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1.9. Using the property that the determinant of the matrix $C=A \cdot B$, which represents the product of square matrices $A$ and $B$ of the same order, is equal to the product of the determinants of matrices $A$ and $B$, i.e., $\operatorname{det} C=\operatorname{det} A \cdot \operatorname{det} B$, calculate the determ... | Solution. We will find the determinants of matrices $A$ and $B$ and then multiply them. We have
$$
\operatorname{det} A=\left|\begin{array}{lll}
3 & 2 & 5 \\
0 & 2 & 8 \\
4 & 1 & 7
\end{array}\right|=\left|\begin{array}{ccc}
3 & 2 & -3 \\
0 & 2 & 0 \\
4 & 1 & 3
\end{array}\right|=2 \cdot(-1)^{2+2}\left|\begin{array}{c... | 2142 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,265 |
1.10. Using the method of the adjugate matrix, find $A^{-1}$ if
$$
A=\left(\begin{array}{lll}
2 & 5 & 6 \\
1 & 2 & 5 \\
1 & 2 & 3
\end{array}\right)
$$ | Solution. First, we calculate the determinant of the given matrix to ensure the existence of the inverse matrix. We have
$$
\operatorname{det} A=\left|\begin{array}{ccc}
2 & 5 & 6 \\
1 & 2 & 5 \\
1 & 2 & 3
\end{array}\right|=\left|\begin{array}{lll}
2 & 5 & 6 \\
0 & 0 & 2 \\
1 & 2 & 3
\end{array}\right|=(-2) \cdot\lef... | A^{-1}=(\begin{pmatrix}-2&-3/2&13/2\\1&0&-2\\0&1/2&-1/2\end{pmatrix}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,266 |
1.13. Using the method of bordering minors, find the rank of the matrix
$$
A=\left(\begin{array}{ccccc}
2 & 1 & 2 & 1 & 2 \\
1 & 1 & 5 & -2 & 3 \\
-1 & 0 & -4 & 4 & 1 \\
3 & 3 & 8 & 1 & 9
\end{array}\right)
$$ | Solution. The second-order minor
$$
M_{2}=\left|\begin{array}{ll}
2 & 1 \\
1 & 1
\end{array}\right|=2 \cdot 1-1 \cdot 1=1 \neq 0
$$
is non-zero and located in the upper left corner of matrix $A$. The third-order minor
$$
M_{3}=\left|\begin{array}{ccc}
2 & 1 & 2 \\
1 & 1 & 5 \\
-1 & 0 & 4
\end{array}\right|=\left|\be... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,267 |
1.14. Using elementary transformations, find the rank of the matrix
$$
A=\left(\begin{array}{ccccc}
5 & 7 & 12 & 48 & -14 \\
9 & 16 & 24 & 98 & -31 \\
14 & 24 & 25 & 146 & -45 \\
11 & 12 & 24 & 94 & -25
\end{array}\right)
$$ | Solution. Let's denote the $i$-th row of matrix $A$ by the symbol $\alpha_{i}$. In the first stage, we perform elementary transformations $\alpha_{2}^{\prime}=\alpha_{2}-\alpha_{3}+\alpha_{1}, \alpha_{3}^{\prime}=\alpha_{3}-\alpha_{2}-\alpha_{1} ; \alpha_{4}^{\prime}=\alpha_{4}-\alpha_{3}+\alpha_{1}$.
In the second st... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,268 |
### 1.16. Investigate the system of equations
$$
\left\{\begin{array}{l}
3 x_{1}+4 x_{2}+3 x_{3}=0 \\
x_{1}-x_{2}+x_{3}=0 \\
x_{1}+3 x_{2}-x_{3}=-2 \\
x_{1}+2 x_{2}+3 x_{3}=2
\end{array}\right.
$$ | $$
\begin{aligned}
& \tilde{A}-\left(\begin{array}{ccc|c}
3 & 4 & 3 & 0 \\
1 & -2 & 1 & 0 \\
1 & 3 & -1 & -2 \\
1 & 2 & 3 & 2
\end{array}\right) \sim\left[\begin{array}{l}
\text { swap } \\
1 \text { and } 2 \text { rows }
\end{array}\right] \sim\left(\begin{array}{ccc|c}
1 & -2 & 1 & 0 \\
3 & 4 & 3 & 0 \\
1 & 3 & -1 &... | x_1=2,x_2=0,x_3=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,269 |
1.18. Solve the system of equations using the matrix method
$$
\left\{\begin{array}{l}
x_{1}+2 x_{2}+4 x_{3}=5 \\
2 x_{1}+x_{2}+5 x_{3}=7 \\
3 x_{1}+2 x_{2}+6 x_{3}=9
\end{array}\right.
$$ | Solution. Rewrite the system in matrix form: $A \cdot X=B$, where
$$
A=\left(\begin{array}{lll}
1 & 2 & 4 \\
2 & 1 & 5 \\
3 & 2 & 6
\end{array}\right), X=\left(\begin{array}{l}
x_{1} \\
x_{2} \\
x_{3}
\end{array}\right), B=\left(\begin{array}{l}
5 \\
7 \\
9
\end{array}\right)
$$
The solution is represented as $X=A^{-... | (\begin{pmatrix}1\\0\\1\end{pmatrix}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,270 |
1.20. Solve the system of equations from example 1.18 using the Gaussian method. | Solution. We have
$$
\begin{aligned}
& \tilde{A}=\left(\begin{array}{ccc|c}
1 & 2 & 4 & 5 \\
2 & 1 & 5 & 7 \\
3 & 2 & 6 & 9
\end{array}\right) \sim\left[\begin{array}{l}
(2)-(1) \times 2 ; \\
(3)-(1) \times 3
\end{array}\right] \sim\left(\begin{array}{ccc|c}
1 & 2 & 4 & 5 \\
0 & -3 & -3 & -3 \\
0 & -4 & -6 & -6
\end{a... | x_1=1,x_2=0,x_3=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,271 |
1.21. Solve the system from example 1.17 using the Gaussian method. | Solution. As a result of elementary transformations over the augmented matrix of the system, we obtained the following system at the end of the forward pass of the Gaussian method:
$$
\left\{\begin{array}{r}
x_{1}+2 x_{2}+x_{3}-x_{4}=1 \\
-x_{2}-5 x_{3}-x_{4}=-1 \\
6 x_{3}+5 x_{4}=1
\end{array}\right.
$$
Here, the fr... | {\begin{pmatrix}\forallx_{4}\in\mathbb{R}\\x_{1}=\frac{1}{2}+\frac{73}{6}x_{4}\\x_{2}=\frac{1}{6}-\frac{31}{6}x_{4}\\x_{3}=-\frac{5}{6}x_{4}+\frac{1}{6}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,272 |
1.22. Solve the system of equations from example 1.15 using the Gaussian method. | Solution. Let's write down the system of equations corresponding to the final form of the augmented matrix of the system from Example 1.15:
$$
\tilde{A} \sim\left(\begin{array}{ccc|c}
1 & 4 & 10 & 1 \\
0 & -5 & -13 & -1.25 \\
0 & 0 & 0 & 1.25
\end{array}\right), \quad\left\{\begin{array}{r}
x_{1}+4 x_{2}+10 x_{3}=1 \\... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,273 |
1.23. Find the fundamental system of solutions for the system of equations and write the general solution of this system:
$$
\left\{\begin{array}{l}
x_{1}+7 x_{2}-8 x_{3}+9 x_{4}=0 \\
2 x_{1}-3 x_{2}+3 x_{3}-2 x_{4}=0 \\
4 x_{1}+11 x_{2}-13 x_{3}+16 x_{4}=0 \\
7 x_{1}-2 x_{2}+x_{3}+3 x_{4}=0
\end{array}\right.
$$ | Solution. The main matrix of the system has the form
$$
\tilde{\boldsymbol{A}}=\left(\begin{array}{cccc}
1 & 7 & -8 & 9 \\
2 & -3 & 3 & -2 \\
4 & 11 & -13 & 16 \\
7 & -2 & 1 & 3
\end{array}\right)
$$
The rank of this matrix is 2, and the minor of the second order located in the upper left corner of this matrix is not... | \alpha(\begin{pmatrix}\frac{3}{17}\\\frac{19}{17}\\1\\0\end{pmatrix})+\beta(\begin{pmatrix}-\frac{13}{17}\\-\frac{20}{17}\\0\\1\end{pmatrix}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,274 |
1.24. Find the general solution of the non-homogeneous system
$$
x_{1}+7 x_{2}-8 x_{3}+9 x_{4}=4 \text {. }
$$ | Solution. The ranks of the main matrix of the system and the extended matrix of the system are equal to one. Since the number of unknowns is four, the number of basic unknowns is one (let the basic unknown be $x_{1}$), and the number of free unknowns is three (let the free unknowns be $x_{2}, x_{3}, x_{4}$). Rewrite th... | \begin{aligned}X^{\text{(generalnon-hom.)}}&=\alpha_{1}(\begin{pmatrix}-7\\1\\0\\0\end{pmatrix})+\alpha_{2}(\begin{pmatrix}8\\0\\1\\0\end{pmatrix})+\alpha_{3}(\begin{pmatrix}9\\0\\0\\1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,275 |
1.25. Find the general solution of the non-homogeneous system
\[
\left\{\begin{array}{l}
x_{1}+7 x_{2}-8 x_{3}+9 x_{4}=1 \\
2 x_{1}-3 x_{2}+3 x_{3}-2 x_{4}=1 \\
4 x_{1}+11 x_{2}-13 x_{3}+16 x_{4}=3 \\
7 x_{1}-2 x_{2}+x_{3}+3 x_{4}=4
\end{array}\right.
