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Example 9. Find the residue of the function
$$
f(z)=e^{1 / z^{2}} \cos z
$$
at the point $z=0$. | Solution. Since the residue at the point $z=0$ is equal to the coefficient of $z^{-1}$, we immediately obtain that in this case the residue is zero, since the function $f(z)$ is even and its expansion in the neighborhood of the point $z=0$ cannot contain odd powers of $z$.
## Problems for Independent Solution
Find th... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 4. Compute the integral
$$
\int_{|x|=2} \frac{1}{z-1} \sin \frac{1}{z} d z
$$ | Solution. In the circle $|z| \leqslant 2$, the integrand has two singular points $z=1$ and $z=0$. It is easy to establish that $z=1$ is a simple pole, therefore
$$
\operatorname{res}\left(\frac{1}{z-1} \sin \frac{1}{z}\right)=\left.\frac{\sin \frac{1}{z}}{(z-1)^{\prime}}\right|_{z=1}=\sin 1
$$
To determine the nature... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 6. Compute the integral
$$
I=\int_{|z|=2} \frac{d z}{1+z^{4}}
$$ | The poles (finite) of the integrand
$$
f(z)=\frac{1}{1+z^{4}}
$$
are the roots $z_{1}, z_{2}, z_{3}, z_{4}$ of the equation $z^{4}=-1$, which all lie inside the circle $|z|=2$. The function $f(z)=\frac{1}{1+z^{4}}$ has an expansion in the neighborhood of the infinitely distant point
$$
f(z)=\frac{1}{1+z^{4}}=\frac{1... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 2. Find the logarithmic residue of the function
$$
f(z)=\frac{\operatorname{ch} z}{e^{i z}-1}
$$
with respect to the contour $C:|z|=8$. | Solution. We find the zeros $z_{k}$ of the function $f(z)$. For this, we solve the equation $\cosh z=0$ or $e^{z}+e^{-z}=0$. Writing the last equation as $e^{2 z}=-1$, we find
$2 z=\operatorname{Ln}(-1)=(2 k+1) \pi i$, so $z_{k}=\frac{2 k+1}{2} \pi i(k=0, \pm 1, \pm 2, \ldots)$ (all zeros are simple). To find the poles... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 3. Find the logarithmic residue of the function
$$
f(z)=\frac{1+z^{2}}{1-\cos 2 \pi z}
$$
with respect to the circle $|z|=\pi$. | Solution. Setting $1+z^{2}=0$, we find two simple zeros of the function $f(z): a_{1}=-i, a_{2}=i$. Setting $1-\cos 2 \pi z=0$, we find the poles of the function $f(z): z_{n}=n, n=0, \pm 1, \pm 2, \ldots$. The multiplicity of the poles is $k=2$.
In the circle $|z|<\pi$, the function has two simple zeros $a_{1}=-i, a_{2... | -12 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 4. Find the number of roots in the right half-plane $\operatorname{Re} z>0$ of the equation
$$
Q_{5}(z) \equiv z^{5}+z^{4}+2 z^{3}-8 z-1=0
$$ | Solution. By the argument principle, the number of zeros inside the contour $C$ is
$$
N=\frac{1}{2 \pi} \Delta_{C} \operatorname{Arg} Q_{5}(z)
$$
where the contour $C$ consists of the semicircle $C_{R}:|z|=R, \operatorname{Re} z>0$, and its diameter on the imaginary axis; the radius $R$ is taken to be so large that a... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 5. Find the number of roots of the equation
$$
Q_{7}(z) \equiv z^{7}-2 z-5=0
$$
in the right half-plane. | Solution. We choose the contour $C$ as indicated in Example 4. Then $\Delta_{C_{R}} \operatorname{Arg} Q_{7}(z)=\Delta_{C_{R}} \operatorname{Arg}\left(z^{7}-2 z-5\right)=$
$$
\begin{aligned}
& =\Delta_{C_{R}} \operatorname{Arg}\left[z^{7}\left(1-\frac{2}{z^{6}}-\frac{5}{z^{7}}\right)\right]=7 \Delta_{C_{R}} \operatorn... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 6. Find the number of zeros of the function
$$
F(z)=z^{8}-4 z^{5}+z^{2}-1
$$
inside the unit circle $|z|<1$. | Solution. Let us represent the function $F(z)$ as the sum of two functions $f(z)$ and $\varphi(z)$, which we choose, for example, as follows:
$$
f(z)=-4 z^{5}, \quad \varphi(z)=z^{8}+z^{2}-1
$$
Then on the circle $|z|=1$ we will have
$$
\begin{aligned}
& |f(z)|=\left|-4 z^{5}\right|=4 \\
& |\varphi(z)|=\left|z^{8}+z... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 7. Determine the number of roots of the equation
$$
z^{6}-6 z+10=0
$$
inside the circle $|z|<1$. | Solution. Let, for example, $f(z)=10$ and $\varphi(z)=z^{6}-6 z$. On the circle $|z|=1$ we have
$$
|f(z)|=10, \quad|\varphi(z)|=\left|z^{6}-6 z\right| \leqslant\left|z^{6}\right|+6|z|=7
$$
Thus, in all points of the circle $|z|=1$, the inequality $|f(z)|>|\varphi(z)|$ holds. The function $f(z)=10$ has no zeros inside... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 8. How many roots of the equation
$$
z^{4}-5 z+1=0
$$
lie in the annulus $1<|z|<2 ?$ | Solution. Let $N$ be the number of roots of equation (4) in the ring $1<|\varphi(z)|$, since $|f(z)|=|-5 z|=5,|\varphi(z)|=\left|z^{4}+1\right| \leqslant$ $\left|z^{4}\right|+1=2$. The function $f(z)=-5 z$ has one root in the circle $|z|<1$, and thus $N_{1}=1$.
In the circle $|z|<2$, $|f(z)|>\left|\varphi(z)\right|$, ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 9. Find the number of roots of the equation
$$
z^{2}-a e^{z}=0, \quad \text { where } \quad 0<a<e^{-1}
$$
in the unit circle $|z|<1$. | Solution. Let $f(z)=z^{2}$ and $\varphi(z)=-a e^{z}$. On the circle $|z|=1$ we have
$$
\begin{aligned}
& |f(z)|=\left|z^{2}\right|=1 \\
& |\varphi(z)|=\left|-a e^{z}\right|=a\left|e^{z}\right|=a\left|e^{x+i y}\right|=a e^{x} \leqslant a e|\varphi(z)|$, if $|z|=1$. The function $f(z)=z^{2}$ in the circle $|z|0, \quad \... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 10. Find the number of roots of the equation
$$
\lambda-\boldsymbol{z}-e^{-z}=0, \quad \lambda>1
$$
in the right half-plane $\operatorname{Re} z>0$. | Solution. Consider the contour composed of the segment $[-i R, i R]$ and the right semicircle $|z|=R$. Let $f(z)=z-\lambda$ and $\varphi(z)=e^{-z}$. On the segment $[-i R, i R]$, where $z=i y$, we have
$$
\begin{aligned}
& |f(z)|=|i y-\lambda|=\sqrt{\lambda^{2}+y^{2}} \geqslant \sqrt{\lambda^{2}}=\lambda>1 \\
& |\varp... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 1. Find the approximate value of the smallest characteristic number of the kernel by the Ritz method
$$
K(x, t)=x t ; \quad a=0, b=1
$$ | Solution. As the coordinate system of functions $\psi_{n}(x)$, we choose the system of Legendre polynomials: $\psi_{n}(x)=P_{n}(2 x-1)$. In formula (1), we limit ourselves to two terms, so that
$$
\varphi_{2}(x)=a_{1} \cdot P_{0}(2 x-1)+a_{2} \cdot P_{1}(2 x-1) .
