Using mean value theorem, prove that there is a point on the curve `y = 2x^(2) - 5x+3` between the points `A(1,0)` and `B(2,1)`, where tangent is parallel to the chord `AB`. Also, find that point.

Explanation

To solve the problem using the Mean Value Theorem, we will follow these steps:

### Step 1: Verify the conditions of the Mean Value Theorem
The Mean Value Theorem states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one point c in (a, b) such that:

\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

In our case, the function is \( y = 2x^2 - 5x + 3 \). Since this is a polynomial function, it is continuous and differentiable everywhere, including the interval [1, 2].

### Step 2: Calculate \( f(1) \) and \( f(2) \)
Now we will calculate the values of the function at the endpoints of the interval.

\[ f(1) = 2(1)^2 - 5(1) + 3 = 2 - 5 + 3 = 0 \]

\[ f(2) = 2(2)^2 - 5(2) + 3 = 8 - 10 + 3 = 1 \]

### Step 3: Calculate the slope of the chord \( AB \)
Now we can find the slope of the chord connecting points A(1, 0) and B(2, 1):

\[ \text{slope of } AB = \frac{f(2) - f(1)}{2 - 1} = \frac{1 - 0}{2 - 1} = 1 \]

### Step 4: Find the derivative of the function
Next, we need to find the derivative of the function:

\[ f'(x) = \frac{d}{dx}(2x^2 - 5x + 3) = 4x - 5 \]

### Step 5: Set the derivative equal to the slope of the chord
According to the Mean Value Theorem, we need to find \( c \) such that:

\[ f'(c) = 1 \]

So we set the derivative equal to the slope of the chord:

\[ 4c - 5 = 1 \]

### Step 6: Solve for \( c \)
Now we solve for \( c \):

\[ 4c = 1 + 5 \]
\[ 4c = 6 \]
\[ c = \frac{6}{4} = \frac{3}{2} \]

### Step 7: Find the corresponding \( y \) value
Now we find the corresponding \( y \) value when \( x = \frac{3}{2} \):

\[ f\left(\frac{3}{2}\right) = 2\left(\frac{3}{2}\right)^2 - 5\left(\frac{3}{2}\right) + 3 \]
\[ = 2 \cdot \frac{9}{4} - \frac{15}{2} + 3 \]
\[ = \frac{18}{4} - \frac{30}{4} + \frac{12}{4} \]
\[ = \frac{18 - 30 + 12}{4} = \frac{0}{4} = 0 \]

### Conclusion
Thus, the point on the curve where the tangent is parallel to the chord \( AB \) is:

\[ \left(\frac{3}{2}, 0\right) \]