If `x^(3) + 2x-3 = 0` then then number of real values of 'x' satisfying the equation is :

Options

A. 0
B. 1
C. 2
D. 3

Answer

To find the number of real values of \( x \) satisfying the equation \( x^3 + 2x - 3 = 0 \), we can follow these steps:

### Step 1: Define the function
Let \( f(x) = x^3 + 2x - 3 \).

### Step 2: Analyze the function
We need to determine how many times the function \( f(x) \) crosses the x-axis, which corresponds to finding the real roots of the equation \( f(x) = 0 \).

### Step 3: Calculate the derivative
To understand the behavior of the function, we calculate its derivative:
\[ f'(x) = 3x^2 + 2 \]

### Step 4: Analyze the derivative
Since \( 3x^2 + 2 \) is always positive (as \( 3x^2 \geq 0 \) and \( 2 > 0 \)), we conclude that \( f'(x) > 0 \) for all \( x \). This means that \( f(x) \) is a strictly increasing function.

### Step 5: Evaluate the function at specific points
To find the number of real roots, we can evaluate \( f(x) \) at a couple of points:
- For \( x = 1 \):
\[ f(1) = 1^3 + 2(1) - 3 = 1 + 2 - 3 = 0 \]
- For \( x = 0 \):
\[ f(0) = 0^3 + 2(0) - 3 = -3 \]
- For \( x = 2 \):
\[ f(2) = 2^3 + 2(2) - 3 = 8 + 4 - 3 = 9 \]

### Step 6: Analyze the function's values
From the evaluations:
- \( f(0) = -3 \) (negative)
- \( f(1) = 0 \) (zero)
- \( f(2) = 9 \) (positive)

Since \( f(x) \) is continuous and strictly increasing, it can only cross the x-axis once. Therefore, there is exactly one real root.

### Conclusion
The number of real values of \( x \) satisfying the equation \( x^3 + 2x - 3 = 0 \) is \( \boxed{1} \).

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