Republish the benchmark with one unified id namespace and one headline total

ANSWER_AUDIT.md

@@ -506,15 +506,15 @@ answer.
 | `dgn.trigonometry.p158.26.6` | wrong answer | (d) | `\left(\dfrac{\sqrt{3}}{3}; 1\right]` | `\left(\dfrac{\sqrt{3}}{2}; 1\right]` | the exercise asks to solve arccos x < pi/6, whose solution is x in (cos(pi/6), 1] = (sqrt(3)/2, 1]; the source gives (sqrt(3)/3, 1] |
 | `dgn.core.2.205` | misprinted statement | second equation of the system | `\begin{cases}x+y+\frac{x^2}{y^2}=7,\\ \frac{(x-y)x^2}{y^2}=12.\end{cases}` | `\begin{cases}x+y+\frac{x^2}{y^2}=7,\\ \frac{(x+y)x^2}{y^2}=12.\end{cases}` | the source answer does not satisfy the second equation: (2; 1) gives (x-y)x^2/y^2 = 4, not 12, and (6; -3) gives 36, not 12. All four source answers satisfy x+y+x^2/y^2 = 7 and (x+y)x^2/y^2 = 12 exactly, so the source misprinted x+y as x-y. The system as written has exactly one real solution, (1.3863205; -0.5580341), whose coordinates are roots of a quintic and are therefore certainly not the intended answer. |
 | `dgn.core.1.206` | wrong answer |  | `\frac{a}{a+1}` | `\frac{a(a+1)}{2a+1}` | the expression (a+1)/(2*cuberoot(sqrt(3)-sqrt(2))*6throot(5+2sqrt(6)) + 1/a) equals a(a+1)/(2a+1), because 2*cuberoot(sqrt(3)-sqrt(2))*6throot(5+2sqrt(6)) = 2 exactly; the difference between the source answer and the true value is a^3/((a+1)(2a+1)), which is not an identity |
-| `dgn.core.1.202` | wrong answer |  | `\sqrt{mn}` | `the expression has no closed form in m and n; its value at m = 8, n = 1 is exactly 40/3, so the source answer cannot be right.  The cube root of mn is not the value either (it would be 2 at that point)` | the expression (cube root of m^2 + n*cube root of m + n^2) * (cube root of m^4 - n^3 + n^2*cube root of (m - m*n)) / (m*n^-1 + n - n^4*m^-1 - n^2) does not have the square root of mn as its value.  At m = 8 and n = 1 the expression is exactly 40/3, a rational number, while the source answer is sqrt(8) = 2*sqrt(2), an irrational number, so the two can never be equal.  The denominator is not zero there (it is 63/8), so the point lies inside the domain of the expression.  The first factor and the denominator do factor neatly - the first one is (m - n^3)/(cube root of m - cube root of n) and the denominator is (m + n^2)(m - n^3)/(m*n) - which is why the statement was read carefully before the answer was called wrong |
-| `dgn.core.8.276` | wrong answer |  | `(\sqrt[3]{5},5)` | `(\sqrt[5]{5},5)` | the source answer is (cuberoot(5), 5) = (5^(1/3), 5), but the answer of the inequality log_x^2 sqrt(5) - log_x 5 sqrt(5) + 1,25 < 0 is (5^(1/5), 5).  With t = log_x 5 the inequality becomes t^2/4 - 3t/2 + 5/4 < 0, i.e. (t-1)(t-5) < 0, so 1 < t < 5 and x = 5^(1/t) runs from 5^(1/5) to 5; the cube root 5^(1/3) is only an interior point of that interval, not its lower endpoint.  The root index in the source answer is a 3, not a 5: it was read from the page image at 800 dpi and the index glyph was compared pixel by pixel with the index of the cube root in 3sqrt35 printed on the same answer page; it matches the 3 far better than the 5 (IoU 0.61 against 0.51, and a 5 would have a horizontal top bar reaching the left edge and a narrow middle stem, neither of which the printed glyph has). |
-| `dgn.core.1.111` | wrong answer |  | `100` | `the expression is negative at the stated values a = 25/7 and x = 7/25; under the transcribed structure its value is exactly -31250*sqrt(7)/343 - 84/25` | the exercise asks for the value of the expression at a = 3 4/7 = 25/7 and x = 0,28 = 7/25.  There ax = 1, so the dominant term -10a sqrt(x) sqrt((a x^-1)^2) is exactly -31250*sqrt(7)/343 (about -241.1), and the remaining terms are bounded by 229 and 3, so the value is negative in every faithful reading of the printed expression (eight readings were evaluated exactly: the transcribed structure gives -31250*sqrt(7)/343 - 84/25, the structure in which the first group is printed as ( ... )^3 gives about -1.39*10^7).  A positive answer of 100 is therefore impossible. |
-| `dgn.algebra.777` | wrong answer |  | `$-\frac{19}{4} < x < -4$` | `$\left(-\frac{11}{4};\, -2\right) \cup \left(-2;\, +\infty\right)$` | the source answer of log_{x+3}(1/2) > 1/2 lies entirely outside the domain of the logarithm, which needs x > -3 and x != -2.  Solving the inequality gives x in (-11/4, -2) union (-2, +infinity): for 0 < x+3 < 1 the logarithm decreases, so 1/2 < (x+3)^(1/2) and x > -11/4, and for x+3 > 1 the inequality holds everywhere.  The point x = -19/8 satisfies the inequality and is missing from the source answer.  Page 99 was read at 400 dpi and shows exactly -19/4 < x < -4. |
-| `dgn.algebra.788` | wrong answer |  | `$-1 < x < -\frac{5}{6}$` | `$-1 < x < \frac{1}{\sqrt[3]{36}} - 1$` | the source answer of log_6 log_{1/6th-root-of-6}(x+1) > 1 is too narrow.  Writing u = log_{6^(-1/9)}(x+1), the condition is u > 6, and because the base is below 1 this gives 0 < x+1 < 6^(-2/3), so the solution is -1 < x < 6^(-2/3) - 1 = -0.6972....  The source upper end -5/6 = -0.8333 excludes the whole stretch from -0.8333 to -0.6972; the point x = -0.8 satisfies the inequality (log_6 log_{6^(-1/9)}(0.2) = 1.166 > 1) and is missing from the source answer.  Page 100 was read at 400 dpi. |
-| `dgn.algebra.926` | wrong answer |  | `$[-18;-3] \cup (0;2)$` | `$[-18;-3) \cup (0;2)$` | in sqrt((x^2+9x-162)/(x-2)) > x+9 equality holds at x = -3, because the radicand is 36 there and x+9 is 6 as well, so the strict inequality fails and -3 must not belong to the solution.  The source closes the bracket at -3. The difference of the two sides is -x(x+3)(x+12)/(x-2), which changes sign exactly at -3.  Page 116 was read at 400 dpi and shows the closed bracket. |
-| `dgn.core.2.343` | wrong answer | first root | `$x_1=3,\ x_2=\sqrt{3}/3,\ x_3=-2\sqrt{3}$` | `$x_1=\sqrt{3},\ x_2=\sqrt{3}/3,\ x_3=-2\sqrt{3}$` | the source's own solution derives x_1 = sqrt(3) and x_1' = sqrt(3)/3, but its answer line prints x_1 = 3.  The value 3 leaves the exact residual 18 + 24*sqrt(3) in 3x^3 + 2*sqrt(3)x^2 - 21x + 6*sqrt(3), while sqrt(3), sqrt(3)/3 and -2*sqrt(3) are exact roots |
-| `dgn.core.2.356` | wrong answer |  | `$x_1=2,\ x_2=-2$` | `$x_1=\sqrt{2},\ x_2=-\sqrt{2}$` | the printed pair is not a root pair of the equation: 2 leaves the exact residual 2 and -2 leaves the residual 34 in x^4 - 4x^3 + 3x^2 + 8x - 10.  The equation factors as (x^2 - 2)(x^2 - 4x + 5), so its two real roots are sqrt(2) and -sqrt(2); the source appears to have lost the radical signs of its own answer |
