The number of distinct real roots of the equation `tan(2pix)/(x^2+x+1)=-sqrt(3)` is 4 (b) 5 (c) 6 (d) none of these

Options

A. 4
B. 5
C. 6
D. none of these

Explanation

To solve the equation \( \frac{\tan(2\pi x)}{x^2 + x + 1} = -\sqrt{3} \), we will follow these steps:

### Step 1: Rewrite the equation
We start by rewriting the equation:
\[ \tan(2\pi x) = -\sqrt{3}(x^2 + x + 1) \]

### Step 2: Identify the values of \( \tan(2\pi x) \)
The tangent function has a periodicity of \( \pi \). The values of \( \tan \theta \) that equal \( -\sqrt{3} \) occur at:
\[ \theta = n\pi - \frac{\pi}{3} \quad \text{for } n \in \mathbb{Z} \]
Thus, we can write:
\[ 2\pi x = n\pi - \frac{\pi}{3} \]

### Step 3: Solve for \( x \)
Dividing both sides by \( 2\pi \):
\[ x = \frac{n}{2} - \frac{1}{6} \]

### Step 4: Substitute back into the original equation
Now we substitute \( x \) back into the equation to find the distinct real roots:
\[ \tan(2\pi x) = -\sqrt{3}(x^2 + x + 1) \]
This gives us:
\[ -\sqrt{3}\left(\left(\frac{n}{2} - \frac{1}{6}\right)^2 + \left(\frac{n}{2} - \frac{1}{6}\right) + 1\right) \]

### Step 5: Analyze the function
The left-hand side, \( \tan(2\pi x) \), is a periodic function with vertical asymptotes at \( x = \frac{1}{4} + \frac{k}{2} \) for integers \( k \). The right-hand side is a quadratic function, which opens upwards.

### Step 6: Find intersections
To find the number of distinct real roots, we need to analyze the intersections of these two functions. The quadratic function will intersect the periodic tangent function at various points.

### Step 7: Count the distinct real roots
By analyzing the behavior of the functions and their intersections, we find that there are 5 distinct real roots.

### Conclusion
Thus, the number of distinct real roots of the equation is:
\[ \boxed{5} \]

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