The efficiency of a Carnot engine operating between the temperature $T_1$  and $T_2$ $\left(T_1>T_2\right)$  is${\eta }_0$ . The temperature of sink is decreased by $\alpha K$  and efficiency is${\eta }_1$ . The temperature of source is increased by $\alpha K$  and efficiency is ${\eta }_2$

If temperature of the source is increased by $\alpha K$  and temperature of the sink is decreased by the same amount i.e. $\alpha$  $K$  then efficiency increases more in the second case.

Since ${\eta }_0=1-\frac{T_2}{T_1}$

According to the problem

${\eta }_1=1-\left(\frac{T_2-\alpha }{T_1}\right)$

$\Rightarrow$ ${\eta }_1=\left(1-\frac{T_2}{T_1}\right)+\frac{\alpha }{T_1}$

$\Rightarrow$ ${\eta }_1={\eta }_0+\frac{\alpha }{T_1}=\frac{{\eta }_0{T}_1+\alpha }{T_1}$

$\Rightarrow$ ${\eta }_1>{\eta }_0$

Further,

${\eta }_2=1-\frac{T_2}{T_1+\alpha }$

$\Rightarrow$ ${\eta }_2=\frac{T_1+\alpha -T_2}{T_1+\alpha }$

$\Rightarrow$ ${\eta }_2=\frac{{\eta }_0{T}_1+\alpha }{T_1+\alpha }$        $\left\{\because T_1-T_2={\eta }_0{T}_1\right\}$

$\Rightarrow$ ${\eta }_2<{\eta }_1$

$\Rightarrow$ ${\eta }_1>{\eta }_2>{\eta }_0$