problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
735. Find $r^{\prime}$, if
$$
r=(\sin \varphi)^{\cos 2 \varphi}
$$ | Solution. $\ln r=\cos 2 \varphi \ln \sin \varphi ; \frac{r^{\prime}}{r}=(\cos 2 \varphi)^{\prime} \ln \sin \varphi+$ $+\cos 2 \varphi(\ln \sin \varphi)^{\prime}=-2 \sin 2 \varphi \ln \sin \varphi+\cos 2 \varphi \cdot \frac{\cos \varphi}{\sin \varphi}$, hence
$$
r^{\prime}=(\sin \varphi)^{\cos 2 \varphi}(\operatorname{... | r^{\}=(\sin\varphi)^{\cos2\varphi}(\operatorname{ctg}\varphi\cos2\varphi-2\sin2\varphi\ln\sin\varphi) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,540 |
736. Find $z^{\prime}$, if
$$
z=\frac{4 x^{2}}{\sqrt[5]{(2-x)^{3}}}
$$ | Solution. $\ln z=\ln 4+2 \ln x-\frac{3}{5} \ln (2-x)$;
$$
\frac{z^{\prime}}{z}=\frac{2}{x}-\frac{3(2-x)^{\prime}}{5(2-x)}=\frac{2}{x}+\frac{3}{5(2-x)}=\frac{20-7 x}{5 x(2-x)}
$$
from this
$$
z^{\prime}=\frac{20-7 x}{5 x(2-x)} \cdot \frac{4 x^{2}}{\sqrt{(2-x)^{3}}}=\frac{4 x(20-7 x)}{5(2-x)^{5}(2-x)^{3}}
$$
Using th... | \frac{4x(20-7x)} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,541 |
745. Find the derivatives of the following functions:
1) $\left.y^{3}-3 y+2 a x=0 ; 2\right) x^{2}+3 x y+y^{2}+1=0$. Calculate $y^{\prime}$ at the point $(2 ;-1) ;$; 3) $\sin \varphi+r \varphi-5 r=0$. Calculate $\frac{d r}{d \psi}$; 4) $e^{y}+x y=e$. Calculate $y^{\prime}$ at the point $(0 ; 1)$. | Solution. 1) Differentiating both sides of the equality with respect to $x$, taking into account that $y$ is a function of $x$, we get
$$
3 y^{2} y^{\prime}-3 y^{\prime}+2 a=0
$$
from which
$$
y^{\prime}=\frac{2 a}{3\left(1+y^{2}\right)}
$$
2) Differentiating with respect to $x$, we get
$$
2 x+3 y+3 x y^{\prime}+2... | y^{\}=\frac{2}{3(1+y^{2})},\;y^{\}(2)=-\frac{1}{4},\;\frac{}{\varphi}=\frac{r+\cos\varphi}{5-\varphi},\;y^{\}(0)=-\frac{1}{e} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,542 |
794. Find the derivatives of the specified order of the given functions:
1) $y=x^{3}+2 x^{2}-4 x+1 ; y^{\prime \prime \prime}=$ ?
2) $y=\ln x ; y^{(4)}=$ ?
3) $s=t^{2}-t+1 ; s^{\prime \prime}(0)=$ ?
4) $f(x)=\sin 2 x ; f^{(5)}\left(\frac{\pi}{2}\right)=$ ? | Solution. 1) Find the first derivative
$$
y^{\prime}=3 x^{2}+4 x-4
$$
Now find the second derivative
$$
y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(3 x^{2}+4 x-4\right)^{\prime}=6 x+4
$$
Differentiate one more time to get the third-order derivative
$$
y^{\prime \prime \prime}=\left(y^{\prime \prime}\r... | 6,-\frac{6}{x^{4}},2,-32 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,543 |
795. Show that the function $y=\cos 2 x$ satisfies the differential equation $y^{\prime \prime}+4 y=0$. | Solution. We determine $y^{\prime}$ and $y^{\prime \prime}$ :
$$
y^{\prime}=-2 \sin 2 x ; y^{\prime \prime}=-4 \cos 2 x
$$
Substituting $y^{\prime}$ and $y^{\prime \prime}$ into the given equation, we obtain the identity $-4 \cos 2 x+4 \cos 2 x=0 ; 0=0$. | proof | Calculus | proof | Yes | Yes | olympiads | false | 30,544 |
801. Write the equation of the tangent and normal to the parabola $f(x)=x^{2}-2 x+5$ at the point where the abscissa is $x=2$. | Solution. We will use formulas (1) and (2).
Given $x_{0}=2$, substituting this value into the given parabola equation, we find $y_{0}: y_{0}=2^{2}-2 \cdot 2+5, y_{0}=5$.
Therefore, the point of tangency is $M_{0}(2 ; 5)$.
Now let's find the slope of the tangent to the given curve at the point $x=2$:
$$
f^{\prime}(x... | 2x-y+1=0x+2y-12=0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,545 |
802. Write the equation of the tangent and normal to the curve $x^{2}-2 x y+3 y^{2}-2 y-16=0$ at the point $(1 ; 3)$. | Solution. Substituting the coordinates of the given point into the equation, we verify that the point $(1 ; 3)$ lies on the curve
$$
1-2 \cdot 1 \cdot 3+3 \cdot 3^{2}-2 \cdot 3-16=0 \text { or } 0=0
$$
Differentiating with respect to $x$, we find
$$
2 x-2 y-2 x y^{\prime}+6 y y^{\prime}-2 y^{\prime}=0
$$
$. Let's find the slopes of the tangents to the curves ... | 554^{\} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,547 |
804. A point moves in a straight line according to the law $s=4 t+t^{3}$. Find the velocity and acceleration of the point at the moments of time $t_{0}=0, t_{1}=1, t_{2}=2$ ( $s$ is given in centimeters, $t$ - in seconds). | Solution. The velocity $v=s_{t}^{\prime}=\left(4 t+t^{3}\right)^{\prime}=4+3 t^{2}$; acceleration $a=s_{t}^{\prime \prime}=\left(4+3 t^{2}\right)^{\prime}=6 t$. Therefore, at the moment $t_{0}=0 \quad v_{0}=4 \mathrm{~cm} /$ sec, $a_{0}=0 \mathrm{~cm} /$ sec $^{2}$. At $t_{1}=1 \quad v_{1}=s^{\prime}(1)=$
$=7 \mathrm{c... | v_{0}=4\mathrm{~}/,a_{0}=0\mathrm{~}/^{2},v_{1}=7\mathrm{~}/,a_{1}=6\mathrm{~}/^{2},v_{2}=16\mathrm{~}/,a_{2}=1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,548 |
815. Find the differentials of the functions:
1) $y=x^{3}$ at the point $x=0$, if $\left.\Delta x=0.3 ; 2\right) y=x^{3}-x^{2}+$ $+3 x-1$; 3) $r=\varphi^{4}+2^{\sin 3 \varphi}$; 4) $y=\ln \left(x^{2}+1\right)+\operatorname{arctg} \sqrt{x}$ at the point $x=1$, if $\Delta x=0.1$. | Solution. 1) Let's find the increment of the given function at the point $x$:
$$
\Delta y=(x+\Delta x)^{3}-x^{3}=3 x^{2} \Delta x+3 x \Delta x^{2}+\Delta x^{3}
$$
The first term of the function's increment contains $\Delta x$ in the first degree, i.e., it is the linear part of the function's increment with respect to... | 0.125 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,549 |
3. Given the function $f(x)=1-\sqrt[3]{x^{2}}$.
Does this function satisfy the conditions of Rolle's Theorem on the segment $[-1,1]$? If the theorem is not applicable, why not? | Solution. The function is continuous on the segment $[-1,1]$ and at the ends of this segment it takes equal values $f(1)=$ $=f(-1)=0$. However, inside the segment there is a point $x=0$, where the derivative does not exist. Therefore, the conditions of Rolle's theorem are not satisfied, and thus Rolle's theorem is not ... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,551 |
840. Does the function $f(x)=x^{2}+3$ satisfy the conditions of Lagrange's theorem on the segment $[-1,2]$? If the theorem is applicable, then find the point $c$ where $f(b)-f(a)=f^{\prime}(c)(b-a)$, where $a=$ $=-1, b=2$. | Solution. The function satisfies the conditions of Lagrange's theorem, as it is continuous on the segment $[-1,2]$ and has
a finite derivative in the interval ( $-1,2$ ). The point $c$ will be found from the equation
$$
f(2)-f(-1)=f^{\prime}(c)[2-(-1)]
$$
hence
$$
7-4=2 c \cdot 3, c=\frac{1}{2}
$$ | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,552 |
847. Find the limits:
1) $\lim _{x \rightarrow 0} x^{x}$; 2) $\lim _{x \rightarrow 0}\left(\frac{1}{x}\right)^{\sin x}$; 3) $\lim _{x \rightarrow 1} x^{\frac{1}{x-1}}$ | Solution. In the given examples, we have indeterminate forms of the type $0^{0}, \infty^{0}, 1^{\infty}$.
