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How many distinct positive integer-valued solutions exist to the equation $\left ( x^{2}-7x+11 \right )^{(x^{2}-13x+42)} =1$?

  1. $6$
  2. $8$
  3. $2$
  4. $4$
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Given that, equation $(x^{2} – 7x + 11)^{x^{2} – 13x + 42} = 1 \quad \longrightarrow (1)$

We know that $a^{b} = 1$

  1. If $ a = 1$
  2.  If $ b = 0, a \neq 0$
  3. 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$: \text{A}$

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