The sum of all real values of x satisfying the equation $(x^{2} - 5 x + 5)^{x^{2} + 4 x - 60} = 1$ is :

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Solution

The correct option is C 3
$(x^{2} - 5 x + 5)^{x^{2} + 4 x - 60} = 1$
Case I
$x^{2} - 5 x + 5 = 1$ and $x^{2} + 4 x - 60$ can be any real number
$\Rightarrow x = 1, 4$
Case II
$x^{2} - 5 x + 5 = - 1$ and $x^{2} + 4 x - 60$ can be an even number
$\Rightarrow x = 2, 3$
where 3 is rejected because for $x = 3, x^{2} + 4 x - 60$ is odd.
Case III
$x^{2} - 5 x + 5$ can be any real number and $x^{2} + 4 x - 60 = 0$
$\Rightarrow x = - 10, 6$
$\Rightarrow$ Sum of all values of $x = - 10 + 6 + 2 + 1 + 4 = 3$

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