Let f(x)=e(p+1)x−ex for real number p>0 , then
a. The value of x=sp for which f(x) is minimum, is
b. Let g(t)=∫tt+1f(x)et−xdx . The value of t=tp , for which g(t) is minimum, is
c. Use the fact that 1+2p≤pep−1≤1+2p+p2 ( 0<p≤1 ), the value of limp→0+(sp−tp) is