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CaptainLevi
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Given f(x) = a|x| − bx^2, where a and b are constants. Then at x = 0, 25 Nov 2019, 07:24
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D
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26% (02:07) correct 74% (01:52) wrong based on 262 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.
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D
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ManjariMishra
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Given f(x) = a|x| − bx^2, where a and b are constants. Then at x = 0, 25 Nov 2019, 09:04
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Bunuel chetan2u Can you please explain this
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Cheeku2023
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Given f(x) = a|x| − bx^2, where a and b are constants. Then at x = 0, 25 Nov 2019, 09:16
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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
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Given f(x) = a|x| − bx^2, where a and b are constants. Then at x = 0, 25 Nov 2019, 09:49
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CaptainLevi
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.
Show more

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

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ManjariMishra
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Given f(x) = a|x| − bx^2, where a and b are constants. Then at x = 0, 25 Nov 2019, 11:52
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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
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Given f(x) = a|x| − bx^2, where a and b are constants. Then at x = 0, 01 Dec 2019, 03:21
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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
CaptainLevi
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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Given f(x) = a|x| − bx^2, where a and b are constants. Then at x = 0, 07 Aug 2020, 05:59
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Kindly see the attachment.

I tried the 2 approaches and man they are difficult. This question involves a lot of thinking.
I preferred the Plotting method.
Attachments

File comment: Method 1. Graph and Plotting
Method 1.png
Method 1.png [ 592.84 KiB | Viewed 6311 times ]

File comment: Method 2. Determinants
Method 2.png
Method 2.png [ 624.81 KiB | Viewed 6306 times ]

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Given f(x) = a|x| − bx^2, where a and b are constants. Then at x = 0, 15 Dec 2021, 05:21
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Hi,

here is a simple solution to this:

Check the function for x>=0 and x<0. You can see directly that the graph is mirrored on the Y-Axis. So we can continue to work on this problem just by looking at x>=0. Now we know that f(0)=0.

To know if f(0) is a min / max, we have to know if there is any point x in the graph with f(x) smaller than or greater than 0.

If we take the first and second derivatives, we get f_1(x)=a-2bx and f_2=-2b.

Now, plug in the conditions from (A) to (E):

(A) It follows f_1(x)=lal-2lblx. For x=0, f_1(x)=lal>0, so f(x)=0 can't be a maximum.
(B) We have a downwards sloping parabola, so 0 can't be a minimum.
(E) It follows f_1(x)=-lal+2lblx. For x=0 we get a negative slope, so 0 can't be the minimum.
(C) and (D) It follows f_1(x)=-lal+2lblx. Plug in 0, and see again that we get a negative first derivative. So 0 can't be a minimum. This leaves us with only (D) as possible choice.

Please correct me if I'm wrong.

Thanks
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Given f(x) = a|x| − bx^2, where a and b are constants. Then at x = 0, 14 Jan 2022, 11:39
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chetan2u
CaptainLevi
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

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chetan2u, thank you very much for your explanation.

A quick question please. Could you kindly highlight why we chose D over A. This is my confusion: given the coefficient of x^2 is negative, I thought the vertex is at a/-2b, which equals - a/2b, which holds when a and b are both positive . Why then do we choose the case where a<0 and b>0?

Thanks in advance.

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