Arcs and angles, signs, extreme values, domain, period and graphs of circular functions (chapters I and II).
Teste 1ITA-72
The convex angle formed by the hour and minute hands at 10 hours and 15 minutes is:
At 10 h 15 min, the minute hand is at $90^\circ$ (from 12) and the hour hand is at $300^\circ + 15 \cdot 0{,}5^\circ = 307^\circ 30'$. The difference is $217^\circ 30'$, whose complement is $142^\circ 30'$.
Teste 2FUVEST-77
The acute angle formed by the hands of a clock at 1 hour and 12 minute is:
The minute hand moves $6^\circ$ per minute and the hour hand moves $0{,}5^\circ$. At 1 h 12 min: minutes in $72^\circ$, hours in $30^\circ + 6^\circ = 36^\circ$; the angle is $36^\circ$.
Teste 3ITA-73
Between 4 and 5 hours, the hour hand of a clock is twice at right angles to the minute hand. The times of these occurrences are:
At 4 h $m$ min, the hands are at $6m$ and $120 + 0{,}5m$ degrees. Right angle: $120 - 5{,}5m = 90 \Rightarrow m = \frac{60}{11} = 5\frac{5}{11}$; or $5{,}5m - 120 = 90 \Rightarrow m = \frac{420}{11} = 38\frac{2}{11}$.
Teste 4PUC-70
Since $\theta$ is an acute angle, then $\left(\frac{5\pi}{2} - \theta\right)$ belongs to:
$\frac{5\pi}{2} - \theta$ is congruous with $\frac{\pi}{2} - \theta$. The template assumes $\theta$ acute (the original statement just says "positive"): so $\frac{\pi}{2} - \theta$ is between $0$ and $\frac{\pi}{2}$, in the 1st quadrant. For any positive $\theta$, nothing could be said.
Teste 5UDESC-74
Arcs whose cosine is $\sqrt{2}$ can be in the quadrants:
There is no arc with cosine $\sqrt{2}$, because $\sqrt{2} > 1$.
Teste 6PUC-76
All values of $x$ for which the expression $\operatorname{sen}\theta = \dfrac{2x - 1}{3}$ exists are:
It takes $-1 \le \frac{2x - 1}{3} \le 1$, that is, $-3 \le 2x - 1 \le 3$, or $-1 \le x \le 2$.
Teste 7MACK-73
The set of real numbers $a$ for which the equation $\operatorname{sen} x = a + a^{-1}$ has a real solution in $x$ is:
For $a > 0$, $a + \frac{1}{a} \ge 2$; to $a < 0$, $a + \frac{1}{a} \le -2$. In no case does the value stay at $[-1, 1]$, so the equation never has a solution.
Teste 8CESCEM-77
If $x \in \left]\pi, \frac{3\pi}{2}\right[$ and $\cos x = 2k - 1$, then $k$ varies in the range:
In the 3rd quadrant, $-1 < \cos x < 0$: $-1 < 2k - 1 < 0 \Rightarrow 0 < k < \frac{1}{2}$.
Teste 9PUC-75
The numerical value of the expression $y = \cos 4x + \operatorname{sen} 2x + \operatorname{tg} 2x - \sec 8x$ to $x = \frac{\pi}{2}$ is:
The smallest value assumed by the expression $6 - \operatorname{sen} x$, for $x$ ranging from $0^\circ$ to $360^\circ$, is:
The largest value of $\operatorname{sen} x$ is 1, therefore the smallest value of $6 - \operatorname{sen} x$ is 5.
Teste 11MACK-76
The maximum value of $y = 2\operatorname{sen} x + \cos 2x$, for $0 \le x < \frac{\pi}{2}$, is:
With $s = \operatorname{sen} x \in [0, 1[$: $y = 2s + 1 - 2s^2$, vertex parabola in $s = \frac{1}{2}$, where $y = 1{,}5$.
