Given that, equation $(x^{2} – 7x + 11)^{x^{2} – 13x + 42} = 1 \quad \longrightarrow (1)$
We know that $a^{b} = 1$
- If $ a = 1$
- If $ b = 0, a \neq 0$
- If $ a = -1, b = \text{even}$
$\textbf{Case 1:}\; x^{2} – 7x + 11 = 1$
$ \Rightarrow x^{2} – 7x + 10 = 0$
$ \Rightarrow x^{2} – 5x -2x + 10 = 0$
$ \Rightarrow x (x-5) -2 (x-5) = 0$
$ \Rightarrow (x-5)(x-2) = 0$
$ \Rightarrow \boxed {x = 2,5}$
$\textbf{Case 2:} \;x^{2} – 13x + 42 = 0 $
$ \Rightarrow x^{2} -6x -7x + 42 = 0$
$ \Rightarrow x(x-6) – 7(x-6) = 0$
$ \Rightarrow (x-6)(x-7) = 0 $
$ \Rightarrow \boxed{ x= 6,7}$
$\textbf{Case 3:}\; x^{2} – 7x + 11 = – 1$
$ \Rightarrow x^{2} – 7x + 12 = 0 $
$ \Rightarrow x^{2} – 4x – 3x + 12 = 0$
$ \Rightarrow x(x-4) – 3(x-4) = 0$
$ \Rightarrow (x-4)(x-3) = 0$
$ \Rightarrow \boxed{x = 3,4}$
And, $x^{2} – 13x + 42 = \text{even}$
$x^{2} – 13x + 42 = 0$
This is the same as case $2.$
$\therefore$ There are $6$ distinct positive integer-valued solutions possible.
Correct Answer$: \text{A}$