Question

If domain of $f(x) = \sqrt{\frac{(x-1)(7-x)(x-5)}{x(x-5)^2}}$ is D, then number of positive integers in D is

• A. 5
• B. 6
• C. 7
• D. 8

Answer 1

3 (None of the given options are correct.)

Answer 2

We must have

$$\frac{(x-1)(7-x)(x-5)}{x(x-5)^2}\ge0,\quad \text{with } x\neq0,\; x\neq5.$$

For $x\neq5$ we can cancel one factor of $(x-5)$ (since it’s nonzero) so that the inequality is equivalent to

$$\frac{(x-1)(7-x)}{x(x-5)}\ge0,\quad x\neq0,\,5.$$

The critical points are:

• $x=0$ (denom. zero),
• $x=1$ (numerator zero),
• $x=5$ (denom. zero),
• $x=7$ (numerator zero).

Analyzing the sign in intervals:

Thus, the expression is non‐negative on:

• $(0,1]$ (including $x=1$ where the value is 0)
• $(5,7]$ (including $x=7$ where the value is 0)

Now, checking for positive integers in the domain:

• In $(0,1]$: only $x=1$ qualifies.
• In $(5,7]$: $x=6$ and $x=7$ qualify.

Total positive integers in the domain = 3.