In each of these cases, the logarithm of the function takes an indeterminate form of the type $0 \cdot \infty$. By resolving this indeterminate form, we find the limit of the logarithm of the function, from which ... | 1,1,e | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,555 |
872. Find the intervals of increase and decrease for the following functions:
1) $y=x^{2}$; 2) $y=x^{3}$; 3) $f(x)=x^{3}+2 x-5$; 4) $y=\ln \left(x^{2}+2 x+3\right)$; 5) $y=2 x^{2}-\ln x$; 6) $y=\frac{1}{x-2}$. | Solution. 1) The given function is defined for all real numbers. Let's find the derivative of the function
$$
y^{\prime}=2 x
$$
The function decreases if $2 x < 0$, i.e., $x < 0$. It increases if $2 x > 0$, i.e., $x > 0$. Therefore, the function decreases in the interval $(-\infty, 0)$ and increases in the interval $... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,556 |
904. Find the points of inflection and intervals of concavity (convexity) of the graphs of the following functions:
$$
\text { 1) } y=x^{4}-6 x^{2}+5 ; \text { 2) } y=x^{\frac{4}{3}} ; \text { 3) } y=\sqrt[8]{x}
$$ | Solution. 1) Find the second derivative:
$$
y^{\prime}=4 x^{3}-12 x ; y^{\prime \prime}=12 x^{2}-12=12\left(x^{2}-1\right)
$$
Solve the equation $12\left(x^{2}-1\right)=0 ; x= \pm 1$.
Investigate the sign of the second derivative in the neighborhood of the points
$$
x_{1}=-1, x_{2}=1
$$
For $x < -1, y^{\prime \pri... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,558 |
916. Find the asymptotes of the curves:
1) $y=\frac{1}{x-2}$;
2) $y=\operatorname{arctg} x$
3) $f(x)=\frac{2 x}{x-1}$;
4) $y=-e^{\frac{1}{x}}$. | Solution. 1) The curve has a horizontal asymptote $Y=0$, i.e., the $O x$ axis, since $\lim _{x \rightarrow \infty} \frac{1}{x-2}=0$, and a vertical asymptote $x=2$, since $\lim _{x \rightarrow 2} \frac{1}{x-2}=\infty$.
2) The curve has two horizontal asymptotes $Y=\frac{\pi}{2}$ and $Y=-\frac{\pi}{2}$, since $\lim _{x... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,559 |
2. Find the greatest and least values of the given functions in the specified intervals:
1) $y=x^{4}-8 x^{2}+3 \quad$ on the segment $[-2,2]$;
2) $y=\tan x-x$
on the segment $\left.\left\lvert\,-\frac{\pi}{4}\right., \frac{\pi}{4}\right]$. | Solution. 1) Find the critical points:
$$
\begin{gathered}
f^{\prime}(x)=4 x^{3}-16 x, 4 x\left(x^{2}-4\right)=0, x_{1}=0, x_{2}=-2, x_{3}=2 \\
f^{\prime \prime}(x)=4\left(3 x^{2}-4\right) ; f^{\prime \prime}(0)=-160 \text{ in } \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] \text{ (except for the point } x=0 \text{, wher... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,561 |
3. Make a box (without a lid) of the greatest capacity from a piece of cardboard $30 \times 14$ (cm²) by cutting out equal squares from the corners and then folding up the sides of the cardboard to form the sides of the box (Fig. 70). | Solution. The area of the base of the box $S=(14-2 x)(30-2 x)$, the height $H=x$. The volume of the box $V'=S H=4 x(7-x)(15-x)$.
Let's investigate this function: $V'(x)=4\left(3 x^{2}-44 x+105\right)$. The roots of the derivative will be $x_{1}=11 \frac{2}{3}, x_{2}=3$.
Next, $V''(x)=8(3 x-22), V''(3)<0$. At the poin... | 576\,^3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,562 |
942. A wire of length $20 m$ is required to fence a flower bed, which should have the shape of a circular sector. What radius of the circle should be taken so that the area of the flower bed is maximized? | Solution. Let the radius of the circle be $x$, and the length of the arc of the sector be $y$ (Fig. 71). The area of the circular sector is $S=\frac{1}{2} x y$. The function $S$ is to be investigated for a maximum. Note that $S$ depends on two variables, $x$ and $y$. Express $y$ in terms of $x$ (or vice versa). Accordi... | 5\mathrm{~} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,563 |
943. A cylindrical tank must hold $V \Omega$ of water. What should its dimensions be so that the surface area (without the lid) is the smallest? | Solution.
$$
V=\pi x^{2} y \text { (equation of connection), }
$$
where $x$ is the radius of the base of the cylinder, and $y$ is its height (Fig. 72). The surface area
$$
S=\pi x^{2}+2 \pi x y
$$
(function to be minimized). From equation (a), we determine $y=\frac{V}{\pi x^{2}}$. Substituting the value of $y$ into... | \sqrt[3]{\frac{V}{\pi}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,564 |
965. Find the domains of the following functions:
1) $z=x+y-1$
2) $f(x, y)=\frac{x+1}{x^{2}+y^{2}}$
3) $z=\sqrt{4-x^{2}-y^{2}}$
4) $u=\frac{1}{V \sqrt{9-x^{2}-y^{2}}}$
5) $\varphi(x, y)=x-\arcsin y$
6) $z=\frac{1}{\sqrt{x y}}$ | Solution. 1) The function is defined on the entire plane $x O y$. The geometric representation of this function will be a plane in space.
2) The function is defined on the entire plane except for the point $O(0 ; 0)$, where the denominator of the fraction becomes zero.
3) The function is defined for values of $x$ and ... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,565 |
976. Find the following limits:
1) $\lim _{\substack{x \rightarrow 2 \\ y \rightarrow 0}} \frac{\tan x y}{y}$
2) $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x+y}{x}$. | Solution. 1) $\lim _{\substack{x \rightarrow 2 \\ y \rightarrow 0}} \frac{\operatorname{tg} x y}{y}=\lim _{\substack{x \rightarrow 2 \\ y \rightarrow 0}} x \frac{\operatorname{tg} x y}{x y}=2 \cdot 1=2$,
since
$$
\lim _{\alpha \rightarrow 0} \frac{\operatorname{tg} \alpha}{\alpha}=1
$$
2) Let's use the definition of... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,566 |
989. Find the partial derivatives of the following functions:
$$
\text { 1) } z=x^{3}+x^{2} y+y^{3}
$$
$$
\begin{aligned}
& \text { 2) } f(x, y)=\sqrt{x^{2}-y^{2}} \text { at the point }(5 ;-3) \\
& \text { 3) } \varphi(x, y)=2 x y \text { at the point }(0 ; 3)
\end{aligned}
$$ | Solution. 1) Considering $y$ as a constant, we differentiate the function with respect to the variable $x$. Using known formulas and rules from Chapter V, we find
$$
\frac{\partial z}{\partial x}=3 x^{2}+2 x y
$$
Similarly, considering $x$ as a constant, we get
$$
\text { 2) } \begin{aligned}
& \frac{\partial z}{\pa... | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,567 | |
1002. Find the total increment and the total differential of the function $z=x^{2}+x y$ at $x=2 ; y=1 ; \Delta x=0.01 ; \Delta y=0.02$. | Solution. We will use formula (1) and the definition of the total differential of the function $z=f(x, y)$:
$$
\begin{gathered}
\Delta z=(x+\Delta x)^{2}+(x+\Delta x)(y+\Delta y)-\left(x^{2}+x y\right)= \\
=(2 x+y) \Delta x+x \Delta y+\Delta x^{2}+\Delta x \Delta y
\end{gathered}
$$
from which
$$
d z=(2 x+y) \Delta ... | 0.0903for\Deltaz,\0.09for | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,568 |
1003. Find the total differentials of the functions:
$$
\text { 1) } z=5 x^{2} y^{3} ; \quad \text { 2) } z=\operatorname{arctg}\left(x^{2}+3 y\right)
$$ | Solution. 1) First, we find the partial derivatives:
$$
\frac{\partial z}{\partial x}=10 x y^{3}, \frac{\partial z}{\partial y}=15 x^{2} y^{2}
$$
By formula (2) we get
$$
\begin{aligned}
& d z=10 x y^{3} d x+15 x^{2} y^{2} d y=5 x y^{2}(2 y d x+3 x d y) \\
& \text { 2) } \frac{\partial z}{\partial x}=\frac{\left(x^{... | =5xy^{2}(2y+3x)=\frac{1}{1+(x^{2}+3y)^{2}}(2x+3) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,569 |
1004. Calculate approximately:
1) $(0.96)^{2}(1.02)^{3}$;
2) $(1.02)^{3.05}$. | Solution. 1) $(0.96)^{2}(1.02)^{3}$ is a particular value of the function $f(x, y)=x^{2} y^{3}$. Using formula (3), we get
$$
(x+\Delta x)^{2}(y+\Delta y)^{3} \approx x^{2} y^{3}+2 x y^{3} \Delta x+3 x^{2} y^{2} \Delta y
$$
Let $x=1, y=1$, then $\Delta x=-0.04$, since $1+\Delta x=0.96$, $\Delta y=0.02$, since $1+\Del... | 1.06 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,570 |
1020. Investigate the function for extremum
$$
f(x, y)=x^{2}+y^{2}-4 y+4
$$ | Solution. Let's find the partial derivatives:
$$
f_{x}^{\prime}(x, y)=2 x ; f_{y}^{\prime}(x, y)=2 y-4
$$
Set them equal to zero, we get a system of two equations with two unknowns:
$$
\left\{\begin{array}{l}
2 x=0 \\
2 y-4=0
\end{array}\right.
$$
from which $x=0, y=2$.
