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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