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 (MVT), we need to follow these steps:

### Step 1: Verify the conditions of the Mean Value Theorem
The Mean Value Theorem states that if a function \( f(x) \) is continuous on the 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 \( f(x) = 2x^2 - 5x + 3 \), and we are considering the interval from \( A(1, 0) \) to \( B(2, 1) \).

### Step 2: Check continuity and differentiability
The function \( f(x) = 2x^2 - 5x + 3 \) is a polynomial, which is continuous and differentiable everywhere. Therefore, it is continuous on \([1, 2]\) and differentiable on \((1, 2)\).

### Step 3: Calculate \( f(1) \) and \( f(2) \)
Now we need to evaluate the function at the endpoints:

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

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

### Step 4: Apply the Mean Value Theorem
Now, we can apply the MVT:

\[ f'(c) = \frac{f(2) - f(1)}{2 - 1} = \frac{1 - 0}{1} = 1 \]

### Step 5: Find \( f'(x) \)
Next, we find the derivative of \( f(x) \):

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

### Step 6: Set \( f'(c) \) equal to 1
Now we set the derivative equal to 1:

\[ 4c - 5 = 1 \]

### Step 7: Solve for \( c \)
Solving for \( c \):

\[ 4c = 6 \implies c = \frac{6}{4} = \frac{3}{2} \]

### Step 8: Find the corresponding \( y \)-coordinate
Now we find the \( y \)-coordinate of the point on the curve at \( x = \frac{3}{2} \):

\[ f\left(\frac{3}{2}\right) = 2\left(\frac{3}{2}\right)^2 - 5\left(\frac{3}{2}\right) + 3 \]

Calculating this:

\[ = 2 \cdot \frac{9}{4} - \frac{15}{2} + 3 = \frac{18}{4} - \frac{30}{4} + \frac{12}{4} = \frac{18 - 30 + 12}{4} = \frac{0}{4} = 0 \]

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

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