The function at the critical point $M_{0}(0 ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,571 |
1021. Among all rectangular parallelepipeds having a given volume $\mathrm{V}$, find the one with the smallest total surface area. | Solution. Let $x, y, z$ be the dimensions of the given parallelepiped. Then its volume is $\mathrm{V}=x y z$ (the constraint equation). The total surface area of the parallelepiped is
$$
S=2 x y+2(x+y) z
$$
From the constraint equation, we find $z=\frac{V}{x y}$ and substitute this value into the expression for the f... | \sqrt[3]{V} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,572 |
1036. Find the following integrals and verify the results by differentiation:
1) $\left.\int\left(x^{2}-2 \sin x+3 e^{x}\right) d x ; 2\right) \int\left(\sec ^{2} x-3 \cos x+1\right) d x$;
2) $\int\left(\csc ^{2} x+7 \sin x-2\right) d x$.[^8] | Solution. 1) $\int\left(x^{2}-2 \sin x+3 e^{x}\right) d x=\frac{x^{3}}{3}+2 \cos x+$ $+3 e^{x}+C$ [I, II; (3), (7), (6)].
Verification. We will find the derivative of the obtained function $F(x)+C$ and ensure that it equals the integrand $f(x):$
$\left(\frac{x^{3}}{3}+2 \cos x+3 e^{x}+C\right)^{\prime}=\left(\frac{x^... | 1)\frac{x^{3}}{3}+2\cosx+3e^{x}+C\\2)\operatorname{tg}x-3\sinx+x+C\\3)-\operatorname{ctg}x-7\cosx-2x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,574 |
1037. Find the integrals:
1) $\int \frac{d x}{2 x^{2}+9}$
2) $\int \frac{d x}{\sqrt{5-4 x^{2}}}$
3) $\int \frac{d x}{\sqrt{2 x^{2}-3}}$
4) $\int \frac{d x}{4 x^{2}-3}$
5) $\int \cos ^{2} \frac{x}{2} d x$
6) $\int \operatorname{tg}^{2} x d x$ | Solution. 1) We will transform the given integral to a tabular form as follows:
$$
I=\int \frac{d x}{2 x^{2}+9}=\int \frac{d x}{2\left(x^{2}+\frac{9}{2}\right)}=\frac{1}{2} \int \frac{d x}{x^{2}+\frac{9}{2}}
$$
here $a^{2}=\frac{9}{2}$ (see formula 12), therefore,
$$
I=\frac{1}{2} \cdot \frac{\frac{1}{3}}{\sqrt{\ove... | -\frac{\sqrt{2}}{6}\operatorname{arctg}\frac{x\sqrt{2}}{3}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,575 |
1120. Find the integrals:
1) $\int(x+2) \ln x \, dx$
2) $\int x^{2} \arcsin x \, dx$
3) $\int \arccos x \, dx$
4) $\int e^{x} \sin x \, dx$
5) $\int\left(x^{2}+2 x+1\right) \sin 3 x \, dx$
6) $\int x^{2} e^{3 x} \, dx$ | Solution. 1) Using formula (1). Let $u=\ln x$, $d v=(x+2) d x$, then $d u=\frac{d x}{x} ; \quad v=\int(x+2) d x=\frac{x^{2}}{2}+2 x=$ $=x\left(\frac{x}{2}+2\right)$, therefore,
$\int(x+2) \ln x d x=x\left(\frac{x}{2}+2\right) \ln x-\int x\left(\frac{x}{2}+2\right) \frac{d x}{x}=$ $=\frac{x}{2}(x+4) \ln x-\frac{x^{2}}{... | \begin{aligned}1)&\quad\frac{x}{2}(x+4)\lnx-\frac{x^{2}}{4}-2x+C\\2)&\quad\frac{1}{3}(x^{3}\arcsinx+\sqrt{1-x^{2}}-\frac{1}{3}\sqrt{(1-x^{2})^{3}})+C\\3)&\ | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,577 |
1141. Find the integrals:
1) $\int \frac{d x}{3 x^{2}-6 x+5}$
2) $\int \frac{x+3}{x^{2}+4 x-1} d x$
3) $\int \frac{x^{3}-2 x^{2}+x-1}{3+2 x-x^{2}} d x$
4) $\int \frac{2 x-1}{x^{2}-2 x} d x$ | Solution. 1) $\int \frac{d x}{3 x^{2}-6 x+5}=\frac{1}{3} \int \frac{d x}{x^{2}-2 x+\frac{5}{3}}=$ $=\frac{1}{3} \int \frac{d x}{(x-1)^{2}+\frac{2}{3}}=\frac{1}{3} \int \frac{d t}{t^{2}+\frac{2}{3}}=\frac{\sqrt{3}}{3 \sqrt{2}} \operatorname{arctg} \frac{t \sqrt{3}}{\sqrt{2}}+C=$ $=\frac{\sqrt{6}}{6} \operatorname{arctg}... | \begin{aligned}1)&\quad\frac{\sqrt{6}}{6}\operatorname{arctg}\frac{\sqrt{6}(x-1)}{2}+C\\2)&\quad\frac{1}{2}\ln|x^{2}+4x-1|+\frac{1}{2\sqrt{5}}\ln|\frac{x+2-\sqrt{5} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,578 |
1142. Find the integrals:
1) $\int \frac{3 x-5}{\sqrt{x^{2}-4 x+5}} d x$
2) $\int \frac{d x}{\sqrt{1+2 x-3 x^{2}}}$ | Solution. 1) $\int \frac{3 x-5}{\sqrt{x^{2}-4 x+5}} d x=\int \frac{3 x-5}{\sqrt{(x-2)^{2}+1}} d x=$ $=\int \frac{3 t+1}{\sqrt{t^{2}+1}} d t=3 \int \frac{t}{\sqrt{t^{2}+1}} d t+\int \frac{d t}{\sqrt{t^{2}+1}}=3 \sqrt{t^{2}+1}+$
$$
\begin{gathered}
+\ln \left(t+\sqrt{t^{2}+1}\right)+C=3 \sqrt{x^{2}-4 x+5}+ \\
\quad+\ln ... | \frac{1}{\sqrt{3}}\arcsin\frac{3x-1}{2}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,579 |
1143. Find the integrals:
1) $\int \sqrt{x^{2}-2 x-1} d x$
2) $\int \sqrt{5+4 x-x^{2}}$. | Solution. The integral $\int \sqrt{a x^{2}+b x+c} d x$ depending on the sign of the coefficient $a$ can be reduced to the integral $\int \sqrt{t^{2}+\sigma} d t$ or $\int \sqrt{\beta^{2}-t^{2}} d t$, each of which is calculated using trigonometric substitutions (see Remark § 3 and Example 15) or the method of integrati... | \begin{aligned}&1)\frac{x-1}{2}\sqrt{x^{2}-2x-1}-\ln|x-1+\sqrt{x^{2}-2x-1}|+C\\&2)\frac{1}{2}[9\arcsin\frac{x-2}{3}+(x-2)\sqrt{5+4x-x^{2}}]+C\ | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,580 |
1166. Find the integrals:
1) $\int\left(5 x^{3}-4 x^{2}+2\right) d x$
2) $\int \frac{d x}{x-2}$
3) $\int \frac{5 d x}{(x+3)^{7}}$
4) $\int \frac{2 x-5}{x^{2}+4 x+8} d x$ | Solution. 1) An entire rational function is integrated directly:
$$
\begin{gathered}
\int\left(5 x^{3}-4 x^{2}+2\right) d x=5 \int x^{3} d x-4 \int x^{2} d x+2 \int d x= \\
=\frac{5}{4} x^{4}-\frac{4}{3} x^{3}+2 x+C
\end{gathered}
$$
2) Integral of a simple fraction
$$
