The given system of equations
4x2 f(–1) + 4x f(1) + f(2) = 3x2 + 3x is satisfied for 3 distinct real numbers a, b and c
Comparing the co-efficients of x2, x and constant terms, we get
4f(–1) = 3, 4f(1) = 3, f(2) = 0
Let f(x) = ax2 + bx + c
Given f(–1) = 3/4
a – b + c = 3/4
f(1) = 3/4
a + b + c = 3/4
Thus, b = 0
And f(2) = 0
4a + 2b + c = 0
c = – 4a
Solving, we get
a = – (1/4) , b = 0, c = 1
Thus, f(1) = – (1/4) x2 + 1
Let the co-ordinates of B be B (t, l – (t2/4))
Let
t = – 8
Therefore, the co-ordinates of B = (– 8, –15).
Hence, the required area
= 125/3