Injective function for a given 'a' If a function is given whose domain and codomain are all real numbers such that
$$f(x) = x^3 + ax^2 + 3x + 100$$ then we have to find the value of $a$ for which the function is injective .
For that first I prove it as increasing or decreasing function . By doing this i got that the function is increasing when $a$ belongs to $[-3,3]$ and the function cannot be decreasing for any value of $a$.
But then I thought of second method by doing $f(x) =f(y)$. Solving this I got a equation of hyperbola i.e
$$x^2 +xy +y^2 + ax + ay +3 =0$$ 
But now how to proceed ?
 A: I would prefer your first method, but the second one can work too:
Actually your "hyperbola" is an ellipse when $|a|>3$ and nothing when $|a|<3$.
The function $$g(x,y) := \frac{f(x)-f(y)}{x-y} = x^2+xy+y^2+ax+ay+3$$ is an elliptic paraboloid opening upwards. You can find its bottom point by solving $\frac{\partial g}{\partial x} = \frac{\partial g}{\partial y}=0$.
If the value of $g$ at the bottom is less than zero, the locus of $g(x,y)=0$ must be a proper ellipse, which means there are nontrivial solutions to $f(x)=f(y)$.
On the other hand if the value of $g$ at the bottom is postive, then $g(x,y)=0$ can have no real solutions, and $f$ must be injective.
A: $f$ is injective $\iff$ $\forall x,y \in D, f(x)=f(y) \Rightarrow x=y$
Assume $f$ is injective.
$$f(x)-f(y)=0 \Rightarrow (x-y)[x^2+xy+y^2+ax+ay+3]=0$$
As f is injective $x=y$ hence $3x^2+2ax+3$ is either equal to $0$ or not equal to $0$ for all $x$ in $\mathbb{R}$
The only possible case is $3x^2+2ax+3$ doesn't have any root in $\mathbb{R}$
This imples $$D<0 \Rightarrow 4(a^2-9)<0$$
A: $$f(x)=f(y) \Rightarrow x^3+ax^2+3x+100=y^3+ay^2+3y+100$$
$$(x^3-y^3)+a(x^2-y^2)+3(x-y)=0$$
$$(x-y)(x^2+xy+y^2+ax+ay+3)=0$$
Then $x-y=0$ and $x^2+xy+y^2+ax+ay+3 \not =0$
$$f(x)=x^2+x(y+a)+y^2+ay+3\not =0$$
$$D=(y+a)^2-4(y^2+ay+3)<0$$
$$-3y^2-2ay+a^2-12<0$$
$$3y^2+2ay-a^2+12>0$$
$$\frac D4=a^2+3(a^2-12)<0$$
$$4a^2-36<0$$
$$a^2<9 \Rightarrow a\in(-3;3)$$
A: For your second method, you can write
$$f(x)-f(y) = (x-y)\left[\frac12(x+y+\tfrac23a)^2+\frac12(x+\tfrac13a)^2+\frac12(y+\tfrac13a)^2 + \color{red}{\left(3-\tfrac13a^2\right)}\right]$$
Notice only if the last term in red is non-negative, the entire portion in square braces is never zero, so $f(x)-f(y) \implies x=y$.  Thus our condition is
$$a^2 \leqslant 9 \iff a \in [-3, 3]$$