$$
Noting that
$$
\psi_{1} \equiv P_{0}(2 x-1)=1 ; \q... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 3. Using the Kellogg method, calculate the smallest characteristic number of the kernel $K(x, t)=x^{2} t^{2}, 0 \leqslant x, t \leqslant 1$. | Solution. Let $\omega(x)=x$. Then
$$
\begin{aligned}
& \omega_{1}(x)=\int_{0}^{1} x^{2} t^{2} t d t=\frac{x^{2}}{4} \\
& \omega_{2}(x)=\int_{0}^{1} x^{2} t^{4} \frac{1}{4} d t=\frac{1}{4} x^{2} \cdot \frac{1}{5} \\
& \omega_{3}(x)=\int_{0}^{1} \frac{1}{4 \cdot 5} x^{2} t^{4} d t=\frac{1}{4 \cdot 5^{2}} x^{2} \\
& \ldo... | 5 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 6. Find the length of the arc of the curve $y=\arcsin \sqrt{x}-\sqrt{x-x^{2}}$ from $x_{1}=0$ to $x_{2}=1$. | Solution. We have $y^{\prime}=\frac{1}{2 \sqrt{x} \sqrt{1-x}}-\frac{1-2 x}{2 \sqrt{x-x^{2}}}=\frac{x}{\sqrt{x-x^{2}}}=$ $=\sqrt{\frac{x}{1-x}}$ $L=\int_{0}^{1} \sqrt{1+y^{\prime 2}} d x=\int_{0}^{1} \sqrt{1+\frac{x}{1-x}} d x=\int_{0}^{1} \frac{1}{\sqrt{1-x}} d x=-\left.2 \sqrt{1-x}\right|_{0} ^{1}=2$.
Answer. $L=2$. | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 6. Calculate the areas of figures bounded by the curves:
a) $\left\{\begin{array}{l}y=x \sqrt{9-x^{2}} \\ y=0(0 \leqslant x \leqslant 3)\end{array}\right.$
b) $\left\{\begin{array}{l}y=2 x-x^{2}+3 \\ y=x^{2}-4 x+3\end{array}\right.$ | Solution. We will construct the corresponding regions (Fig. 3.8) and determine the appropriate limits of integration from them, omitting the systems of inequalities. Thus:
a) $S=\int_{D} \int d x d y=\int_{0}^{3} d x \int_{0}^{x \sqrt{9-x^{2}}} d y=$
$=\left.\int_{0}^{3} d x \cdot y\right|_{0} ^{x \sqrt{9-x^{2}}}=\in... | 9 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 3. Calculate the mass of the surface $z=x y$, located inside the cylinder $x^{2}+\frac{y^{2}}{4}=1$, if the density is $\rho=\frac{|z|}{\sqrt{1+x^{2}+y^{2}}}$. | Solution. Given the symmetries of the integration region $\sigma$: $x^{2}+\frac{y^{2}}{4} \leqslant 1$, the equation of the surface, and the density function, it is sufficient to compute the integral over one quarter of the region and multiply the result by 4: $m=4 \int_{D} \rho(x, y, z) d \sigma$, where $D$ is the qua... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 1. Calculate the mass of the tetrahedron bounded by the planes $x=0, y=0, z=0$ and $x / 10+y / 8+z / 3=1$, if the density distribution of mass at each point is given by the function $\rho=(1+x / 10+y / 8+z / 3)^{-6}$. | Solution. We have $m=\iint_{W} \int \rho d V$. The triple integral is reduced to a double and a definite one (see point $4^{\circ}$):
$$
m=\iint_{D} d x d y \int_{0}^{z} \frac{d z}{\left(1+\frac{x}{10}+\frac{y}{8}+\frac{z}{3}\right)^{6}}
$$
The upper limit, or the exit point from the region, is the ordinate of the po... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 8. Check the conditions of Green's theorem for the line integral $\int_{L} 2 x y d x + x^{2} d y$ and compute this integral along the parabola $y=\frac{x^{2}}{4}$ from the origin to the point $A(2,1)$. | Solution. We have $P(x, y)=2 x y, Q(x, y)=x^{2}$. These functions are defined, continuous, and differentiable at any point $(x, y)$ in the plane. We have $\frac{\partial P}{\partial y}=2 x, \frac{\partial Q}{\partial x}=2 x$. The conditions of Green's theorem are satisfied. Therefore, the given integral is independent ... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 10. Calculate the integral $I=\oint_{L}\left(x^{2}-y^{2}\right) d x+2 x y d y$, where $L-$ is the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. | Solution. We apply Green's formula and compute the double integral, transitioning to "generalized" polar coordinates. We have: $P=x^{2}-y^{2}, Q=2 x y, \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=2 y+2 y=4 y$. Therefore,
$$
\begin{aligned}
& \int\left(x^{2}-y^{2}\right) d x+2 x y d y=4 \iint_{D} y d x ... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 1. Investigate the convergence of the series $\sum_{n=2}^{\infty} \frac{18}{n^{2}+n-2}$ and, if possible, find its sum. | Solution. We have $\frac{18}{n^{2}+n-2}=\frac{6}{n-1}-\frac{6}{n+2}$. Let's form the partial sum and find its limit. Notice which terms cancel each other out (!). We have:
$$
\begin{aligned}
& S_{n}=u_{2}+u_{3}+\ldots+u_{n}= \\
& =6\left[\left(\frac{1}{1}-\frac{1}{A}\right)+\left(\frac{1}{2}-\frac{1}{b}\right)+\left(\... | 11 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. Find the most probable number of hits in the ring in five throws, if the probability of hitting the ring with the ball in one throw is $p=0.6$. | Solution. We have $n=5 ; p=0.6 ; q=0.4$. For the number $k_{0}$, we obtain the estimate: $5 \cdot 0.6-0.4 \leqslant k_{0} \leqslant 5 \cdot 0.6+0.6$, i.e., $2.6 \leqslant k_{0} \leqslant 3.6$. Since $k_{0}$ is an integer, then $k_{0}=3$.
Direct calculations of $p_{5}(k)$ lead to the values: $p_{5}(0)=$ $=0.01024 ; p_{... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 1. Given a statistical series - the number of days missed due to illness by employees of a laboratory.
| Number of days | 0 | 2 | 3 | 4 | 5 | 7 | 10 | Total |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Number of employees | 7 | 3 | 5 | 2 | 5 | 6 | 2 | 30 |
Determine the average n... | Solution. Let's determine the sample mean of size $n=30(k=7)$ using formula (1):
$$
\bar{x}_{3}=\frac{1}{30}(7 \cdot 0+3 \cdot 2+5 \cdot 3+2 \cdot 4+5 \cdot 5+6 \cdot 7+2 \cdot 10)=\frac{116}{30}=3.87
$$
We will calculate the variance and standard deviation using formula (2):
$$
\begin{aligned}
\overline{x_{\mathrm{... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example. Compute the limit
$$
\lim _{n \rightarrow \infty} \frac{(2 n+1)^{2}-(n+1)^{2}}{n^{2}+n+1}
$$ | Solution. Here $(2 n+1)^{2}-(n+1)^{2}=3 n^{2}+2 n-$ is a polynomial of the second degree (an infinitely large sequence of order $n^{2}$) and $n^{2}+n+1$ is a polynomial of the second degree (an infinitely large sequence of order $n^{2}$).
