Consider the function, f(x)=(2k+1)x−3−ke−x+2ex since f(x) is monotonically increasing for all x∈R so, f′(x)≥0 (2k+1)+ke−x+2ex≥0 e−x(2k+1)ex+k+2e2x≥0 2e2x(ex+k)+1(ex+k)≥0 Further simplify the above, (2ex+1)(ex+k)≥0 ex+k≥0 k≥0 So, the least value of k is 0