\] | Solution. The main matrix of the system coincides with the one considered in Example 1.23. It is not difficult to show that the ranks of the augmented matrix of the system and the main matrix of the system are equal to 2, and the system is consistent. The system is equivalent to the following:
$$
\left\{\begin{array}{... | (\begin{pmatrix}\alpha\cdot\frac{3}{17}+\beta\cdot(-\frac{13}{17})+\frac{10}{17}\\\alpha\cdot(\frac{19}{17})+\beta\cdot(-\frac{20}{17})+\frac{1}{17}\\\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,276 |
### 1.26. Find the coordinates of the vector
$$
X=\left(\begin{array}{l}
x_{1} \\
x_{2} \\
x_{3}
\end{array}\right)
$$
if the linear transformation that maps it to the vector
$$
Y=\left(\begin{array}{l}
1 \\
2 \\
3
\end{array}\right)
$$
is given by the system of equations
$$
\left\{\begin{array}{l}
x_{1}+2 x_{2}+3... | Solution. We find the inverse matrix for the main matrix of the system
$$
A=\left(\begin{array}{ccc}
1 & 2 & 3 \\
-1 & 3 & 4 \\
2 & -3 & 5
\end{array}\right)
$$
We compute the determinant of this matrix to ensure the existence of the inverse matrix. We have
 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,277 |
1.27. Given a linear transformation using a system of equations
$$
\left\{\begin{array}{l}
g_{1}=-x_{1}+0 \cdot x_{2}+0 \cdot x_{3} \\
g_{2}=0 \cdot x_{1}+1 \cdot x_{2}+0 \cdot x_{3} \\
g_{3}=0 \cdot x_{1}+0 \cdot x_{2}+1 \cdot x_{3}
\end{array}\right.
$$
Find the transformation matrix, will it be singular? | Solution. With the matrix representation of the transformation, we get:
$$
\left(\begin{array}{l}
g_{1} \\
g_{2} \\
g_{3}
\end{array}\right)=\left(\begin{array}{ccc}
-1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right) \cdot\left(\begin{array}{l}
x_{1} \\
x_{2} \\
x_{3}
\end{array}\right)
$$
Here, the transformati... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,278 |
1.28. There are linear transformations $A x=\left\{x_{2}-x_{3}\right.$, $\left.x_{1}, x_{1}+x_{3}\right\}$ and $B x=\left\{x_{2}, 2 x_{2}, x_{1}\right\}$ of the vector $x=\left\{x_{1}, x_{2}, x_{3}\right\}$. Find the mapping $B \cdot(2 A-B) x$. | Solution. Let's write down the matrices of the linear mappings. For this, we will write the given mappings in matrix form:
$$
\left(\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right) \cdot\left(\begin{array}{l}
x_{1} \\
x_{2} \\
x_{3}
\end{array}\righ... | (\begin{pmatrix}2&-2&0\\0&-4&0\\1&1&-2\end{pmatrix}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,279 |
1.29. Let the matrix $A=\left(\begin{array}{ccc}1 & 1 & 2 \\ 3 & -1 & 0 \\ 1 & 1 & -2\end{array}\right)$ be used to
transform the vector $x$ in the basis $\left\{e_{1}, e_{2}, e_{3}\right\}$. Determine the transformation matrix $A^{\prime}$ in the basis $\left\{e_{1}^{\prime}, e_{2}^{\prime}, e_{3}^{\prime}\right\}$,... | Solution. Let's write down the transformation matrix from the old basis to the new one:
$$
B=\left(\begin{array}{ccc}
1 & -1 & 1 \\
-1 & 1 & -2 \\
-1 & 2 & 1
\end{array}\right)
$$
Find the inverse matrix $B^{-1}$, first ensuring that it is not singular:
$$
\begin{aligned}
& \operatorname{det} B=\left|\begin{array}{c... | (\begin{pmatrix}9&-1&27\\5&-1&14\\-3&1&-8\end{pmatrix}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,280 |
### 1.30. Applying the Gram-Schmidt orthogonalization process, construct the orthonormal basis $\left\{e_{1}, e_{2}, e_{3}\right\}$ from the given basis $\left\{g_{1}, g_{2}, g_{3}\right\}:$
$$
g_{1}=(1 ; 1 ; 0), \quad g_{2}=(2 ; 0 ; 1), \quad g_{3}=(0 ; 1 ;-2)
$$ | Solution.
\[
\begin{aligned}
& f_{1}=g_{1}=(1 ; 1 ; 0), \\
& C_{1}^{2}=\left(g_{2}, f_{1}\right) /\left(f_{1}, f_{1}\right)=(2 \cdot 1+0 \cdot 1+1 \cdot 0) /\left(1^{2}+1^{2}+0^{2}\right)=1, \\
& f_{2}=g_{2}-C_{1}^{2} f_{1}=(2 ; 0 ; 1)-1 \cdot(1 ; 1 ; 0)=(1 ;-1 ; 1), \\
& C_{1}^{3}=\left(g_{3}, f_{1}\right) /\left(f_{... | \begin{aligned}&e_{1}=(\frac{\sqrt{2}}{2};\frac{\sqrt{2}}{2};0),\\&e_{2}=(\frac{\sqrt{3}}{3};-\frac{\sqrt{3}}{3};\frac{\sqrt{3}}{3}),\\&e_{3}=(\frac{\sqrt{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,281 |
1.31. Find the eigenvalues and eigenvectors of the linear operator $\hat{A}$ with the matrix
$$
A=\left(\begin{array}{ll}
1 & 2 \\
3 & 2
\end{array}\right)
$$
Will the eigenvectors be orthogonal? | Solution. Let's form the characteristic equation
$$
\left|\begin{array}{cc}
1-\lambda & 2 \\
3 & 2-\lambda
\end{array}\right|=0 \Rightarrow(1-\lambda)(2-\lambda)-6=0 \Rightarrow \lambda^{2}-3 \lambda-4=0
$$
Solving the quadratic equation, we find the eigenvalues $\lambda_{1}=-1, \lambda_{2}=4$. Next, we will find the... | x^{(1)}\cdotx^{(2)}=-0.5^2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,282 |
1.33. Reduce the quadratic form, which defines the equation of a second-order curve $5 x^{2}+6 x y+5 y^{2}-5 x-5 y=2$, to its canonical form. | Solution. The quadratic form defining the given equation is as follows:
$$
\begin{gathered}
F(x, y)=5 x^{2}+6 x y+5 y^{2}=5 x^{2}+3 x y+3 y x+5 y^{2}, \\
a_{11}=5, \quad a_{12}=a_{21}=3, \quad a_{22}=5
\end{gathered}
$$
The corresponding matrix $\tilde{A}$ of the quadratic form is:
$$
A=\left(\begin{array}{ll}
5 & 3... | \frac{(x^{\\})^{2}}{1^{2}}+\frac{(y^{\\})^{2}}{(0.5)^{2}}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,284 |
1.34. Reduce the equation of the surface to canonical form
$$
3 x^{2}+4 y^{2}+5 z^{2}+4 x y-4 y z+2 x-8 y+14 z=16
$$ | Solution. The quadratic form defining the given equation has the form
$$
F(x, y)=3 x^{2}+4 y^{2}+5 z^{2}+4 x y-4 y z
$$
The corresponding matrix $\tilde{A}$ of the quadratic form is:
$$
\tilde{A}=\left(\begin{array}{ccc}
3 & 2 & 0 \\
2 & 4 & -2 \\
0 & -2 & 5
\end{array}\right)
$$
To find the eigenvalues, we form th... | \frac{(x^{\\})^{2}}{28}+\frac{(y^{\\})^{2}}{7}+\frac{(z^{\\})^{2}}{4}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,285 |
2.1. The segment $A B$ is divided into three equal parts by points $C(3,4)$ and $D(5,6)$. Find the coordinates of points $A$ and $B$. | Solution. Let the coordinates of points $A$ and $B$ be: $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. For segment $A D$, point $C$ is the midpoint, so $\lambda=A C / C D=1$ and by the formulas for dividing a segment in a given ratio (2.6), we get
$$
x_{C}=\frac{x_{1}+\lambda x_{D}}{1+\lambda}=\frac{x_{1}+x... | A(1,2),B(7,8) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 33,286 |
2.2. The coordinate system $X O Y$ was first translated in parallel so that the new origin became located at point $O^{\prime}(2,3)$, and then the axes were rotated by an angle $\varphi=30^{\circ}$. Determine the new coordinates of point $A$, if the old coordinates in the system $X O Y$ are: $A(-1,5)$. Determine the co... | Solution:
a) when the coordinate system is translated, the coordinates of point $A$ will be as follows:
$$
\left\{\begin{array}{l}
x_{A}^{\prime}=x_{A}-a=-1-2=-3 \\
y_{A}^{\prime}=y_{A}-b=5-3=2
\end{array}\right.