1. Factor out $n^{2}$ in the numerator, we get
$$
(2 n+1)^{2}-(n+1)^{2}=n^{2}\l... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example. Compute the limit
$$
\lim _{n \rightarrow \infty} \frac{n \sqrt[6]{n}+\sqrt[5]{32 n^{10}+1}}{(n+\sqrt[4]{n}) \sqrt[3]{n^{3}-1}}
$$ | Solution. The numerator $n \sqrt[6]{n}+\sqrt[5]{32 n^{10}+1}$ is an infinitely large sequence of order $n^{2}$, and the denominator $(n+\sqrt[4]{n}) \sqrt[3]{n^{3}-1}$ is an infinitely large sequence of order $n^{2}$.
1. Factor out $n^{2}$ in the numerator, we get
$$
n \sqrt[6]{n}+\sqrt[5]{32 n^{10}+1}=n^{2}\left(\fr... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
## Example. Compute the limit
$$
\lim _{x \rightarrow 0} \frac{2 x \sin x}{1-\cos x}
$$ | Solution. The expression under the limit sign is the ratio of two infinitesimals at the point $x=0$, since
$$
\lim _{x \rightarrow 0}(2 x \sin x)=0, \quad \lim _{x \rightarrow 0}(1-\cos x)=0
$$
The infinitesimals in the numerator and denominator are replaced by equivalent ones:
$$
\begin{array}{ll}
2 x \sin x \sim 2... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example. Calculate the limit
$$
\lim _{x \rightarrow \pi} \frac{\cos 3 x-\cos x}{\tan^{2} 2 x}
$$ | Solution.
1. Since
$$
\lim _{x \rightarrow \pi}[\cos 3 x-\cos x]=0, \quad \lim _{x \rightarrow \pi} \operatorname{tg}^{2} 2 x=0
$$
the expression under the limit sign is a ratio of two infinitesimal functions as $x \rightarrow \pi$. We need to replace these infinitesimal functions with equivalent ones. For this, we ... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example. Compute the limit
$$
\lim _{x \rightarrow 0} \sqrt[3]{x\left(2+\sin \frac{1}{x}\right)+8 \cos x}
$$ | Solution.
1. Since the function $y=\sqrt[3]{x}$ is continuous for all $x$, by passing to the limit under the sign of a continuous function, we get
$$
\lim _{x \rightarrow 0} \sqrt[3]{x\left(2+\sin \frac{1}{x}\right)+8 \cos x}=\sqrt[3]{\lim _{x \rightarrow 0}\left[x\left(2+\sin \frac{1}{x}\right)+8 \cos x\right]}
$$
... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example. According to the definition, find the derivative of the function
$$
f(x)=\left[\begin{array}{ll}
1-\cos \left(x \sin \frac{1}{x}\right), & x \neq 0 \\
0, & x=0
\end{array}\right.
$$
at the point $x=0$. | Solution.
1. By definition
$$
f^{\prime}(0)=\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x}=\lim _{x \rightarrow 0} \frac{1-\cos (x \sin (1 / x))-0}{x}
$$
2. Since $\sin (1 / x)$ is a bounded function and $x$ is an infinitesimal function as $x \rightarrow 0$, by the theorem on the product of an infinitesimal function an... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example. Find the area of the region bounded by the graphs of the functions
$$
y=x^{2}-4 x+3, \quad y=-x^{2}+2 x+3
$$ | ## Solution.
1. Find the abscissas $a$ and $b$ of the points of intersection of the graphs. For this, solve the equation
$$
x^{2}-4 x+3=-x^{2}+2 x+3
$$
We get $a=0, \quad b=3$.
2. Investigate the sign of the function $\varphi=x^{2}-4 x+3-\left(-x^{2}+2 x+3\right)$ on the interval $[a, b]=[0,3]$. For this, assign $x... | 9 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example. Evaluate the double integral
$$
\iint_{D}\left(54 x^{2} y^{2}+150 x^{4} y^{4}\right) d x d y
$$
where the region $D$ is bounded by the lines $x=1, y=x^{3}$ and $y=-\sqrt{x}$. | Solution.
1. Let's define the region $D$ by inequalities. It is obvious that $-\sqrt{x} \leq x^{3}$. Therefore, $-\sqrt{x} \leq y \leq x^{3}$. Since $x$ appears under the square root, $x \geq 0$. For $x$, the possible inequalities are $0 \leq x \leq 1$ or $1 \leq x$. In the second case, the region is unbounded, which ... | 11 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example. Evaluate the double integral
$$
\iint_{D} \frac{x}{y^{5}} d x d y
$$
where the region $D$ is defined by the inequalities
$$
1 \leq \frac{x^{2}}{16}+y^{2} \leq 3, \quad y \geq \frac{x}{4}, \quad x \geq 0
$$ | SOLUTION.
1. The region $D$ is defined by inequalities in the Cartesian coordinate system:
$$
D=\left\{(x, y): \begin{array}{c}
1 \leq \frac{x^{2}}{16}+y^{2} \leq 3 \\
\\
y \geq \frac{x}{4}, \quad x \geq 0
\end{array}\right\}
$$
2. Since the region $D$ is bounded by ellipses and lines passing through the origin, it ... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 1. Find the volume of the body bounded by the surfaces
$$
x=17 \sqrt{2 y}, \quad x=2 \sqrt{2 y}, \quad z=1 / 2-y, \quad z=0
$$ | ## Solution.
1. By formula (1) with $f_{2}=1 / 2-y$ and $f_{1}=0$, the desired volume is
$$
V=\iint_{D}\left(\frac{1}{2}-y\right) d x d y
$$
where $D$ is the projection of the body onto the $X O Y$ plane.
2. To find $D$, we define the body using inequalities and eliminate $z$ from them. In this case, the body is de... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 1. Find the mass of the plate $D$ with surface density $\mu=16 x+9 y^{2} / 2$, bounded by the curves
$$
x=\frac{1}{4}, \quad y=0, \quad y^{2}=16 x \quad(y \geq 0)
$$ | Solution.
1. The mass of the plate $D$ with surface density $\mu=16 x+9 y^{2} / 2$ is determined by the formula
$$
m=\iint_{D}\left(16 x+\frac{9 y^{2}}{2}\right) d x d y
$$
2. We compute the obtained double integral in Cartesian coordinates:
a) define the region $D$ by a system of inequalities:
$$
\left\{\begin{ar... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 3. Find the mass of the plate $D$ with surface density $\mu=x / y^{5}$, bounded by the curves
$$
\frac{x^{2}}{16}+y^{2}=1, \quad \frac{x^{2}}{16}+y^{2}=3, \quad y=\frac{x}{4}, \quad x=0 \quad\left(y \geq \frac{x}{4}, x \geq 0\right)
$$ | Solution.
1. The mass of the plate $D$ with surface density $\mu=x / y^{5}$ is determined by the formula
$$
m=\iint_{D} \frac{x}{y^{5}} d x d y
$$
2. We calculate the obtained double integral:
a) define the region $D$ by inequalities in Cartesian coordinates
$$
D=\left\{(x, y): \begin{array}{c}
1 \leq \frac{x^{2}}... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 1. Find the volume of the body $\Omega$, bounded by the surfaces
$$
x=17 \sqrt{2 y}, \quad x=2 \sqrt{2 y}, \quad z=\frac{1}{2}-y, \quad z=0
$$ | Solution.
