I have few doubts here.
There are two cases at first.
b>0 parabola opens in downward direction. Means expression will have maximum value.
b<0 parabola opens in upward direction. Means expression will have minimum value.
From above two points I choose a, d, e,.
Now |x| = a/2b
If b<0, a<0 or if b>0, a>0
after this point, I am stuck.
Now need your help!
chetan2u wrote:
CaptainLevi wrote:
Given f(x) = \(a|x| − bx^2\), where a and b are constants. Then at x = 0, f (x) is
a) maximized whenever a > 0, b > 0.
b) minimized whenever a > 0, b > 0.
c) minimized whenever a < 0, b > 0.
d) maximized whenever a < 0, b > 0.
e) minimized whenever a < 0 , b < 0.
This can be done in two ways..
(I) Parabola..f(x) = \(a|x| − bx^2\)
Since we have a NEGATIVE coefficient of x^2, the parabola will open downwards and the value of y will be at the vertex, that is \(\frac{-a}{2b}\)
This will be maximum when a is negative and b is positive
D
(II) DeterminantsDeterminant of f(x) is -2bx+a=0.......x=-a/2b
Next determinant .. -2b, As this is negative, the value will be maximized at x=-a/2b
Again D as above
D
Not likely to be tested in GMAT