The set of all continuous functions $f:[0,1]\rightarrow R$ satisfying $ \int_{0}^{1}t^{n}f(t)dt=0$ I was thinking about the problem:
The set of all continuous functions $f:[0,1]\rightarrow R$ satisfying $$\int_0^1 t^n f(t) \, dt=0,\qquad n=1,2,\ldots$$ 

(a) $\text{is empty},$
  (b) $\text{contains a single element},$
  (c) $\text{is countably infinite},$
  (d) $\text{is uncountably infinite}.$

I am stuck on it and do not know how to proceed.Please help.Thanks in advance for your time.

Consider the map $\varphi:C[0,1]\to\mathbb{R}$ given by 
$$g\mapsto \int_0^1 g(t)f(t)\; dt$$
It's easy to see that this map is a continuous linear functional when $C[0,1]$ is given the $\|\cdot\|_\infty$ norm. Note then that by applying the hypothesis you can show that $\mathbb{R}[x]\subseteq\ker\varphi$, but since $\mathbb{R}[x]$ is dense in $C[0,1]$ this implies that $\varphi=0$. Take $g=f$ then to conclude that $f=0$.
EDIT: As Michael Hardy points out, I should probably mention that the density of $\mathbb{R}[x]$ in $(C[0,1],\|\cdot\|_\infty)$ is precisely the statement of the Stone-Weierstrass theorem. 
EDIT EDIT: As Pete. L Clark points out, we don't need the complexity of the Stone-Weierstrass theorem (which deals with $C(X)$ for $X$ l.c. Hausdorff) we really only need the special case of $[a,b]$ and the subalgebra being polynomials--this is the Weierstrass Approximation Theorem. This particular case is much easier than the general Stone-Weierstrass theorem. There is a constructive proof using Bernstein polynomials--a Numerical Analysis approximant that has the advantage of approximating a huge class of functions (at least the continuous ones) but with the downside of the rate of convergence being sublinear. 
