If `f(x)=1/((x-1)(x-2))` and `g(x)=1/x^2` then points of discontinuity of `f(g(x))` are

Options

A. `{--1,0,1,1/(sqrt(2))}`
B. `{-1/(sqrt(2)),-1,0,1,1/(sqrt(2))}`
C. `{0,1}`
D. `{0,1,1/(sqrt(2))}`

Answer

To find the points of discontinuity of the function \( f(g(x)) \), where \( f(x) = \frac{1}{(x-1)(x-2)} \) and \( g(x) = \frac{1}{x^2} \), we will follow these steps:

### Step 1: Define the composition \( f(g(x)) \)

Given:
- \( g(x) = \frac{1}{x^2} \)

Now, substituting \( g(x) \) into \( f(x) \):

\[ f(g(x)) = f\left(\frac{1}{x^2}\right) = \frac{1}{\left(\frac{1}{x^2} - 1\right)\left(\frac{1}{x^2} - 2\right)} \]

### Step 2: Simplify the expression

We need to simplify the expression inside the function:

1. Calculate \( \frac{1}{x^2} - 1 \):
\[ \frac{1}{x^2} - 1 = \frac{1 - x^2}{x^2} \]

2. Calculate \( \frac{1}{x^2} - 2 \):
\[ \frac{1}{x^2} - 2 = \frac{1 - 2x^2}{x^2} \]

Now substituting these back into \( f(g(x)) \):

\[ f(g(x)) = \frac{1}{\left(\frac{1 - x^2}{x^2}\right)\left(\frac{1 - 2x^2}{x^2}\right)} = \frac{x^4}{(1 - x^2)(1 - 2x^2)} \]

### Step 3: Identify points of discontinuity

The function \( f(g(x)) \) will be discontinuous where the denominator is zero. Therefore, we need to solve:

\[ (1 - x^2)(1 - 2x^2) = 0 \]

This gives us two equations to solve:

1. \( 1 - x^2 = 0 \)
\[ x^2 = 1 \implies x = \pm 1 \]

2. \( 1 - 2x^2 = 0 \)
\[ 2x^2 = 1 \implies x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}} \]

### Step 4: Check for additional points of discontinuity

Since \( g(x) = \frac{1}{x^2} \) is also undefined at \( x = 0 \), we need to include this point as well.

### Final List of Points of Discontinuity

Thus, the points of discontinuity for \( f(g(x)) \) are:
- \( x = -1 \)
- \( x = 1 \)
- \( x = -\frac{1}{\sqrt{2}} \)
- \( x = \frac{1}{\sqrt{2}} \)
- \( x = 0 \)

### Conclusion

The points of discontinuity of \( f(g(x)) \) are:
\[ \{-1, 1, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\} \]