If $kx^2-4x+3k+1>0$ for at least one $x>0$ and if $k\in S$, find $S$ 
Options ; $A)~ (1,\infty)~~ B)~(0,\infty)~~C)~(-1,\infty)~~D)~(-\frac 14 , \infty)$

Obviously
$$16-4(k)(3k+1)<0$$
$$k\in (-\infty, -\frac 43)\cup (1,\infty)$$
And also $k>0$ so the answer should be A)
The answer is, however, A, B, D
I think it has something to do with  ‘ least one $x>0$’ but I don’t know what the question exactly means by that
 A: Case 1, when $k>0$
Then $f(x)=kx^2-4x+3k+1$ is vertical upward parabola which will be positive definite for infinitely many values of $x$. So $k>0$ will enable $f(x)$ to become positive for many values of $x$.
Case 2: when $k<0$
$f(x)$ will be a downward parabola and it will be positive for at least one value of $x>0$ if at least one of its roots is positive and this happens when  $B^2 \ge 4AC \implies k[-4/3. 1]$ and when the product of the roots is negative  $\implies \frac{3k+1}{k} \implies -1/3<k<0$.  The overlap of the previous two domains is $\implies -1/3<k <0.$
If $k<0$, this equation cannot have both roots positive because sum of the roots $\alpha+\beta=4/k<0.$
Final answer is the union of these two cases: $k\in(-1/3, \infty).$
A: If $k\ge 0$, then the said condition will always be true, as we get an upward facing parabola. (or a line, for the $k=0$ case)
If $k\lt 0$, then we need the discriminant to be $\gt 0$, for otherwise the quadratic always assumes negative values, i.e. $$16-4k(3k+1) \gt 0 \\ \implies k\in \left(-\frac 43, 1\right ) $$
Now, since the quadratic must be positive for at least one positive $x$, at least one of the roots must be positive (can you see why?), so $$\frac{4+\sqrt{16-4k(3k+1)}}{2k} \gt 0 $$
But this is never true for $k\lt 0$. Although $$\frac{4-\sqrt{16-4k(3k+1)}}{2k} \gt 0 $$ is true for $$k\in \left(-\frac 13,0\right)$$
Taking the union of the two cases, we must have that $$\boxed{k\in \left(-\frac 13, \infty\right)} $$ Only options (A),(B),(D) are subsets of this set.
A: $\lim_{x\to0^+}f(x)=3k+1$ which is positive for $k\in(-1/3,\infty)$ so the condition holds in this interval. For any other values of $k$ the condition is false because $f'(x)=2kx-4$ is negative for $x\gt2/k$ and in particular this is true for $x\ge0$ so the function is globally maximised (over the positive reals) at $\lim_{x\to0^+}f(x)=3k+1\le0$.
