The number of values of $x$ satisfying the equation $(x^{2} + 7 x + 11)^{(x^{2} - 4 x - 21)} = 1$ is

6

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4

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2

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5

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Solution

The correct option is B $4$
$(x^{2} + 7 x + 11)^{(x^{2} - 4 x - 21)} = 1$
Let $a = x^{2} + 7 x + 11$ and $b = x^{2} - 4 x - 21$

Case $1 :$ $a^{b} = 1$ when $a = 1$
$x^{2} + 7 x + 11 = 1 \Rightarrow x^{2} + 7 x + 10 = 0 \Rightarrow (x + 2) (x + 5) = 0 \Rightarrow x = - 5, - 2$

Case $2 :$ $a^{b} = 1$ when $b = 0$ and $a \neq 0$
$x^{2} - 4 x - 21 = 0 \Rightarrow (x + 3) (x - 7) = 0 \Rightarrow x = - 3, 7$

Case $3 :$ $a^{b} = 1$ when $a = - 1$ and $b$ is an even number.
$x^{2} + 7 x + 11 = - 1 \Rightarrow x^{2} + 7 x + 12 = 0 \Rightarrow (x + 3) (x + 4) = 0 \Rightarrow x = - 3, - 4$
$x = - 3 \Rightarrow b = 0$ which is even.
$x = - 4 \Rightarrow b = 11$ which is odd.
So, $x = - 4$ doesn't satisfy the assumed conditions.

So, the solutions are $- 5, - 2, - 3, 7$

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