$$
when the axes are rotated by an angle of $30^{\circ}$, the coordinates of point $A$ are transformed a... | x_{A}^{\\}\approx0.722;y_{A}^{\\}\approx3.598;x_{B}\approx5.732;y_{B}\approx5.464 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 33,287 |
2.3. The Cartesian coordinates of point $A(4,3)$ are known. Find the coordinates of this point in the polar coordinate system, if the pole is located at the origin of the Cartesian coordinate system, and the polar axis coincides in direction with the $O X$ axis. | Solution. Let's use the formulas connecting polar coordinates with Cartesian ones:
$$
\begin{gathered}
\rho=\sqrt{x^{2}+y^{2}}=\sqrt{4^{2}+3^{2}}=\sqrt{25}=5 \\
\tan \varphi=3 / 4 ; \varphi=\arctan 0.75 \approx 36^{\circ} 53'
\end{gathered}
$$
, $0 \leqslant \varphi \leqslant 2 \pi$, by giving $\varphi$ values at intervals of $\pi / 4$. | Solution. Let's create a table:
Connecting the obtained points with a smooth line, we will construct the desired curve (Fig. 2.10). | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 33,289 |
2.5. Given points $A(1,2,-2)$ and $B(3,1,4)$. Find the coordinates of vectors $\overrightarrow{A B}$ and $\overline{B A}$: Find $|\overrightarrow{A B}|$ and $|\overrightarrow{B A}|$. | Solution. The coordinates of vectors $\overline{A B}$ and $\overrightarrow{B A}$ are found as the differences of the corresponding coordinates of the end and start points of the vectors, i.e., the differences of the corresponding coordinates of points $A$ and $B$. Thus, we have
$$
\overrightarrow{A B}=\{3-1,1-2,4-(-2)... | \sqrt{41} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,290 |
2.6. Calculate the direction cosines of the vector $\overrightarrow{A B}$, if $A(-3,2,0)$ and $B(3,-3,1)$. | The solution is as follows. The direction cosines of vector $\overrightarrow{A B}$ are determined by its coordinates using formulas (2.28). We find the coordinates of vector $\overrightarrow{A B}$ in the same way as it was done in problem (2.5). We have $\bar{A} B=\{3+3,-3+2,1-0\}=\{6,-1,1\}$. The modulus of vector $\o... | \begin{aligned}&\cos\alpha=\frac{6}{\sqrt{38}}\\&\cos\beta=\frac{-1}{\sqrt{38}}\\&\cos\gamma=\frac{1}{\sqrt{38}}\end{aligned} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,291 |
2.7. Vector $\vec{a}$ forms equal angles with the coordinate axes. Find its coordinates if $|\bar{a}|=\sqrt{3}$. | The problem is solved. The direction cosines satisfy condition (2.29), and since the angles formed by vector $\vec{a}$ with the coordinate axes are equal, i.e., $\alpha=\beta=\gamma$, taking this into account, (2.29) in the considered case will take the form: $3 \cos ^{2} \alpha=1$. From this, it follows that
$$
\cos ... | \vec{}={1,1,1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,292 |
2.9. Write the decomposition of vector $\vec{a}$ in terms of vectors
$$
\vec{p}=\{1,2,4\} ; \vec{q}=\{1,-1,1\} ; \vec{r}=\{2,2,4\} ; \vec{a}=\{-1,-4,-2\} .
$$ | Solution. According to (2.20), we need to find such numbers $\lambda, \mu, v$ that
$$
\ddot{a}=\lambda \cdot \vec{p}+\mu \cdot \vec{q}+v \cdot \vec{r}
$$
Substituting into this equality the expansions of vectors $\vec{a}, \vec{p}, \vec{q}, \vec{r}$ in terms of the basis $\vec{i}, \vec{j}, \vec{k}$, as given in the pr... | \vec{}=\vec{p}+2\vec{q}-2\vec{r} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,293 |
2.10. Are the vectors $\vec{a}$ and $\vec{b}$, constructed from vectors $\vec{p}$ and $\vec{q}$, collinear?
$$
\vec{a}=3 \vec{p}-4 \vec{q} ; \vec{b}=2 \vec{p}+\vec{q} ; \vec{p}=\{1,1,2\} ; \vec{q}=\{3,1,0\}
$$ | The problem is solved. To solve this problem, we can use the geometric property of the first vector product, i.e., find the coordinates of vectors $\vec{a}$ and $\vec{b}$ relative to the basis $\vec{i}, \vec{j}, \vec{k}$, and then compute their vector product. We have
$$
\begin{aligned}
& \vec{a}=3(\vec{i}+\vec{j}+2 \... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,294 |
2.11. Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$, where $A(1,2,1), B(3,-1,7), C(7,4,-2)$. | Solution. From the definition of the scalar product (formula (2.30)), it follows that the cosine of the angle between the vectors is determined by the formula
$$
\cos \varphi=\frac{\vec{A} \vec{B} \cdot \overrightarrow{A C}}{|\overrightarrow{A B}| \cdot|\overrightarrow{A C}|}
$$
To find $\cos \varphi$ using this form... | -\frac{12}{49} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,295 |
2.12. Calculate the area of the parallelogram constructed on vectors $\vec{a}$ and $\vec{b}$, the lengths of the diagonals of the parallelogram, $\cos (\widehat{\vec{a}, \vec{p}})$ and $r_{\dot{a}} \ddot{b}$, if
$$
\vec{a}=5 \vec{p}+2 \vec{q}, \vec{b}=\vec{p}-3 \vec{q},|\vec{p}|=4 \sqrt{2},|\vec{q}|=6,(\widehat{\vec{p... | S o l u t i o n. The area $S$ of the parallelogram constructed on vectors $\vec{a}$ and $\vec{b}$ is equal to the magnitude of the vector product of these vectors. Using the algebraic properties of the vector product, we have
$$
\begin{aligned}
\vec{a} \times \vec{b}=(5 \vec{p}+2 \vec{q}) & \times(\vec{p}-3 \vec{q})=5... | S=408,|\vec{}+\vec{b}|=30,|\vec{}-\vec{b}|=2\sqrt{593},\cos(\widehat{\vec{},\vec{p}})=\frac{13}{\sqrt{178}},\Pi_{\vec{}}\vec{b}=-\frac{92}{\sqrt{89}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,296 |
2.13. Are the vectors coplanar
$$
\vec{a}, \vec{b}, \vec{c}: \vec{a}=\{1,-2,1\}, \vec{b}=\{3,1,-2\}, \vec{c}=\{7,14,-13\} ?
$$ | Solution. To answer the given question, it is necessary to calculate the mixed product of these vectors, and if it turns out to be zero, this will indicate that the vectors $\ddot{a}, \vec{b}, \vec{c}$ are coplanar. We find the mixed product $\vec{a} \cdot \vec{b} \cdot \vec{c}$ using formula (2.40).
$$
\begin{aligned... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,297 |
2.14. Given the vertices of a tetrahedron (see Fig. 2.11) \( A(2,-1,2) \), \( B(5,5,5) \), \( C(3,2,0) \), \( D(4,1,4) \). Calculate: 1) the volume of the tetrahedron; 2) its height dropped from vertex \( D \); 3) the angle between edge \( AD \) and its face \( ABC \).
$ through which the line passes, and a vector perpendicular to the line $\vec{n}=\{A ; B\}=\{4 ;-1\}$. | Solution.
1st step. Make a schematic drawing (Fig. 3.1).
2nd step. Write down the coordinates of the vectors defining the given geometric object - the line. This is the normal vector of the line
$$
\vec{n}=\{A ; B\}=\{4 ;-1\}
$$
$ and parallel to the vector $\ddot{a}=\{l ; m\}=\{8 ;-4\}$. | Solution.
