Question

The area enclosed by the curves

$\begin{array}{l} y=\text{log}_e\left(x+e^2\right),\ x=\text{log}_e\left(\frac{2}{y}\right)\text{ and }x=\text{ log }_e\ 2,\end{array}$

above the line y = 1 is

• A. 2 + e - loge 2
• B. 1+ e - loge2
• C. 2 + e + loge 2
• D. 1 - e - loge2

Answer 1

Answer: (B)

1+ e - loge2

Answer 2

The required region A , which is shaded in crossed lines and comes out to be

$\begin{array}{l} A=\displaystyle\int\limits_1^2\left(\text{In }\frac{2}{y}-e^y+e^2\right)dy=1+e-\text{In}2 \end{array}$

But according to us the required region A comes out to be shaded in parallel lines, which can be obtained as

$\begin{array}{l} A=\displaystyle\int\limits_0^{\text{In }2}\left(\text{In}\left(x+e^2\right)-2e^{-x}\right)dx\end{array}$

\begin{array}{l} \left. =\left\\{\left(x+e^2\right)\text{In}\left(x+e^2\right)-x+2e^{-x} \right\\} \right|_0^{\text{In }2}\end{array}

$\begin{array}{l} =\left(\text{In}2+e^2\right)\text{In}\left(\text{In2}+e^2\right)-\text{In}2+1-2e^2 – 2\end{array}$
$\begin{array}{l} =\left(\text{In}2+e^2\right)\text{In}\left(\text{In}2-e^2\right)-\text{In}2-2e^2-1\end{array}$

Not given in any option.

The region asked in the question is bounded by three curves

$\begin{array}{l} y=\text{In}\left(x+e^2\right)\end{array} \begin{array}{l} x=\text{In}\left(\frac{2}{y}\right) \end{array} \begin{array}{l} x=\text{In}2\end{array}$

There is only one region which satisfies above requirement and which also lies above line y = 1

Line y = 1 may or may not be the boundary of the region.