$F=ma$ can only be used if the relativistic mass of the object is known, true? I have received this question 

"Einstein's ideas on mass mean in essence the equation $F=ma$ can only be used if the relativistic mass of the object is known. Describe in your own words to what extent this statement is true." 

for a homework, by my understanding the equation $F=ma$ can't be used at velocities close to the speed of light because mass at those speeds is increased. But I am not sure this answers the question. 
Could someone help me with answering the question, mainly:


*

*Can $F=ma$ be used without knowing the relativistic mass?

*If yes, why/(how)/ if no, why/(how)? 


Thanks in advance!
 A: Skipping the normal caveats about the obsolescence of relativistic mass, $F = ma$ still doesn't work. The more proper expression for force is the rate of change of momentum. So,
\begin{array}{ll}
F &= \frac{dp}{dt} \\
&= \frac{d(\gamma m_0v)}{dt} \\
&= m_0\left(\gamma\frac{dv}{dt} + v\frac{d\gamma}{dt} \right)\\
&= m_0\left(\gamma a + v\frac{-3v/c^2}{(1-v^2/c^2)^{3/2}} \right)\\
\end{array}
where $m_0$ is the rest mass.
If it helps, here's what Einstein had to say in later years:

It is not good to introduce the concept of the mass $M = m/\sqrt{1 - v^2/c^2}$ of a moving body for which no clear definition can be given. It is better to introduce no other mass concept than the ’rest mass’ m. Instead of introducing M it is better to mention the expression for the momentum and energy of a body in motion.
— Albert Einstein in letter to Lincoln Barnett, 19 June 1948 (quote from L. B. Okun (1989), p. 42)

https://en.wikipedia.org/wiki/Mass_in_special_relativity#Relativistic_mass
A: It would be more correct to state Newton's second law as following:
$$\vec F = \frac{d \vec p}{dt}$$
This holds in relativistic mechanics too:
$$\vec F = \frac{d \vec p}{dt} = \frac{d (\gamma m \vec v)}{dt} = \gamma^3 m \vec a_{//}+\gamma m \vec a_{\perp}$$
Where $\vec a_{//}$ is the component of the acceleration which is parallel to $\vec v$ and $\vec a_{\perp}$ is the orthogonal component. So the form $\vec F = m \vec a $ doesn't hold in relativistic mechanics.
Moreover, as AccidentalFourierTransform pointed out, relativistic mass is an obsolete concept. The $m$ I used in the above formula is a constant. It is $\gamma$ that is a function of velocity:
$$\gamma = \frac{1}{\sqrt{1-(v/c)^2}}$$
In old times people used to define the "relativistic mass" as
$$ m_r =\gamma m = \frac{m}{\sqrt{1-(v/c)^2}}$$
But even if we wanted to use this concept, $\vec F = m_r \vec a$ would not hold. You can verify that starting from
$$\vec F = \frac{d \vec p}{dt} = \frac{d (\gamma m \vec v)}{dt}$$
Since $\gamma$ is a function of $\vec v$ and $\vec v$ is a function of time, you cannot take $\gamma$ and $m$ out of the time derivative.
So the answer is that $\vec F = m \vec a$ just doesn't hold in relativistic mechanics.
A: No, because
$$F = \frac{dp}{dt}$$
(p is momentum)
$$\rightarrow F = \frac{d \ m \times v}{dt}$$
$$\rightarrow F = \frac{dm}{dt} + \frac{dv}{dt}$$
cross-product rule
$$\rightarrow F = \frac{dm}{dt} + \frac{dv}{dt}$$
$$\rightarrow F = \frac{d\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}}{dt} + \frac{dv}{dt}$$
which gives the correct relation between Force, mass and acceleration.
