Subject: Mathematics — Trigonometry
Question: Consider the curve $C_{1}$ given by
$$y = e^{- x}\text{~for~}x \in \lbrack 0,10\pi\rbrack$$
and the curve $C_{2}$ given by
$$y = e^{- x}(sinx + cosx)\text{~for~}x \in \lbrack 0,10\pi\rbrack$$
Let $n$ be the total number of points of intersection of the curves
$C_{1}$ and $C_{2}$.\
Suppose that
$\alpha_{1},\alpha_{2},\ldots,\alpha_{n}, \in \lbrack 0,10\pi\rbrack$
are the $x$ - coordinates of the points of intersection of the curves
$C_{1}$ and $C_{2}$ such that
$$\alpha_{1} < \alpha_{2} < \ldots < \alpha_{n}.$$
The value of $n$ is $\_\_\_\_$ .
Answer: 11.00\
Solution: $e^{- x} = e^{- x}(sinx + cosx)$
$${sinx + cosx = 1
}{cos\left( x - \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}
}{x - \frac{\pi}{4} = 2n\pi \pm \frac{\pi}{4}
}$$$x = 2n\pi$ or $2n\pi + \frac{\pi}{2}$\
$$x \in \left\{ 0,\frac{\pi}{2},2\pi,\frac{5\pi}{2},4\pi,\frac{9\pi}{2},6\pi,\frac{13\pi}{2},8\pi,\frac{17\pi}{2},10\pi \right\}
$$so $n = 11$