Question Stem for Question 15 and 16:

Consider the curve $C_1$ given by $y=e^{-x}$ for $x\in [0,10\pi ]$, and the curve $C_2$ given by $y=e^{-x(\sin x+\cos x)}$ for $x\in [0,10\pi ].\; Let$ n$be\; the\; total\; number\; of\; points\; of\; intersection\; of\; the\; curves$ C_1$and$ C_2$.$

Suppose that ${\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n\in [0,10\pi ]$ are the $x$ -coordinates of the points of intersection of the curves $C_1$ and $C_2$ such that ${\alpha }_1<{\alpha }_2<\dots <{\alpha }_n$.

Let β be the area of the region enclosed between the curves C₁, C₂, and the lines x = α₁ and x = α₄. Then the value of $-\frac{1}{\pi }\ln (\beta -2e^{-\pi /2})$ is __________.

Correct Answer: The correct answer is 2.

Step-by-Step Explanation:

Step 1: Determine the points of intersection of the two curves.

The given curves are:

$C_1:y=e^{-x}$

$C_2:y=e^{-x}(\sin x+\cos x)$

To find the x-coordinates of the points of intersection, we equate the two expressions for $y$:

$e^{-x}=e^{-x}(\sin x+\cos x)$

Since $e^{-x}>0$ for all real $x$, we can divide both sides by $e^{-x}$:

$\sin x+\cos x=1$

Rewriting this trigonometric equation:

$\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x=\frac{1}{\sqrt{2}}$

$\cos \left(x-\frac{\pi }{4}\right)=\cos \left(\frac{\pi }{4}\right)$

The general solution is $x-\frac{\pi }{4}=2k\pi \pm \frac{\pi }{4}$ for integer $k$, which gives:

$x=2k\pi \text{ or }x=2k\pi +\frac{\pi }{2}$

Listing the first four points of intersection in ascending order in the interval $[0,10\pi ]$:

${\alpha }_1=0$
${\alpha }_2=\frac{\pi }{2}$
${\alpha }_3=2\pi$
${\alpha }_4=\frac{5\pi }{2}$

Step 2: Calculate the area $\beta$ of the enclosed region.

The area $\beta$ between $x={\alpha }_1=0$ and $x={\alpha }_4=\frac{5\pi }{2}$ is split into three sub-intervals:

1. On $[0,\pi /2]$, $\sin x+\cos x\ge 1$, so $C_2\ge C_1$.

2. On $[\pi /2,2\pi ]$, $\sin x+\cos x\le 1$, so $C_1\ge C_2$.

3. On $[2\pi ,5\pi /2]$, $\sin x+\cos x\ge 1$, so $C_2\ge C_1$.

Notice that $\int e^{-x}(\sin x+\cos x)dx=-e^{-x}\cos x$.

Calculating the area for each interval:

$A_1={\int }_0^{\pi /2}\left(e^{-x}(\sin x+\cos x)-e^{-x}\right)dx={\left[-e^{-x}\cos x+e^{-x}\right]}_{}^{}0\pi /2=e^{-\pi /2}$

$A_2={\int }_{\pi /2}^{2\pi }\left(e^{-x}-e^{-x}(\sin x+\cos x)\right)dx={\left[-e^{-x}+e^{-x}\cos x\right]}_{}^{}\pi /22\pi =e^{-\pi /2}$

$A_3={\int }_{2\pi }^{5\pi /2}\left(e^{-x}(\sin x+\cos x)-e^{-x}\right)dx=e^{-2\pi }$

Summing up all three regions gives the total area $\beta$:

$\beta =A_1+A_2+A_3=2e^{-\pi /2}+e^{-2\pi }$

Step 3: Evaluate the required expression.

We are asked to find the value of:

$-\frac{1}{\pi }\ln (\beta -2e^{-\pi /2})$

Substituting $\beta -2e^{-\pi /2}=e^{-2\pi }$ into the expression:

$-\frac{1}{\pi }\ln \left(e^{-2\pi }\right)=-\frac{1}{\pi }(-2\pi )=2$

Thus, the final answer is 2.