2.The number of ordered pair(s) (x,y) satisfying y = 2 sinx and y = 5x² + 2x + 3 is equal to-

Options

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

Answer

To find the number of ordered pairs \((x, y)\) satisfying the equations \(y = 2 \sin x\) and \(y = 5x^2 + 2x + 3\), we can follow these steps:

### Step 1: Analyze the equations
We have two equations:
1. \(y = 2 \sin x\)
2. \(y = 5x^2 + 2x + 3\)

### Step 2: Determine the range of \(y = 2 \sin x\)
The function \(y = 2 \sin x\) oscillates between -2 and 2, since the sine function oscillates between -1 and 1. Therefore, the range of \(y = 2 \sin x\) is:
\[ -2 \leq y \leq 2 \]

### Step 3: Analyze the quadratic equation \(y = 5x^2 + 2x + 3\)
Next, we analyze the quadratic function \(y = 5x^2 + 2x + 3\). This is a parabola that opens upwards (since the coefficient of \(x^2\) is positive).

### Step 4: Find the minimum value of the quadratic function
To find the minimum value of the quadratic function, we can use the vertex formula. The x-coordinate of the vertex for a quadratic \(ax^2 + bx + c\) is given by:
\[ x = -\frac{b}{2a} \]
Here, \(a = 5\) and \(b = 2\):
\[ x = -\frac{2}{2 \cdot 5} = -\frac{1}{5} \]

### Step 5: Calculate the minimum value of \(y\)
Now, we substitute \(x = -\frac{1}{5}\) back into the quadratic equation to find the minimum value of \(y\):
\[ y = 5\left(-\frac{1}{5}\right)^2 + 2\left(-\frac{1}{5}\right) + 3 \]
Calculating this:
\[ y = 5 \cdot \frac{1}{25} - \frac{2}{5} + 3 = \frac{1}{5} - \frac{2}{5} + 3 = -\frac{1}{5} + 3 = \frac{14}{5} = 2.8 \]

### Step 6: Compare the minimum value of the quadratic with the range of \(y = 2 \sin x\)
The minimum value of the quadratic function \(y = 5x^2 + 2x + 3\) is \(2.8\), which is greater than the maximum value of \(y = 2 \sin x\) (which is 2).

### Conclusion
Since the minimum value of the quadratic function is greater than the maximum value of the sine function, the two graphs do not intersect. Therefore, there are no ordered pairs \((x, y)\) that satisfy both equations.

Thus, the number of ordered pairs \((x, y)\) satisfying the equations is:
\[ \boxed{0} \]