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 - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

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

$∴$ Area $= \int_{0}^{1} (a^{2} x^{2} + a x + 1) d x$

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

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

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

The value of the parameter ‘a’ such that the area bounded by $y = a^{2} x^{2} + a x + 1$ , coordinate axes and the line x = 1 attains its least value, is equal to