-| `dgn.algebra.454` | wrong answer | second root | `$x_2 = \frac{4\log_3 2 - 1}{2\sqrt{3}(\log_3 5 - 1)}$` | `$x_2 = \frac{4\log_3 2 + 1}{2\sqrt{3}(\log_3 5 - 1)}$` | the printed second root leaves the exact value -8/3 in the reduced quadratic r^2 - 5*sqrt(3)*r + 12 with r = (5/3)^(sqrt(3) x), and the exact value -97.610355 in the equation itself, so it is not a solution.  The sign of the 1 is wrong: the exact solution needs 4*log_3(2) + 1 |
+| `dgn.core.1.202` | wrong answer |  | `\sqrt{mn}` | `the expression has no closed form in m and n; its value at m = 8, n = 1 is exactly 40/3, so the source answer cannot be right. The cube root of mn is not the value either (it would be 2 at that point)` | the expression (cube root of m^2 + n*cube root of m + n^2) * (cube root of m^4 - n^3 + n^2*cube root of (m - m*n)) / (m*n^-1 + n - n^4*m^-1 - n^2) does not have the square root of mn as its value. At m = 8 and n = 1 the expression is exactly 40/3, a rational number, while the source answer is sqrt(8) = 2*sqrt(2), an irrational number, so the two can never be equal. The denominator is not zero there (it is 63/8), so the point lies inside the domain of the expression. The first factor and the denominator do factor neatly - the first one is (m - n^3)/(cube root of m - cube root of n) and the denominator is (m + n^2)(m - n^3)/(m*n) - which is why the statement was read carefully before the answer was called wrong |
+| `dgn.core.8.276` | wrong answer |  | `(\sqrt[3]{5},5)` | `(\sqrt[5]{5},5)` | the source answer is (cuberoot(5), 5) = (5^(1/3), 5), but the answer of the inequality log_x^2 sqrt(5) - log_x 5 sqrt(5) + 1,25 < 0 is (5^(1/5), 5). With t = log_x 5 the inequality becomes t^2/4 - 3t/2 + 5/4 < 0, i.e. (t-1)(t-5) < 0, so 1 < t < 5 and x = 5^(1/t) runs from 5^(1/5) to 5; the cube root 5^(1/3) is only an interior point of that interval, not its lower endpoint. The root index in the source answer is a 3, not a 5: it was read from the page image at 800 dpi and the index glyph was compared pixel by pixel with the index of the cube root in 3sqrt35 printed on the same answer page; it matches the 3 far better than the 5 (IoU 0.61 against 0.51, and a 5 would have a horizontal top bar reaching the left edge and a narrow middle stem, neither of which the printed glyph has). |
+| `dgn.core.1.111` | wrong answer |  | `100` | `the expression is negative at the stated values a = 25/7 and x = 7/25; under the transcribed structure its value is exactly -31250*sqrt(7)/343 - 84/25` | the exercise asks for the value of the expression at a = 3 4/7 = 25/7 and x = 0,28 = 7/25. There ax = 1, so the dominant term -10a sqrt(x) sqrt((a x^-1)^2) is exactly -31250*sqrt(7)/343 (about -241.1), and the remaining terms are bounded by 229 and 3, so the value is negative in every faithful reading of the printed expression (eight readings were evaluated exactly: the transcribed structure gives -31250*sqrt(7)/343 - 84/25, the structure in which the first group is printed as ( ... )^3 gives about -1.39*10^7). A positive answer of 100 is therefore impossible. |
+| `dgn.algebra.777` | wrong answer |  | `$-\frac{19}{4} < x < -4$` | `$\left(-\frac{11}{4};\, -2\right) \cup \left(-2;\, +\infty\right)$` | the source answer of log_{x+3}(1/2) > 1/2 lies entirely outside the domain of the logarithm, which needs x > -3 and x != -2. Solving the inequality gives x in (-11/4, -2) union (-2, +infinity): for 0 < x+3 < 1 the logarithm decreases, so 1/2 < (x+3)^(1/2) and x > -11/4, and for x+3 > 1 the inequality holds everywhere. The point x = -19/8 satisfies the inequality and is missing from the source answer. Page 99 was read at 400 dpi and shows exactly -19/4 < x < -4. |
+| `dgn.algebra.788` | wrong answer |  | `$-1 < x < -\frac{5}{6}$` | `$-1 < x < \frac{1}{\sqrt[3]{36}} - 1$` | the source answer of log_6 log_{1/6th-root-of-6}(x+1) > 1 is too narrow. Writing u = log_{6^(-1/9)}(x+1), the condition is u > 6, and because the base is below 1 this gives 0 < x+1 < 6^(-2/3), so the solution is -1 < x < 6^(-2/3) - 1 = -0.6972.... The source upper end -5/6 = -0.8333 excludes the whole stretch from -0.8333 to -0.6972; the point x = -0.8 satisfies the inequality (log_6 log_{6^(-1/9)}(0.2) = 1.166 > 1) and is missing from the source answer. Page 100 was read at 400 dpi. |
+| `dgn.algebra.926` | wrong answer |  | `$[-18;-3] \cup (0;2)$` | `$[-18;-3) \cup (0;2)$` | in sqrt((x^2+9x-162)/(x-2)) > x+9 equality holds at x = -3, because the radicand is 36 there and x+9 is 6 as well, so the strict inequality fails and -3 must not belong to the solution. The source closes the bracket at -3. The difference of the two sides is -x(x+3)(x+12)/(x-2), which changes sign exactly at -3. Page 116 was read at 400 dpi and shows the closed bracket. |
+| `dgn.core.2.343` | wrong answer | first root | `$x_1=3,\ x_2=\sqrt{3}/3,\ x_3=-2\sqrt{3}$` | `$x_1=\sqrt{3},\ x_2=\sqrt{3}/3,\ x_3=-2\sqrt{3}$` | the source's own solution derives x_1 = sqrt(3) and x_1' = sqrt(3)/3, but its answer line prints x_1 = 3. The value 3 leaves the exact residual 18 + 24*sqrt(3) in 3x^3 + 2*sqrt(3)x^2 - 21x + 6*sqrt(3), while sqrt(3), sqrt(3)/3 and -2*sqrt(3) are exact roots |
+| `dgn.core.2.356` | wrong answer |  | `$x_1=2,\ x_2=-2$` | `$x_1=\sqrt{2},\ x_2=-\sqrt{2}$` | the printed pair is not a root pair of the equation: 2 leaves the exact residual 2 and -2 leaves the residual 34 in x^4 - 4x^3 + 3x^2 + 8x - 10. The equation factors as (x^2 - 2)(x^2 - 4x + 5), so its two real roots are sqrt(2) and -sqrt(2); the source appears to have lost the radical signs of its own answer |
+| `dgn.algebra.454` | wrong answer | second root | `$x_2 = \frac{4\log_3 2 - 1}{2\sqrt{3}(\log_3 5 - 1)}$` | `$x_2 = \frac{4\log_3 2 + 1}{2\sqrt{3}(\log_3 5 - 1)}$` | the printed second root leaves the exact value -8/3 in the reduced quadratic r^2 - 5*sqrt(3)*r + 12 with r = (5/3)^(sqrt(3) x), and the exact value -97.610355 in the equation itself, so it is not a solution. The sign of the 1 is wrong: the exact solution needs 4*log_3(2) + 1 |
 | `dgn.core.2.059` | wrong answer |  | `x=2` | `` | Our text matches the print, but the printed answer x=2 gives LHS 5.162 != 6; the printed equation's root is about 3.4017. |
 | `dgn.core.2.295` | wrong answer |  | `x=65` | `` | Our text matches the print, but the printed answer x=65 gives 0.33019 != 1/3; the printed equation's root is x=64 (1/12+1/4=1/3). |
 | `dgn.core.2.298` | wrong answer |  | `x=4` | `` | Print reads 8,4*12th-root(x^-7) - 0,2*4th-root(x^-1)*cuberoot(x^2) = 12th-root(x^11); our text turned the root indices into powers 4^12 and 4^4. (The printed answer x=4 also fails the printed equation, residual -0.4.). After the corrected reading of the printed statement the printed answer still fails: the engine says mismatch. The statement was therefore misread by us and the answer is wrong in the source |
@@ -831,7 +831,7 @@ answer.
