40. The number of solutions of tan x = x − x 3 with −1 ≤ x ≤ 1

Since expansion of Tanx = x + 1/3(x^3) + 2/15(x^5) + ------.Therefore x + 1/3(x^3) + 2/15(x^5) + ------ = x - x^3.Thus (1+ 1/3)x^3 + 2/15(x^5) + ----- = 0Therefore LHS will be 0 only for x = 0. Since for any x less than 0 LHS will be negative and for any x greater than 0 it will be positive.In other words LHS will be 0 only for x = 0.Therefore x=0 is the solution.