1. Define the region $\Omega$ by inequalities. Since $17 \sqrt{2 y} \geq 2 \sqrt{2 y}$, for $x$ we have the inequalities $2 \sqrt{2 y} \leq x \leq 17 \sqrt{2 y}$. Since $y$ appears under the square root, $y \geq 0$. For $z$, the possible inequalities are $0 \leq z \leq 1 / 2-y$ or $1 / 2-y \leq z \leq 0$. In... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 1. Find the mass of the body $\Omega$ with density $\mu=2 x$, bounded by the surfaces
$$
x=2 \sqrt{2 y}, \quad x=\sqrt{2 y}, \quad z=1-y, \quad z=0
$$ | Solution.
1. The mass of the body $\Omega$ with density $\mu=2 x$ is determined by the formula
$$
m=\iiint_{\Omega} 2 x d x d y d z
$$
2. Let's define the region $\Omega$ using inequalities. Since $2 \sqrt{2 y} \geq \sqrt{2 y}$, for $x$ we have the inequalities $\sqrt{2 y} \leq x \leq 2 \sqrt{2 y}$. Since $y$ appear... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 3. Find the derivative of the scalar field $u=x z^{2}+2 y z$ at the point $M_{0}(1,0,2)$ along the circle
$$
\left\{\begin{array}{l}
x=1+\cos t \\
y=\sin t-1 \\
z=2
\end{array}\right.
$$ | Solution. The vector equation of the circle has the form
$$
\mathbf{r}(t)=(1+\cos t) \mathbf{i}+(\sin t-1) \mathbf{j}+2 \mathbf{k} .
$$
We find the vector $T$, tangent to it at any point $M$. We have
$$
\left.T=\frac{d r}{d t}=-\sin t \mathbf{i}+\cos t \mathbf{j}\right]
$$
The given point $M_{0}(1,0,2)$ lies in the... | -4 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 1. Using the invariant definition, calculate the divergence of the vector $a=x \mathbf{i}$ at the point $O(0,0,0)$, choosing as the surface $\sigma$ surrounding the point $O$, a sphere $\sigma_{\varepsilon}$ of radius $\varepsilon$ centered at this point. | Solution. By the definition of divergence at the given point, we have
$$
\operatorname{div} a(0)=\lim _{\left(\sigma_{k}\right) \rightarrow 0} \frac{\int\left(a, n^{0}\right) d \sigma}{v_{\varepsilon}}
$$
where $v_{\varepsilon}$ is the volume of the ball bounded by the sphere $\sigma_{\varepsilon}$, or
$$
\operatorn... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 5. Show that the vector field $a=\frac{2 \cos \theta}{r^{3}} \mathbf{e}_{r}+\frac{\sin \theta}{r^{3}} \mathbf{e}_{\theta}$
is solenoidal. | Solution. Using formula (5), we will have
$$
\begin{aligned}
& \operatorname{div}=\frac{1}{r^{2}} \frac{\theta}{\partial r}\left(r^{2} \frac{2 \cos \theta}{r^{3}}\right)+\frac{1}{r \sin \theta} \frac{\theta}{\partial \theta}\left(\sin \theta \frac{\sin \theta}{r^{3}}\right)+0= \\
&=\frac{1}{r^{2}}\left(-\frac{2 \cos \... | 0 | Calculus | proof | Yes | Yes | olympiads | false |
Example 12. Calculate the circulation of the vector field given in cylindrical coordinates: $2=\rho \sin \varphi \mathrm{e}_{\rho}+\rho z \mathrm{e}_{\varphi}+\rho^{3} \mathrm{e}_{z}$, along the curve L: $\{\rho=\sin \varphi, z=0,0 \leqslant \varphi \leqslant \pi\}$ directly and using Stokes' theorem. | Solution. Coordinates of the given vector
$$
a_{p}=\rho \sin \varphi, \quad a_{v}=\rho z, \quad a_{n}=\rho^{3}
$$
The contour $L$ represents a closed curve located in the plane $z=0$ (Fig. 41).
1) Direct calculation of circulation.
Substituting the coordinates of the vector into formula (13), we get
$$
\boldsymbol... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
45. How many numbers need to be selected from a table of random numbers to be confident with a probability of at least 0.9 that among them there is at least one even number? | Solution. Let $n$ be the required number of random numbers. Consider the events:
$A_{k}$ - one randomly selected number is even $(k=1,2, \ldots, n)$;
$\overline{A_{k}}$ - one randomly selected number is odd;
$B$ - among $n$ random numbers there will be at least one even;
$\bar{B}$ - among $n$ random numbers there w... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
118. The random variable $X$ is given by the distribution function:
$$
F(x)=\left\{\begin{array}{ccc}
0 & \text { if } & x \leq -c \\
\frac{1}{2}+\frac{1}{\pi} \arcsin \frac{x}{c} & \text { if } & -c < x \leq c \\
1 & \text { if } & x > c
\end{array}\right.
$$
( arcsine law ).
Find the mathematical expectation of th... | Solution. Let's find the probability density function of the random variable $X: f(x)=F^{\prime}(x)$. For $x \leq -c$ and for $x > c$, the function $f(x)=0$. For $-c < x \leq c$, we have
$$
f(x)=\left(\frac{1}{2}+\frac{1}{\pi} \arcsin \frac{x}{c}\right)^{\prime}=\frac{1}{\pi \sqrt{c^{2}-x^{2}}}
$$
The expected value ... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
120. Calculate $\left(\frac{9}{16}\right)^{-1 / 10}:\left(\frac{25}{36}\right)^{-3 / 2}-\left[\left(\frac{4}{3}\right)^{-1 / 2}\right]^{-2 / 5}\left(\frac{6}{5}\right)^{-3}$. | Solution. We will perform the actions sequentially:
1) $\left(\frac{9}{16}\right)^{-1 / 10}=\left[\left(\frac{3}{4}\right)^{2}\right]^{-1 / 10}=\left(\frac{3}{4}\right)^{-1 / 5}=\left(\frac{4}{3}\right)^{1 / 5}$;
2) $\left(\frac{25}{36}\right)^{-3 / 2}=\left[\left(\frac{5}{6}\right)^{2}\right]^{-3 / 2}=\left(\frac{5}{... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
137. $5^{x}=625$ | Solution. Writing 625 as $5^{4}$, we get $5^{x}=5^{4}$, hence $x=4$. 138. $8^{x}=32$.
Solution. We have $32=2^{5} ; 8^{x}=\left(2^{3}\right)^{x}=2^{3 x}$. Therefore, $2^{3 x}=2^{5}$, hence $3 x=5$, i.e., $x=5 / 3$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
154. $(0,25)^{2-x}=\frac{256}{2^{x+3}}$.