Teste 12CESCEA-73
Select the true statement:
The tangent takes on all real values. In (b) $a \ge -1$ would be missing; in (c), the correct one is $|a| \ge 1$.
Teste 13CESCEM-72
The quadrants where the angles $\alpha$, $\beta$ and $\gamma$ are such as $\operatorname{sen}\alpha < 0$ and $\cos\alpha < 0$; $\cos\beta < 0$ and $\operatorname{tg}\beta < 0$; $\operatorname{sen}\gamma > 0$ and $\operatorname{cotg}\gamma > 0$ are, respectively:
Negative sine and cosine: 3rd quadrant. Negative cosine and tangent: 2nd. Positive sine and cotangent: 1st.
Teste 14SANTA CASA-77
If $F(x) = \cos x$, then:
The four numbers are in $[0, \pi]$, where the cosine is decreasing: $\frac{\sqrt{3}}{2} \approx 0{,}87 < \sqrt{2} \approx 1{,}41 < 1{,}5 < \frac{\pi}{2} \approx 1{,}57$, so the cosines come in reverse order. The printed template indicates (c), which is precisely the increasing order of the numbers, not the cosines.
Teste 15CESCEM-70
Highlight the true inequality for all $x$ (where the functions are defined):
$(|\cos x| + |\operatorname{sen} x|)^2 = 1 + 2|\operatorname{sen} x\cos x| \ge 1$. The others fail: in (d), for example, $\sec^2 x = 1 + \operatorname{tg}^2 x > \operatorname{tg}^2 x$.
Teste 16CESCEM-73
Among the statements below, one and only one is true. Mark it:
A square is never negative, and $(\operatorname{sen} x - \cos x)^2 = 1 - 2\operatorname{sen} x\cos x$. In (a), in the 1st quadrant the sine increases and the cosine decreases, but in the 4th both increase; (b) failure in $x = \frac{\pi}{2}$; (c): the period of $\operatorname{tg}\frac{x}{4}$ is $4\pi$.
Teste 17CESCEA-73
Let $x$ and $y$ be two real numbers such as $0 \le x < y < \frac{\pi}{2}$. Mark the statement false:
In the 1st quadrant the cosine is decreasing: $x < y \Rightarrow \cos x > \cos y$. The tangent and sine are increasing, and so is $2^t$.
Teste 18GV-70
Function $F(x) = \operatorname{sen} x \cdot \log_{\frac{1}{2}} x$ is:
In $]0, \pi[$ the sine is positive, and $\log_{1/2} x$ is positive for $0 < x < 1$ and negative for $x > 1$.
Teste 19POLI-68
If $x$ and $y$ satisfy $0 < x < y < \frac{\pi}{2}$ and $z = \operatorname{sen} x - \operatorname{tg} y \cdot \cos x$, then:
Fixed $x$, $\cos x > 0$ and $\operatorname{tg} y$ increase with $y$, therefore $z$ decreases. Note also that $z = \frac{\operatorname{sen}(x - y)}{\cos y} < 0$, which eliminates (c) and (d).
Teste 20CESCEM-73
Consider the sequence of real numbers that is obtained by doing $x = \dfrac{2}{\pi + 2n\pi}$ in the expression $y = \operatorname{sen}\dfrac{1}{x}$, $n = 0, 1, 2, \ldots$. It can be stated that:
$\frac{1}{x} = \frac{\pi}{2} + n\pi$, and $y = \operatorname{sen}\left(\frac{\pi}{2} + n\pi\right) = (-1)^n$: the sequence alternates between $1$ and $-1$.
Teste 21FFCLUSP-69
The solution of $\operatorname{sen}^2 x + \operatorname{sen}^4 x + \operatorname{sen}^6 x = 3$ is ($k$ any integer):
As $\operatorname{sen}^2 x \le 1$, each installment is at most 1, and the sum is only worth 3 if $\operatorname{sen}^2 x = 1$, that is, $x = \frac{\pi}{2} + k\pi = (2k + 1)\frac{\pi}{2}$.
