Given a bounded real sequence {an}, for what values of the real exponent p can you be sure that the seriesn=1annpsin(nx)sums to a twice-continuously differentiable function of x? Explain your answer.It is a calculus problem(119) Let f:[0,1]R be a non-negative continuous function, so that f(0,1)>0f'f(0,)>0supxin(0,)f'(x)f(x)=f(0)=0,f(x)>0,0. Assume also fis differentiable on(0,1). Prove that for any >0, the functionf'fis unbounded on(0,). That is, show that for all >0,supxin(0,)f'(x)f(x)=Hint: Argue by contradiction.(119) Let f:[0,1]R be a non-negative continuous function, so that f(0,1)>0f'f(0,)>0supxin(0,)f'(x)f(x)=f(0)=0,f(x)>0,0. Assume also fis differentiable on(0,1). Prove that for any >0, the functionf'fis unbounded on(0,). That is, show that for all >0,supxin(0,)f'(x)f(x)=Hint: Argue by contradiction.