극한·미분 심화 전체 문제

$1$

diverges

$1/2$

$1/2$

diverges; Limit Comparison Test with $1/x$.

converges; Limit Comparison Test with $1/x^{3/2}$.

diverges; Limit Comparison Test with $1/x$.

converges; Direct Comparison Test with $xe^{-x}$.

converges; Direct Comparison Test with $e^{-x}$.

converges; Direct Comparison Test with $xe^{-x}$.

converges; Direct Comparison Test with $1/(x^2-1)$.

diverges; Direct Comparison Test with $x/(x^2+1)$.

converges; Direct Comparison Test with $1/e^x$.

converges; Limit Comparison Test with $1/e^x$.

(a) We can plug any number into the cosine function, and it will return a number in $[-1,1]$. The domain of $\arcsin x$ is $[-1,1]$, so any number we plug into cosine will give us a valid number to plug into arcsine. So, the domain of $f(x)$ is all real numbers.

(b) We can plug any number into the cosine function, and it will return a number in $[-1,1]$. The domain of $\mathop{\mathrm{arccsc}} x$ is $(-\infty,-1] \cup [1,\infty)$, so in order to have a valid number to plug into arccosecant, we need $\cos x = \pm 1$. That is, the domain of $g(x)$ is all values $x=n\pi$ for some integer $n$.

(c) The domain of arccosine is $[-1,1]$. The domain of sine is all real numbers, so no matter what number arccosine spits out, we can safely plug it into sine. So, the domain of $h(x)$ is $[-1,1]$.

False: $\cos t=1$ for infinitely many values of $t$; arccosine gives only the single value $t=0$ for which $\cos t=1$ and $0 \leq t \leq \pi$. The particle does not start moving until $t=10$, so $t=0$ is not in the domain of the function describing its motion.

The particle will have height $1$ at time $2\pi n$, for any integer $n \geq 2$.

First, we restrict the domain of $f$ to force it to be one–to–one. There are many intervals we could choose over which $f$ is one–to–one, but the question asks us to contain $x=0$ and be as large as possible; this leaves us with the following restricted function:

The inverse of a function swaps the role of the input and output; so if the graph of $y=f(x)$ contains the point $(a,b)$, then the graph of $Y=f^{-1}(X)$ contains the point $(b,a)$. That is, the graph of $Y=f^{-1}(X)$ is the graph of $y=f(x)$ with the $x$-coordinates and $y$-coordinates swapped. (So, since $y=f(x)$ crosses the $y$-axis at $y=1$, then $Y=f^{-1}(X)$ crosses the $X$-axis at $X=1$.) This swapping is equivalent to reflecting the curve $y=f(x)$ over the line $y=x$.

Remark: while you're getting accustomed to inverse functions, it is sometimes clearer to consider $y=f(x)$ and $Y=f^{-1}(X)$: using slightly different notations for $x$ (the input of $f$, hence the output of $f^{-1}$) and $X$ (the input of $f^{-1}$, which comes from the output of $f$). However, the convention is to use $x$ for the inputs of both functions, and $y$ as the outputs of both functions, as is written on the graph above.

The tangent line is horizontal when $0=y'=a-\sin x$. That is, when $a=\sin x$.

• If $|a|>1$, then there is no value of $x$ for which $a=\sin x$, so the curve has no horizontal tangent lines.
• If $|a| = 1$, then there are infinitely many solutions to $a=\sin x$, but only one solution in the interval $[-\pi,\pi]$: $x=\arcsin(a)=\arcsin(\pm1)=\pm\frac{\pi}{2}$. Then the values of $x$ for which $a=\sin x$ are $x=2\pi n +a \frac{\pi}{2}$ for any integer $n$.
• If $|a|<1$, then there are infinitely many solutions to $a=\sin x$. The solution in the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ is given by $x=\arcsin(a)$. The other solution in the interval $\left(-\pi,\pi\right)$ is given by $x=\pi-\arcsin(a)$, as shown in the unit circles below.

So, the values of $x$ for which $x=\sin a$ are $x=2\pi n+\arcsin(a)$ and $x=2\pi n + \pi - \arcsin (a)$ for any integer $n$.

Remark: when $a=1$, then $$2\pi n+\arcsin(a) = 2\pi n + \dfrac{\pi}{2}=2\pi n +\pi -\left(\dfrac {\pi}{2}\right)=2\pi n+\pi-\arcsin(a).$$ Similarly, when $a=-1$, $$2\pi n+\arcsin(a) = 2\pi n - \dfrac{\pi}{2}=2\pi (n-1) +\pi -\left(-\dfrac {\pi}{2}\right) =2\pi(n-1)+\pi-\arcsin(a).$$ So, if we try to use the descriptions in the third bullet point to describe points where the tangent line is horizontal when $|a|=1$, we get the correct points but each point is listed twice. This is why we separated the case $|a|=1$ from the case $|a|<1$.

Using the chain rule,

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\left\{\arcsin\left(\frac{x}{3}\right)\right\}&=\frac{1}{\sqrt{1-\left(\frac{x}{3}\right)^2}}\cdot \frac{1}{3}\\ &=\frac{1}{3\sqrt{1-\frac{x^2}{9}}}\\ &=\frac{1}{\sqrt{9-x^2}} \end{aligned}$$

Since the domain of arcsine is $[-1,1]$, and we are plugging in $\dfrac{x}{3}$ to arcsine, the values of $x$ that we can plug in are those that satisfy $-1 \le \dfrac{x}{3} \leq 1$, or $-3\leq x \leq 3$. So the domain of $f$ is $[-3,3]$.

Using the quotient rule,

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\left\{\dfrac{\arccos t}{t^2-1}\right\}&= \frac{(t^2-1)\left(\frac{-1}{\sqrt{1-t^2}}\right)-(\arccos t)(2t)}{(t^2-1)^2} \end{aligned}$$

The domain of arccosine is $[-1,1]$, and since $t^2-1$ is in the denominator, the domain of $f$ requires $t^2-1 \neq 0$, that is, $t \neq \pm 1$. So the domain of $f(t)$ is $(-1,1)$.