Teste 22CESCEM-73
Consider the trigonometric equation $\operatorname{sen} x + \operatorname{sen} 2x = 2$. Then:
It would take $\operatorname{sen} x = 1$ and $\operatorname{sen} 2x = 1$ at the same time; but $\operatorname{sen} x = 1 \Rightarrow x = \frac{\pi}{2} + 2k\pi \Rightarrow \operatorname{sen} 2x = \operatorname{sen}\pi = 0$.
Teste 23CESCEA-72
Let $A \subset B = \{x \in \mathbb{R} \mid 0 \le x \le 2\pi\}$ be the domain of the function $f$ given by $f(x) = \dfrac{1 - \operatorname{sen}^2 x}{1 + \operatorname{sen} x}$. Then $A$ is equal to:
You only need $1 + \operatorname{sen} x \ne 0$, that is, $x \ne \frac{3\pi}{2}$ in the range $[0, 2\pi]$.
Teste 24GV-74
Let $n$ be the number of points in the set $\{x \in \mathbb{R} \mid 0 \le x \le 2\pi\}$ in which $\dfrac{\operatorname{tg} x}{\operatorname{sen} 4x}$ is not defined. Then $n$ is equal to:
The expression is not defined where $\cos x = 0$ or $\operatorname{sen} 4x = 0$. The second condition gives $x = \frac{k\pi}{4}$, $k = 0, 1, \ldots, 8$ (nine points), and it already includes $\frac{\pi}{2}$ and $\frac{3\pi}{2}$.
Teste 25CESCEA-75
Select the correct statement:
The secant is $\frac{1}{\cos x}$ and $|\cos x| \le 1$, therefore $|\sec x| \ge 1$. The rest fail: the tangent is not defined in $\frac{\pi}{2} + k\pi$; the cotangent is decreasing and does not exist in $k\pi$; the cosecant also takes values $\le -1$; the sine is not always increasing.
Teste 26CESCEM-71
Which of the following sets of values of $x$ could constitute a domain for the function $\log(\operatorname{sen} x)$?
The logarithm requires $\operatorname{sen} x > 0$, which happens in $\left]\frac{\pi}{2}, \pi\right[$ (2nd quadrant). In the other sets there are points with $\operatorname{sen} x \le 0$.
Teste 27CESCEM-75
The function that best fits the graph below is:
The graph passes through the origin, goes up to 1 in $x = \pi$ and has period $4\pi$: it is $y = \operatorname{sen}\frac{x}{2}$, whose period is $\frac{2\pi}{1/2} = 4\pi$.
Teste 28CESCEA-73
The figure is a sketch of the graph of the function:
The curve is worth 1 in $x = 0$, zero in $\pm\frac{\pi}{4}$ and $-1$ in $\pm\frac{\pi}{2}$: it is $\cos 2x$.
Teste 29CESCEM-73
Which of the functions below best fits the graph?
The graph vanishes at $x = k\pi$, is always $\ge 0$ and has vertical asymptotes at $\frac{\pi}{2} + k\pi$: it is $|\sec x| - 1$.
Teste 30MACK-77
The graph below may be of the function:
The function is valid 0 in $0$, $\pi$ and $2\pi$ and is valid 1 (with nozzle) in $\frac{\pi}{2}$ and $\frac{3\pi}{2}$: it is $1 - |\cos x|$. (The printed template indicates (c), but $1 - |\operatorname{sen} x|$ is equal to 1 in $x = 0$, and the graph starts at zero.)
Teste 31GV-74
The equations below represent curves in a Cartesian system of axes $x$ and $y$. Only one of these curves does not pass through the point $x = -0{,}5$, $y = 2$:
The period of $a\cos(cx)$ is $\frac{2\pi}{|c|} = \frac{2\pi}{4} = \frac{\pi}{2}$.
Teste 33EAESP-GV-77
The period of the function given by $y = 3\operatorname{sen}\left(2\pi x + \frac{\pi}{2}\right)$ is:
The period is $\frac{2\pi}{2\pi} = 1$.
