Answer: B, C, D
Setting d[B]/dt=0 gives k1[A]=k2[B], so [B]/[A]=k1/k2=0.2, and tmax=ln(k2/k1)/(k2−k1)=1.609/0.40=4.02 min. C forms from B, which is zero at t=0, so its initial rate is zero. The peak [B]/[A]0=(k2/k1)k2/(k1−k2)=0.134, not 0.20 (that is the ratio to the then-current [A]).