\int \frac{d x}{x-2}=\ln |x-2|+C \quad[(1) \S 3... | \begin{aligned}1)&\quad\frac{5}{4}x^{4}-\frac{4}{3}x^{3}+2x+C\\2)&\quad\ln|x-2|+C\\3)&\quad-\frac{5}{6(x+3)^{6}}+C\\4)&\quad\ln(x^{2}+4x+8)-\frac | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,581 |
1184. Find the integrals:
1) $\int \frac{d x}{\sqrt{x}+\sqrt[4]{x}}$
2) $\int \frac{x+\sqrt{1+x}}{\sqrt[3]{1+x}} d x$
3) $\int \frac{\sqrt{x}}{x^{2} \sqrt{x-1}} d x$ | Solution. 1) Apply the substitution (a):
$$
\sqrt[4]{x}=t, \text { from which } x=t^{4}, d x=4 t^{3} d t
$$
$$
\int \frac{d x}{\sqrt{x}+\sqrt[4]{x}}=\int \frac{4 t^{3} d t}{t^{2}+t}=4 \int \frac{t^{2} d t}{t+1}=4 \int\left(t-1+\frac{1}{t+1}\right) d t=4 \times
$$
$$
\times\left[\frac{t^{2}}{2}-t+\ln (t+1)\right]+C=4... | \begin{aligned}1)&\quad4[\frac{\sqrt{x}}{2}-\sqrt[4]{x}+\ln(1+\sqrt[4]{x})]+C\\2)&\quad6\sqrt[3]{(1+x)^{2}}(\frac{1+x}{10}+\frac{\sqrt{1+x}}{7}-\frac{1}{4})+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,583 |
1192. Find the integrals:
1) $\int \frac{\cos x}{\sqrt{3+\sin ^{2} x}} d x$
2) $\int \sin ^{5} x d x$
3) $\int \frac{1}{3 \tan^{2} x+5} \cdot \frac{d x}{\cos ^{2} x}$
4) $\int \cot^{6} x d x$ | Solution. 1) We use the substitution (1):
$$
\begin{gathered}
\int \frac{\cos x}{\sqrt{3+\sin ^{2} x}} d x=\int \frac{d t}{\sqrt{3+t^{2}}}=\ln \left(t+\sqrt{3+t^{2}}\right)+C= \\
=\ln \left(\sin x+\sqrt{3+\sin ^{2} x}\right)+C
\end{gathered}
$$
2) We apply the substitution (2):
$\int \sin ^{5} x d x=\int\left(1-\cos... | \begin{aligned}1)&\ln(\sinx+\sqrt{3+\sin^{2}x})+C\\2)&\frac{2}{3}\cos^{3}x-\frac{1}{5}\cos^{5}x-\cosx+C\\3)&\frac{1}{\sqrt{15}}\operatorname{arctg}\frac{\sqrt{3}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,584 |
1193. Find the integrals:
1) $\int \sin ^{2} x \cos ^{4} x d x ;$ 2) $\int \cos ^{4} x d x ;$ 3) $\int \frac{d x}{2+3 \cos x}$. | Solution. 1) Applying the formulas from point 3, we get
$$
\begin{gathered}
I=\int \sin ^{2} x \cos ^{4} x d x=\frac{1}{4} \int \sin ^{2} 2 x \frac{1+\cos 2 x}{2} d x= \\
=\frac{1}{8} \int \sin ^{2} 2 x d x+\frac{1}{8} \int \sin ^{2} 2 x \cos 2 x d x=\frac{1}{16} \int(1-\cos 4 x) d x+ \\
+\frac{1}{16} \int t^{2} d t
\... | \begin{aligned}1)&\quadI=\frac{x}{16}-\frac{\sin4x}{64}+\frac{\sin^{3}2x}{48}+C0\\2)&\quad\int\cos^{4}x=\frac{3x}{8}+\frac{\sin2x}{4}+\frac{\sin4x}{32}+C0\\ | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,585 |
3. Consider the Newton-Leibniz formula. If the function \( f(x) \) is continuous on the interval \([a, b]\), then
\[
\int_{a}^{b} f(x) \, d x = \left. F(x) \right|_{a}^{b} = F(b) - F(a),
\]
where \( F(x) \) is the antiderivative of \( f(x) \), i.e., \( F'(x) = f(x) \). 1242. Compute the integrals:
1) \(\int_{0}^{8} (... | Solution. Applying the Newton-Leibniz formula (2) and the properties of the definite integral, we get:
1) $\int_{0}^{8}(\sqrt{2 x}+\sqrt[3]{x}) d x=\sqrt{2} \int_{0}^{8} x^{\frac{1}{2}} d x+\int_{0}^{8} x^{\frac{1}{3}} d x=$
$$
=\left.\left(\frac{2 \sqrt{2}}{3} x^{\frac{3}{2}}+\frac{3}{4} x^{\frac{4}{3}}\right)\right... | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,586 | |
1258. Compute the integrals:
1) $\int_{0}^{\frac{\pi}{2}} \sin ^{3} x d x$;
2) $\int_{0}^{\ln 2} \sqrt{e^{x}-1} d x$
3) $\int_{-a}^{a} x^{2} \sqrt{a^{2}-x^{2}} d x$;
4) $\int_{1}^{2} \frac{\sqrt{x^{2}-1}}{x} d x$. | Solution. 1) The given integral can be easily reduced to an integral of the form
$$
\int f(\cos x) \sin x d x
$$
therefore, we apply the substitution $\cos x=t, \sin x d x=-d t$.
Determine the new interval of integration. If $x=0$, then $\cos 0=t$ and $t=1$; if $x=\frac{\pi}{2}$, then $\cos \frac{\pi}{2}=t$ and $t=0... | \frac{2}{3},2-\frac{\pi}{2},\frac{1}{8}\pi^{4},\sqrt{3}-\frac{\pi}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,587 |
1269. Calculate the integrals:
1) $\int_{0}^{1} x e^{-x} d x$
2) $\int_{1}^{2} x \log _{2} x d x$
3) $\int_{1}^{e} \ln ^{2} x d x$ | Solution. 1) Using formula (1). Let $u=x$, $d v=e^{-x} d x$ we have $d u=d x, v=-e^{-x}$, therefore,
$$
\int_{0}^{1} x e^{-x} d x=-\left.x e^{-x}\right|_{0} ^{1}+\int_{0}^{1} e^{-x} d x=\left.\left(-x e^{-x}-e^{-x}\right)\right|_{0} ^{1}=1-\frac{2}{e}
$$
2) Let $u=\log _{2} x, d v=x d x$, we have $d u=\frac{d x}{x \l... | 1-\frac{2}{e},2-\frac{3}{4\ln2},e-2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,588 |
1284. Calculate the area of the figure bounded by the following lines:
1) the coordinate axes, the line $x=3$ and the parabola $y=x^{2}+1$;
2) the ordinate axis, the lines $y=-2, y=3$ and the parabola $x=$ $=\frac{1}{2} y^{2}$
3) the parabolas $y=x^{2}$ and $x=y^{2}$;
4) the parabolas $y=x^{2}+1, y=\frac{1}{2} x^{2}$ a... | Solution. 1) The area of the curvilinear trapezoid $O C B A$ (Fig. $79, a$) will be found using formula (1):
$$
S=\int_{0}^{3}\left(x^{2}+1\right) d x=\left.\left(\frac{x^{3}}{3}+x\right)\right|_{0} ^{3}=9+3=12
$$
2) The area of the curvilinear trapezoid $A B C D$ (Fig. 79, b) will be found using formula (2):
$$
S=\... | 10.41 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,589 |
1285. Find the areas of two figures bounded by the parabola $y^{2}=2 x$ and the circle $y^{2}=4 x-x^{2}$. | Solution. Transforming the equation of the circle, we get
$$
(x-2)^{2}+y^{2}=4
$$
from which it follows that the center of the circle lies at point $C(2 ; 0)$ and its radius $R=2$ (Fig. 80).