154. $(0.25)^{2-x}=\frac{256}{2^{x+3}}$. | Solution. Let's convert all powers to base $2: 0.25=1/4=2^{-2}$; $256=2^{8}$. Therefore, $\left(2^{-2}\right)^{2-x}=\frac{2^{8}}{2^{x+3}}$. Applying the rule of dividing powers, we have
$$
\begin{gathered}
2^{-4+2 x}=2^{8-x-3} ; 2^{-4+2 x}=2^{5-x} ;-4+2 x=5-x ; 2 x+x=5+4 \\
3 x=9 ; x=3
\end{gathered}
$$
=44 ; 2^{x-3}(8+4-1)=44 ; 2^{x-3} \cdot 11=44
$$
Dividing both sides of the equation by 11, we get
$$
2^{x-3}=4 ; 2^{x-3}=2^{2} ; x-3=2 ; x=5
$$ | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
162. $7^{x}-3 \cdot 7^{x-1}+7^{x+1}=371$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
162. $7^{x}-3 \cdot 7^{x-1}+7^{x+1}=371$. | Solution. The least exponent is $x-1$; therefore, we factor out $7^{x-1}$:
$$
\begin{gathered}
7^{x-1} \cdot\left(7^{1}-3 \cdot 1+7^{12}\right)=371 ; 7^{x-1}(7-3+49)=371 \\
7^{x-1} \cdot 53=371 ; 7^{x-1}=7 ; x-1=1 ; x=2
\end{gathered}
$$ | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
171. $7^{2 x}-48 \cdot 7^{x}=49$
171. $7^{2 x}-48 \cdot 7^{x}=49$ | Solution. Let $7^{x}=y$, we get the quadratic equation $y^{2}-$ $-48 y-49=0$. Let's solve it. Here $a=1, b=-48, c=-49 ; \quad D=b^{2}-$ $-4 a c=(-48)^{2}-4 \cdot 1(-49)=2304+196=2500 ; \sqrt{D}=50$. Using the formula $y_{1,2}=\frac{-b \pm \sqrt{D}}{2 a}$, we find
$$
y_{1}=\frac{48-50}{2}=\frac{-2}{2}=-1 ; \quad y_{2}=... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
179. $5^{2}=25$. | Solution. Since the base of the power is 5, the exponent (logarithm) is 2, and the power is 25, then $\log _{5} 25=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
231. $\log _{4}(x+3)-\log _{4}(x-1)=2-3 \log _{4} 2$. | solution. Representing the number 2 as the logarithm of 16 to the base 4, we rewrite the given equation as
$$
\log _{4}(x+3)-\log _{4}(x-1)=\log _{4} 16-3 \log _{4} 2
$$
From this, we obtain
$$
\log _{4} \frac{x+3}{x-1}=\log _{4} \frac{16}{8}, \text { or } \frac{x+3}{x-1}=2
$$
We solve this equation:
$$
x+3=2(x-1)... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
320. $\sqrt{x-1} \cdot \sqrt{2 x+6}=x+3$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
320. $\sqrt{x-1} \cdot \sqrt{2 x+6}=x+3$. | Solution. First, we perform the multiplication of the roots:
$\sqrt{(x-1)(2 x+6)}=x+3 ; \sqrt{2 x^{2}-2 x+6 x-6}=x+3 ; \sqrt{2 x^{2}+4 x-6}=x+3$.
Now, we square both sides of the equation:
$\left(\sqrt{2 x^{2}+4 x-6}\right)^{2}=(x+3)^{2} ; 2 x^{2}+4 x-6=x^{2}+6 x+9 ; x^{2}-2 x-15=0$.
We solve the quadratic equation... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
321. $\sqrt{2 x+5}+\sqrt{x-1}=8$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
321. $\sqrt{2 x+5}+\sqrt{x-1}=8$. | Solution. Isolate one of the radicals and square both sides of the equation:
$$
(\sqrt{2 x+5})^{2}=(8-\sqrt{x-1})^{2} ; 2 x+5=64-16 \sqrt{x-1}+(x-1)
$$
Move $16 \sqrt{x-1}$ to the left side, and all other terms to the right side:
$$
16 \sqrt{x-1}=64+x-1-2 x-5 ; 16 \sqrt{x-1}=58-x
$$
Square both sides of the equatio... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
385. $(\tan \alpha+\cot \alpha)^{2}-(\tan \alpha-\cot \alpha)^{2}$. | Solution. We will use the formulas for the square of the sum and difference of two numbers:
$$
\begin{gathered}
(\operatorname{tg} \alpha+\operatorname{ctg} \alpha)^{2}-(\operatorname{tg} \alpha-\operatorname{ctg} \alpha)^{2}=\operatorname{tg}^{2} \alpha+2 \operatorname{tg} \alpha \operatorname{ctg} \alpha+\operatorna... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
26. Compute the second-order determinants:
a) $\left|\begin{array}{rr}2 & 5 \\ -3 & -4\end{array}\right|$;
b) $\left|\begin{array}{ll}a^{2} & a b \\ a b & b^{2}\end{array}\right|$. | Solution. a) $\left|\begin{array}{rr}2 & 5 \\ -3 & -4\end{array}\right|=2(-4)-5(-3)=-8+15=7$;
b) $\left|\begin{array}{ll}a^{2} & a b \\ a b & b^{2}\end{array}\right|=a^{2} \cdot b^{2}-a b \cdot a b=a^{2} b^{2}-a^{2} b^{2}=0$.
27-32. Calculate the determinants: | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
33. Calculate the determinants of the third order:
a) $\left|\begin{array}{lll}3 & 2 & 1 \\ 2 & 5 & 3 \\ 3 & 4 & 3\end{array}\right|$
b) $\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ | Solution.
a) $\left|\begin{array}{lll}3 & 2 & 1 \\ 2 & 5 & 3 \\ 3 & 4 & 3\end{array}\right|=3 \cdot 5 \cdot 3+2 \cdot 3 \cdot 3+2 \cdot 4 \cdot 1-1 \cdot 5 \cdot 3-2 \cdot 2 \cdot 3-$ $-3 \cdot 3 \cdot 4=45+18+8-15-12-36=71-63=8$;
b) $\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|=a c b+... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
92. Show that as $t \rightarrow \infty$ the limit of the variable
$x=\frac{6 t^{3}-9 t+1}{2 t^{3}-3 t}$ is 3. | Solution. We find the difference between the variable $x$ and the number 3:
$$
\begin{gathered}
x-3=\frac{6 t^{3}+9 t+1}{2 t^{3}+3 t}-3=\frac{6 t^{3}+9 t+1-3\left(2 t^{3}+3 t\right)}{2 t^{3}+3 t}=\frac{6 t^{3}+9 t+1-6 t^{3}-9 t}{2 t^{3}+3 t}= \\
=\frac{1}{2 t^{3}+3 t}
\end{gathered}
$$
If $t \rightarrow \infty$, then... | 3 | Calculus | proof | Yes | Yes | olympiads | false |
94. Find $\lim _{x \rightarrow 2}\left(3 x^{2}-2 x\right)$. | Solution. Using properties 1, 3, and 5 of limits sequentially, we get
$$
\begin{gathered}
\lim _{x \rightarrow 2}\left(3 x^{2}-2 x\right)=\lim _{x \rightarrow 2}\left(3 x^{2}\right)-\lim _{x \rightarrow 2}(2 x)=3 \lim _{x \rightarrow 2} x^{2}-2 \lim _{x \rightarrow 2} x= \\
=3\left(\lim _{x \rightarrow 2} x\right)^{2}... | 8 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
95. Find $\lim _{x \rightarrow 4} \frac{x^{2}-2 x}{x-3}$. | Solution. To apply the property of the limit of a quotient, we need to check whether the limit of the denominator is not zero when $x=4$. Since $\lim _{x \rightarrow 4}(x-3)=$[^3]$=\lim _{x \rightarrow 4} x-\lim 3=4-3=1 \neq 0$, in this case, we can use property 4:
$$
\lim _{x \rightarrow 4} \frac{x^{2}-2 x}{x-3}=\fra... | 8 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
115. Find $\lim _{x \rightarrow 3} \frac{x^{2}-9}{3-x}$. | Solution. A direct transition to the limit is impossible here, since the limit of the denominator is zero: $\lim _{x \rightarrow 3}(3-x)=3-3=0$. The limit of the dividend is also zero: $\lim _{x \rightarrow 3}\left(x^{2}-9\right)=9-9=0$. Thus, we have an indeterminate form of $0 / 0$. However, this does not mean that t... | -6 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
116. Find $\lim _{x \rightarrow \infty} \frac{3}{x+5}$. | Solution. When $x \rightarrow \infty$, the denominator $x+5$ also tends to infinity, and its reciprocal $\frac{1}{x+5} \rightarrow 0$. Therefore, the product $\frac{1}{x+5} \cdot 3=\frac{3}{x+5}$ tends to zero if $x \rightarrow \infty$. Thus, $\lim _{x \rightarrow \infty} \frac{3}{x+5}=0$.
