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A quadratic function f(x) attains a max of 3 at x =1, the

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A quadratic function f(x) attains a max of 3 at x =1, the [#permalink] New post 07 Dec 2009, 10:52
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A quadratic function f(x) attains a max of 3 at x =1, the value of the function at x =0 is 1.the value of f(x) at x =10 is

a)-159,
b)-110,
c)-180,
d)-105,
e)-119

Pls explain
[Reveal] Spoiler: OA

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Last edited by Bunuel on 03 Feb 2012, 04:25, edited 1 time in total.
Added the OA
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Re: function problem [#permalink] New post 07 Dec 2009, 13:13
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xcusemeplz2009 wrote:
A quadratic function f(x) attains a max of 3 at x =1, the value of the function at x =0 is 1.the value of f(x) at x =10 is

a)-159,
b)-110,
c)-180,
d)-105,
e)-119

Pls explain


Nice explanation sriharimurthy.

I did some calculation mistake,
answer is A, -159.

My approach:

f(x)=ax^2+bx+c

Given forx=0, f(x)=1
so c=1
Given for x=1, f(x)=3
so a+b+c=3
a+b=2

now question is 100a+10b+c=?
90a+10(a+b)+c
90a+21=?

a, b, and c are integers

now pick an answer choice for which a is an integer

90a+21=-159
a=-2

so A is the answer.

Last edited by swatirpr on 07 Dec 2009, 14:11, edited 1 time in total.
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Re: function problem [#permalink] New post 07 Dec 2009, 13:32
xcusemeplz2009 wrote:
A quadratic function f(x) attains a max of 3 at x =1, the value of the function at x =0 is 1.the value of f(x) at x =10 is

a)-159,
b)-110,
c)-180,
d)-105,
e)-119

Pls explain



Let the quadratic function be : f(x) = ax^2 + bx + c

Given : f(0) = 1

Substituting this in our equation we get : c = 1

Thus our function becomes : f(x) = ax^2 + bx + 1

Given : f(1) = 3 is the max value of 'x'.

If 'x' is maximum at x = 1, then \frac{d}{dx}f(x) = 0 at x = 1

\frac{d}{dx}f(x) = 2ax + b = 0 Therefore, putting value of x =1 we get b = -2a

Now going back to f(1) = 3, we get a + b + 1 = 3 --> a - 2a + 1 = 3

Thus, a = -2 and b = 4

So our quadratic function is : f(x) = -2x^2 + 4x + 1

Thus f(10) = -159

Answer : A


Ps. I doubt this is a GMAT problem. Please mention the source.
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Re: function problem [#permalink] New post 07 Dec 2009, 19:53
sriharimurthy wrote:
xcusemeplz2009 wrote:
A quadratic function f(x) attains a max of 3 at x =1, the value of the function at x =0 is 1.the value of f(x) at x =10 is

a)-159,
b)-110,
c)-180,
d)-105,
e)-119

Pls explain



Let the quadratic function be : f(x) = ax^2 + bx + c

Given : f(0) = 1

Substituting this in our equation we get : c = 1

Thus our function becomes : f(x) = ax^2 + bx + 1

Given : f(1) = 3 is the max value of 'x'.

If 'x' is maximum at x = 1, then \frac{d}{dx}f(x) = 0 at x = 1

\frac{d}{dx}f(x) = 2ax + b = 0 Therefore, putting value of x =1 we get b = -2a

Now going back to f(1) = 3, we get a + b + 1 = 3 --> a - 2a + 1 = 3

Thus, a = -2 and b = 4

So our quadratic function is : f(x) = -2x^2 + 4x + 1

Thus f(10) = -159

Answer : A


Ps. I doubt this is a GMAT problem. Please mention the source.


wow!!
grt job
i solved lke this and got -159 ,but was not sure abt the ans

let f(x) = ax2 + bx +c

at x = 0 value of f(x) = 1 so equation becomes

1 = c

since it attains max of 3 at x =1

3 = a+b+c put c=1

a+b = 2

now differntiate f(x) we will get 2ax +b =0 since it attains max at x =1 relaion becomes 2a = -b

now u will get values of a=-2,b =4 nd c =1

f(x) = -2x2 + 4x + 1

now put x = 10 nd answer is -159
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Re: function problem [#permalink] New post 08 Dec 2009, 14:39
xcusemeplz2009 wrote:
A quadratic function f(x) attains a max of 3 at x =1, the value of the function at x =0 is 1.the value of f(x) at x =10 is

a)-159,
b)-110,
c)-180,
d)-105,
e)-119

Pls explain


Not sure this would be on the GMAT, seems you should use calculus for this.

