Q.
The integral I=∫xexcos(ex)dx=f(x)+c (where, c is the constant of integration) and f(ln(4π))2=2 . Then, the number of solutions of f(x)=2e(∀x∈R−{0}) is equal to
Given integral is I=∫xexcos(ex)dx
Let ex=t ⇒2xexdx=dt ∴I=2∫costdt =2sint+c =2sin(ex)+c ∴f(x)=2sin(ex)
which has range =[−2,2]
Hence, f(x)=2e has no solution