Identical transformations un... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
117. Find $\lim _{x \rightarrow \infty} \frac{2 x^{3}+x}{x^{3}-1}$. | Solution. Here, the numerator and denominator do not have a limit, as both increase indefinitely. In this case, it is said that there is an indeterminate form of $\infty / \infty$. We will divide the numerator and denominator term by term by $x^{3}$ (the highest power of $x$ in this fraction):
$$
\lim _{x \rightarrow ... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
118. Find $\lim _{\alpha \rightarrow 0} \frac{\sin 2 \alpha}{\alpha}$. | Solution. We will transform this limit into the form (1). For this, we multiply the numerator and the denominator of the fraction by 2, and take the constant factor 2 outside the limit sign. We have
$$
\lim _{\alpha \rightarrow 0} \frac{\sin 2 \alpha}{\alpha}=\lim _{\alpha \rightarrow 0} \frac{2 \sin 2 \alpha}{2 \alph... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
120. Find $\lim _{x \rightarrow 3}\left(x^{2}-7 x+4\right)$. | Solution. To find the limit of the given function, we will replace the argument $x$ with its limiting value:
$$
\lim _{x \rightarrow 3}\left(x^{2}-7 x+4\right)=3^{2}-7 \cdot 3+4=-8
$$ | -8 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
135. Find $\lim _{x \rightarrow \infty} \frac{\sqrt{x^{2}+4}}{x}$. | Solution. As the argument $x$ tends to infinity, we have an indeterminate form of $\infty / \infty$. To resolve it, we divide the numerator and the denominator of the fraction by $x$. Then we get
$$
\lim _{x \rightarrow \infty} \frac{\sqrt{x^{2}+4}}{x}=\lim _{x \rightarrow \infty} \frac{\sqrt{\frac{x^{2}+4}{x^{2}}}}{1... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
136. Find $\lim _{x \rightarrow \infty} \frac{3 x}{\sqrt{x^{2}-2 x+3}}$. | Solution. The limit transition as $x \rightarrow \infty$ can always be replaced by the limit transition as $\alpha \rightarrow 0$, if we set $\alpha=1 / x$ (the method of variable substitution).
Thus, setting $x=1 / \alpha$ in this case, we find that $\alpha \rightarrow 0$ as $x \rightarrow \infty$. Therefore,
$$
\be... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
137. Find $\lim _{x \rightarrow \infty} \frac{3 x^{2}+5 x+1}{x^{2}-2}$. | Solution. Method I. Dividing the numerator and the denominator by $x^{2}$, we find
$$
\lim _{x \rightarrow \infty} \frac{3 x^{2}+5 x+1}{x^{2}-2}=\lim _{x \rightarrow \infty} \frac{3+\frac{5}{x}+\frac{1}{x^{2}}}{1-\frac{2}{x^{2}}}=\frac{3}{1}=3
$$
Method II. Let $x=1 / a$; then $a \rightarrow 0$ as $x \rightarrow \inf... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
142. Find $\lim _{x \rightarrow \infty}(\sqrt{x+1}-\sqrt{x})$. | Solution. Here, it is required to find the limit of the difference of two quantities tending to infinity (indeterminate form of $\infty-\infty$). By multiplying and dividing the given expression by its conjugate, we get
$$
\sqrt{x+1}-\sqrt{x}=\frac{(\sqrt{x+1}-\sqrt{x})(\sqrt{x+1}+\sqrt{x})}{\sqrt{x+1}+\sqrt{x}}=\frac... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
168. Find the derivative of the function $y=5x$. | Solution. $1^{0} . y_{\mathrm{H}}=5(x+\Delta x)=5 x+5 \Delta x$.
$2^{0} . \Delta y=y_{\mathrm{k}}-y=(5 x+5 \Delta x)-5 x=5 \Delta x$.
$3^{0} . \frac{\Delta y}{\Delta x}=\frac{5 \Delta x}{\Delta x}=5$. | 5 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
329. Find the value of the arbitrary function $f(x)=\sin ^{4} x-$ $-\cos ^{4} x$ at $x=\pi / 12$. | Solution. First, we transform the given function:
$$
f(x)=\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{2} x-\cos ^{2} x\right)=-\cos 2 x
$$
(since $\sin ^{2} \alpha+\cos ^{2} \alpha=1, \cos ^{2} \alpha-\sin ^{2} \alpha=\cos 2 \alpha$). Using the rule for differentiating a composite function, we get
$$
f^{\prime}... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
374. Find the slope of the tangent line drawn to the curve $y=x^{3}$ at the point $C(-2;-8)$. | Solution. Let's find the derivative of the function $y=x^{3}$ at the point $x=-2$:
$$
y^{\prime}=\left(x^{3}\right)^{\prime}=3 x^{2} ; \quad y_{x=-2}^{\prime}=3(-2)^{2}=12
$$
Thus, the slo... | 12 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
461. The law of temperature change $T$ of a body depending on time $t$ is given by the equation $T=0.2 t^{2}$. At what rate is this body heating at the moment of time $10 \mathrm{c}$? | Solution. The heating rate of a body is the derivative of temperature $T$ with respect to time $t$:
$$
\frac{d T}{d t}=\left(0.2 t^{2}\right)^{\prime}=0.4 t
$$
Determine the heating rate of the body at $t=10$:
$$
\left(\frac{d T}{d t}\right)_{t=10}=0.4 \cdot 10=4(\text{ deg } / \mathrm{s})
$$ | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
564. $y=x^{2}+2$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
564. $y=x^{2}+2$. | Solution. $1^{0}$. Find the derivative: $y^{\prime}=\left(x^{2}+2\right)^{\prime}=2 x$.
$2^{0}$. Set it to zero; $2 x=0$, from which $x=0$ - the critical point.
$3^{\circ}$. Determine the sign of the derivative at the value $x0$, for example at $x=1$: $y_{x=1}^{\prime}=2 \cdot 1=2$. Since the derivative changes sign ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
582. $y=4x-x^{2}$. | Solution. $1^{0}$. Find the first derivative: $y^{\prime}=4-2 x$. $2^{0}$. Solving the equation $4-2 x=0$, we get the critical point $x=2$. $3^{0}$. Calculate the second derivative: $y^{\prime \prime}=-2$.
$4^{0}$. Since $y^{\prime \prime}$ is negative at any point, then $y^{\prime \prime}(2)=-2<0$. This means that th... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
585. $y=x^{6}$. | Solution. $1^{0} . y^{\prime}=6 x^{5}$.
