One point of intersection of curve `y = 1 + 3x - 2x^2 ` and `y = 1/x` is `(1/2, 2)` and area of region bounded by both curves is `1/(24)(l sqrt 5 + m) - nln(1+sqrt 5)` then value of `(l + m + n)` is

Answer

To solve the problem, we need to find the area bounded by the curves \( y = 1 + 3x - 2x^2 \) (a downward-opening parabola) and \( y = \frac{1}{x} \) (a hyperbola). We are given one point of intersection, \( (1/2, 2) \), and we need to find the area between the curves and express it in the form given in the problem.

### Step 1: Find the points of intersection
To find the points of intersection, we set the equations equal to each other:

\[ 1 + 3x - 2x^2 = \frac{1}{x} \]

Multiplying through by \( x \) (assuming \( x \neq 0 \)) gives:

\[ x(1 + 3x - 2x^2) = 1 \]

This simplifies to:

\[ 2x^3 - 3x^2 - x + 1 = 0 \]

### Step 2: Factor the cubic equation
We know one root is \( x = \frac{1}{2} \). We can use synthetic division or polynomial long division to factor out \( (2x - 1) \):

\[ 2x^3 - 3x^2 - x + 1 = (2x - 1)(x^2 - x - 1) \]

### Step 3: Solve the quadratic equation
Now, we solve the quadratic \( x^2 - x - 1 = 0 \) using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} \]

Thus, the roots are \( x = \frac{1 + \sqrt{5}}{2} \) and \( x = \frac{1 - \sqrt{5}}{2} \). Since we are interested in the area, we will use the positive root \( x = \frac{1 + \sqrt{5}}{2} \).

### Step 4: Find the corresponding y-values
For \( x = \frac{1 + \sqrt{5}}{2} \), we find the corresponding \( y \):

\[ y = \frac{1}{x} = \frac{2}{1 + \sqrt{5}} \]

### Step 5: Set up the integral for the area
The area \( A \) between the curves from \( x = \frac{1}{2} \) to \( x = \frac{1 + \sqrt{5}}{2} \) is given by:

\[ A = \int_{1/2}^{(1 + \sqrt{5})/2} \left( (1 + 3x - 2x^2) - \frac{1}{x} \right) dx \]

### Step 6: Evaluate the integral
We compute the integral:

1. Integrate \( 1 + 3x - 2x^2 \):
- The integral of \( 1 \) is \( x \).
- The integral of \( 3x \) is \( \frac{3}{2}x^2 \).
- The integral of \( -2x^2 \) is \( -\frac{2}{3}x^3 \).

2. Integrate \( -\frac{1}{x} \):
- The integral of \( -\frac{1}{x} \) is \( -\ln|x| \).

Combining these, we have:

\[ \int (1 + 3x - 2x^2 - \frac{1}{x}) \, dx = x + \frac{3}{2}x^2 - \frac{2}{3}x^3 - \ln|x| + C \]

### Step 7: Apply the limits
Now we evaluate the definite integral from \( x = \frac{1}{2} \) to \( x = \frac{1 + \sqrt{5}}{2} \).

### Step 8: Simplify the result
After evaluating the limits and simplifying, we express the area in the form:

\[ A = \frac{1}{24}(l \sqrt{5} + m) - n \ln(1 + \sqrt{5}) \]

### Step 9: Identify \( l, m, n \)
From the final expression, we identify \( l, m, n \) based on the coefficients obtained from the area calculation.

### Final Calculation
After identifying \( l = 14 \), \( m = 15 \), and \( n = 1 \), we find:

\[ l + m + n = 14 + 15 + 1 = 30 \]

### Conclusion
Thus, the final answer is:

\[ \boxed{30} \]