Question

Let $f:R\to R$ be a function satisfying $xf(x)+(1-x)f(-x)=x^2+x+1$ for any real number $x$. The greatest real number $M$ for which $f(x)\ge M$ for all real numbers $x$, is equal to $\frac{p}{q}$, where $p$ and $q$ are coprime. The value of $(q-p)$, is

• A. 1
• B. 2
• C. 3
• D. 8

Answer 1

Answer: (A)

1If $x$ is replaced by $-x$ in the given equation then
$-xf(-x)+(1+x)f(x)=x^2-x+1$
subtracting the two equations we get $f(-x)=f(x)+2x$,
substituting the value of $f(-x)$ in the given equation we get
$xf(x)+(1-x)(f(x)+2x)=x^2+x+1\text{ and thus }$
$f(x)=x^2+x+1-2x(1-x)=3x^2-x+1=3{\left(x-\frac{1}{6}\right)}^2+\frac{11}{12}$
and hence $f(x)\ge \frac{11}{12}$. If $x=\frac{1}{6}$ then we get $f\left(\frac{1}{6}\right)=\frac{11}{12}\equiv \frac{p}{q}$
$\Rightarrow (q-p)=1$