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For what value of 'a' is the area bounded by the curve y=a2 x 2+ax+1 and
the straight line y = 0, x = 0 & x = 1 the least ?
- a)-(3/4)
- b)-(3/5)
- c)-(3/2)
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
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For what value of a is the area bounded by the curve y=a2 x 2+ax+1 and...
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For what value of a is the area bounded by the curve y=a2 x 2+ax+1 and...
Understanding the problem:
The area bounded by the curve \(y=a^2x^2+ax+1\) and the straight lines \(y=0\), \(x=0\), and \(x=1\) can be calculated by finding the definite integral of the curve between the x-values of 0 and 1.
Finding the area:
To find the area, we need to calculate the integral of the curve with respect to x between 0 and 1. The integral can be calculated as follows:
\[
\int_{0}^{1} (a^2x^2 + ax + 1) dx
\]
Integrating the above expression, we get:
\[
\left[\frac{a^2x^3}{3} + \frac{ax^2}{2} + x\right]_0^1
\]
Substitute the limits of integration (0 and 1) into the expression and simplify:
\[
\left(\frac{a^2}{3} + \frac{a}{2} + 1\right) - \left(0 + 0 + 0\right)
\]
This simplifies to:
\[
\frac{a^2}{3} + \frac{a}{2} + 1
\]
Minimizing the area:
To find the least value of the area, we need to minimize the expression \(\frac{a^2}{3} + \frac{a}{2} + 1\). This can be done by finding the critical points of the expression. Taking the derivative and setting it equal to zero:
\[
\frac{d}{da}(\frac{a^2}{3} + \frac{a}{2} + 1) = 0
\]
Solving the above equation gives:
\[
\frac{2a}{3} + \frac{1}{2} = 0
\]
\[
a = -\frac{3}{4}
\]
Therefore, the least value of the area is obtained when \(a = -\frac{3}{4}\), which corresponds to option 'A'.
The area bounded by the curve \(y=a^2x^2+ax+1\) and the straight lines \(y=0\), \(x=0\), and \(x=1\) can be calculated by finding the definite integral of the curve between the x-values of 0 and 1.
Finding the area:
To find the area, we need to calculate the integral of the curve with respect to x between 0 and 1. The integral can be calculated as follows:
\[
\int_{0}^{1} (a^2x^2 + ax + 1) dx
\]
Integrating the above expression, we get:
\[
\left[\frac{a^2x^3}{3} + \frac{ax^2}{2} + x\right]_0^1
\]
Substitute the limits of integration (0 and 1) into the expression and simplify:
\[
\left(\frac{a^2}{3} + \frac{a}{2} + 1\right) - \left(0 + 0 + 0\right)
\]
This simplifies to:
\[
\frac{a^2}{3} + \frac{a}{2} + 1
\]
Minimizing the area:
To find the least value of the area, we need to minimize the expression \(\frac{a^2}{3} + \frac{a}{2} + 1\). This can be done by finding the critical points of the expression. Taking the derivative and setting it equal to zero:
\[
\frac{d}{da}(\frac{a^2}{3} + \frac{a}{2} + 1) = 0
\]
Solving the above equation gives:
\[
\frac{2a}{3} + \frac{1}{2} = 0
\]
\[
a = -\frac{3}{4}
\]
Therefore, the least value of the area is obtained when \(a = -\frac{3}{4}\), which corresponds to option 'A'.
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