How does $f^{(3)} (0) \geq 3\ $? 
Let $f(x)$ be a real polynomial of degree $4.$ Suppose $f(-1) = 0,f(0) = 0, f(1) = 1$ and $f^{(1)} (0) = 0,$ where $f^{(k)} (a)$ is the value of $k^{\text {th}}$ derivative of $f(x)$ at $x = a.$ Which of the following statements are true?


$(1)$ There exists $a \in (-1,1)$ such that $f^{(3)} (a) \geq 3$


$(2)$ $f^{(3)} (a) \geq 3,$ for all $a \in (-1,1)$


$(3)$ $0 \lt f^{(3)} (0) \leq 2$


$(4)$ $f^{(3)} (0) \geq 3$

My attempt $:$ Since $f$ is a real polynomial of degree $4$ having $-1$ and $0$ as roots it follows that there exist $p,q,r \in \Bbb R$ with $p \neq 0$ such that $$f(x) = x(x+1)(px^2 + qx + r).$$ By the given hypothesis $f(1) = 1$ which yields $p + q+ r = \frac {1} {2}.$ Now by expanding $f(x)$ we get $$f(x) = p x^4 + q x^3 + (p + r) x^2 + qx  + r.$$ So $$f^{(1)} (x) = 4 p x^3 + 3 q x^2 + 2 (p + r) x + q.$$ Since $f^{(1)} (0) = 0$ it follows that $q = 0.$ Hence we have $p + r = \frac {1} {2}.$ Therefore $f^{(1)} (x) = 4px^3 + x.$ Thus $f^{(3)} (x) = 24 p x.$ So $f^{(3)} (0) = 0,$ which violates $(3)$ and $(4).$ Also if we take $p = \frac {1} {24}$ then for all $a \in (-1,1)$ we have $f^{(3)} (a) \lt 1,$ which violates $(1)$ and $(2)$ as well. So neither of the given options happens to be true. But the answer key to this question suggests that $(1)$ and $(4)$ are the only correct options. Am I doing something wrong here? Can anybody please check my solution?
Thanks for your time.
Source $:$ CSIR NTA NET DECEMBER $2019.$

The conditions $f(0)=0$ and $f'(0)=0$ tell us that $f(x)=mx^4+ax^3+bx^2$ for some $a,b\in\Bbb R$. Furthermore, $f(-1)=0$ implies $b-a=-m$ and $f(1)=1$ implies $a+b=1-m$. Thus the polynomial is $$f(x)=mx^4+\frac12x^3+\left(\frac12-m\right)x^2.$$
