The value of the parameter ‘a’ such that the area bounded by  y=a2x2+ax+1, coordinate axes and the line x = 1 attains its least value, is equal to

# 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

1. A

$-\frac{1}{4}$

2. B

$-\frac{1}{2}$

3. C

$-\frac{3}{4}$

4. D

$-1$

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### Solution:

${a}^{2}{x}^{2}+ax+1$ is positive for all real values of x.

$\therefore$  Area $=\underset{0}{\overset{1}{\int }}\left({a}^{2}{x}^{2}+ax+1\right)dx$

$=\frac{{a}^{2}}{3}+\frac{a}{2}+1=\frac{1}{6}\left(2{a}^{2}+3a+6\right)$

$\frac{1}{6}\left(2\left({a}^{2}+\frac{3}{2}a+\frac{a}{16}\right)+6-\frac{18}{16}\right)$

$=\frac{1}{6}\left(2{\left(a+\frac{3}{4}\right)}^{2}+\frac{39}{8}\right)$,which is minimum for $a=-\frac{3}{4}$

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