 1

Here it is not specified in question in which domain p belongs.

So, Let's take $p\in R^{+}$.

$p\left(x\right)=\frac{5-2^{2-x}}{2^x}$

To have a unique solution for $p\left(x\right)$,

$p\left(x_0\right)=p\left(x_1\right)\Rightarrow x_0=x_1$

ie, $\frac{5-2^{2-x_0}}{2^{x_0}}=\frac{5-2^{2-x_1}}{2^{x_1}}$

$\Rightarrow 5\times 2^{x_1}-2^{2+x_1-x_0}=5\times 2^{x_0}-2^{2-(x_1-x_0)}$

$\Rightarrow 2^{x_1-x_0}-2^{x_0-x_1}=\frac{5}{4}(2^{x_1}-2^{x_0})\Rightarrow (2^{x_1}-2^{x_0})(\frac{2^{x_1}+2^{x_0}}{2^{x_1+x_0}}-\frac{5}{4})=0$

Case 1: $x_1=x_0=x$ and $(\frac{2^{x_1}+2^{x_0}}{2^{x_1+x_0}}-\frac{5}{4})=0$ (both solutions are equal)

ie, $\frac{2^{x+1}}{2^{2x}}=\frac{5}{4}$

$\Rightarrow x=1-\log \frac{5}{4}$

Case 2: $p\left(x_0\right)=p\left(x_1\right)=0$ (just have to consider numerators)

$\Rightarrow 5-2^{2-x}=0\Rightarrow x=2-\log 5$

So, $a=2$

ie, $\frac{1}{{\alpha }_n}=\frac{1}{2}-n\frac{\frac{1}{2}-\frac{1}{6}}{21}=\frac{1}{2}-\frac{n}{63}$

and, ${\beta }_n=1+n\frac{6-1}{21}=1+\frac{5n}{21}$

So, ${\alpha }_{18}{\beta }_3=(\frac{1}{2}-\frac{18}{63}{)}^{-1}(1+\frac{15}{21})=\frac{14}{3}\frac{12}{7}=8$