 | `dgn.parameters.p120.44` | answer recovered |  | `` | `x = \sin(\pi/12) = (\sqrt{6} - \sqrt{2})/4` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: §1.1. Simplest equations and inequalities with a parameter; answer block on page 121) |
 | `dgn.parameters.p121.45` | answer recovered |  | `` | `(x; y; z) = (\tg \alpha; \tg 2\alpha; \tg 4\alpha), where \alpha = \pm\pi/7, \pm2\pi/7, \pm3\pi/7.` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: §1.1. Simplest equations and inequalities with a parameter; answer block on page 122) |
 | `dgn.parameters.242` | answer recovered |  | `` | `[0; (3 - \sqrt{5})/6)` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: §1.1. Simplest equations and inequalities with a parameter; answer block on page 124) |
-| `dgn.parameters.243` | wrong answer |  | `page 124, item 243 of the 'Ответы' block: x₁ = −1/√2, x₂ = √2 − √3/2  (printed problem, page 122: 'Задача 243. Найдите все решения уравнения √((1 + 2x√(1−x²))/2) + 2x² = 1.')` | `x_1 = -1/\sqrt{2},\quad x_2 = (\sqrt{6}-\sqrt{2})/4` | With x = sin t, \|x\| ⩽ 1, the equation becomes \|sin t + cos t\|/√2 = cos 2t; it has exactly two real roots. x₁ = −1/√2 checks out (radicand 0, 2x² = 1). The printed x₂ = √2 − √3/2 = 0.5481882 does NOT: 2x² + √((1+2x√(1−x²))/2) − 1 = +0.5800404 ≠ 0. The true second root is (√6 − √2)/4 = sin(π/12) = 0.2588190451, for which 2x√(1−x²) = 1/2, the radicand is 3/4 and the left side is √3/2 + (2−√3)/2 = 1. A sign-change scan over [−1, 1] returns only −0.7071067812 and 0.2588190451, matching the corrected pair. The engine's complaint is valid. |
+| `dgn.parameters.243` | wrong answer |  | `page 124, item 243 of the 'Ответы' block: x₁ = −1/√2, x₂ = √2 − √3/2 (printed problem, page 122: 'Задача 243. Найдите все решения уравнения √((1 + 2x√(1−x²))/2) + 2x² = 1.')` | `x_1 = -1/\sqrt{2},\quad x_2 = (\sqrt{6}-\sqrt{2})/4` | With x = sin t, \|x\| ⩽ 1, the equation becomes \|sin t + cos t\|/√2 = cos 2t; it has exactly two real roots. x₁ = −1/√2 checks out (radicand 0, 2x² = 1). The printed x₂ = √2 − √3/2 = 0.5481882 does NOT: 2x² + √((1+2x√(1−x²))/2) − 1 = +0.5800404 ≠ 0. The true second root is (√6 − √2)/4 = sin(π/12) = 0.2588190451, for which 2x√(1−x²) = 1/2, the radicand is 3/4 and the left side is √3/2 + (2−√3)/2 = 1. A sign-change sweep over [−1, 1] returns only −0.7071067812 and 0.2588190451, matching the corrected pair. The engine's complaint is valid. |
 | `dgn.parameters.244` | answer recovered |  | `` | `\pm\sqrt{2}/6, (\sqrt{2} + \sqrt{6})/12` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: §1.1. Simplest equations and inequalities with a parameter; answer block on page 124) |
 | `dgn.parameters.245` | answer recovered |  | `` | `x_1 = \cos(\pi/9), x_2 = \cos(2\pi/7), x_3 = \cos(\pi/3) = 1/2` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: §1.1. Simplest equations and inequalities with a parameter; answer block on page 124) |
 | `dgn.parameters.246` | answer recovered |  | `` | `-10` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: §1.1. Simplest equations and inequalities with a parameter; answer block on page 124) |
@@ -867,15 +867,7 @@ answer.
 | `dgn.parameters.275` | answer recovered |  | `` | `a > 3/82` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: §1.2. Simplest problems with a modulus; answer block on page 139) |
 | `dgn.parameters.276` | answer recovered |  | `` | `\|a\| \leqslant 1` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: §1.2. Simplest problems with a modulus; answer block on page 139) |
 | `dgn.parameters.277` | answer recovered |  | `` | `1) a \in [-\sqrt{26} - 1; \sqrt{26} + 1], 2) a \in [-\sqrt{26} + 1; \sqrt{26} - 1]` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: §1.2. Simplest problems with a modulus; answer block on page 139) |
-| `dgn.parameters.278` | mispaired answer |  | `Continuation of a previous variant's solutions (Problems 6–7) and answers to problems 1–7, followed by a new header.
-
-Problem 6. The bisector $CD$ of angle $ACB$ at the base of an isosceles triangle $ABC$ ($AB = AC$) divides the side $AB$ in such a way that $AD = BC$. Find the length of the bisector $CD$ and the area of triangle $ABC$, if its base $BC$ is equal to $2$.
-
-Problem 7. For which integers $a$ does the inequality
-$$2\log_{1/2} a - 3 + 2x\log_{1/2} a - x^2 < 0$$
-hold for every value of $x$?
-
-Answers. 1. $0, \pm 1, \pm 2, \pm 3$. 2. $240$. 3. $(-3/4; 0) \cup (3; +\infty)$. 4. $2\pi$. 5. $\pi/2 + 2\pi n, \ n \in \mathbb{Z}$. 6. $l = 2, \ S = \operatorname{tg}(2\pi/5) = \sqrt{5 + 2\sqrt{5}}$. 7. $1, 2, 3, 4, 5, 6, 7$.` | `b = -1` | the answer was taken from the answer block of the neighbouring exam variant printed on the same page; the page was read again and the printed answer of this problem is published instead (variant: §1.2. Simplest problems with a modulus; answer block on page 139) |
+| `dgn.parameters.278` | mispaired answer |  | `Continuation of a previous variant's solutions (Problems 6–7) and answers to problems 1–7, followed by a new header. Problem 6. The bisector $CD$ of angle $ACB$ at the base of an isosceles triangle $ABC$ ($AB = AC$) divides the side $AB$ in such a way that $AD = BC$. Find the length of the bisector $CD$ and the area of triangle $ABC$, if its base $BC$ is equal to $2$. Problem 7. For which integers $a$ does the inequality $$2\log_{1/2} a - 3 + 2x\log_{1/2} a - x^2 < 0$$ hold for every value of $x$? Answers. 1. $0, \pm 1, \pm 2, \pm 3$. 2. $240$. 3. $(-3/4; 0) \cup (3; +\infty)$. 4. $2\pi$. 5. $\pi/2 + 2\pi n, \ n \in \mathbb{Z}$. 6. $l = 2, \ S = \operatorname{tg}(2\pi/5) = \sqrt{5 + 2\sqrt{5}}$. 7. $1, 2, 3, 4, 5, 6, 7$.` | `b = -1` | the answer was taken from the answer block of the neighbouring exam variant printed on the same page; the page was read again and the printed answer of this problem is published instead (variant: §1.2. Simplest problems with a modulus; answer block on page 139) |
 | `dgn.parameters.279` | answer recovered |  | `` | `a \geqslant 0` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: §1.2. Simplest problems with a modulus; answer block on page 139) |
 | `dgn.parameters.280` | answer recovered |  | `` | `a \in (15/2; 8) \cup (12; +\infty)` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: §1.2. Simplest problems with a modulus; answer block on page 139) |
 | `dgn.parameters.281` | answer recovered |  | `` | `a = 4/3` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: §1.2. Simplest problems with a modulus; answer block on page 139) |
@@ -926,7 +918,7 @@ Answers. 1. $0, \pm 1, \pm 2, \pm 3$. 2. $240$. 3. $(-3/4; 0) \cup (3; +\infty)$
