Question

Let $f\left(x\right)=a{x}^2+bx+c$ where a,b, and c are constants. If f(x) takes it maximum value at $x=\frac{1}{3},$ then which of the following is necessarily true?

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

$f\left(x\right)=a{x}^2+bx+c$
Since maxima of f(x) occurs at $x=\frac{1}{3},$f(x) can be rewritten as $f\left(x\right)=a{(x-\frac{1}{3})}^2+k$ (where $f\left(\frac{1}{3}\right)=k$)
It is obvious that a$<$ 0.
Assuming f(n)=f(0) where n is any constant.
$f\left(0\right)=9\left(\frac{1}{9}\right)+k=f\left(n\right)=a{(n-\frac{1}{3})}^2+k$
$\Rightarrow {(n-\frac{1}{3})}^2=\frac{1}{9}$
$\Rightarrow n=\frac{2}{3}$
Therefore,f(0)=$f\left(\frac{2}{3}\right)$

Alternative Method:
The maximum value of $f\left(x\right)=a{x}^2+bx+c$ occurs at $x=-\frac{b}{2a}.$
$-\frac{b}{2a}=\frac{1}{3}\Rightarrow b=\frac{-2}{3a}$
Also,f(0)=c. Substituting $b=\frac{-2}{3a}$in f(x),we can easily observe that $f\left(0\right)=f\left(\frac{2}{3}\right).$