1st step. Make a schematic drawing (Fig. 3.2).
2nd step. Write down the coordinates of the vectors defining the given geometric object - the line. These are the direction vector of the line $\vec{a}=\{l ; m\}=\{8 ;-4\}$ and the current vector $\widehat{P M}=\{x+1 ; y-2\}$.
3rd step. Write down one of the c... | \frac{x+1}{8}=\ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,300 |
3.3. Obtain the equation of the line passing through two given points $P_{0}(1 ; 2)$ and $P_{1}(3 ; 2)$. | Solution.
1st step. Make a schematic drawing (Fig. 3.3).
2nd step. Write down the coordinates of the vectors defining the given geometric object - the line. This is the current vector lying on the line $\overrightarrow{P_{0} M}=\{x-1 ; y-2\}$ and the vector formed by two points on the line
$$
\overrightarrow{P_{0} P... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,301 |
3.4. Based on the equation of a straight line in general form $3 x-4 y+$ $+20=0$, transform this equation: 1 ) into the slope-intercept form; 2) into the intercept form; 3 ) into the canonical form; 4) into the normal form; 5) into the parametric equations; 6) into the polar equation. | Solution.
1) to obtain the equation with the slope coefficient, solve the original equation for $y$:
$$
y=\frac{3}{4} x+5
$$
2) determine the segments that the given line intercepts on the coordinate axes. Setting $x_{0}=0$, we get from the general equation $3 \cdot 0-4 \cdot y_{0}+20=0$ the ordinate of the point of... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,302 |
3.6. A line is given by the equation with the slope $y=3 x-4$
1) write the equation of the line parallel to the given one and passing through the point $\left.P_{0}=(1 ; 2) ; 2\right)$ write the equation of the line perpendicular to the given one and passing through the point $P_{0}=(1 ; 2)$. | Solution.
1) Based on the condition of parallelism (3.14) of two lines given by equations with angular coefficients $k_{2}=k_{1}=3$, using formula (3.5), we obtain that the required equation has the form $y-2=3(x-1)$ or $y=3 x-1$;
2) Based on the condition (3.15) of perpendicularity of two lines, we get $k_{2}=-1 / k_... | 3x-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,303 |
3.8. Find the equation of the line passing through the point $P(2, -3)$ and the intersection point of the lines $3x + 2y - 4 = 0$ and $x - y + 5 = 0$. | Solution.
Let's use the equation (3.21) of the pencil of lines $3 x+2 y-4+\lambda(x-y+5)=0$. Since the desired line passes through the point $P(2, -3)$, we can find the parameter $\lambda$ by substituting the coordinates of this point into the equation of the pencil of lines:
$$
3 \cdot 2 + 2 \cdot (-3) - 4 + \lambda... | 3.4x+1.6y-2=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,304 |
3.10. Obtain the equations of the median, bisector, and altitude, emanating from the vertex of the triangle, which is the intersection of two sides of the triangle $3 x-4 y+$ $+8=0$ and $12 x+5 y-73=0$, if the equation of the third side is $x+y+12=0$. | S o l u t i o n. We will make a schematic drawing (Fig. 3.4).
1. We will find the coordinates of the vertex from which the required median, bisector, and height originate. For this, we will solve the system of equations $3 x-4 y+8=0,12 x+5 y-$ $-73=0$ simultaneously. From the first equation, multiplied by 4, we subtra... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 33,305 | |
3.11. The equation of one side of the square \(ABCD-AB\) is \(x + 3y - 5 = 0\). Form the general equation of the other three sides of the square, given that \(P(-1; 0)\) is the point of intersection of its diagonals.
.
1. Find the coordinates of point $K$, the intersection of $AB$ with the perpendicular $PK$ dropped from point $P$ to $AB$, i.e., having the current vector $\overrightarrow{PM}=\{x+1 ; y-0\}$ and perpendicular to the line $AB$ (parallel to its norma... | \begin{aligned}&EquationofAD:&3x-y+9=0\\&EquationofBC:&3x-y-3=0\\&EquationofCD:&x+3y+7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 33,306 |
3.12. Given the equations of one side $A B$ of the rhombus $A B C D$: $x-3 y+10=0$, and one of its diagonals: $x+4 y-4=0$. The diagonals of the rhombus intersect at point $P(0 ; 1)$. Find the equations of the other sides of the rhombus, the distance from point $P$ to side $A B$, and the internal angles of the rhombus. | Solution. Let's make a schematic drawing (Fig. 3.6).
1. The equation of the diagonal $AC$ can be obtained from the condition of orthogonality of $AC$ and $BD$ and the fact that $BD$ passes through the point $P$,
$$
(x-0) / 1=(y-1) / 4, 4 x-y+1=0
$$
 \\
-x+y-1 \leqslant 0 \quad(AB) \\
y \geqslant-1 \quad(AC)
\end{array}\right.
$$ | The solution is as follows. We will choose a test point that does not belong to any of the lines obtained from the given inequalities
Fig. 3.7 by replacing the inequality signs with equalit... | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 33,308 |
3.15. Form the equation of a circle with center at point $P_{0}(1; -2)$ and radius $R=3$. | S o l u t i o n. According to formula (3.25), the equation of the circle will be written as: $(x-1)^{2}+(y+2)^{2}=3^{2}=9$. | (x-1)^{2}+(y+2)^{2}=9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 33,310 |
3.16. Reduce the equation of the circle $x^{2}-2 x+y^{2}+$ $+6 y=6$ to its canonical form. Determine the coordinates of the center and the radius. | Solution. We use the formulas for the complete square of the sum and difference, and complete the square of the expressions
a) $x^{2}-2 x=x^{2}-2 \cdot x \cdot 1+1^{2}-1^{2}=(x-1)^{2}-1$;
b) $y^{2}+6 y=y^{2}-2 \cdot y \cdot 3+3^{2}-3^{2}=(y+3)^{2}-9$.
The equation of the circle can now be rewritten as follows:
$$
(... | (x-1)^{2}+(y+3)^{2}=4^{2},centerO(1,-3),radiusR=4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 33,311 |
3.17. Form the equation of the geometric locus of points on the plane, each of which is twice as close to point $A(4 ; 0)$ as to point $B(1 ; 0)$. | Solution. The distance from the current point $M(x ; y)$ of the desired curve to the point $A(4 ; 0)$ is denoted by
$$
r_{1}=\sqrt{(x-4)^{2}+(y-0)^{2}},
$$
to the point $B(1 ; 0)$ - by
$$
r_{2}=\sqrt{(x-1)^{2}+(y-0)^{2}}
$$
From the condition of the problem, it follows that $r_{1}=2 r_{2}$, from which we obtain
$$... | x^{2}+y^{2}=(2\sqrt{3})^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 33,312 |
3.18. Find the eccentricity, coordinates of the foci, and vertices of the ellipse $x^{2} / 25+y^{2} / 9=1$. | Solution. According to the condition $a^{2}=25, b^{2}=9$, hence $a=5$, $b=3$. In accordance with the notations in Figure 3.10, we get $c=\sqrt{25-9}=4, \varepsilon=4 / 5=0.8, A_{1,2}(\mp 5 ; 0), B_{1,2}(0 ; \mp 3)$, $F_{1,2}(\mp 4 ; 0)$. | =4,\varepsilon=0.8,A_{1,2}(\5;0),B_{1,2}(0;\3),F_{1,2}(\4;0) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 33,313 |
3.19. To construct the geometric locus of points on the plane, the ratio of whose distances to the point \( F(4.5; 0) \) and to the line \( x=0.5 \) is three. | Solution. Let's make a schematic
Fig. 3.13 diagram (Fig. 3.13).
Let the coordinates of the current point of the desired curve be \( M(x ; y) \). Then the distance from the point \( M(x ; ... | \frac{x^{2}}{2.25}-\frac{y^{2}}{18}=1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 33,314 |
3.20. To construct the geometric locus of points equidistant from the point $A(4 ; 1)$ and from the y-axis. | Solution. From the definition of a parabola, we conclude that the sought geometric locus of points is a parabola. Let's choose a coordinate system in such a way that the equation of the parabola has a canonical form. The origin should be at the midpoint of the perpendicular dropped from the focus to the directrix, i.e.... | (y-1)^2=8(x-2) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 33,315 |
3.22. Determine that the equation $\rho=36 /(4-5 \cos \varphi)$ defines a hyperbola, and find its semi-axes. | Solution. Divide both the numerator and the denominator by 4. Transform the given equation to the form
$$
\rho=\frac{9}{\left(1-\frac{5}{4} \cos \varphi\right)}
$$
Since $\varepsilon=c / a=5 / 4>1$, the found equation defines the right branch of a hyperbola. Using formulas (3.50), $b^{2}=c^{2}-a^{2}$, and the equatio... | =16,b=12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,316 |
3.23. Derive the polar equation of the ellipse $x^{2} / a^{2}+$ $+y^{2} / b^{2}=1$ under the condition that the direction of the polar axis coincides with the positive direction of the x-axis, and the pole is at the center of the ellipse. | Solution. Substitute into the ellipse equation the expressions for $x$ and $y$ in terms of the polar radius $\rho$ and the angle $\varphi$.