 | `dgn.parameters.2003 Mechanics` | mispaired answer |  | `1. 24. 2. x = 3, y = z = 1; no. 3. n = 0, \,-5. 4. The minor arc AB (without the endpoints) of the circle tangent to the sides of angle \angle AOB at the points A and B; 8\pi/(3\sqrt{3}).` | `1. 24; 2. x = 3, y = z = 1; no; 3. n = 0,-5; 4. The minor arc AB (without endpoints) of the circle tangent to the sides of angle AOB at points A and B; 8\pi/(3\sqrt{3})` | the answer was taken from the answer block of the neighbouring exam variant printed on the same page; the page was read again and the printed answer of this problem is published instead (variant: Variant 2003, Mechanics and Mathematics Faculty, oral exam problems.; answer block on page 156) |
 | `dgn.parameters.p157.1` | mispaired answer |  | `$a \in (44 - 24\sqrt{2}; 44 + 24\sqrt{2})$` | `\{-1; \pm \sqrt[6]{(\sqrt{17} - 1)/2}\}` | the answer was taken from the answer block of the neighbouring exam variant printed on the same page; the page was read again and the printed answer of this problem is published instead (variant: Variant 2003 (April), Faculty of Computational Mathematics and Cybernetics, 2.; answer block on page 158) |
 | `dgn.parameters.p157.2` | mispaired answer |  | `$\angle A = \angle C = \alpha, \angle B = \pi - 2\alpha$, where $\alpha = 2\arcsin \dfrac{\sqrt{8k^2+1}-1}{4k}; \ k > 0$` | `\angle A = \alpha, \angle B = \frac{\pi}{2} - \alpha, where \alpha = 2\operatorname{arctg}\frac{1}{n\sqrt{2 + 1}}; n > 0` | the answer was taken from the answer block of the neighbouring exam variant printed on the same page; the page was read again and the printed answer of this problem is published instead (variant: Variant 2003 (April), Faculty of Computational Mathematics and Cybernetics, 2.; answer block on page 158) |
-| `dgn.parameters.p157.3` | wrong answer |  | `page 158, item 3 of the 'Ответы' block for ВАРИАНТ 2003 (апрель), факультет вычислительной математики и кибернетики, 2.: {π} ∪ {3π} ∪ [π/2; 3π/4) ∪ [5π/2; 11π/4 − arcsin((1−√2)/√2)]  (printed problem, page 157: 'Задача 3. Найдите все решения системы неравенств √(−cos x) ⩽ √(sin³x + cos²x), π/4 < \|x − π\| ⩽ 2π.')` | `{-\pi} \cup \{3\pi\} \cup [\pi/2; 3\pi/4) \cup [5\pi/2; 11\pi/4 - \arcsin((1-\sqrt{2})/\sqrt{2})]` | The printed system is satisfied at x = −π (constraint \|x−π\| = 2π is allowed, and the first inequality is an equality: sin³(−π)+cos²(−π) = 1 = −cos(−π)), but NOT at x = π, where the constraint π/4 < \|x−π\| fails (\|π−π\| = 0). The book printed {π} where the answer key must print {−π}; every other element of the printed answer is correct. With F(x) = sin³x + cos²x + cos x: F(8.936631158693137) = 3.3e−16 (the printed endpoint 11π/4 − arcsin((1−√2)/√2)), F(π/2) = 1, F(5π/2) = 1, F(3π) = 0, F(−π) = F(π) = 0; 3π/4 is open because \|3π/4−π\| = π/4 is excluded by the strict constraint, not by the inequality. The engine's complaint is false: the radicand printed on the page is sin³x + cos²x (only this reading reproduces the printed endpoints 5π/2 and 8.9366; the engine's sin(x³) reading gives F = −1.756 at the printed root), and for the printed reading F(1.9635) = +0.5523 > 0, i.e. x = 1.9635 satisfies the inequality and in fact lies inside the printed interval [π/2; 3π/4). |
+| `dgn.parameters.p157.3` | wrong answer |  | `page 158, item 3 of the 'Ответы' block for ВАРИАНТ 2003 (апрель), факультет вычислительной математики и кибер��етики, 2.: {π} ∪ {3π} ∪ [π/2; 3π/4) ∪ [5π/2; 11π/4 − arcsin((1−√2)/√2)] (printed problem, page 157: 'Задача 3. Найдите все решения системы неравенств √(−cos x) ⩽ √(sin³x + cos²x), π/4 < \|x − π\| ⩽ 2π.')` | `{-\pi} \cup \{3\pi\} \cup [\pi/2; 3\pi/4) \cup [5\pi/2; 11\pi/4 - \arcsin((1-\sqrt{2})/\sqrt{2})]` | The printed system is satisfied at x = −π (constraint \|x−π\| = 2π is allowed, and the first inequality is an equality: sin³(−π)+cos²(−π) = 1 = −cos(−π)), but NOT at x = π, where the constraint π/4 < \|x−π\| fails (\|π−π\| = 0). The source printed {π} where the answer key must print {−π}; every other element of the printed answer is correct. With F(x) = sin³x + cos²x + cos x: F(8.936631158693137) = 3.3e−16 (the printed endpoint 11π/4 − arcsin((1−√2)/√2)), F(π/2) = 1, F(5π/2) = 1, F(3π) = 0, F(−π) = F(π) = 0; 3π/4 is open because \|3π/4−π\| = π/4 is excluded by the strict constraint, not by the inequality. The engine's complaint is false: the radicand printed on the page is sin³x + cos²x (only this reading reproduces the printed endpoints 5π/2 and 8.9366; the engine's sin(x³) reading gives F = −1.756 at the printed root), and for the printed reading F(1.9635) = +0.5523 > 0, i.e. x = 1.9635 satisfies the inequality and in fact lies inside the printed interval [π/2; 3π/4). |
 | `dgn.parameters.p157.5` | answer recovered |  | `` | `a \in (44 - 24\sqrt{2}; 44 + 24\sqrt{2})` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2003 (April), Faculty of Computational Mathematics and Cybernetics, 1.; answer block on page 157) |
 | `dgn.parameters.p158.2003.1` | answer recovered |  | `` | `(1; -1) \cup [2; 3)` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2003 (July), Faculty of Computational Mathematics and Cybernetics (Specialist Division), 1.; answer block on page 159) |
 | `dgn.parameters.p158.2003.2` | answer recovered |  | `` | `\{\pm\frac{\pi}{4} + 2\pi k, k \in \mathbb{Z}\}` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2003 (July), Faculty of Computational Mathematics and Cybernetics (Specialist Division), 1.; answer block on page 159) |
@@ -935,16 +927,16 @@ Answers. 1. $0, \pm 1, \pm 2, \pm 3$. 2. $240$. 3. $(-3/4; 0) \cup (3; +\infty)$
 | `dgn.parameters.p159.2` | answer recovered |  | `` | `\{\pi/8 + 2\pi k; 5\pi/8 + 2\pi n; k, n \in \mathbb{Z}\}` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2003 (July), Faculty of Computational Mathematics and Cybernetics (Specialist Division), 2.; answer block on page 160) |
 | `dgn.parameters.p159.3` | answer recovered |  | `` | `(-\infty; -2] \cup (1/2; 1]` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2003 (July), Faculty of Computational Mathematics and Cybernetics (Specialist Division), 2.; answer block on page 160) |
 | `dgn.parameters.p159.5` | mispaired answer |  | `$\alpha\in\left(\dfrac{3}{2};+\infty\right)$` | `\alpha \in (\frac{3}{2}; +\infty)` | the answer was taken from the answer block of the neighbouring exam variant printed on the same page; the page was read again and the printed answer of this problem is published instead (variant: Variant 2003 (July), Faculty of Computational Mathematics and Cybernetics (Specialist Division), 1.; answer block on page 159) |