Solving the system of equations
$$
\left\{\begin{array}{l}
y^{2}=4 x-x^{2} \\
y^{2}=2 x
\end{array}\right.
$$
we obtain the ... | 2(\pi-\frac{8}{3})2(\pi+\frac{8}{3}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,590 |
1286. Find the area of the figure bounded by the line
$$
y^{2}=x(x-1)^{2}
$$ | Solution. The function $y^{2}=x(x-1)^{2}$ is even with respect to the variable $y$, therefore, the figure bounded by this line is symmetric with respect to the $O x$ axis (Fig. 81). Let's find the interval of integration. Let $y=0$, then $x_{1}=0$, $x_{2}=1$. Using formula (1), we get
$$
\frac{1}{2} S=\int_{0}^{1} V \... | \frac{8}{15} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,591 |
1287. Calculate the area of the figure bounded by the parabola $y=-x^{2}+$ $+6 x-5$ and the coordinate axes. | Solution. If the figure is located on opposite sides of the $O x$ axis (Fig. $82, a$), then the area $S$ should be calculated using the formula
$$
S=\int_{a}^{c} f(x) d x+\left|\int_{c}^{b} f(x) d x\right|
$$
, we find the length of the arc $AB$, i.e., $\frac{1}{4}$ of the circumference (Fig. 83):
$$
\begin{aligned}
& \frac{1}{4} s=\int_{0}^{R} \sqrt{1+\frac{x^{2}}{y^{2}}} d x=\int_{0}^{R} \frac{\s... | 2\piR | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,593 |
1296. Calculate the length of the arc of the parabola $y^{2}=4 x$ from the vertex to the point $M(1 ; 2)$. | Solution. Apply formula (2). From the parabola equation $x=\frac{y^{2}}{4}, x^{\prime}=\frac{y}{2}$, by the condition $y$ varies from 0 to 2, therefore,
$$
s=\int_{0}^{2} \sqrt{1+\frac{y^{2}}{4}} d y=\frac{1}{2} \int_{0}^{2} \sqrt{4+y^{2}} d y
$$
It is known that $\int \sqrt{x^{2}+a} d x=\frac{x}{2} \sqrt{x^{2}+a}+\f... | \sqrt{2}+\ln(1+\sqrt{2}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,594 |
1298. Find the length of the astroid
$$
x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}
$$ | Solution. Applying formula (1), we will compute the length of the arc $AB$, i.e., $\frac{1}{4}$ of the astroid (Fig. 84). We find the derivative of the implicit function $y$ with respect to $x$, obtaining
$$
\frac{2}{3} x^{-\frac{1}{3}}+\frac{2}{3} y^{-\frac{1}{3}} y_{x}^{\prime}=0
$$
from which
$$
y_{x}^{\prime}=-\... | 6a | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,596 |
1307. An ellipse with a major axis of $2a$ and a minor axis of $2b$ ($a > b$) rotates: 1) around the major axis; 2) around the minor axis. Find the volumes of the resulting ellipsoids of revolution. In a special case, determine the volume of a sphere. | Solution. Let's write the equation of the ellipse:
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1
$$
Using formula (1), we will find the volume of the body formed by rotating the ellipse around the $O x$ axis. From the equation of the ellipse,
$$
y^{2}=\frac{b^{2}}{a^{2}}\left(a^{2}-x^{2}\right)
$$
By the condition, ... | \frac{4}{3}\pi^{2}b | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,597 |
1308. Calculate the volume of the body obtained by rotating around the $O x$ axis the figure bounded by the parabolas $y=x^{2}$ and $x=y^{2}$. | Solution. Solving the system of equations
$$
\left\{\begin{aligned}
y & =x^{2} \\
y^{2} & =x
\end{aligned}\right.
$$
we get \(x_{1}=0, x_{2}=1, y_{1}=0, y_{2}=1\), from which the points of intersection of the curves are \(O(0 ; 0)\) and \(B(1 ; 1)\). As can be seen (Fig. 88), the volume of the solid of revolution is ... | \frac{3}{10}\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,598 |
1309. Calculate the volume of the body obtained by rotating around the $O y$ axis the curvilinear trapezoid bounded by the hyperbola $x y=4$, the lines $y=1, y=2$ and the $O y$ axis. | Solution. We apply formula (2). From the equation of the curve, we find $x=\frac{4}{y}$, and the limits of integration are given by the condition: $c=1, d=2$. Therefore,
$$
\begin{gathered}
\mathrm{V}=\pi \int_{c}^{d} x^{2} d y=\pi \int_{1}^{2} \frac{16}{y^{2}} d y=\left.\pi\left(-\frac{16}{y}\right)\right|_{1} ^{2}= ... | 8\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,599 |
1323. Determine the force of water pressure on a vertical wall having the shape of a semicircle with radius $R=6$ m, the diameter of which is at the water surface. | Solution. The pressure force of a liquid on an area $S$ at a depth of immersion $x$ is equal to
$$
P=\gamma x S
$$
where $\gamma$ is the specific weight of the liquid.
Fig. 90
. The radius of its base $R=3$ m, depth $H=5$ m. The boiler is filled with a liquid of specific weight $0.8 \Gamma /$ cm $^{3}$. Calculate the work required to pump the liquid out of the boiler. | Solution. In the section plane $x O y A O B$, there is a parabola with the equation $y=a x^{2}$. Let's find the parameter $a$. The coordinates of point $B$ must satisfy this equation, i.e., $H=a R^{2}$, hence $a=\frac{H}{R^{2}}$, so $y=\frac{H}{R^{2}} x^{2}$.
Divide the paraboloid into layers by planes parallel to the... | 294300\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,602 |
1334. Find the improper integrals:
1) $\int_{0}^{+\infty} e^{-x} d x$
2) $\int_{-\infty}^{+\infty} \frac{d x}{1+x^{2}}$
3) $\int_{1}^{+\infty} \frac{d x}{x}$
4) $\int_{1}^{+\infty} \frac{d x}{x^{2}}$ | Solution. 1) The integrand is continuous on the interval $[0,+\infty)$, hence by definition $\int_{0}^{+\infty} e^{-x} d x=\lim _{b \rightarrow+\infty} \int_{0}^{b} e^{-x} d x=\left.\lim _{b \rightarrow+\infty}\left(-e^{-x}\right)\right|_{0} ^{b}=\lim _{b \rightarrow+\infty}\left(-e^{-b}+1\right)=1$.
It is known that ... | 1,\pi,+\infty,1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,603 |
1348. Calculate the integral $\int_{0}^{6}\left(x^{2}+3\right) d \bar{x}$ using the Newton-Leibniz formula and approximate formulas of rectangles and trapezoids, dividing the interval of integration into 6 equal parts $(n=6)$. Find the absolute and relative errors of the results obtained by the approximate formulas. | Solution. By the Newton-Leibniz formula
$$
I=\int_{0}^{6}\left(x^{2}+3\right) d x=\left.\left(\frac{x^{3}}{3}+3 x\right)\right|_{0} ^{6}=90
$$
The interval $[0,6]$ is divided into 6 equal parts, and we construct a table of values of the integrand function $y=x^{2}+3$ :
$$
\begin{array}{lllr}
x_{0}=0 & y_{0}=3 & x_{4... | 90 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,605 |
1349. Knowing that $I=\int_{1}^{2} \frac{d x}{x}=\ln 2$, compute approximately the value of $\ln 2$ using the rectangle and trapezoidal formulas (for $n=10$). Estimate the absolute error of the obtained results. | Solution. The interval [1, 2] is divided into 10 equal parts, and we construct a table of values of the integrand function $y=\frac{1}{x}$
$$
\begin{array}{llll}
x_{0}=1,0 & y_{0}=1,0000 & x_{6}=1,6 & y_{6}=0,6250 \\
x_{1}=1,1 & y_{1}=0,9091 & x_{7}=1,7 & y_{7}=0,5882 \\
x_{2}=1,2 & y_{2}=0,8333 & x_{8}=1,8 & y_{8}=0,... | 0,6938 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,606 |
1356. Check that the following functions are solutions to the differential equations:
1) $x y^{\prime}-y=0, y=C x$
2) $y^{\prime \prime}-\frac{2}{x} y^{\prime}+\frac{2 y}{x^{2}}=0, y=C_{1} x+C_{2} x^{2}$
3) $v^{\prime \prime}+\frac{2}{r} v^{\prime}=0, v=\frac{C_{1}}{r}+C_{2}$ | Solution. 1) The equation is of the first order, as it contains the first-order derivative. Let's find the derivative of the given function $y^{\prime}=C$. Substituting into the given equation $y=$ $=C x$ and $y^{\prime}=C$, we get the identity
$$
x C-C x=0,0=0
$$
2) The equation is of the second order, as it contain... | proof | Calculus | proof | Yes | Yes | olympiads | false | 30,607 |
1357. Check that the function $y=x^{2}+C$ is a solution to the differential equation $\frac{d y}{d x}=2 x$. Construct the family of integral curves and highlight the integral curve passing through the point $(1 ; 2)$. | Solution. Let's find the derivative $\frac{d y}{d x}=\frac{d}{d x}\left(x^{2}+C\right)=2 x$ and substitute it into the given equation. We obtain the identity
$$
2 x=2 x
$$
Therefore, $y=x^{2}+C$ is a solution to the given differential equation and, moreover, a general solution since it contains an arbitrary constant.... | x^{2}+1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,608 |
1364. Integrate the following equations and find the particular solutions that satisfy the given initial conditions:
1) $(x+2) d y+(y-1) d x=0, y(1)=2$
2) $y^{\prime}=(2 y-1) \operatorname{ctg} x, y\left(\frac{\pi}{6}\right)=5$. | Solution. 1) First, we find the general integral:
$$
\begin{aligned}
& \frac{d x}{x+2}+\frac{d y}{y-1}=0, \int \frac{d x}{x+2}+\int \frac{d y}{y-1}=\ln |C|(C \neq 0) \\
& \ln |x+2|+\ln |y-1|=\ln |C|, \ln |(x+2)(y-1)|=\ln |C|
\end{aligned}
$$
from which
$$
|(x+2)(y-1)|=|C| \text { and }(x+2)(y-1)= \pm C
$$
or
$$
(x... | 18\sin^{2}x+\frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,610 |
1373. Integrate the equation
$$
x d y=(x+y) d x
$$
and find the particular solution that satisfies the initial conditions $y=2$ when $x=-1$. | Solution. We will solve the given equation with respect to the derivative. We obtain an equation of the form (6):
$$
y^{\prime}=\frac{x+y}{x} \text { . }
$$
The function on the right-hand side of this equation is homogeneous of degree zero, hence the original equation is homogeneous. We apply the substitution $y=u x$... | x\ln|x|-2x | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,611 |
1376. Integrate the equation
$$
x^{2} y^{\prime}+x y=1 \quad(x \neq 0)
$$
and find the particular solutions satisfying the initial conditions: $y=3$ when $x=1 ; y=-5$ when $x=-e$. | Solution. We will transform the given equation to the form (7):
$$
y^{\prime}+\frac{1}{x} y=\frac{1}{x^{2}}
$$
## Apply the substitution
$$
y=u v .
