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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 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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Originally posted by xcusemeplz2009 on 07 Dec 2009, 11:52.
Last edited by Bunuel on 03 Feb 2012, 05:25, edited 1 time in total.
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Re: function problem
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11 Dec 2009, 03:32
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=a2a > 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 \(\frac{b}{2a}=1\) > \(b=2a\) > \(a+b=a2a=a=2\) > \(a=2\) and \(b=4\). \(f(x)=2x^2+4x+1\) > \(f(10)=200+40+1=159\) Answer: A.
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Re: function problem
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07 Dec 2009, 14:32
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
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Updated on: 07 Dec 2009, 15:11
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.
Originally posted by swatirpr on 07 Dec 2009, 14:13.
Last edited by swatirpr on 07 Dec 2009, 15:11, edited 1 time in total.



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Re: function problem
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07 Dec 2009, 20:53
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
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08 Dec 2009, 15:39
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=a2a > 2=a f(10) = 2*100+10(4)+1 > 159



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Re: function problem
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11 Dec 2009, 16:32
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
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19 Dec 2009, 09:33
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
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19 Dec 2009, 11:00
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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