-| `dgn.parameters.p160.1` | wrong answer |  | `page 161, item 1 of the 'Ответы' block for ВАРИАНТ 2003 (июль), отделение бакалавров, 1.: [−8/7; 0]  (printed problem, page 160: 'Задача 1. Решите неравенство 3\|x + 2\| − 4\|x + 1\| ⩽ 2.')` | `(-\infty; -8/7] \cup [0; +\infty)` | sympy: solve(3\|x+2\|-4\|x+1\| <= 2, x) returns (0 <= x) \| (x <= -8/7), i.e. (-inf; -8/7] u [0; +inf). The printed answer [-8/7; 0] is exactly the solution set of the REVERSED inequality (sympy: solve(3\|x+2\|-4\|x+1\| >= 2, x) = (-8/7 <= x) & (x <= 0)), so the book's answer key is inconsistent with the printed sign. The engine's probe is legitimate: x=-12/11 lies inside [-8/7;0] and 3\|x+2\|-4\|x+1\|-2 = 4/11 = 0.3636 > 0, so it violates '⩽ 2'. |
-| `dgn.parameters.p160.2` | wrong answer |  | `page 161, item 2 of the 'Ответы' block for ВАРИАНТ 2003 (июль), факультет вычислитель��ой математики и кибернетики (отделение бакалавров), 1.: (−4/3; −1) ∪ [0; 1/2)  (printed problem, page 160: 'Задача 2. Решите неравенство log₍(1−2x)/3₎ (12/(3x+4)) ⩽ −1.')` | `(-1; 0]` | Domain: base (1−2x)/3 > 0 and ≠ 1, argument 12/(3x+4) > 0, i.e. x ∈ (−4/3, 1/2), x ≠ −1. For x < −1 the base is > 1 (log increasing): log ⩽ −1 ⟺ 12/(3x+4) ⩽ 3/(1−2x) ⟺ 12−24x ⩽ 9x+12 ⟺ x ⩾ 0, impossible in (−4/3, −1) — the whole branch is empty. For −1 < x < 1/2 the base is in (0,1) (log decreasing): log ⩽ −1 ⟺ 12/(3x+4) ⩾ 3/(1−2x) ⟺ x ⩽ 0, giving (−1; 0]. The printed answer (−4/3; −1) ∪ [0; 1/2) is exactly the solution set of the REVERSED inequality '⩾ −1', so the key contradicts the printed problem. The engine's probe is legitimate: x = −1.31667 lies inside the printed (−4/3; −1) and gives log₍(1−2x)/3₎(12/(3x+4)) = 28.6146 > −1, violating '⩽ −1'. |
+| `dgn.parameters.p160.1` | wrong answer |  | `page 161, item 1 of the 'Ответы' block for ВАРИАНТ 2003 (июль), отделение бакалавров, 1.: [−8/7; 0] (printed problem, page 160: 'Задача 1. Решите неравенство 3\|x + 2\| − 4\|x + 1\| ⩽ 2.')` | `(-\infty; -8/7] \cup [0; +\infty)` | sympy: solve(3\|x+2\|-4\|x+1\| <= 2, x) returns (0 <= x) \| (x <= -8/7), i.e. (-inf; -8/7] u [0; +inf). The printed answer [-8/7; 0] is exactly the solution set of the REVERSED inequality (sympy: solve(3\|x+2\|-4\|x+1\| >= 2, x) = (-8/7 <= x) & (x <= 0)), so the source's answer key is inconsistent with the printed sign. The engine's probe is legitimate: x=-12/11 lies inside [-8/7;0] and 3\|x+2\|-4\|x+1\|-2 = 4/11 = 0.3636 > 0, so it violates '⩽ 2'. |
+| `dgn.parameters.p160.2` | wrong answer |  | `page 161, item 2 of the 'Ответы' block for ВАРИАНТ 2003 (июль), факультет вычислительной математики и кибернетики (отделение бакалавров), 1.: (−4/3; −1) ∪ [0; 1/2) (printed problem, page 160: 'Задача 2. Решите неравенство log₍(1−2x)/3₎ (12/(3x+4)) ⩽ −1.')` | `(-1; 0]` | Domain: base (1−2x)/3 > 0 and ≠ 1, argument 12/(3x+4) > 0, i.e. x ∈ (−4/3, 1/2), x ≠ −1. For x < −1 the base is > 1 (log increasing): log ⩽ −1 ⟺ 12/(3x+4) ⩽ 3/(1−2x) ⟺ 12−24x ⩽ 9x+12 ⟺ x ⩾ 0, impossible in (−4/3, −1) — the whole branch is empty. For −1 < x < 1/2 the base is in (0,1) (log decreasing): log ⩽ −1 ⟺ 12/(3x+4) ⩾ 3/(1−2x) ⟺ x ⩽ 0, giving (−1; 0]. The printed answer (−4/3; −1) ∪ [0; 1/2) is exactly the solution set of the REVERSED inequality '⩾ −1', so the key contradicts the printed problem. The engine's probe is legitimate: x = −1.31667 lies inside the printed (−4/3; −1) and gives log₍(1−2x)/3₎(12/(3x+4)) = 28.6146 > −1, violating '⩽ −1'. |
 | `dgn.parameters.p160.3` | answer recovered |  | `` | `\{5\pi/4 + 2\pi k; 7\pi/4 + 2\pi n; k, n \in \mathbb{Z}\}` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2003 (July), Faculty of Computational Mathematics and Cybernetics (Bachelor's Division), 1.; answer block on page 161) |
 | `dgn.parameters.p160.4` | answer recovered |  | `` | `9\sqrt{3}` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2003 (July), Faculty of Computational Mathematics and Cybernetics (Specialist Division), 2.; answer block on page 160) |
 | `dgn.parameters.p160.5` | answer recovered |  | `` | `\beta \in (2/3; +\infty)` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2003 (July), Faculty of Computational Mathematics and Cybernetics (Specialist Division), 2.; answer block on page 160) |
 | `dgn.parameters.p160.6` | answer recovered |  | `` | `5\pi/12` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2003 (July), Faculty of Computational Mathematics and Cybernetics (Specialist Division), 2.; answer block on page 160) |
 | `dgn.parameters.320` | answer recovered |  | `` | `6\sqrt{3}` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2003 (July), Faculty of Computational Mathematics and Cybernetics (Bachelor's Division), 1.; answer block on page 161) |
 | `dgn.parameters.321` | mispaired answer |  | `$\frac{5\pi}{12}$` | `5\pi/12` | the answer was taken from the answer block of the neighbouring exam variant printed on the same page; the page was read again and the printed answer of this problem is published instead (variant: Variant 2003 (July), Faculty of Computational Mathematics and Cybernetics (Bachelor's Division), 1.; answer block on page 161) |
-| `dgn.parameters.p161.322` | wrong answer |  | `Item 1 of the «Ответы» block on p. 162: «1. [1; 17/7].»` | `(-\infty;\, -\frac{53}{3}] \cup [\frac{73}{7};\, +\infty)` | The printed problem (p. 161, confirmed at 600 dpi: «5\|x − 2\| − 2\|x − 31\| ⩾ 1») has solution set x ≤ −53/3 or x ≥ 73/7, so the printed key «[1; 17/7]» does not answer it (at x=1: 5·\|1−2\| − 2·\|1−31\| = 5−60 = −55 < 1, i.e. F−1 = −56, exactly the engine's number). The printed key is the solution of the older-looking statement 5\|x−2\| − 2\|x−3\| ⩽ 1, so the book is internally inconsistent; the answer that matches the printed problem is (−∞;−53/3] ∪ [73/7;+∞). |
-| `dgn.parameters.p161.323` | wrong answer |  | `page 162, item 2 of the 'Ответы' block for ВАРИАНТ 2003 (июль), отделение бакалавров, 2.: (5/2; 8/3) ∪ [6; +∞)  (printed problem, page 161: 'Задача 2. Решите неравенство log₍₄/(3x−4)₎ (2/(2x−5)) ⩽ 1.')` | `(8/3; 6]` | Domain: x > 5/2 (argument > 0), x ≠ 8/3 (base = 1). For x in (5/2, 8/3) the base is > 1, so log ≤ 1 requires 2/(2x-5) ≤ 4/(3x-4), i.e. x ≥ 6 — empty; for x > 8/3 the base is < 1, so log ≤ 1 requires 2/(2x-5) ≥ 4/(3x-4), i.e. x ≤ 6. Hence the solution is (8/3; 6] (at x = 6 the argument equals the base, log = 1, included). The printed answer (5/2; 8/3) ∪ [6; +∞) is exactly the solution set of the reversed inequality '⩾ 1', so the key does not match the printed problem. |
+| `dgn.parameters.p161.322` | wrong answer |  | `Item 1 of the «Ответы» block on p. 162: «1. [1; 17/7].»` | `(-\infty;\, -\frac{53}{3}] \cup [\frac{73}{7};\, +\infty)` | The printed problem (p. 161, confirmed at 600 dpi: «5\|x − 2\| − 2\|x − 31\| ⩾ 1») has solution set x ≤ −53/3 or x ≥ 73/7, so the printed key «[1; 17/7]» does not answer it (at x=1: 5·\|1−2\| − 2·\|1−31\| = 5−60 = −55 < 1, i.e. F−1 = −56, exactly the engine's number). The printed key is the solution of the older-looking statement 5\|x−2\| − 2\|x−3\| ⩽ 1, so the source is internally inconsistent; the answer that matches the printed problem is (−∞;−53/3] ∪ [73/7;+∞). |
