JEE Main 2022 — Mathematics Calculus

The area enclosed by the curves $y={\mathrm{log}}_{e}(x+{e}^{2}),x={\mathrm{log}}_{e}(\frac{2}{y})$y=loge​(x+e2),x=loge​(y2​) and $x={\mathrm{log}}_{e}2$x=loge​2, above the line $y=1$y=1 is

Now plotting the diagram of given functions we get,

According to JEE Mains, we have to calculate the required region ${A}_{2}$A2​ which is shaded in crossed lines and comes out to be

$${A}_{2}={\int }_{1}^{2}(\mathrm{ln}\frac{2}{y}-{e}^{y}+{e}^{2})dy=1+e-\mathrm{ln}2$$A2​=∫12​(lny2​−ey+e2)dy=1+e−ln2

But according to the question the required region ${A}_{1}$A1​ comes out to be shaded in parallel lines, which can be obtained as

$${A}_{1}={\int }_{0}^{\mathrm{ln}2}(\mathrm{ln}(x+{e}^{2})-2{e}^{-x})dx$$A1​=∫0ln2​(ln(x+e2)−2e−x)dx

$${={(x+{e}^{2})\mathrm{ln}(x+{e}^{2})-x+2{e}^{-x}}}_{0}^{\text{ln}2}$$=(x+e2)ln(x+e2)−x+2e−x0ln2​

$$=(\mathrm{ln}2+{e}^{2})\mathrm{ln}(\mathrm{ln}2+{e}^{2})-\mathrm{ln}2+1-2{e}^{2}-2$$=(ln2+e2)ln(ln2+e2)−ln2+1−2e2−2

$$=(\mathrm{ln}2+{e}^{2})\mathrm{ln}(\mathrm{ln}2+{e}^{2})-\mathrm{ln}2-2{e}^{2}-1$$=(ln2+e2)ln(ln2+e2)−ln2−2e2−1

Not given in any option.