We have
$$
\frac{\rho^{2} \cos ^{2} \varphi}{a^{2}}+\frac{\rho^{2} \sin ^{2} \varphi}{b^{2}}=1, \rho^{2} \frac{b^{2} \cos ^{2} \varphi+a^{2} \sin ^{2} \varphi}{a^{2} b^{2}}=1
$$
... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 33,317 |
4.1. Form the equation of the plane that is perpendicular to the vector $\vec{n}=(8,2,1)$ and passes through the point $P_{0}(3,-5,4)$. | Solution. Let's write down the coordinates of the current vector: $P_{0} M=(x-3, y+5, z-4)$. We will write the condition of orthogonality of vectors $\vec{n}$ and $\overline{P_{0} M}$ in coordinate form:
$$
\begin{aligned}
& 8(x-3)+2(y+5)+1(z-4)=0 \\
& 8 x+2 y+z-18=0
\end{aligned}
$$ | 8x+2y+z-18=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,318 |
4.2. Two planes are given by their general equations:
$$
\begin{aligned}
& 5 x-4 y+z-1=0 \\
& 10 x+2 y+4 z-7=0
\end{aligned}
$$
Determine the magnitude of the dihedral angle between these planes. | Solution. The dihedral angle between planes is measured, as is known, by a linear angle, and the latter is equal to the angle between vectors perpendicular to the planes. The vector perpendicular to the first plane has coordinates $\vec{n}_{1}=(5,-4,1)$, to the second plane $\ddot{n}_{2}=(10,2,4)$. Let's use the formul... | 52 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 33,319 |
4.3. Form the equation of the plane passing through the points $P_{0}(2,-1,2), P_{1}(4,3,0), P_{2}(5,2,1)$. | Solution. Let's form the coordinates of the current vector of the plane $\overline{P_{0}} M=(x-2, y+1, z-2)$ and the vectors
$$
\overrightarrow{P_{0} P_{1}}=(2,4,-2), \overline{P_{0} P_{2}}=(3,3,-1) .
$$
For these vectors, the condition of coplanarity is valid:
$$
\left|\begin{array}{ccc}
x-2 & y+1 & z-2 \\
2 & 4 & ... | x-2y-3z+2=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,320 |
4.4. Given the equation of a plane in general form $2 x-2 y+$ $+z-20=0$.
Transform this equation: 1) into the equation of a plane in intercept form; 2) into the normal equation of the plane. | Solution.
1) Let's calculate the lengths of the segments cut off by the plane on the coordinate axes, counting from the origin. Assuming $y_{0}=0, z_{0}=0$, we get $2 x_{0}-2 \cdot 0+0-20=0, x_{0}=10=a$. Now let $x_{1}=0, z_{1}=0$, then $2 \cdot 0-2 \cdot y_{1}+0-20=0$, $y_{1}=-10=b$. Finally, we find the segment cut ... | x/10+y/(-10)+z/20=1-2/3\cdotx+2/3\cdoty-1/3\cdotz+20/3=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 33,321 |
4.6. A line is given by its general equations:
$$
\left\{\begin{array}{l}
3 x-4 y-z-1=0 \\
x+2 y+4 z-5=0
\end{array}\right.
$$
From these equations, obtain: 1) the canonical equations; 2) the parametric equations; 3) find the distance from the point $A(3,2 ; 4,4 ;-1)$ to the given line. | Solution.
1) To find the coordinates of two points belonging to the line, we first set $z_{0}=0$ and substitute it into the equation of the line:
$$
\left\{\begin{array} { l }
{ 3 x _ { 0 } - 4 y _ { 0 } - 0 - 1 = 0 , } \\
{ x _ { 0 } + 2 y _ { 0 } + 4 \cdot 0 - 5 = 0 , }
\end{array} \Rightarrow \left\{\begin{array}... | 1.57 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 33,322 |
4.7. Form the equation of the line passing through the given point $P_{0}(3,4,2)$ and perpendicular to the plane $8 x-4 y+5 z-4=0$ | Solution. Let's find the coordinates of the current vector of the desired line $\widehat{P}_{0} M=(x-3, y-4, z-5)$. Since the line is perpendicular to the plane, the direction vector of the line is the normal vector of the plane $\vec{n}=(8,4,5)$. We write the condition of collinearity of vectors $\overrightarrow{P_{0}... | (x-3)/8=(y-4)/(-4)=(z-2)/5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,323 |
4.8. Form the equation of the line passing through the point $P_{0}(4,-3,2)$ and parallel to the given line:
$$
\left\{\begin{array}{l}
2 x-y-z+1=0 \\
3 x+2 y+z-8=0
\end{array}\right.
$$ | Solution. Let's determine the coordinates of the direction vector of the line by finding the vector product of the vectors perpendicular to the planes $2 x-y-z+1=0$ and $3 x+2 y+z-8=0$. We have $\vec{n}_{1}=(2,-1,-1), \vec{n}_{2}=(3,2,1)$ :
$$
\begin{aligned}
\ddot{n}_{1} \times \vec{n}_{2}=\left|\begin{array}{ccc}
\v... | (x-4)/1=(y+3)/(-5)=(z-2)/7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,324 |
4.9. Form the equation of the plane passing through the line $(x-1) / 2=(y-3) / 4=z /(-1)$ and the point $P_{1}(1,5,2)$. | Solution. Let's write the condition of coplanarity of three vectors: the current vector $\overrightarrow{P_{0} M}=(x-1, y-3, z)$, the vector $\overrightarrow{P_{0} P_{1}}=(0,2,2)$; belonging to the plane, and the direction vector of the given line $\vec{a}=(2,4,-1)$, parallel to the given plane:
$$
\begin{aligned}
& \... | 5x-2y+2z+1=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,325 |
4.10. Given the coordinates of the vertices of a pyramid $A_{1}(4,2,5)$, $A_{2}(0,7,2)$, $A_{3}(0,2,7)$, $A_{4}(1,5,0)$. Find: 1) the equations of the planes $A_{1} A_{2} A_{3}$ and $A_{1} A_{2} A_{4}$; 2) the angle between the edge $A_{1} A_{3}$ and the face $A_{1} A_{2} A_{4}$; 3) the distance from the vertex $A_{3}$... | Solution.
1) The equation of the plane $A_{1} A_{2} A_{3}$ is obtained from the condition of coplanarity of three vectors: the current
$$
\overrightarrow{A_{1} M}=(x-4, y-2, z-5)
$$
and two vectors belonging to the given plane,
$$
\begin{gathered}
\overline{A_{1} A_{2}}=(-4,5,-3) \text { and } \overline{A_{1} A_{3}... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 33,326 | |
4.11. Form the equation of the plane equidistant from the points $B(4,1,0)$ and $C(2,0,3)$. | Solved. We will use the distance formula between points to find the coordinates of point $A(0,0, z)$, equidistant from the two given points, and set the magnitudes $|A B|=|A C|$.
We have
$$
\begin{aligned}
& |A B|=\sqrt{(4-0)^{2}+(1-0)^{2}+(0-z)^{2}} \\
& |A C|=\sqrt{(2-0)^{2}+(0-0)^{2}+(3-z)^{2}} \\
& 17+z^{2}=4+(3-... | -2x-y+3z+2=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 33,327 |
4.12. Find the point of intersection of the line
$$
(x-2) / 1=(y-3) / 1=(z+1) / 4
$$
and the plane
$$
x+2 y+3 z-10=0
$$ | Solution. Let's write down the parametric equations of the given line (formula (4.14)):
$$
x=2+t, y=3+t, z=-1+4 t
$$
Substitute the coordinates of the points on the line into the equation of the plane:
$$
(2+t)+2(3+t)+3(-1+4 t)-10=0, t=1 / 3
$$
The desired coordinates of the intersection point are obtained by subst... | P(7/3,10/3,1/3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,328 |
4.14. Find the coordinates of the point $A^{\prime}$, symmetric to the point $A(2,0,2)$ with respect to the plane $4 x+6 y+4 z-50=0$. | Solution. Let's write the equation of the line passing through point $\boldsymbol{A}$ and perpendicular to the given plane:
$$
(x-2) / 4=(y-0) / 6=(z-2) / 4
$$
Convert the equation of the line to parametric form:
$$
\begin{aligned}
& (x-2) / 4=(y-0) / 6=(z-2) / 4=t \\
& x=2+2 t, \quad y=3 t, \quad z=2+2 t
\end{align... | (6,6,6) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 33,329 |
4.15. Let $k=2$ be the coefficient of homothety with the center at the origin. Is it true that the point $A(2$, $1,-1)$ belongs to the image of the plane $2 x+3 y+z-3=0$? | Solution.