+| `dgn.parameters.p161.323` | wrong answer |  | `page 162, item 2 of the 'Ответы' block for ВАРИАНТ 2003 (июль), отделение бакалавров, 2.: (5/2; 8/3) ∪ [6; +∞) (printed problem, page 161: 'Задача 2. Решите неравенство log₍₄/(3x−4)₎ (2/(2x−5)) ⩽ 1.')` | `(8/3; 6]` | Domain: x > 5/2 (argument > 0), x ≠ 8/3 (base = 1). For x in (5/2, 8/3) the base is > 1, so log ≤ 1 requires 2/(2x-5) ≤ 4/(3x-4), i.e. x ≥ 6 — empty; for x > 8/3 the base is < 1, so log ≤ 1 requires 2/(2x-5) ≥ 4/(3x-4), i.e. x ≤ 6. Hence the solution is (8/3; 6] (at x = 6 the argument equals the base, log = 1, included). The printed answer (5/2; 8/3) ∪ [6; +∞) is exactly the solution set of the reversed inequality '⩾ 1', so the key does not match the printed problem. |
 | `dgn.parameters.p161.324` | answer recovered |  | `` | `\{\frac{\pi}{8} + 2\pi k; \frac{13\pi}{8} + 2\pi n; k, n \in \mathbb{Z}\}` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2003 (July), Faculty of Computational Mathematics and Cybernetics (Bachelor's Division), 2.; answer block on page 162) |
 | `dgn.parameters.p161.325` | answer recovered |  | `` | `If d \leqslant 0, then x = \log_{2/7}(2 + \sqrt{4 - d}); if 0 < d < 4, then x = \log_{2/7}(2 \pm \sqrt{4 - d}); if d = 4, then x = \log_{2/7} 2; if d > 4, then there are no solutions` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2003 (July), Faculty of Computational Mathematics and Cybernetics (Bachelor's Division), 2.; answer block on page 162) |
 | `dgn.parameters.p161.326` | answer recovered |  | `` | `(-\infty; -1) \cup (1; 3]` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2003 (July), Faculty of Computational Mathematics and Cybernetics (Bachelor's Division), 2.; answer block on page 162) |
@@ -1032,7 +1024,7 @@ Answers. 1. $0, \pm 1, \pm 2, \pm 3$. 2. $240$. 3. $(-3/4; 0) \cup (3; +\infty)$
 | `dgn.parameters.183 b` | answer recovered |  | `` | `1. x = 1; 2. \pi/6 + \pi n/2, n \in \mathbb{Z}; 3. 41; 4. (2k; 7k; 3k), k \in \mathbb{Z}` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2003 (May), Faculty of Soil Science, 2.; answer block on page 184) |
 | `dgn.parameters.p184.1` | answer recovered |  | `` | `\pi/4 + \pi n/2, n \in \mathbb{Z}` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2003 (July), Faculty of Soil Science, 1.; answer block on page 184) |
 | `dgn.parameters.p184.2` | answer recovered |  | `` | `(0; -1); (1; 0)` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2003 (July), Faculty of Soil Science, 1.; answer block on page 184) |
-| `dgn.parameters.p184.3` | wrong answer |  | `page 184, item 3 of the SECOND 'Ответы' block (the one following ВАРИАНТ 2003 (июль), факультет почвоведения, 1.): 3. (−∞; 2/3] ∪ {0} ∪ [2/3; +∞)  (printed problem, page 184: 'Задача 3. Решите неравенство 3x²/2 − \|x\| ⩾ 0.')` | `(-\infty; -2/3] \cup \{0\} \cup [2/3; +\infty)` | Putting u = \|x\| ≥ 0 the inequality 3u²/2 − u ≥ 0 becomes u(3u/2 − 1) ≥ 0, i.e. u = 0 or u ≥ 2/3 (sympy: roots of 6u²+u−5 are −1 and 5/6), so x = 0 or \|x\| ≥ 2/3, i.e. (−∞; −2/3] ∪ {0} ∪ [2/3; +∞). The book prints '−∞; 2/3]' — the minus sign of the first endpoint is missing on the page — so the printed answer wrongly includes e.g. x = −1/3, where 3x²/2 − \|x\| = −1/6 < 0 (exactly the engine's counterexample). |
+| `dgn.parameters.p184.3` | wrong answer |  | `page 184, item 3 of the SECOND 'Ответы' block (the one following ВАРИАНТ 2003 (июль), факультет почвоведения, 1.): 3. (−∞; 2/3] ∪ {0} ∪ [2/3; +∞) (printed problem, page 184: 'Задача 3. Решите неравенство 3x²/2 − \|x\| ⩾ 0.')` | `(-\infty; -2/3] \cup \{0\} \cup [2/3; +\infty)` | Putting u = \|x\| ≥ 0 the inequality 3u²/2 − u ≥ 0 becomes u(3u/2 − 1) ≥ 0, i.e. u = 0 or u ≥ 2/3 (sympy: roots of 6u²+u−5 are −1 and 5/6), so x = 0 or \|x\| ≥ 2/3, i.e. (−∞; −2/3] ∪ {0} ∪ [2/3; +∞). The source prints '−∞; 2/3]' — the minus sign of the first endpoint is missing on the page — so the printed answer wrongly includes e.g. x = −1/3, where 3x²/2 − \|x\| = −1/6 < 0 (exactly the engine's counterexample). |
 | `dgn.parameters.p184.4` | answer recovered |  | `` | `4\sqrt{2} - 3\sqrt{3}` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2003 (July), Faculty of Soil Science, 1.; answer block on page 184) |
 | `dgn.parameters.p184.5` | answer recovered |  | `` | `2\operatorname{arctg}(1/4)` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2003 (July), Faculty of Soil Science, 1.; answer block on page 184) |
 | `dgn.parameters.p184.6` | answer recovered |  | `` | `(-\infty; -6) \cup (-1/3; +\infty)` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2003 (July), Faculty of Soil Science, 1.; answer block on page 184) |
@@ -1190,14 +1182,7 @@ Answers. 1. $0, \pm 1, \pm 2, \pm 3$. 2. $240$. 3. $(-3/4; 0) \cup (3; +\infty)$
 | `dgn.parameters.p221.8` | answer recovered |  | `` | `\operatorname{arctg} \frac{2}{3}` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2004 (March), Physics Faculty, 1.; answer block on page 221) |
 | `dgn.parameters.2004` | mispaired answer |  | `1. $\frac{\pi}{10}\pm \arccos\frac{2}{5}+2\pi n,\ n\in \mathbb{Z}$. 2. $1-\sqrt{2}<x<0,\ 2<x<1+\sqrt{2}$. 3. $-\sqrt{13}<x<-3,\ -2<x<0,\ 0<x<2,\ 3<x<\sqrt{13}$. 4. $\frac{60}{143}$. 5. $x_1=\frac{1}{2},\ y_1=\frac{3}{2};\ x_2=\frac{1}{2}+\log_5 3,\ y_2=\frac{3}{2}-\log_5 3$. 6. $\frac{35}{11}$. 7. If $a=0$, then $x>0$; if $a\ne 0$, then $x=2a+\sqrt{4a^2+1}$. 8. $\arctg 2$.` | `1. \frac{\pi}{10} \pm \arccos \frac{2}{5} + 2\pi n, n \in \mathbb{Z}; 2. 1 - \sqrt{2} < x < 0, 2 < x < 1 + \sqrt{2}; 3. -\sqrt{13} < x < -3, -2 < x < 0, 0 < x < 2, 3 < x < \sqrt{13}; 4. \frac{60}{143}; 5. x_{1} = \frac{1}{2}, y_{1} = \frac{3}{2}; x_{2} = \frac{1}{2} + \log_{5} 3, y_{2} = \frac{3}{2} - \log_{5} 3; 6. \frac{35}{11}; 7. If a = 0, then x > 0; if a \neq 0, then x = 2a + \sqrt{4a^{2} + 1}; 8. \operatorname{arctg} 2` | the answer was taken from the answer block of the neighbouring exam variant printed on the same page; the page was read again and the printed answer of this problem is published instead (variant: Variant 2004 (March), Physics Faculty, 2.; answer block on page 222) |
 | `dgn.parameters.Variant 2004 (iyul), physics Faculty, 1` | answer recovered |  | `` | `1. \frac{\pi}{12} + \frac{\pi n}{6}, \frac{\pi}{2} + \pi n, n \in \mathbb{Z}; 2. -5 < x < -1, 3 < x < 7; 3. -2 < x < -1 + \sqrt{2}; 4. -3 + \sqrt{24}; 5. x = \frac{1 + \sqrt{13}}{2}, y = \frac{-5 + \sqrt{13}}{4}; 6. 96; 7. -\frac{13}{30} \leqslant a < -\frac{9}{30}, \frac{11}{30} < a \leqslant \frac{15}{30}; 8. \frac{\sqrt{14}}{3}` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2004 (July), Physics Faculty, 1.; answer block on page 223) |
-| `dgn.parameters.2004 (iyul), physics Faculty, 2` | mispaired answer |  | `1. $\frac{\pi n}{6}$, $\frac{\pi}{2} + \pi n$, $n \in \mathbb{Z}$.
