In photoelectric emission, "The number of photoelectrons emitted per unit area per unit time is directly proportional to the intensity of light used." Is this true if the frequency is not held constant? If yes, please explain how. If a light of frequency f and intensity I emits n photoelectrons from a metal surface(of threshold frequency f/4) per unit area per unit time, how many photoelectrons will be emitted from the same metal surface per unit area per unit time when a lights of frequency and intensity 1)f/2, I 2)2f, 2I 3)f/2, 2I are separately used.
Yes, the number of photoelectrons is proportional to the light intensity. At a fixed frequency.
The rest of this question is impossible to answer. The workfunction threshold is given by the OP as $f/4$, so $f$ should be in the far UV. It is then likely that the photoelectric yield is higher at $f/2$. And that it would be lower at $2f$ (likely above the plasmon cutoff).
But photoemission intensities depend on details of the electronic structure of the material: its density of occupied states and also on the final states reached by vertical transitions in $k$-space. And at $2f$ (eight times the work function) there might be emission from a shallow core level.
The number of photoelectrons emitted per unit area per unit time is directly proportional to the intensity of light used.
is a correct statement. The frequency of the light is proportional to the energy of the light (through $E=hf$) and therefore only affects the kinetic energy of the photoelectrons coming out from the surface not their numbers.
For example, if you use light with low energy (that is low frequency) than the work function of the metal then you can increase the intensity as much as you want; no photoelectrons would be emitted. On the other hand, if you have light with energy higher than the work function of the metal even a very dim light source would emit photoelectrons.
So, the number of electrons emitted is only proportional to the intensity of the light not the energy (or frequency) as long as the energy of the photons is higher then the work function of the metal.