$$
Find the derivative
$$
y^{\prime}=u^{\prime} v+u v^{\prime}
$$
Substituting these values of $y$ and $y^{\prime}$ into the given equation, we get
$$
u^{\prime} v+u... | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,612 | |
1379. Find the curve passing through the point $(2 ; 3)$ and having the property that the segment of any of its tangents, enclosed between the coordinate axes, is bisected at the point of tangency. | Solution. Let $M(x ; y)$ be an arbitrary point on the desired curve, then by condition $A M=M B$ (Fig. 98), hence $O C=C B, O B=2 O C=$ $=2 x$ and similarly $O A=2 O D=2 y$.
Further, since $\varphi=\pi-\alpha$, then $\operatorname{tg} \varphi=\operatorname{tg}(\pi-\alpha)=-\operatorname{tg} \alpha=\frac{O A}{O B}=\fra... | x6 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,613 |
1380. The rate of decay of radium is proportional to its current amount $R$. Find the dependence of $R$ on $t$; formulate the differential equation and determine the proportionality coefficient from experimental data, according to which after 1600 years half of the current amount of radium will remain. | Solution. According to the problem, $\frac{d R}{d t}=k R$ (the rate of change is the derivative with respect to time), from which $\frac{d R}{R}=k d t$,
$$
\begin{gathered}
\ln R=k t+\ln C, \quad \ln R=\ln e^{k t}+\ln C= \\
=\ln C e^{k t}, \quad R=C e^{k t} .
\end{gathered}
$$
Let the amount of radium at the initial ... | R_{0}e^{-0.00043} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,614 |
1388. Write the first five terms of the series given the general term $a_{n}=\frac{1}{n(n+1)}$. | Solution. Setting $n=1$, we get $a_{1}=\frac{1}{1 \cdot 2}$. If $n=2$, then $a_{2}=\frac{1}{2 \cdot 3}$, if $n=3$, then $a_{3}=\frac{1}{3 \cdot 4}$, and further (for $n=4 ; 5$) $a_{4}=\frac{1}{4 \cdot 5}, a_{5}=\frac{1}{5 \cdot 6}$. Therefore,
$$
\sum_{n=1}^{\infty} \frac{1}{n(n+1)}=\frac{1}{1 \cdot 2}+\frac{1}{2 \cdo... | \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,615 |
1389. Write the first five terms of the series $\sum_{n=1}^{\infty}(-1)^{n-1} \frac{n}{n+1}$. | Solution. The general term $a_{n}=(-1)^{n-1} \frac{n}{n+1}$. Setting $n=1$, we get $a_{1}=(-1)^{0} \frac{1}{1+1}=\frac{1}{2}$. If $n=2$, then $a_{2}=$
$=(-1) \frac{2}{2+1}=-\frac{2}{3} \quad$ and $\quad$ further $\quad$ (for $\quad n=3 ; 4 ; 5$ ) $a_{3}=$
$=(-1)^{2} \frac{3}{3+1}=\frac{3}{4}, \quad a_{4}=(-1)^{3} \frac... | \frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}+\frac{5}{6}-\ldots | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,616 |
1390. Write the formula for the general term of each series:
1) $1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\ldots$,
2) $\frac{2}{5}-\frac{4}{8}+\frac{6}{11}-\frac{8}{14}+\ldots$ | Solution. 1) The denominators of the terms of this series are squares of natural numbers, hence the general term of the series is
$$
a_{n}=\frac{1}{n^{2}}
$$
2) The numerators of the terms of this series are even numbers of the form $2 n$, and the denominators are numbers obtained by the formula $3 n+2$ $(n=1,2,3, \l... | a_{n}=\frac{1}{n^{2}},a_{n}=(-1)^{n-1}\frac{2n}{3n+2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,617 |
1391. Find for each series the partial sum of the first $n$ terms $\left(S_{n}\right)$; show, using the definition, the convergence (divergence) of the series; find the sum of the series ( $S$ ):
1) $a+a q+a q^{2}+\ldots+a q^{n-1}+\ldots$
$$
\text { 2) } \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldo... | Solution. Let's consider several cases.
1) Let $|q|<1$, then the partial sum $S_{n}$ is found using the formula for the sum of a decreasing geometric progression
$$
S=\frac{a-a q^{n}}{1-q}
$$
Therefore, the partial sum of the given series is
$$
S_{n}=\frac{a-a q^{n}}{1-q}
$$
The sum of the series is
$$
S=\lim _{n... | \frac{}{1-q} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,618 |
1392. Check if the necessary condition for convergence is satisfied for the series:
1) $\sum_{n=1}^{\infty} \frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}+\ldots$;
2) $\sum_{n=1}^{\infty} \frac{2 n+1}{5 n-3}=\frac{3}{2}+\frac{5}{7}+\frac{7}{12}+\ldots+\frac{2 n+1}{5 n-3}+\ldots$ | Solution. 1) The given series is called harmonic. The necessary condition for convergence of this series is satisfied:
$$
\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} \frac{1}{n}=0
$$
nevertheless, the harmonic series diverges. Let's verify this. We will find the sequence of partial sums:
$$
\begi... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,619 |
1393. Investigate the convergence of the series using the necessary condition and the comparison test:
1) $1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+\ldots$;
2) $\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\ldots$
3) $\sum_{n=2}^{\infty} \frac{1}{\ln n}=\frac{1}{\ln 2}+\frac{1}{\ln 3}+\ldots$;
4) $\sum_{n=1}^{... | Solution. 1) The general term of the given series $a_{n}=\frac{1}{\sqrt{n}}$ tends to zero as $n \rightarrow \infty$, i.e., the necessary condition for convergence is satisfied. Let's compare series 1 with the harmonic series, whose general term is $a_{n}=\frac{1}{n}$. It is easy to see that $\frac{1}{\sqrt{n}} \geqsla... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,620 |
1394. Investigate the convergence of the series by the D'Alembert's criterion:
1) $\sum_{n=1}^{\infty} \frac{2^{n}}{n!}$
2) $\sum_{n=1}^{\infty} \frac{3^{n}}{n \cdot 2^{n}}$
3) $\sum_{n=1}^{\infty} \frac{1}{n^{3}}$. | Solution. 1) $a_{n}=\frac{2^{n}}{n!}$, the next term in the series $a_{n+1}=$ $=\frac{2^{n+1}}{(n+1)!}$. We divide the subsequent term by the previous one and take the limit
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}}=\lim _{n \rightarrow \infty}\left(\frac{2^{n+1}}{(n+1)!}: \frac{2^{n}}{n!... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,621 |
1395. Investigate the convergence of the series by the Cauchy criterion:
1) $\sum_{n=1}^{\infty} \frac{1}{n^{n}}$
2) $\sum_{n=1}^{\infty}\left(\frac{n}{2 n+1}\right)^{n}$ | Solution. 1) The general term of the given series is $a_{n}=\frac{1}{n^{n}}$. Let's find the limit
$$
\lim _{n \rightarrow \infty} \sqrt[n]{a_{n}}=\lim _{n \rightarrow \infty} \sqrt[n]{\frac{1}{n^{n}}}=\lim _{n \rightarrow \infty} \frac{1}{n}=0
$$
We obtained $l=0(l<1)$, therefore, series 1 converges.