$2^{0} .6 x^{5}=0$, i.e. $x=0$.
$3^{0} \cdot y^{\prime \prime}=30 x^{4}$.
$4^{0}$. Since $y^{\prime \prime}(0)=0$, the method of investigation using the second derivative is not applicable.
We will investigate the function for an extremum using the first derivative. If $x<0$,... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
614. There is a square sheet of tin, the side of which $a=$ $=60 \mathrm{~cm}$. By cutting out equal squares from all its corners and folding up the remaining part, a box (without a lid) needs to be made. What should be the dimensions of the squares to be cut out so that the box has the maximum volume? | Solution. According to the condition, the side of the square $a=60$. Let the side of the squares cut out from the corners be denoted by $x$. The bottom of the box is a square with side $a-2 x$, and the height of the box is equal to the side $x$ of the cut-out square. Therefore, the volume of the box can be expressed by... | 10 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
228. Given the function $f(x)=2 x+4$. Find the increment of any of its antiderivatives when $x$ changes from -2 to 0. | Solution. Let's find the antiderivative of the given function:
$$
F=\int(2 x+4) d x=x^{2}+4 x+C
$$
Consider, for example, the antiderivatives $F_{1}=x^{2}+4 x, F_{2}=x^{2}+4 x+2$, $F_{3}=x^{2}+4 x-1$, and compute the increment of each of them on the interval $[-2,0]: \quad F_{1}(0)-F_{1}(-2)=0-(-4)=4 ; \quad F_{2}(0)... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
245. Find $\int_{\pi / 2}^{\pi} \frac{2 \sin x d x}{(1-\cos x)^{2}}$. | Solution. Let's use the substitution $u=1-\cos x$, from which $d u=$ $=\sin x d x$. Then we will find the new limits of integration; substituting into the equation $u=1-\cos x$ the values $x_{1}=\pi / 2$ and $x_{2}=\pi$, we will respectively obtain $u_{1}=1-\cos (\pi / 2)=1$ and $u_{2}=1-\cos \pi=2$. The solution is wr... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
294. $\int_{1}^{e} \frac{3 \ln ^{2} x d x}{x}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
294. $\int_{1}^{e} \frac{3 \ln ^{2} x \, dx}{x}$. | Solution. $\int_{1}^{e} \frac{3 \ln ^{2} x d x}{x}=\left|\begin{array}{l}\ln x=t, t_{1}=\ln 1=0, \\ \frac{d x}{x}=d t ; t_{2}=\ln e=1\end{array}\right|=3 \int_{0}^{1} t^{2} d t=\left.t^{3}\right|_{0} ^{1}=1$. | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
380. Two bodies simultaneously started linear motion from a certain point in the same direction with speeds $v_{1}=$ $=\left(6 t^{2}+4 t\right) \mathrm{m} /$ s and $v_{2}=4 t \mathrm{m} / \mathrm{s}$. After how many seconds will the distance between them be 250 m? | Solution. Let $t_{1}$ be the moment of the meeting. Then
$$
s_{1}=\int_{0}^{t_{1}}\left(6 t^{2}+4 t\right) d t=\left.\left(2 t^{3}+2 t^{2}\right)\right|_{0} ^{t_{1}}=2 t_{1}^{3}+2 t_{1}^{2} ; \quad s_{2}=\int_{0}^{t_{1}} 4 t d t=2 t_{1}^{2}
$$
Since $s_{1}-s_{2}=250$, we get the equation $2 t_{1}^{3}+2 t_{1}^{2}-2 t_... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
195. In a motorboat with an initial speed of $v_{0}=$ $=5 \mathrm{m} / \mathrm{s}$, the engine was turned off. During movement, the boat experiences water resistance, the force of which is proportional to the square of the boat's speed, with a proportionality coefficient of $m / 50$, where $m$ is the mass of the boat. ... | Solution. $1^{0}$. According to the condition, we can take the function as the path $s$, and the argument as time $t$. Using Newton's second law $F=m a$, we obtain the equation
$$
m s_{(t)}^{\prime \prime}=-\frac{m}{50}\left(s_{(t)}^{\prime}\right)^{2}, \text { or } s_{(t)}^{\prime \prime} \doteq-\frac{1}{50}\left(s_{... | 10 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
150. Two shooters shoot at a target. The probability of a miss with one shot for the first shooter is 0.2, and for the second shooter it is 0.4. Find the most probable number of volleys in which there will be no hits on the target, if the shooters will make 25 volleys. | Solution. Misses by the shooters are independent events, so the multiplication theorem of probabilities of independent events applies. The probability that both shooters will miss in one volley, $p=0.2 \cdot 0.4=0.08$.
Since the product $n p=25 \cdot 0.08=2$ is an integer, the most probable number of volleys in which ... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
172. After the student answers the questions in the examination ticket, the examiner asks the student additional questions. The teacher stops asking additional questions as soon as the student shows ignorance of the given question. The probability that the student will answer any additional question is 0.9. Required: a... | Solution. a) The discrete random variable $X$ - the number of additional questions asked - has the following possible values: $x_{1}=1, x_{2}=2, x_{3}=3, \ldots, x_{k}=k, \ldots$ Let's find the probabilities of these possible values.
The variable $X$ will take the possible value $x_{i}=1$ (the examiner will ask only o... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
189. Find the mathematical expectation of the random variable $Z$, if the mathematical expectations of $X$ and $Y$ are known:
a) $Z=X+2 Y, M(X)=5, M(Y)=3 ;$ b) $Z=3 X+4 Y$, $M(X)=2, M(Y)=6$ | Solution. a) Using the properties of mathematical expectation (the expectation of a sum is equal to the sum of the expectations of the terms; a constant factor can be factored out of the expectation), we get
$$
\begin{aligned}
M(Z)=M(X+2 Y)= & M(X)+M(2 Y)=M(X)+2 M(Y)= \\
& =5+2 \cdot 3=11 .
\end{aligned}
$$ | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
277. The random variable $X$ in the interval (-c, c) is given by the probability density function $f(x)=1 /\left(\pi \sqrt{c^{2}-x^{2}}\right)$; outside this interval, $f(x)=0$. Find the mathematical expectation of the variable $X$. | Solution. We use the formula $M(X)=\int_{a} x f(x) \mathrm{d} x$. Substituting $a=-c, b=c, f(x)=1 /\left(\pi \sqrt{c^{2}-x^{2}}\right)$, we get
$$
M(X)=\frac{1}{\pi} \int_{-c}^{c} \frac{x d x}{\sqrt{c^{2}-x^{2}}}
$$
Considering that the integrand is an odd function and the limits of integration are symmetric with res... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
280. Find the mathematical expectation of the random variable $X$, given by the distribution function
$$
F(x)=\left\{\begin{array}{ccr}
0 & \text { for } & x \leqslant 0 \\
x / 4 & \text { for } & 0 < x \leqslant 4 \\
1 & \text { for } & x > 4
\end{array}\right.
$$ | Solution. Find the density function of the random variable $\boldsymbol{X}$:
$$
f(x)=F^{\prime}(x)=\left\{\begin{array}{ccc}
0 & \text { for } & x4 .
