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

Question

MARKS Content Team · Accepted Answer

Correct answer: (C) -3/4. a^2x^2+ax+1 is positive for all real values of x. ∴ Area=∫01a^2x^2+ax+1dx =(a^2)/3+a/2+1 =1/62a^2+3a+6 =1/62a^2+3/2a+9/16+6-18/16 =1/62a+3^42+39/8 , which is minimum for a=-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

Solution

More Area Under Curves Questions

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

Solution

More Area Under Curves Questions

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