The formulas for similarity transformation with a coefficient $k=2$ are
$$
x^{\prime}=2 x, \quad y^{\prime}=2 y, \quad z^{\prime}=2 z
$$
To obtain the image of the plane $2 x+3 y+z-3=0$, we make the substitution
$$
x=x^{\prime} / 2, \quad y=y^{\prime} / 2, \quad z=z^{\prime} / 2
$$
After this, we get
$$... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 33,330 |
4.17. A sphere passes through the point $A(4,2,2)$ and has its center at the point $C(1,-1,-1)$. Write its equation. | Solution. The equation of a sphere centered at the origin has the form
$$
\left(x^{\prime}\right)^{2}+\left(y^{\prime}\right)^{2}+\left(z^{\prime}\right)^{2}=R^{2}
$$
Since the center of the given sphere is located at the point with coordinates $C(1,-1,-1)$, it is natural to perform a parallel translation of the coor... | (x-1)^2+(y+1)^2+(z+1)^2=27 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 33,331 |
4.18. What surface is defined by the equation
$$
x^{2}-2 x-3 y^{2}+12 y+2 z^{2}+12 z-11=0 ?
$$ | Solution. Let's complete the squares for $x, y, z$:
$$
\begin{aligned}
& x^{2}-2 x+1^{2}-1^{2}-3\left(y^{2}-4 y+2^{2}-2^{2}\right)+ \\
& \quad+2\left(z^{2}+6 z+3^{2}-3^{2}\right)-11=0 \\
& (x-1)^{2}-3(y-2)^{2}+2(z+3)^{2}=18 \\
& (x-1)^{2} / 18-(y-2)^{2} / 6+(z+3)^{2} / 9=1
\end{aligned}
$$
- The equation represents a... | (x-1)^{2}/18-(y-2)^{2}/6+(z+3)^{2}/9=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,332 |
4.20. Given the equation of a line lying in the XY plane, $4 x^{2}+p y^{2}=1$ and the coordinates of point $A(3 / 2,1,1)$. It is required: 1) to derive the equation of the surface formed by rotating this line around the $O X$ axis; 2) to select the value of the parameter $p$ so that point $A$ lies on the surface of rot... | Solution.
1) Since the line rotates around the axis $O X$, to obtain the equation of the surface of revolution, it is necessary to replace $y$ with $\pm \sqrt{y^{2}+z^{2}}$ in the relation $4 x^{2}+p y^{2}=1$. We have
$$
4 x^{2}+p\left( \pm \sqrt{y^{2}+z^{2}}\right)^{2}=1, \quad 4 x^{2}+p\left(y^{2}+z^{2}\right)=1
$$... | 4x^{2}-4\cdot(y^{2}+z^{2})=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,333 |
4.21. Determine that the plane $z-2=0$ intersects the ellipsoid $x^{2} / 12+y^{2} / 4+z^{2} / 16=1$ along an ellipse. Formulate its equation and find the semi-axes and vertices. | Solution. Substitute the number 4 for $z^{2}$ in the surface equation,
$$
\begin{aligned}
& x^{2} / 12+y^{2} / 4+4 / 16=1 \\
& x^{2} / 12+y^{2} / 4=3 / 4 \\
& x^{2} / 3^{2}+y^{2} / 3=1
\end{aligned}
$$
from which the ellipse has semi-axes $a=3, b=\sqrt{3}$. | =3,b=\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 33,334 |
4.22. Form the equation of the geometric locus of points equidistant from the point $A(-2,0,0)$ and the plane $x=2$. | Solution. Let's equate the distances from the point to the plane and between two points:
$$
\begin{aligned}
& |1 \cdot x+0 \cdot y+0 \cdot z-2| / \sqrt{1^{2}+0^{2}+0^{2}}= \\
& \quad=\sqrt{(x+2)^{2}+(y-0)^{2}+(z-0)^{2}}
\end{aligned}
$$
Square both sides of the equation:
$$
(x-2)^{2}=(x+2)^{2}+y^{2}+z^{2}, x=-\left(... | -(y^{2}+z^{2})/8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 33,335 |
5.1. Find the limit of the sequence
$$
\lim _{n \rightarrow \infty} \frac{3 n^{5}-6 n^{4}+3 n+5}{(n+1)^{6}-(n-2)^{6}}
$$ | Solution. First, we transform the denominator of this sequence using the formulas for abbreviated multiplication:
$$
\begin{gathered}
(n+1)^{6}-(n-2)^{6}=\left((n+1)^{3}-(n-2)^{3}\right)\left((n+1)^{3}+(n-2)^{3}\right)= \\
=3\left((n+1)^{2}+(n+1)(n-2)+(n-2)^{2}\right)\left((n+1)^{2}-(n+1) \times\right. \\
\left.\times... | \frac{1}{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,336 |
5.2. Find the limit of the sequence
$$
\lim _{n \rightarrow \infty} \frac{\sqrt{2 n-1}-\sqrt[3]{n^{3}+10}}{\sqrt{n^{2}+1}-\sqrt[4]{n+1}}
$$ | Solution. Divide the numerator and the denominator of the fraction by $n$. After transformations and discarding infinitely small quantities, we obtain the required result:
$$
\begin{gathered}
\lim _{n \rightarrow \infty} \frac{\sqrt{2 n-1}-\sqrt[3]{n^{3}+10}}{\sqrt{n^{2}+1}-\sqrt[4]{n+1}}=\lim _{n \rightarrow \infty} ... | -1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,337 |
5.3. Find the limit of the sequence:
$$
\lim _{n \rightarrow \infty}(\sqrt{(n+2)(n+7)}-n)
$$ | Solution. Multiply and divide the general term of the sequence by $\sqrt{(n+2)(n+7)}+n$ and use the formula for the difference of squares:
$$
\begin{gathered}
\lim _{n \rightarrow \infty}\left(\frac{(\sqrt{(n+2)(n+7)}-n) \cdot(\sqrt{(n+2)(n+7)}+n)}{(\sqrt{(n+2)(n+7)}+n)}\right)= \\
=\lim _{n \rightarrow \infty} \frac{... | \frac{9}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,338 |
5.5. Find the limit of the sequence
$$
\lim _{n \rightarrow \infty}\left(\frac{2 n^{2}+2 n+3}{2 n^{2}-2 n+5}\right)^{5-7 n}
$$ | Solution. In this case, we have the indeterminate form "1":
$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{2 n^{2}+2 n+3}{2 n^{2}-2 n+5}\right)^{5-7 n}=\lim _{n \rightarrow \infty}\left(1+\frac{4 n-2}{2 n^{2}-2 n+5}\right)^{5-7 n}= \\
&=\lim _{n \rightarrow \infty}\left(\left(1+\frac{4 n-2}{2 n^{2}-2 n+5... | e^{-14} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,339 |
## 5.6. Find the limit of the function
$$
\lim _{x \rightarrow \infty} \frac{3 x^{4}-2 x^{3}+6 x^{2}+x-2}{9 x^{4}+5 x^{3}+7 x^{2}+x+1}
$$ | Solution.
$$
\begin{gathered}
\lim _{x \rightarrow \infty} \frac{3 x^{4}-2 x^{3}+6 x^{2}+x-2}{9 x^{4}+5 x^{3}+7 x^{2}+x+1}=\lim _{x \rightarrow \infty} \frac{\frac{3 x^{4}-2 x^{3}+6 x^{2}+x-2}{x^{4}}}{\frac{9 x^{4}+5 x^{3}+7 x^{2}+x+1}{x^{4}}}= \\
=\frac{3-2 \lim _{x \rightarrow \infty} \frac{1}{x}+6 \lim _{x \rightar... | \frac{1}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,340 |
5.9. Calculate the limit
$$
\lim _{x \rightarrow \pm \infty}\left(\sqrt{x^{2}+1}-\sqrt{x^{2}-1}\right)
$$ | Solution. In this case, we have an indeterminate form of type «$\infty - \infty$». To resolve it, we multiply and divide the given expression by its conjugate. We get
$$
\begin{gathered}
\lim _{x \rightarrow \pm \infty}\left(\sqrt{x^{2}+1}-\sqrt{x^{2}-1}\right)=\lim _{x \rightarrow \pm \infty} \frac{x^{2}+1-x^{2}+1}{\... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,341 |
5.10. Calculate the limit
$$
\lim _{x \rightarrow 0} \frac{\tan 6 x}{\sin 3 x}
$$ | Solution. To calculate this limit, which represents an indeterminate form of the type "0/0", it is best to use the theorem on the replacement of infinitesimal functions with equivalent quantities.