ax^2+bx+c=0

f(1)=3=a+b+c
f(0)=1=c

f'(x)= 2xa+b.

since the max is 3, f'(1)=0. Thus, 0=2a+b--> b=-2a --> 2=a-2a --> -2=a

f(10) = -2*100+10(4)+1 --> -159
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Re: function problem [#permalink] New post 11 Dec 2009, 02:32
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Expert's post
GMATBLACKBELT wrote:
xcusemeplz2009 wrote:
A quadratic function f(x) attains a max of 3 at x =1, the value of the function at x =0 is 1.the value of f(x) at x =10 is

a)-159,
b)-110,
c)-180,
d)-105,
e)-119

Pls explain


Not sure this would be on the GMAT, seems you should use calculus for this.

ax^2+bx+c=0

f(1)=3=a+b+c
f(0)=1=c

f'(x)= 2xa+b.

since the max is 3, f'(1)=0. Thus, 0=2a+b--> b=-2a --> 2=a-2a --> -2=a

f(10) = -2*100+10(4)+1 --> -159


This can be solved in the following way too:

We have f(x)=ax^2+bx+c.

f(0)=c=1 --> f(x)=ax^2+bx+1

We are told that f_{max}(1)=a+b+1=3, --> a+b=2.

f_{max} is vertex of parabola and the x coordinate of vertex is -\frac{b}{2a}=1 --> b=-2a --> a+b=a-2a=-a=2 --> a=-2 and b=4.

f(x)=-2x^2+4x+1 --> f(10)=-200+40+1=-159

Answer: A.
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Re: function problem [#permalink] New post 11 Dec 2009, 15:32
You've got 3 points and 3 equations, so you could use a system of equations to solve.
Coordinates: (1,3) Given, (0,1) Given, and (2,1) Implied. I got (2,1) by doing a really rough sketch and seeing that (2,1) is a mirror of (0,1) with (1,3) as the max.

3 systems in quadratic form:
a(1)^2 + b(1) + c = 3
a(0)^2 + b(0) + c = 1
a(2)^2 + b(2) + c = 1

a + b + c = 3
c = 1
4a + 2b + c = 1

Solving this would yield a = -2, b = 4, c = 1.
f(x) = -2x^2 + 4x + 1

Then, just plug in 10 for x and solve.
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Re: function problem [#permalink] New post 19 Dec 2009, 08:33
Hey guys,

I understand how you arrived at the f(x)=ax^2+bx+c, but you I don't fully understand how you arrived at 2x^2+4x+1. Could you please explain a little more?

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Re: function problem [#permalink] New post 19 Dec 2009, 10:00
Expert's post
CoIIegeGrad09 wrote:
Hey guys,

I understand how you arrived at the f(x)=ax^2+bx+c, but you I don't fully understand how you arrived at 2x^2+4x+1. Could you please explain a little more?

Posted from my mobile device Image


We have the function f(x)=ax^2+bx+c.

Stem: the value of the function at x =0 is 1.

Substitute x by 0 --> f(0)=1=a*0^2+b*0+c --> c=1, hence the function becomes: f(x)=ax^2+bx+1.


Stem: A quadratic function f(x) attains a max of 3 at x =1.

f_{max}(1)=3=ax^2+bx+1=a*1^2+b*1+1=a+b+1, --> a+b+1=3 --> a+b=2.

Also the property of quadratic function:

The maximum (or minimum if a>0) of the quadratic function f(x)=ax^2+bx+c is the y coordinate (the value of f(x)) of the vertex of the given quadratic function, which is parabola.

The vertex of the parabola is located at the point (-\frac{b}{2a}, c-\frac{b^2}{4a}).

Given: A quadratic function f(x) attains a max of 3 at x =1 --> coordinates of the vertex are (1, 3) --> -\frac{b}{2a}=1, c-\frac{b^2}{4a}=3

So we have:
a+b=2
and:
-\frac{b}{2a}=1

From this a=-2 and b=4. so the function is: f(x)=-2x^2+4x+1

Hope it's clear.
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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: function problem [#permalink] New post 12 Oct 2010, 22:10
Is it REALLY gmatprep's task? I've never seen this in gmatprep. Or the "tag" is wrong?
Re: function problem   [#permalink] 12 Oct 2010, 22:10
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