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

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Bunuel on 03 Feb 2012, 04:25, edited 1 time in total.

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Re: function problem [#permalink ]
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 for

x=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 ]
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 ]
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 ]
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 ]
11 Dec 2009, 02:32
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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 ]
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 ]
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?

Posted from my mobile device

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Re: function problem [#permalink ]
19 Dec 2009, 10:00
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 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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Re: function problem [#permalink ]
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
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12 Oct 2010, 22:10