According to this theorem, as $x \rightarrow 0, \operatorname{tg} 6 x \sim 6 x$, $\sin 3 x \sim 3 x$. Taking this into acc... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,342 |
### 5.11. Compute the limit
$$
\lim _{x \rightarrow 0} \frac{1+\sin 2 x-\cos 2 x}{1-\sin 2 x-\cos 2 x}
$$ | Solution. Under the limit sign, we have an indeterminate form of type "0/0". To resolve this, we will use the known trigonometric formulas:
$$
\sin 2 x=2 \sin x \cos x, \quad 1-\cos 2 x=2 \sin ^{2} x
$$
Taking this into account, we get
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{1+\sin 2 x-\cos 2 x}{1-\sin 2 ... | -1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,343 |
### 5.13. Calculate
$$
\lim _{x \rightarrow \pi}(\pi-x) \cot x
$$ | Solution. The given limit represents an indeterminate form of type «0 $\cdot \infty »$. To resolve this, we introduce a new variable $\pi-x=t$. Then
$\lim _{x \rightarrow \pi}(\pi-x) \operatorname{ctg} x=\lim _{t \rightarrow 0} t \operatorname{ctg}(\pi-t)=-\lim _{t \rightarrow 0} t \operatorname{ctg} t=-\lim _{t \righ... | -1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,344 |
5.14. Calculate the limit
$$
\lim _{x \rightarrow 0} \frac{\sqrt[3]{1+\operatorname{arctan} 4 x}-\sqrt[3]{1-\operatorname{arctan} 4 x}}{\sqrt{1-\arcsin 3 x}-\sqrt{1+\operatorname{arctan} 3 x}}
$$ | Solution. Here we have an indeterminate form of type "0/0". To resolve it, we replace the functions under the limit sign with their equivalent functions.
$$
A=\lim _{x \rightarrow 0} \frac{\sqrt[3]{1+\operatorname{arctg} 4 x}-\sqrt[3]{1-\operatorname{arctg} 4 x}}{\sqrt{1-\arcsin 3 x}-\sqrt{1+\operatorname{arctg} 3 x}}... | -\frac{8}{9} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,345 |
### 5.17. Compute the limit
$$
\lim _{x \rightarrow \infty}\left(\frac{x^{3}+1}{x^{3}-1}\right)^{x^{3}}
$$ | Solution. The calculation of this limit is also related to the resolution of an indeterminate form of type $1^{x}$. We will carry out the transformations recommended in section 5.7.5 (point 5). We have
$$
\begin{gathered}
\lim _{x \rightarrow \infty}\left(\frac{x^{3}+1}{x^{3}-1}\right)^{x^{3}}=\lim _{x \rightarrow \in... | e^2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,346 |
5.18. Calculate the limit
$$
\lim _{x \rightarrow \infty}\left(\frac{4 x^{2}-x+1}{2 x^{2}+x+1}\right)^{\frac{x^{3}}{2-x}}
$$ | Solution. This limit is not an indeterminate form, since when
$$
x \rightarrow \infty \frac{4 x^{2}-x+1}{2 x^{2}+x+1} \rightarrow 2, \text { and } \frac{x^{3}}{2-x} \sim -x^{2} \rightarrow -\infty
$$
Therefore:
$$
\begin{gathered}
\lim _{x \rightarrow \infty}\left(\frac{4 x^{2}-x+1}{2 x^{2}+x+1}\right)^{\frac{x^{3}}... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,347 |
5.20. Calculate the limits
$$
\lim _{x \rightarrow 0 \pm 0} \frac{1}{1+3^{2 / x}}
$$ | Solution. When $x \rightarrow 0+0 (x \rightarrow +0)$, the quantity $2 / x \rightarrow +\infty$. Then $3^{2 / x} \rightarrow 0$. Therefore, $1 /\left(1+3^{2 / x}\right) \rightarrow 0$. Thus,
$$
\lim _{x \rightarrow 0+0} \frac{1}{1+3^{2 / x}}=0
$$
Reasoning similarly, when $x \rightarrow 0-0 (x \rightarrow -0)$, the q... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,349 |
5.22. Let
$$
f(x)=\left\{\begin{array}{l}
e^{3 x}, \text { if } x<0, \\
a+5 x, \text { if } x \geqslant 0
\end{array}\right.
$$
For which choice of the number $a$ will the function $f(x)$ be continuous? | The problem is solved. Let's write down the condition for the continuity of the function at the point \( x=0 \), where it is currently not continuous due to the arbitrariness of the number \( a \). We have
\[
\lim _{x \rightarrow 0-0} f(x)=\lim _{x \rightarrow 0+0} f(x)=f(0)
\]
In this case,
\[
\begin{aligned}
& \li... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,350 |
5.23. Investigate the function
$$
f(x)=\left\{\begin{array}{l}
x^{3}, \text { if }|x| \leqslant 1 \\
1, \text { if }|x|>1
\end{array}\right.
$$
for continuity and determine the nature of the points of discontinuity. | Solution. A discontinuity in this case may occur at the points $x= \pm 1$. At the point $x=1$ we have
$$
\lim _{x \rightarrow 1-0} x^{3}=1, \lim _{x \rightarrow 1+0} 1=1, f(1)=1 \Rightarrow
$$
the function is continuous at the point $x=1$. Let's examine the point $x=-1$. We have
$$
\lim _{x \rightarrow-1-0} f(x)=\li... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,351 |
### 5.24. Investigate the function
$$
f(x)=\frac{5^{1 / x}}{1+5^{1 / x}}
$$
for continuity and determine the nature of the points of discontinuity. | Solution. The function is defined everywhere except at the point $x=0$. Let's investigate the behavior of the function in the neighborhood of the point $x=0$. We will find the one-sided limits
$$
\lim _{x \rightarrow 0-0} f(x)=\lim _{x \rightarrow 0-0} \frac{5^{1 / x}}{1+5^{1 / x}}=0
$$
since as $x \rightarrow 0-0$, ... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,352 |
5.25. Investigate the function
$$
f(x)=\frac{\sin 3 x}{x}
$$
for continuity and determine the nature of the points of discontinuity. | Solution. The one-sided limits of this function exist and are equal to 3:
$$
\lim _{x \rightarrow 0} \frac{\sin 3 x}{x}=3 \lim _{x \rightarrow 0} \frac{\sin t}{t}=3
$$
(a substitution $3 x=t$ has been made).
At the point $x=0$, the function is undefined. Thus, at the point $x=0$, we have a removable discontinuity. T... | (x)={\begin{pmatrix}\frac{\sin3x}{x},x\neq0\\3,0\end{pmatrix}.} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,353 |
5.26. Investigate the function $f(x)=e^{-\frac{2}{(x-1)}}$ for continuity and determine the nature of the discontinuity. | Solution. The function is defined everywhere except at the point $x=1$. At this point, the one-sided limits are
$$
\lim _{x \rightarrow 1-0} f(x)=\lim _{x \rightarrow 1-0} e^{-\frac{2}{x-1}}=\infty
$$
since as $x \rightarrow 1-0-2 /(x-1) \rightarrow+\infty$ and
$$
e^{-\frac{2}{x-1}} \rightarrow \infty \lim _{x \righ... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,354 |
5.27. The function $f(x)$ is not defined at $x=0$. Determine the value of $f(0)$ so that $f(x)$ is continuous at $x=0$, if
$$
f(x)=\frac{\sqrt[3]{1+x}-1}{\sqrt{4+x}-2}
$$ | Solution. Let's find the limit of $f(x)$ at the point $x=0$. We have
$$
\begin{gathered}
\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\sqrt[3]{1+x}-1}{\sqrt{4+x}-2}= \\
=\lim _{x \rightarrow 0} \frac{(\sqrt[3]{1+x}-1)(\sqrt{4+x}+2)\left(\sqrt[3]{(1+x)^{2}}+\sqrt[3]{1+x}+1\right)}{(\sqrt{4+x}-2)(\sqrt{4+x... | \frac{4}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,355 |
6.2. Prove that if $f(x)=0$ and
$$
\lim _{x \rightarrow 0} \frac{f(x)}{x}=0 \text {, then } \lim _{x \rightarrow 0} \frac{f(x)}{x}=f^{\prime}(0)
$$ | Solution. We have
$$
\begin{aligned}
\left.f^{\prime}(x)\right|_{x=0}= & \left.\lim _{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}\right|_{x=0}=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}= \\
& =\left.\lim _{\Delta x \rightarrow 0} \frac{f(\Delta x)-0}{\Delta x}\right|_{x=0}=\lim ... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,357 |
6.4. Using the definitions of the left and right derivatives of a function $f(x)$ at a point $x_{0}$, find them for
$$
f(x)=\sqrt{1-e^{-x^{2}}} \text { and } x_{0}=0 \text {. }
$$ | Solution.
