The value of the parameter a such that the area bounded by `y=a^(2)x^(2)+ax+1,` coordinate axes, and the line x=1 attains its least value is equal to

Options

A. `(1)/(4)` sq. units
B. `-(1)/(2)` sq. units
C. `(3)/(4)` sq. units
D. `-1` sq. units

Explanation

To find the value of the parameter \( a \) such that the area bounded by the curve \( y = a^2 x^2 + ax + 1 \), the coordinate axes, and the line \( x = 1 \) attains its least value, we can follow these steps:

### Step 1: Set up the area integral
The area \( A \) bounded by the curve, the x-axis, and the lines \( x = 0 \) and \( x = 1 \) can be expressed as:
\[ A = \int_0^1 (a^2 x^2 + ax + 1) \, dx \]

### Step 2: Calculate the integral
Now, we compute the integral:
\[ A = \int_0^1 (a^2 x^2 + ax + 1) \, dx = \left[ \frac{a^2 x^3}{3} + \frac{ax^2}{2} + x \right]_0^1 \]
Evaluating this from 0 to 1 gives:
\[ A = \left( \frac{a^2}{3} + \frac{a}{2} + 1 \right) - \left( 0 \right) = \frac{a^2}{3} + \frac{a}{2} + 1 \]

### Step 3: Simplify the area expression
To simplify, we can find a common denominator:
\[ A = \frac{2a^2 + 3a + 6}{6} \]

### Step 4: Find the minimum value of the area
To find the minimum value of \( A \), we can differentiate \( A \) with respect to \( a \) and set the derivative to zero:
\[ \frac{dA}{da} = \frac{1}{6}(4a + 3) \]
Setting the derivative equal to zero:
\[ 4a + 3 = 0 \implies a = -\frac{3}{4} \]

### Step 5: Verify that this is a minimum
To confirm that this value of \( a \) gives a minimum area, we can check the second derivative:
\[ \frac{d^2A}{da^2} = \frac{4}{6} = \frac{2}{3} > 0 \]
Since the second derivative is positive, \( a = -\frac{3}{4} \) is indeed a minimum.

### Final Result
Thus, the value of the parameter \( a \) such that the area attains its least value is:
\[ \boxed{-\frac{3}{4}} \]