The efficiency of a Carnot engine operating between the temperature T1  and T2 (T1>T2)  isη0 . The temperature of sink is decreased by αK  and efficiency isη1 . The temperature of source is increased by αK  and efficiency is η2

If temperature of the source is increased by αK  and temperature of the sink is decreased by the same amount i.e. α  K  then efficiency increases more in the second case. Since η0=1−T2T1 According to the problemη1=1−(T2−αT1) ⇒ η1=(1−T2T1)+αT1 ⇒ η1=η0+αT1=η0T1+αT1 ⇒ η1>η0 Further,η2=1−T2T1+α ⇒ η2=T1+α−T2T1+α ⇒ η2=η0T1+αT1+α        {∵T1−T2=η0T1} ⇒ η2<η1 ⇒ η1>η2>η0