Teste 34SANTA CASA-73
Regarding the function $y = 2\operatorname{sen} x + 3\cos\frac{x}{2}$, it can be stated that:
$2\operatorname{sen} x$ has period $2\pi$ and $3\cos\frac{x}{2}$ has period $4\pi$; the sum is repeated every $4\pi$ (but not every $2\pi$, because $\cos\frac{x + 2\pi}{2} = -\cos\frac{x}{2}$).
Teste 35CESCEA-74
The domain, range and period of the function $f(x) = \operatorname{tg}\left(x - \frac{\pi}{4}\right)$ are, respectively:
It takes $x - \frac{\pi}{4} \ne \frac{\pi}{2} + k\pi$, that is, $x \ne \frac{3\pi}{4} + k\pi$ — the same set as $x \ne -\frac{\pi}{4} - k\pi$. The image is $\mathbb{R}$ and the period is $\pi$ (shifting does not change either).
Teste 36CESCEM-73
A straight line through the origin, with a negative angular coefficient, has three and only three points in common with the graph of the function $y = \operatorname{sen} x$. The smallest of the three corresponding abscissa:
For $x > 0$ the line is negative and can only meet the sine in the negative arcs ($\pi < x < 2\pi$, …), where the sine begins and ends above the line: the encounters come in pairs, unless the line is tangent. With exactly three points (the origin and two symmetrical ones, as the two functions are odd), the line is tangent at a point $x_0$ with $\operatorname{tg} x_0 = x_0$, $x_0 \approx 4{,}49$, between $\pi$ and $\frac{3\pi}{2}$. The smallest abscissa is $-x_0$.
Teste 37MACK-74
The intersection of the graphs of the sine and tangent functions for $0 < x < \pi$:
$\operatorname{sen} x = \operatorname{tg} x \iff \operatorname{sen} x\left(1 - \frac{1}{\cos x}\right) = 0$. In $]0, \pi[$, $\operatorname{sen} x \ne 0$ and $\cos x \ne 1$; there is no solution.
Teste 38CESCEA-75
Given the curves $y = x^2$ and $y = \cos x$, select the true one among the statements below:
$x^2 \ge 0$ and $\cos x \le 1$: there can only be a meeting with $|x| \le 1$. In this interval $x^2$ increases from 0 to 1 and $\cos x$ decreases from 1 to $\cos 1 \approx 0{,}54$ (to $x \ge 0$): one point on each side, for symmetry. There are two points ($x \approx \pm 0{,}82$).
Teste 39MACK-75
The number of points of intersection of the graphs of the functions $f$ and $g$ given by $f(x) = -|\cos x|$ and $g(x) = \cos\left(\frac{\pi}{2} + x\right)$, with $-\pi < x < \pi$, is:
$g(x) = \cos\left(\frac{\pi}{2} + x\right) = -\operatorname{sen} x$. Equality $-|\cos x| = -\operatorname{sen} x$ requires $\operatorname{sen} x = |\cos x| \ge 0$; in $]-\pi, \pi[$ this gives $x = \frac{\pi}{4}$ or $x = \frac{3\pi}{4}$.
Teste 40FAAP-74
For each $t \in [0, \pi]$, $A(t) = 1 - \cos t$ represents the area under the graph of $f(x) = \operatorname{sen} x$, above the $x$ axis, between $0$ and $t$. Based on this, the area under the graph of $f$ (above the $x$ axis) with $x \in \left[\frac{\pi}{4}, \frac{\pi}{2}\right]$ is:
The area between $\frac{\pi}{4}$ and $\frac{\pi}{2}$ is the difference $A\left(\frac{\pi}{2}\right) - A\left(\frac{\pi}{4}\right) = (1 - 0) - \left(1 - \frac{\sqrt{2}}{2}\right) = \frac{\sqrt{2}}{2}$.
