The number of solutions of the equation `x^(3)+x^(2)+4x+2sinx=0` in `0 le x le 2pi` is

Options

A. 0
B. 1
C. 2
D. 4

Explanation

To find the number of solutions of the equation \( x^3 + x^2 + 4x + 2\sin x = 0 \) in the interval \( 0 \leq x \leq 2\pi \), we can follow these steps:

### Step 1: Define the Functions
Let \( f(x) = x^3 + x^2 + 4x \) and \( g(x) = -2\sin x \). We need to find the number of intersections between the curves \( f(x) \) and \( g(x) \).

### Step 2: Analyze \( f(x) \)
We first analyze the function \( f(x) \):
\[ f(x) = x^3 + x^2 + 4x \]
To understand the behavior of \( f(x) \), we can find its derivative:
\[ f'(x) = 3x^2 + 2x + 4 \]
Since \( 3x^2 + 2x + 4 \) is a quadratic function with a positive leading coefficient, it opens upwards. We can find the discriminant to check if it has any real roots:
\[ D = b^2 - 4ac = 2^2 - 4 \cdot 3 \cdot 4 = 4 - 48 = -44 \]
Since the discriminant \( D < 0 \), \( f'(x) \) has no real roots, which means \( f'(x) > 0 \) for all \( x \). Thus, \( f(x) \) is a strictly increasing function.

### Step 3: Evaluate \( f(x) \) at the Endpoints
Next, we evaluate \( f(x) \) at the endpoints of the interval:
- At \( x = 0 \):
\[ f(0) = 0^3 + 0^2 + 4 \cdot 0 = 0 \]
- At \( x = 2\pi \):
\[ f(2\pi) = (2\pi)^3 + (2\pi)^2 + 4(2\pi) = 8\pi^3 + 4\pi^2 + 8\pi \]
Since \( 8\pi^3 + 4\pi^2 + 8\pi > 0 \), we have \( f(2\pi) > 0 \).

### Step 4: Analyze \( g(x) \)
Now, we analyze \( g(x) = -2\sin x \):
- The function \( g(x) \) oscillates between -2 and 0 for \( 0 \leq x \leq 2\pi \).

### Step 5: Finding Intersections
Since \( f(x) \) is strictly increasing from \( f(0) = 0 \) to \( f(2\pi) > 0 \), and \( g(x) \) oscillates between -2 and 0, we can conclude:
- At \( x = 0 \), \( f(0) = 0 \) and \( g(0) = 0 \), so they intersect at this point.
- As \( x \) increases, \( f(x) \) will continue to increase while \( g(x) \) will oscillate. The graphs will intersect again when \( g(x) \) reaches its maximum value of 0.

### Step 6: Counting Solutions
Since \( f(x) \) is strictly increasing and \( g(x) \) oscillates, there will be exactly one intersection in the interval \( (0, 2\pi) \) after the initial intersection at \( x = 0 \). Therefore, the total number of solutions to the equation \( x^3 + x^2 + 4x + 2\sin x = 0 \) in the interval \( 0 \leq x \leq 2\pi \) is:

\[ \text{Number of solutions} = 1 \]