2) The general... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,622 |
1427. Investigate the convergence of the alternating series:
1) $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+(-1)^{n-1} \frac{1}{n}+\ldots$;
2) $1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\ldots+(-1)^{n-1} \frac{1}{2 n-1}+\ldots$;
3) $\frac{\sin \alpha}{1}+\frac{\sin 2 \alpha}{2^{2}}+\frac{\sin 3 \alpha}{3^{2}}+\ldots$ ( $... | Solution. 1) The terms of the given series decrease in absolute value, the signs alternate, and the general term tends to zero as $n$ increases. Therefore, according to the Leibniz criterion, series 1 converges. The series formed from the absolute values of the terms of the given series
$$
1+\frac{1}{2}+\frac{1}{3}+\l... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,623 |
1428. Given a convergent alternating series $\sum_{n=1}^{\infty}(-1)^{n-1} \frac{1}{n^{2}}$. Estimate the error made when replacing the sum of this series: 1) by the sum of its first three terms; 2) by the sum of its first four terms. | Solution. 1) $S=S_{n}+r_{n}$, where $S$ is the sum of the series; $S_{n}$ is the partial sum; $r_{n}$ is the remainder of the series. We find $S=S_{3}+r_{3}$ or $S=1-\frac{1}{2^{2}}+\frac{1}{3^{2}}+r_{3}$, from which $S=\frac{31}{36}+r_{3}$. By the Leibniz theorem, the absolute error $\left|r_{n}\right|<\frac{1}{(n+1)^... | \frac{115}{144} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,624 |
1429. Check that the series $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{\sqrt{n}}$ converges. How many terms of this series need to be taken to compute its sum with an accuracy of 0.01? | Solution. For this series, all conditions of the Leibniz criterion are satisfied, so the series converges. We can write the sum of the series in the form: $S=S_{n}+r_{n},\left|r_{n}\right|<\left|a_{n+1}\right|$. We need to find such an $n$ that $\left|a_{n+1}\right|=\frac{1}{\sqrt{n+1}} \leqslant 0.01$. Solving the equ... | 9999 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,625 |
1443. Find the domain of convergence of the series:
1) $\sum_{n=1}^{\infty} \frac{x^{n}}{n!}=\frac{x}{11}+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\ldots+\frac{x^{n}}{n!}+\ldots$;
2) $\sum_{n=1}^{\infty} n$ $n!x^{n}=1!x+2!x^{2}+3!x^{3}+\ldots+n!x^{n}+\ldots$
3) $\sum_{n=1}^{\infty}(-1)^{n-1} \frac{x^{n}}{n^{2}}=\frac{x}{1^{2}... | Solution. 1) $a_{n}=\frac{1}{n!}, a_{n+1}=\frac{1}{(n+1)!}$. By formula (3) $R=\lim _{n \rightarrow \infty}\left|\frac{a_{n}}{a_{n+1}}\right|=\lim _{n \rightarrow \infty} \frac{(n+1)!}{n!}=\lim _{n \rightarrow \infty} \frac{1 \cdot 2 \cdot 3 \ldots n(n+1)}{1 \cdot 2 \cdot 3 \ldots n}=\lim _{n \rightarrow \infty}(n+$
$+... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,627 |
1454. Expand the function $\sin x$ into a Taylor series at $x_{0}=\frac{\pi}{4}$. | Solution. We find the values of the function and its derivatives at $x=\frac{\pi}{4}$:
$$
\begin{gathered}
f(x)=\sin x, f\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} \\
f^{\prime}(x)=\cos x, f^{\prime}\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} \\
f^{\prime \prime}(x)=-\sin x, f^{\prime \prime}\left(\frac{\pi}{4}\r... | \sin\frac{\sqrt{2}}{2}[1+\frac{x-\frac{\pi}{4}}{1!}-\frac{(x-\frac{\pi}{4})^{2}}{2!}-\frac{(x-\frac{\pi}{4})^{3}}{3!}+\ldots] | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,629 |
1464. Expand $\ln \frac{1+x}{1-x}$ into a power series in $x$. | Solution. $\ln \frac{1+x}{1-x}=\ln (1+x)-\ln (1-x)$.
We already know that
$$
\ln (1+x)=\frac{x}{1}-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\ldots(-1<x \leqslant 1)
$$
Therefore,
$\ln (1-x)=-\frac{x}{1}-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\frac{x^{4}}{4}-\ldots(-1 \leqslant x<1)$.
Subtracting the second series ... | 2(x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\ldots+\frac{x^{2n-1}}{2n-1}+\ldots)(-1<x<1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,630 |
1465. Expand $\operatorname{arctg} x$ into a Maclaurin series. | Solution. It is known that $\operatorname{arctg} x=\int_{0}^{x} \frac{d x}{1+x^{2}}$.
We expand the integrand into a series using the binomial series (4) § 4:
$$
\frac{1}{1+x^{2}}=\left(1+x^{2}\right)^{-1}=1-x^{2}+x^{4}-x^{6}+\ldots
$$
Integrating this series within its interval of convergence $(-1,1)$, we obtain
$... | x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\ldots(-1\leqslantx\leqslant1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,631 |
1466. Calculate $\int_{0}^{1} e^{-x^{2}} d x$ with an accuracy of 0.001. | Solution. As mentioned above, this integral cannot be expressed in a closed form through elementary functions. Nevertheless, it can be calculated with any degree of accuracy using power series.
Let's expand the integrand into a power series. We will use the known series
$$
e^{x}=1+\frac{x}{1!}+\frac{x^{2}}{2!}+\frac{... | 0.747 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,632 |
1467. Calculate $\sin 10^{\circ}$ with an accuracy of 0.0001. | Solution. We will use series (6) of § 4, expressing degrees in radians:
$$
\sin 10^{\circ}=\sin \frac{\pi}{18}=\frac{\pi}{18}-\left(\frac{\pi}{18}\right)^{3} \frac{1}{3!}+\left(\frac{\pi}{18}\right)^{5} \frac{1}{5!}-\ldots
$$
$\frac{\pi}{18}<0.2$, therefore, the third term of the expansion
$$
a_{3}=\left(\frac{\pi}{... | 0.1736 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,633 |
1469. Calculate $\sqrt{1.004}$ with an accuracy of 0.0001. | Solution. We will use formula (4) §4:
$$
\sqrt{1.004}=\sqrt{1+0.004}=(1+0.004)^{\frac{1}{2}}
$$
Here $m=\frac{1}{2}, x=0.004$
$$
\begin{aligned}
\sqrt{1.004}=1 & +\frac{0.004}{2}+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2!}(0.004)^{2}+\ldots= \\
& =1+0.002-\frac{(0.004)^{2}}{2!4}+\ldots
\end{aligned}
$$
Notice ... | 1.002 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,634 |
1481. A randomly taken part can be of the first grade (event $A$), the second grade (event $B$), or the third grade (event $C$). What do the following events represent: $A+B$; $\overline{A+C} ; A C ; A B+C$? | Solution. $A+B$ is the event that will occur if at least one of the events $A$ and $B$ occurs. Therefore, $A+B$ in our case is a part of the first or second grade. Since $A+C$ is a part of the first or third grade, the opposite of this event $\overline{A+C}$ is a part of the second grade. $A C$ is an impossible event, ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 30,635 |
1489. Out of 5000 randomly taken parts, 32 were defective. Find the frequency of defective parts in this batch. | Solution. In this problem, we are interested in the event $A$ - the appearance of a defective part. $n=5000$ trials were conducted, and the event $A$ occurred $m=32$ times. Therefore, the sought frequency is
$$
w_{5000}(A)=\frac{32}{5000}=0.0064
$$ | 0.0064 | Other | math-word-problem | Yes | Yes | olympiads | false | 30,636 |
1494. In the book "An Introduction to Probability Theory and Its Applications" by W. Feller, the 1964 edition has 500 pages. What is the probability that a randomly opened page will have an ordinal number that is a multiple of 7? | Solution. From the condition of the problem, it follows that in this case, a certain event decomposes into $n=500$ equally likely cases. Of these, $m=71$ cases are favorable to the occurrence of the event of interest, since a number divisible by 7 has the form $7k$, where $k$ is an integer, and $0<7k \leqslant 500$ $\l... | 0.142 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 30,637 |
1495. A randomly selected phone number consists of 5 digits. What is the probability that in it: 1) all digits are different; 2) all digits are odd? | Solution. 1) Since any of the digits: $0,1,2,3,4,5,6,7,8$, 9 can stand in each of the five positions in a five-digit number, there will be $10^{5}$ different five-digit numbers (all of which can be enumerated as follows: number 00000 - 1st, $00001-2$-nd, $00002-3$-rd, ..., $99998-99999$-th and, finally, 99999 - 100,000... | 0.3024 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 30,638 |
