Question

Let $f(x)=x^2+a x+b$. If $\forall x \in R$, there exist a real value of $y$ such that $f(y)=f(x)+y$, then find the maximum value of $100 a$.

Answer 1

Given, $f ( y )= f ( x )+ y$
$\Rightarrow y^2+a y+b=x^2+a x+b+y \Rightarrow y^2+y(a-1)-x^2-a x=0$
As, $y \in R$, so $D \geq 0 \forall x \in R$
$\Rightarrow (a-1)^2+4\left(x^2+a x\right) \geq 0 \forall x \in R$
$\Rightarrow 4 x^2+4 a x+a^2-2 a+1 \geq 0 \forall x \in R$
Now, $D \leq 0$
$\Rightarrow 16 a^2-16\left(a^2-2 a+1\right) \leq 0 \Rightarrow 2 a-1 \leq 0 \Rightarrow a \leq \frac{1}{2}$
So, $a _{\max .}=\frac{1}{2}$
Hence, maximum value of $100 a =50$