$$
\begin{aligned}
f_{ \pm}(x)= & \lim _{\Delta x \rightarrow \pm 0} \frac{\sqrt{1-e^{-(0+\Delta x)^{2}}}-\sqrt{1-e^{-0^{2}}}}{\Delta x} \\
& =\lim _{\Delta x \rightarrow \pm 0} \frac{|\Delta x|}{\Delta x} \sqrt{\frac{1-e^{-(\Delta x)^{2}}}{(\Delta x)^{2}}}= \\
& = \pm \lim _{\Delta x \rightarrow \pm 0} \sqr... | \1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,358 |
6.5. Using the definition of the derivative, prove that the derivative of an even (odd) differentiable function is an odd (even) function. | Solution.
$$
\begin{gathered}
f^{\prime}(-x)=\lim _{\Delta x \rightarrow 0} \frac{f(-x+\Delta x)-f(-x)}{\Delta x}= \\
=\left\{\begin{array}{l}
\lim _{\Delta x \rightarrow 0} \frac{-f(x-\Delta x)+f(x)}{-(-\Delta x)}=f^{\prime}(x), \text { if } f(-x)=-f(x) \\
\lim _{\Delta x \rightarrow 0}-\left(\frac{f(x-\Delta x)-f(-x... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,359 |
6.6. Using the definition of the derivative, prove that the derivative of a periodic function is a periodic function. | Solution.
$$
\begin{gathered}
f^{\prime}(x \pm T)=\lim _{\Delta x \rightarrow 0} \frac{f((x \pm T)+\Delta x)-f(x \pm T)}{\Delta x}= \\
=\lim _{\Delta x \rightarrow 0} \frac{f((x+\Delta x) \pm T)-f(x \pm T)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}=f^{\prime}(x)
\end{gathered}
$$
CO... | proof | Calculus | proof | Yes | Yes | olympiads | false | 33,360 |
6.7. $f(x)=4 x^{3}-\frac{5}{x^{7}}-8 \sqrt[6]{x^{5}}+10$.
6.7. $f(x)=4 x^{3}-\frac{5}{x^{7}}-8 \sqrt[6]{x^{5}}+10$. | Solution.
$$
\begin{aligned}
& f^{\prime}(x)=\left(4 x^{3}-5 x^{-7}-8 x^{\frac{5}{6}}+10\right)^{\prime}=\left[\left((x)^{n}\right)^{\prime}=n x^{n-1},(\text { const })^{\prime}=0\right]= \\
& =4\left(x^{3}\right)^{\prime}-5\left(x^{-7}\right)^{\prime}-8\left(x^{\frac{5}{6}}\right)^{\prime}+(10)^{\prime}=4 \cdot 3 x^{... | 12x^{2}+\frac{35}{x^{8}}-\frac{20}{3\sqrt[6]{x}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,361 |
6.8. $f(x)=3 x^{2}-4 x^{-0.5}+7 \sqrt[3]{x^{2}}-\frac{8}{\sqrt[5]{x^{4}}}$.
6.8. $f(x)=3 x^{2}-4 x^{-0.5}+7 \sqrt[3]{x^{2}}-\frac{8}{\sqrt[5]{x^{4}}}$.
The function is given by:
$f(x)=3 x^{2}-4 x^{-0.5}+7 \sqrt[3]{x^{2}}-\frac{8}{\sqrt[5]{x^{4}}}$.
This function includes:
- A quadratic term: $3 x^{2}$
- An inverse s... | Solution.
Let's rewrite the function as follows:
$$
\begin{gathered}
f(x)=3 x^{2}-4 x^{-0.5}+7 x^{2 / 3}-8 x^{-4 / 5} \\
f^{\prime}(x)=3 \cdot 2 x^{2-1}-4 \cdot(-0.5) x^{-0.5-1}+7 \cdot \frac{2}{3} x^{\frac{2}{3}-1}-8 \cdot\left(-\frac{4}{5}\right) x^{-\frac{4}{5}-1}= \\
=6 x+\frac{2}{\sqrt{x^{3}}}+\frac{32}{5 \sqrt[... | 6x+\frac{2}{\sqrt{x^{3}}}+\frac{32}{5\sqrt[5]{x^{9}}}+\frac{14}{3}\cdot\frac{1}{\sqrt[3]{x}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 33,362 |
6.9. \( f(x)=4 x \ln x+5 \frac{e^{x}}{\cos x} \). | Solution.
It is recommended to use the rules of differentiation for the sum, product, and quotient of functions.
$$
\begin{gathered}
(f(x))^{\prime}=\left(x \ln x-5 \frac{e^{x}}{\cos x}\right)^{\prime}= \\
=\left[(u v)^{\prime}=u^{\prime} v+u v^{\prime},(u / v)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}},(x)^{\p... | \lnx+1-5\cdot\frac{e^x\cdot\cosx+e^x\cdot\sinx}{\cos^2x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,363 |
6.10. $f(x)=\sin \left(x^{2}\right)$. | Solution. This function is a complex two-link function: the first outer link is the sine, and the second inner link, the argument of the first link, is a quadratic function. The derivative of a composite function is the derivative of the first link with respect to its argument, keeping this argument, multiplied by the ... | (f(x))^{\}=\cos(x^{2})\cdot2x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,364 |
6.11. $f(x)=\tan\left(2^{x} \cdot \sqrt{x}\right)$. | Solution.
$$
\begin{aligned}
f^{\prime}(x)= & \frac{1}{\cos ^{2}\left(2^{x} \cdot \sqrt{x}\right)}\left(2^{x} \cdot \sqrt{x}\right)^{\prime}=\frac{1}{\cos ^{2}\left(2^{x} \cdot \sqrt{x}\right)}\left(\left(2^{x}\right)^{\prime} \cdot \sqrt{x}+2^{x} \cdot(\sqrt{x})^{\prime}\right)= \\
& =\frac{1}{\cos ^{2}\left(2^{x} \c... | \frac{1}{\cos^{2}(2^{x}\cdot\sqrt{x})}(2^{x}\ln2\cdot\sqrt{x}+2^{x}\cdot\frac{1}{2\sqrt{x}}),x\in(0;+\infty) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,365 |
6.12. Find the derivative of the complex three-link function
$$
f(x)=\arcsin \left(\sqrt{1-e^{x}}\right), x \leqslant 0
$$ | Solution.
Here, the first outer link is the arcsine, the second is the square root, and the third is the sum of one and the exponent. We differentiate sequentially:
$$
\begin{aligned}
& f^{\prime}(x)=\frac{1}{\sqrt{1-\left(\sqrt{1-e^{x}}\right)^{2}}} \cdot\left(\sqrt{1-e^{x}}\right)^{\prime}=\frac{1}{\sqrt{1-\left(1-... | \frac{-e^{x/2}}{2\sqrt{1-e^{x}}},x<0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,366 |
6.13. $f(x)=\operatorname{arctg}\left(\frac{\log _{3} x}{x^{5}}\right)$.
6.13. $f(x)=\arctan\left(\frac{\log _{3} x}{x^{5}}\right)$. | Solution.
$$
\begin{aligned}
& f^{\prime}(x)=\frac{1}{1+\left(\frac{\log _{3} x}{x^{5}}\right)^{2}} \cdot \frac{\left(\log _{3} x\right)^{\prime} \cdot x^{5}-\log _{3} x \cdot\left(x^{5}\right)^{\prime}}{\left(x^{5}\right)^{2}} \\
& =\frac{1}{1+\left(\frac{\log _{3} x}{x^{5}}-\right)^{2}} \cdot \frac{-\frac{1}{\ln 3} ... | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,367 | |
6.14. Find the derivative of the function
$$
f(x)=\left\{\begin{array}{l}
\operatorname{arctg}\left(x^{2} \cdot \sin (1 /(5 x))\right), \quad x \neq 0, \\
0, x=0
\end{array}\right.
$$
and compute its value at the point $x_{0}=0$. | Solution. Let's find the derivatives for $x \neq 0$. We have
$$
\begin{gathered}
f^{\prime}(x)=\frac{1}{1+\left(x^{2} \cdot \sin (1 / 5 x)\right)^{2}} \cdot\left(x^{2} \cdot \sin (1 /(5 x))\right)^{\prime}= \\
=\frac{1}{1+\left(x^{2} \cdot \sin (1 /(5 x))\right)^{2}} \cdot\left(\left(x^{2}\right)^{\prime} \cdot \sin (... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 33,368 |
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