1496. In a railway carriage compartment, two sofas stand opposite each other, each with five seats. Out of ten passengers, four wish to sit facing the locomotive, and three wish to sit with their backs to it. What is the probability that the two passengers from the three who are indifferent about where they sit will en... | Solution. First, we find the number of all possible arrangements of passengers in the carriage, taking into account the restrictions stated in the problem (four passengers wish to sit facing the locomotive, and three - on the opposite bench). Let $A, B, C$ be passengers who are indifferent to where they sit. If $A$ sit... | \frac{2}{15} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 30,639 |
1523. The master services 5 machines. $20 \%$ of the working time he spends at the first machine, 10 - at the second, 15 - at the third, $25-$ at the fourth and, finally, 30 - at the fifth. Find the probability that at a randomly chosen moment in time he is: 1) at the first or third machine; 2) at the second or fifth; ... | Solution. Let's introduce the notation. Let $A, B, C, D, E$ be the events that occur if the master is at the first, second, third, fourth, and finally, the fifth machine at a randomly chosen moment in time, respectively. From the problem statement, it follows that $A, B, C, D, E$ are pairwise mutually exclusive events ... | 0.35,0.40,0.45,0.45,0.55 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 30,640 |
1532. Find the probability that a randomly taken two-digit number will be divisible by either 2, or 5, or both at the same time. | Solution. Let $A$ be the event that a randomly chosen two-digit number is divisible by 2, and $B$ be the event that it is divisible by 5. We need to find $P(A+B)$. Since $A$ and $B$ are compatible events, we have
$$
P(A+B)=P(A)+P(B)-P(A B)
$$
Two-digit numbers are $10,11, \ldots, 98,99$. There are 90 of them in total... | 0.6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 30,641 |
1536. In the workshop, two motors operate independently of each other. The probability that the first motor will not require the master's attention during an hour is 0.9, and for the second motor, this probability is 0.85. Find the probability that during the hour neither motor will require the master's attention. | Solution. Let $A$ be the event that the first motor will not require attention within an hour, and $B$ be the event that the second motor will not require attention. We need to find $P(A B)$. Since $A$ and $B$ are independent events, $P(A B)=P(A) P(B)=0.9 \cdot 0.85=0.765$. | 0.765 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,642 |
1537. In an urn, there are 5 white and 4 black balls. Two balls are drawn in succession. Find the probability that both balls are white. | Solution. Let $A_{1}$ be the appearance of a white ball on the first draw, and $A_{2}$ be the appearance of a white ball on the second draw. We are interested in the event $A_{1} A_{2}$. By the multiplication theorem of probabilities,
$$
P\left(A_{1} A_{2}\right)=P\left(A_{1}\right) P\left(A_{2} / A_{1}\right)=\frac{5... | \frac{5}{18} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 30,643 |
1538. Three shooters shoot at a target independently of each other. The probability of hitting the target for the first shooter is 0.6, for the second - 0.7, and for the third - 0.75. Find the probability of at least one hit on the target if each shooter makes one shot. | Solution. Let $A, B, C$ be the events that the first, second, and third shooter hit the target, respectively. From the problem, we have $P(A)=0.6, P(B)=0.7$,
$P(C)=0.75$. We need to find the probability of at least one hit, i.e., $P(A+B+C)$:
$$
\begin{aligned}
P(A+B+C) & =1-P(\bar{A}) P(\bar{B}) P(\bar{C})=1-(1-0.6)(... | 0.97 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 30,644 |
1557. Electric lamps are manufactured at 3 plants. The first plant produces $45 \%$ of the total number of lamps, the second - 40, the third - 15. The production of the first plant contains $70 \%$ standard lamps, the second - 80, the third - 81. Products from all three plants are supplied to stores. What is the probab... | Solution. Let's introduce the following notations. Let event $A_{1}$ be that the purchased lamp is manufactured at the first factory, $A_{2}$ - the lamp is from the second factory, $A_{3}$ - the lamp is from the third factory, and finally, event $B$ - the lamp turns out to be standard. From the problem statement, it fo... | 0.7565 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,645 |
1565. There are 12 urns, of which 6 urns (composition $A_{1}$) contain 3 white and 4 black balls, 3 urns (composition $A_{2}$) contain 2 white and 8 black balls, 2 urns (composition $A_{3}$) contain 6 white and 1 black ball, and 1 urn (composition $A_{4}$) contains 4 white and 3 black balls. A ball is drawn from a rand... | Solution. Let's introduce notation. Let event $B$ be the event that the ball drawn is white. According to the condition, $P\left(A_{1}\right)=\frac{6}{12}=\frac{1}{2}$, $P\left(A_{2}\right)=\frac{3}{12}=\frac{1}{4}, \quad P\left(A_{3}\right)=\frac{2}{12}=\frac{1}{6}, \quad P\left(A_{4}\right)=\frac{1}{12} ; \quad P\lef... | 0.314 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 30,646 |
1571. A regular coin was tossed 4 times. What is the probability of the coat of arms landing twice? | Solution. In this case, $A$ is the occurrence of heads in a separate trial; $p=P(A)=0.5$ and, consequently, $q=1-p=0.5$. Using the Bernoulli formula, we find the probability
$$
P_{2,4}=C_{4}^{2} p^{2} q^{4-2}=\frac{4 \cdot 3}{1 \cdot 2} \cdot(0.5)^{2} \cdot(0.5)^{2}=0.375
$$ | 0.375 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 30,647 |
1572. Let the germination rate of rye seeds be $90 \%$. What is the probability that out of 7 sown seeds, 5 will germinate? | Solution. The probability of germination of a single seed $p=$ $=0.9$. Therefore, $q=1-p=0.1$. By the Bernoulli formula, we determine
$$
P_{5,7}=C_{7}^{5} p^{5} q^{7-5}=C_{7}^{7-5} p^{5} q^{2}=C_{7}^{2} \cdot(0.9)^{5} \cdot(0.1)^{2}=0.124
$$ | 0.124 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 30,648 |
1586. The share of premium quality items in this enterprise is $31 \%$. What is the most probable number of premium quality items in a randomly selected batch of 75 items? | Solution. It is known that $p=0.31, n=75$, and $q=1-p=$ $=0.69$. We form the double inequality (1):
$$
\begin{gathered}
75 \cdot 0.31 - 0.69 \leqslant m_{0} \leqslant 75 \cdot 0.31 + 0.31 ; \\
22.56 \leqslant m_{0} \leqslant 23.56 .
\end{gathered}
$$
From this, it follows that $m_{0}=23$. | 23 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 30,649 |
1587. How many times should a die be thrown so that the most probable number of times a two appears is 32? | Solution. In this case, $p=\frac{1}{6}, m_{0}=32$. It is required to find the number of independent trials $n$. The quantities $p, q=1-p$, $m_{0}$, and $n$ are related by inequality (1): $n p-q \leqslant m_{0} \leqslant n p+$ $+p$, from which
$$
n \cdot \frac{1}{6}-\frac{5}{6} \leqslant 32 ; n \cdot \frac{1}{6}+\frac{... | 191to197 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 30,650 |
1588. What is the probability of event $A$ occurring in each trial, if the most probable number of occurrences of event $A$ in 120 trials is 32? | Solution. Here $n=120, m_{0}=32$. According to inequality (1),
$$
120 p-(1-p) \leqslant 32 ; 120 p+p \geqslant 32
$$
Solving the obtained system, we find that $\frac{32}{121} \leqslant p \leqslant \frac{33}{121}$. | \frac{32}{121}\leqslantp\leqslant\frac{33}{121} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 30,651 |
1598. The probability of manufacturing a part of the highest grade on a given machine is 0.4. Find the probability that among 26 randomly selected parts, half will be of the highest grade. | Solution. It is required to find the probability $P_{13}$ 26. Knowing the Laplace theorem, the necessary calculations are performed according to the following scheme:
$$
\begin{array}{cc}
p=0.4 ; & n p=26 \cdot 0.4=10.4 ; \\
q=1-0.4=0.6 ; & n p q=10.4 \cdot 0.6=6.24 ; \\
n=26 ; & \sqrt{n p q}=V \overline{6.24}=2.50 ; ... | 0.093 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 30,652 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.