-2. $-10 < x < -1$, $5 < x < 14$.
-3. $-3 < x < -2 + \sqrt{5}$.
-4. $-2 + \sqrt{20}$.
-5. $x = \frac{1-\sqrt{13}}{6}$, $y = \frac{-5 - \sqrt{13}}{4}$.
-6. $80$.
-7. $-\frac{11}{30} \leqslant a < -\frac{7}{30}$, $\frac{5}{30} < a \leqslant \frac{9}{30}$.
-8. $\frac{1}{2}$.` | `1. \frac{\pi n}{6}, \frac{\pi}{2} + \pi n, n \in \mathbb{Z}; 2. -10 < x < -1, 5 < x < 14; 3. -3 < x < -2 + \sqrt{5}; 4. -2 + \sqrt{20}; 5. x = \frac{1 - \sqrt{13}}{6}, y = \frac{-5 - \sqrt{13}}{4}; 6. 80; 7. -\frac{11}{30} \leqslant a < -\frac{7}{30}, \frac{5}{30} < a \leqslant \frac{9}{30}; 8. \frac{1}{2}` | the answer was taken from the answer block of the neighbouring exam variant printed on the same page; the page was read again and the printed answer of this problem is published instead (variant: Variant 2004 (July), Physics Faculty, 2.; answer block on page 224) |
+| `dgn.parameters.2004 (iyul), physics Faculty, 2` | mispaired answer |  | `1. $\frac{\pi n}{6}$, $\frac{\pi}{2} + \pi n$, $n \in \mathbb{Z}$. 2. $-10 < x < -1$, $5 < x < 14$. 3. $-3 < x < -2 + \sqrt{5}$. 4. $-2 + \sqrt{20}$. 5. $x = \frac{1-\sqrt{13}}{6}$, $y = \frac{-5 - \sqrt{13}}{4}$. 6. $80$. 7. $-\frac{11}{30} \leqslant a < -\frac{7}{30}$, $\frac{5}{30} < a \leqslant \frac{9}{30}$. 8. $\frac{1}{2}$.` | `1. \frac{\pi n}{6}, \frac{\pi}{2} + \pi n, n \in \mathbb{Z}; 2. -10 < x < -1, 5 < x < 14; 3. -3 < x < -2 + \sqrt{5}; 4. -2 + \sqrt{20}; 5. x = \frac{1 - \sqrt{13}}{6}, y = \frac{-5 - \sqrt{13}}{4}; 6. 80; 7. -\frac{11}{30} \leqslant a < -\frac{7}{30}, \frac{5}{30} < a \leqslant \frac{9}{30}; 8. \frac{1}{2}` | the answer was taken from the answer block of the neighbouring exam variant printed on the same page; the page was read again and the printed answer of this problem is published instead (variant: Variant 2004 (July), Physics Faculty, 2.; answer block on page 224) |
 | `dgn.parameters.Variant 2003 (aprel), Faculty Sciences o Materials, 1, Task 1` | mispaired answer |  | `2.5 L` | `2,5 \text{ l` | the answer was taken from the answer block of the neighbouring exam variant printed on the same page; the page was read again and the printed answer of this problem is published instead (variant: Variant 2003 (April), Faculty of Materials Science, 1.; answer block on page 225) |
 | `dgn.parameters.Variant 2003 (aprel), Faculty Sciences o Materials, 1, Task 4` | mispaired answer |  | `$\dfrac{16}{3}$ and $\dfrac{21}{25}$ or $3$ and $\dfrac{112}{75}$` | `\frac{16}{3} \text{ and } \frac{21}{25} \text{ or } 3 \text{ and } \frac{112}{75}` | the answer was taken from the answer block of the neighbouring exam variant printed on the same page; the page was read again and the printed answer of this problem is published instead (variant: Variant 2003 (April), Faculty of Materials Science, 1.; answer block on page 225) |
 | `dgn.parameters.Variant 2003 (aprel), Faculty Sciences o Materials, 1, Task 5` | mispaired answer |  | `$\pi k;\ (-1)^m \arcsin\left(\dfrac{1-\sqrt{5}}{4}\right)+\pi m;\ (-1)^{n+1}\dfrac{\pi}{6}+\pi n,\ k,m,n\in\mathbb{Z}$` | `\pi k; (-1)^{m} \arcsin\left(\frac{1 - \sqrt{5}}{4}\right) + \pi m; (-1)^{n+1}\frac{\pi}{6} + \pi n, k, m, n \in \mathbb{Z}` | the answer was taken from the answer block of the neighbouring exam variant printed on the same page; the page was read again and the printed answer of this problem is published instead (variant: Variant 2003 (April), Faculty of Materials Science, 1.; answer block on page 225) |
@@ -1219,7 +1204,7 @@ Answers. 1. $0, \pm 1, \pm 2, \pm 3$. 2. $240$. 3. $(-3/4; 0) \cup (3; +\infty)$
 | `dgn.parameters.334` | answer recovered |  | `` | `(x; y) = (2a; a), (-2a; a), a \in \mathbb{R}` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2004 (2 July), exam for the winners of the olympiads at the faculties of chemistry, biology, materials science and fundamental medicine, 1.; answer block on page 228) |
 | `dgn.parameters.335` | answer recovered |  | `` | `DC = 2, DC = 4` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2004 (2 July), exam for the winners of the olympiads at the faculties of chemistry, biology, materials science and fundamental medicine, 1.; answer block on page 228) |
 | `dgn.parameters.p228.1` | answer recovered |  | `` | `x = \pi n/3, n \in \mathbb{Z}` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2004 (2 July), exam for the winners of the olympiads at the faculties of chemistry, biology, materials science and fundamental medicine, 2.; answer block on page 229) |
-| `dgn.parameters.p228.2` | wrong answer |  | `page 229, item 2 of the 'Ответы' block for ВАРИАНТ 2004 (2 июля), экзамен для победителей олимпиад на факультетах: химическом, биологическом, наук о материалах и фундаментальной медицины, 2.: x ∈ (1; 7)  (printed problem, page 228: 'Задача 2. Решите неравенство log₂(7−x) − log₂(x+1) > log₍√2₎√3.')` | `x \in (-1; 1)` | log₍√2₎√3 = ln√3/ln√2 = log₂3 = 1.5849625, so the printed inequality is log₂((7−x)/(x+1)) > log₂3 on the domain −1 < x < 7, i.e. (7−x)/(x+1) > 3 ⟺ 7−x > 3x+3 ⟺ x < 1. The solution is therefore (−1; 1). The printed answer (1; 7) is exactly the solution set of the REVERSED inequality '<', so the key contradicts the printed problem. The engine's probe is legitimate and reproduces exactly: F(1.3) = log₂(5.7/2.3) − log₍√2₎√3 = 1.308939 − 1.584963 = −0.275634 (the engine's value), i.e. x = 1.3 violates '>'. |
