GMATSUCKER wrote:A quadratic function f(x) attains a maximum value of 3 at x=1. The value of the function at x=0 is 1.what is the value of f(x) at x =10 ?

a)-105

b)-119

c)-159

d)-110

e)-180

Let f (x) = a x^2 + b x + c is the quadratic function, such that

At x = 1, a + b + c = 3, and at x = 0, c = 1

Then, at x = 10, f (x) = a (10) ^2 + b (10) + c = 100 a + 10 b + c = 90 a + 21 (?);

but f (x) cannot exceed 3, hence a is a negative integer.

Under these restrictions, no choice other than [spoiler]C[/spoiler] can take 90 a + 21, which happens at a = -2.

[spoiler]C[/spoiler]


I preferred to EXPLOIT the choices in hand, instead of going through the ways which are either long or beyond the GMAT specifications.

Any other approach to solve this problem ?

sanju09 wrote: Under these restrictions, no choice other than [spoiler]C[/spoiler] can take 90 a + 21, which happens at a = -2.

[spoiler]C[/spoiler]

I preferred to EXPLOIT the choices in hand, instead of going through the ways which are either long or beyond the GMAT specifications.

nice approach!

GMATSUCKER wrote:Any other approach to solve this problem ?

as said by sanju, the other methods to solve this problem will probably employ techniques that are beyond GMAT.

For e.g.

generic quad eq f(x) = ax^2 + bx + c (observe that there are 3 variables)

now based on the question, you can easily get value of C =1 and another eq. which gives a+b=2.

what about the third eq ?

you will have to differentiate the quadratic eq. with respect to x ( a concept beyond the scope), which will give you :

maximum value of this quad eq. is at x= (-b/2a). which on further solving will give a=-2 and b=4.

Hence the quad function is y= -2x^2 + 4x +1

at x=10, the value will be -159.

But, as a point to remember you can keep this in mind :

maximum/minimum value of a 2nd degree equation (quad function) in x represented by ax^2 + bx + c is at x= (-b)/2a

would like to see any simpler methods.