Question Statement

Find the points of discontinuity of the function:

$f(x)=\frac{x-1}{\sin 2x}$

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Background and Explanation

A rational function involving trigonometric terms is discontinuous where its denominator equals zero. For this function, we need to find where $\sin 2x=0$ since division by zero is undefined.

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Solution

The function $f(x)=\frac{x-1}{\sin 2x}$ is a rational function with a trigonometric denominator.

Step 1: Identify where the function is undefined

The function $f(x)$ is not defined when the denominator equals zero:

$\sin 2x=0$

Step 2: Solve for $x$

To find where $\sin 2x=0$, we take the inverse sine:

$2x={\sin }^{-1}(0)$

The general solution for $\sin \theta =0$ is $\theta =n\pi$ where $n\in \mathbb{Z}$ (any integer). Therefore:

$2x=n\pi ,(n\in \mathbb{Z})$

Step 3: Solve for $x$

Dividing both sides by 2:

$x=\frac{n\pi }{2}$

Conclusion:

The function $f(x)$ is discontinuous at all points where:

$x=\frac{n\pi }{2},n\in \mathbb{Z}$

That is, at $x=0,\pm \frac{\pi }{2},\pm \pi ,\pm \frac{3\pi }{2},\dots$

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Key Formulas or Methods Used

• Condition for discontinuity of rational functions: Denominator equals zero
• General solution of $\sin \theta =0$: $\theta =n\pi$ where $n\in \mathbb{Z}$

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Summary of Steps

1. Set the denominator equal to zero: $\sin 2x=0$
2. Apply the general solution for sine: $2x=n\pi$ where $n$ is any integer
3. Solve for $x$: $x=\frac{n\pi }{2}$
4. Conclude that discontinuities occur at all points $x=\frac{n\pi }{2}$ for $n\in \mathbb{Z}$