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# Given f(x) = a|x| − bx^2, where a and b are constants. Then at x = 0,

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Manager
Joined: 18 Jul 2019
Posts: 54
Given f(x) = a|x| − bx^2, where a and b are constants. Then at x = 0,  [#permalink]

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25 Nov 2019, 07:24
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20% (02:08) correct 80% (01:38) wrong based on 49 sessions

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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.
Manager
Joined: 10 May 2018
Posts: 57
Given f(x) = a|x| − bx^2, where a and b are constants. Then at x = 0,  [#permalink]

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25 Nov 2019, 09:04
1
Bunuel chetan2u Can you please explain this
Intern
Joined: 13 Oct 2019
Posts: 5
Concentration: Technology, Finance
Re: Given f(x) = a|x| − bx^2, where a and b are constants. Then at x = 0,  [#permalink]

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25 Nov 2019, 09:16
I am not clear on how to solve. If x=0, then a and b will also become 0. Do we substitute +ve and -ve values to figure out which way the function goes? That would require 6 combinations of a and b as +ve and -ve. Please help
Math Expert
Joined: 02 Aug 2009
Posts: 8284
Re: Given f(x) = a|x| − bx^2, where a and b are constants. Then at x = 0,  [#permalink]

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25 Nov 2019, 09:49
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) Determinants
Determinant 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
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Manager
Joined: 10 May 2018
Posts: 57
Given f(x) = a|x| − bx^2, where a and b are constants. Then at x = 0,  [#permalink]

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25 Nov 2019, 11:52
chetan2u Thanks for the explanation .

But I am unable to understand why the max value is at -a/2b

for ax^2 +bx +c when a<0 the max value will be at x=-b/2a

as per this f(x) = -bx^2 + ax

the max value will be at x= -a/2*-b
x= a/2b

Also for the second method
II) Determinants
Determinant of f(x) is -2bx+a=0.......x=-a/2b ---> Isn't it should be -2bx +a= 0

-2bx=-a
x=a/2b

Could you please guide where I am going wrong
Director
Joined: 09 Aug 2017
Posts: 604
Re: Given f(x) = a|x| − bx^2, where a and b are constants. Then at x = 0,  [#permalink]

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01 Dec 2019, 03:21
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.

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) Determinants
Determinant 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
Re: Given f(x) = a|x| − bx^2, where a and b are constants. Then at x = 0,   [#permalink] 01 Dec 2019, 03:21
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