Find the number of values of $x$ of the form $6 n$ , where $n$ is an integer , in the domain of the function $f (x) = x ln | x - 1 | + \frac{\sqrt{64 - x^{2}}}{sin x}$

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Solution

Considering the domain, we get
$| x - 1 | > 0$
$\Rightarrow x \neq 1$
And
$64 - x^{2} \geq 0$
$x^{2} \leq 64$
$x \leq | 8 |$
$- 8 \leq x \leq 8$
Hence
domain is
$x ϵ [- 8, 1) \cup (1, 8]$
Also,
$s i n x \neq 0$
Hence, $x \neq k π$
Hence the number of the form $6 n$ are
$- 6, 6$
Hence there will be two numbers of the form 6n.
Hence answer is 2.

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