\end{array}\right.
$$
Find the required expected value:
$$
M(X)=\int_{0}^{4} x f \cdot(x) \mathrm{d} x=\int_{0}^{4} x \cdot(1 / 4) \mathrm{d} x=2
$$ | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
647. Why, when checking the hypothesis of exponential distribution of the population by the Pearson criterion, is the number of degrees of freedom determined by the equation $k=s-2$, where $s$ is the number of sample intervals? | Solution. When using the Pearson criterion, the number of degrees of freedom is $k=s-1-r$, where $r$ is the number of parameters estimated from the sample. The exponential distribution is defined by one parameter $\lambda$. Since this parameter is estimated from the sample, $r=1$, and therefore, the number of degrees o... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
657. Why, when testing the hypothesis of a uniform distribution of the population $X$ using the Pearson's criterion, is the number of degrees of freedom determined by the equation $k=s-3$, where $s$ is the number of intervals in the sample? | Solution. When using the Pearson criterion, the number of degrees of freedom $k=s-1-r$, where $r$ is the number of parameters estimated from the sample. The uniform distribution is defined by two parameters $a$ and $b$. Since these two parameters are estimated from the sample, $r=2$, and therefore, the number of degree... | -3 | Other | math-word-problem | Yes | Yes | olympiads | false |
Example 1. Find $\lim _{x \rightarrow 2}\left(4 x^{2}-6 x+3\right)$. | Solution. Since the limit of the algebraic sum of variables is equal to the same algebraic sum of the limits of these variables (formula (1.42)), a constant multiplier can be factored out of the limit sign (formula (1.44)), the limit of an integer positive power is equal to the same power of the limit (formula (1.45)),... | 7 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 3. Find $\lim _{x \rightarrow 4} \frac{x^{2}-6 x+8}{x-4}$. | Solution. When $x=4$, both the numerator and the denominator of the given function become zero. This results in an indeterminate form $\frac{0}{0}$, which needs to be resolved. We will transform the given function by factoring the numerator using the formula
$$
x^{2}+p x+q=\left(x-x_{1}\right)\left(x-x_{2}\right)
$$
... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 6. Find $\lim _{x \rightarrow+\infty} \frac{6 x^{2}+5 x+4}{3 x^{2}+7 x-2}$. | Solution. As $x \rightarrow+\infty$, the numerator and denominator increase without bound (we get an indeterminate form of $\frac{\infty}{\infty}$). To find the limit, we transform the given fraction by dividing its numerator and denominator by $x^{2}$, i.e., by the highest power of $x$. Using the properties of limits,... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 7. Find $\lim _{x \rightarrow+\infty} \frac{7 x^{2}+6 x-3}{9 x^{3}+8 x^{2}-2}$. | Solution. By dividing the numerator and the denominator of the fraction by $x^{3}$, i.e., by the highest degree, we get
$$
\begin{aligned}
\lim _{x \rightarrow+\infty} \frac{7 x^{2}+6 x-3}{9 x^{3}+8 x^{2}-2}= & \lim _{x \rightarrow+\infty} \frac{\frac{7}{x}+\frac{6}{x^{2}}-\frac{3}{x^{3}}}{9+\frac{8}{x}-\frac{2}{x^{3}... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 11. Find $\lim _{n \rightarrow \infty} \frac{1+3+5+\ldots+(2 n-1)}{1+2+3+\ldots+n}$. | Solution. The numerator and denominator of the fraction are the sums of $n$ terms of the corresponding arithmetic progressions. Finding these sums using the known formula, we get
\[
\begin{gathered}
\lim _{n \rightarrow \infty} \frac{1+3+5+\ldots+(2 n-1)}{1+2+3+\ldots+n}=\lim _{n \rightarrow \infty} \frac{\frac{1+(2 n... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 14. Find $\lim _{n \rightarrow \infty} \frac{12 n+5}{\sqrt[3]{27 n^{3}+6 n^{2}+8}}$. | Solution. As $n \rightarrow \infty$, the numerator and denominator also tend to infinity, resulting in an indeterminate form $\frac{\infty}{\infty}$. To find the limit, we divide the numerator and denominator by $\boldsymbol{n}$ and bring $\boldsymbol{n}$ under the root sign:
$$
\lim _{n \rightarrow \infty} \frac{12 n... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 2. Find $\lim _{x \rightarrow 0} \frac{\ln (1+x)}{x}$. | Solution. Since
$$
\frac{\ln (1+x)}{x}=\frac{1}{x} \ln (1+x)=\ln (1+x)^{\frac{1}{x}}
$$
then based on formula (1.58) we find
$$
\lim _{x \rightarrow 0} \frac{\ln (1+x)}{x}=\lim _{x \rightarrow 0}\left[\ln (1+x)^{\frac{1}{x}}\right]=\ln \left[\lim _{x \rightarrow 0}(1+x)^{\frac{1}{x}}\right]=\ln e=1
$$
Therefore,
$... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 6. Find $\lim _{x \rightarrow 0} \frac{\tan x}{x}$. | Solution. Taking into account that $\operatorname{tg} x=\frac{\sin x}{\cos x}$ and $\lim _{x \rightarrow 0} \cos x=\cos 0=1$, based on the properties of limits (1.43) and (1.46) and formula (1.59), we obtain
$$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\operatorname{tg} x}{x} & =\lim _{x \rightarrow 0}\left(\frac{... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 9. Find $\lim _{x \rightarrow 0} \frac{\sin x}{\sqrt{x+9}-3}$. | Solution. When $x=0$, both the numerator and the denominator become zero. The denominator contains an irrationality. We will eliminate the irrationality and use formula (1.59)
$$
\begin{gathered}
\lim _{x \rightarrow 0} \frac{\sin x}{\sqrt{x+9}-3}=\lim _{x \rightarrow 0} \frac{\sin x(\sqrt{x+9}+3)}{(\sqrt{x+9}-3)(\sqr... | 6 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 10. Find $\lim _{x \rightarrow 0}\left(\frac{\sin 3 x}{x}\right)^{x+2}$. | Solution. This is a limit of the form (1.60), where $\varphi(x)=\frac{\sin 3 x}{x}, \psi(x)=x+2$.
From (1.67)
$$
\lim _{x \rightarrow 0} \frac{\sin 3 x}{x}=3, \lim _{x \rightarrow 0}(x+2)=2
$$
According to formula (1.62), we get
$$
\lim _{x \rightarrow 0}\left(\frac{\sin 3 x}{x}\right)^{x+2}=3^{2}=9
$$ | 9 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 11. Find $\lim _{x \rightarrow \infty}\left(\frac{2 x-1}{3 x+4}\right)^{x^{2}}$. | Solution. This is also a limit of the form $(1.60)$, where $\varphi(x)=\frac{2 x-1}{3 x+4}, \psi(x)=x^{2}$
Since
$$
\lim _{x \rightarrow \infty} \frac{2 x-1}{3 x+4}=\lim _{x \rightarrow \infty} \frac{2-\frac{1}{x}}{3+\frac{4}{x}}=\frac{2}{3}, \lim _{x \rightarrow \infty} x^{2}=\infty
$$
then according to formula (1.... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 13. Find $\lim _{x \rightarrow 0}(\cos x)^{\frac{1}{x}}$. | Solution. By adding and subtracting 1 from $\cos x$ and applying the corresponding formula, we get
$$
\begin{aligned}
& \lim _{x \rightarrow 0}(\cos x)^{\frac{1}{x}}=\lim _{x \rightarrow 0}[1-(1-\cos x)]^{\frac{1}{x}}=\lim _{x \rightarrow 0}\left(1-2 \sin ^{2} \frac{x}{2}\right)^{\frac{1}{x}}= \\
& =\lim _{x \rightarr... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
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