Question 15.

$(x^2 - 7x + 11)^{(x^2 - 13x + 42)} = 1$

This holds if  

$(x^2 - 13x + 42) = 0$  

or $(x^2 - 7x + 11) = 1$  

or $(x^2 - 7x + 11) = -1$ and $(x^2 - 13x + 42)$ is even.

For $x = 6,7$, the value $(x^2 - 13x + 42) = 0$.

$(x^2 - 7x + 11) = 1$ for $x = 5,2$.

$(x^2 - 7x + 11) = -1$ for $x = 3,4$ and for $x = 3$ or $4$, $(x^2 - 13x + 42)$ is even.

∴ $\{2,3,4,5,6,7\}$ is the solution set of $x$.

x can take 6 values.