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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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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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.

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

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

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.

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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