+| `dgn.parameters.p228.2` | wrong answer |  | `page 229, item 2 of the 'Ответы' block for ВАРИАНТ 2004 (2 июля), экзамен для победителей олимпиад на факультетах: химическом, биологическом, наук о материалах и фундаментальной медицины, 2.: x ∈ (1; 7) (printed problem, page 228: 'Задача 2. Решите неравенство log₂(7−x) − log₂(x+1) > log₍√2₎√3.')` | `x \in (-1; 1)` | log₍√2₎√3 = ln√3/ln√2 = log₂3 = 1.5849625, so the printed inequality is log₂((7−x)/(x+1)) > log₂3 on the domain −1 < x < 7, i.e. (7−x)/(x+1) > 3 ⟺ 7−x > 3x+3 ⟺ x < 1. The solution is therefore (−1; 1). The printed answer (1; 7) is exactly the solution set of the REVERSED inequality '<', so the key contradicts the printed problem. The engine's probe is legitimate and reproduces exactly: F(1.3) = log₂(5.7/2.3) − log₍√2₎√3 = 1.308939 − 1.584963 = −0.275634 (the engine's value), i.e. x = 1.3 violates '>'. |
 | `dgn.parameters.p228.3` | answer recovered |  | `` | `(x; y) = (3b; b), (-3b; b), b \in \mathbb{R}` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2004 (2 July), exam for the winners of the olympiads at the faculties of chemistry, biology, materials science and fundamental medicine, 2.; answer block on page 229) |
 | `dgn.parameters.p228.4` | answer recovered |  | `` | `KN = 1, KN = 5` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2004 (2 July), exam for the winners of the olympiads at the faculties of chemistry, biology, materials science and fundamental medicine, 2.; answer block on page 229) |
 | `dgn.parameters.p228.5` | mispaired answer |  | `$\mathbf{5}.$ $(54;60)$, $(54;63)$, $(55;60)$, $(56;63)$.` | `(48; 54), (48; 56), (49; 56), (50; 55)` | the answer was taken from the answer block of the neighbouring exam variant printed on the same page; the page was read again and the printed answer of this problem is published instead (variant: Variant 2004 (2 July), exam for the winners of the olympiads at the faculties of chemistry, biology, materials science and fundamental medicine, 2.; answer block on page 229) |
@@ -1267,12 +1252,12 @@ Answers. 1. $0, \pm 1, \pm 2, \pm 3$. 2. $240$. 3. $(-3/4; 0) \cup (3; +\infty)$
 | `dgn.parameters.p237.3` | answer recovered |  | `` | `\{-17\} \cup [0; 17]` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2004 (July), Geology Faculty, 2.; answer block on page 238) |
 | `dgn.parameters.p237.4` | answer recovered |  | `` | `\operatorname{arctg} 4; 17` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2004 (July), Geology Faculty, 2.; answer block on page 238) |
 | `dgn.parameters.p237.5` | answer recovered |  | `` | `[3; 4)` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2004 (July), Geology Faculty, 2.; answer block on page 238) |
-| `dgn.parameters.p237.7` | wrong answer |  | `page 237, item 7 of the 'Ответы' block for ВАРИАНТ 2004 (июль), геологический факультет, 1.: { (π/4 + 2πm; 3π/4 + 2πk); (−π/4 + 2πm; π/4 + 2πk); k, m ∈ ℤ }  (printed problem, page 237: 'Задача 7. Решите систему уравнений sin x/cos(x+y) = −1/√2, cos x/cos(x+y) = 1/√2.')` | `\{ (-\pi/4 + 2\pi m;\ \pi/4 + 2\pi k);\ (3\pi/4 + 2\pi m;\ \pi/4 + 2\pi k);\ k, m \in \mathbb{Z} \}` | Adding and subtracting the two equations gives sin x + cos x = 0 (so tan x = −1) and cos x − sin x = √2·cos(x+y), hence cos(x+y) = ±1 and x+y ∈ πℤ; the parity of the branch fixes y ≡ π/4 (mod 2π), while x ≡ −π/4 or 3π/4 (mod 2π). The book's FIRST family (π/4 + 2πm; 3π/4 + 2πk) has the two coordinates swapped: substituting x = π/4, y = 3π/4 gives cos(x+y) = −1, so eq1 = 0 (satisfied) but eq2 = −√2 (violated); the corrected first family is (3π/4 + 2πm; π/4 + 2πk), which satisfies both equations with residual (0, 0). The book's SECOND family (−π/4 + 2πm; π/4 + 2πk) is valid, so the engine's claim that 'no assignment of the source tuple satisfies the system' is false — at x = −π/4, y = π/4 the residuals are (0, 1.1e−16). Verdict is still source_error because the printed key contains one invalid family. |
+| `dgn.parameters.p237.7` | wrong answer |  | `page 237, item 7 of the 'Ответы' block for ВАРИАНТ 2004 (июль), геологический факультет, 1.: { (π/4 + 2πm; 3π/4 + 2πk); (−π/4 + 2πm; π/4 + 2πk); k, m ∈ ℤ } (printed problem, page 237: 'Задача 7. Решите систему уравнений sin x/cos(x+y) = −1/√2, cos x/cos(x+y) = 1/√2.')` | `\{ (-\pi/4 + 2\pi m;\ \pi/4 + 2\pi k);\ (3\pi/4 + 2\pi m;\ \pi/4 + 2\pi k);\ k, m \in \mathbb{Z} \}` | Adding and subtracting the two equations gives sin x + cos x = 0 (so tan x = −1) and cos x − sin x = √2·cos(x+y), hence cos(x+y) = ±1 and x+y ∈ πℤ; the parity of the branch fixes y ≡ π/4 (mod 2π), while x ≡ −π/4 or 3π/4 (mod 2π). The source's FIRST family (π/4 + 2πm; 3π/4 + 2πk) has the two coordinates swapped: substituting x = π/4, y = 3π/4 gives cos(x+y) = −1, so eq1 = 0 (satisfied) but eq2 = −√2 (violated); the corrected first family is (3π/4 + 2πm; π/4 + 2πk), which satisfies both equations with residual (0, 0). The source's SECOND family (−π/4 + 2πm; π/4 + 2πk) is valid, so the engine's claim that 'no assignment of the source tuple satisfies the system' is false — at x = −π/4, y = π/4 the residuals are (0, 1.1e−16). Verdict is still source_error because the printed key contains one invalid family. |
 | `dgn.parameters.p238.1` | answer recovered |  | `` | `(-\infty; -5) \cup (-\frac{9}{7}; +\infty)` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2004 (July), Philology Faculty, 1.; answer block on page 239) |
 | `dgn.parameters.p238.2` | answer recovered |  | `` | `(-1)^{n}\frac{\pi}{18} + \frac{\pi n}{3}, n \in \mathbb{Z}` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2004 (July), Philology Faculty, 1.; answer block on page 239) |
 | `dgn.parameters.p238.3` | answer recovered |  | `` | `-4` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2004 (July), Philology Faculty, 1.; answer block on page 239) |
 | `dgn.parameters.p238.6` | answer recovered |  | `` | `\frac{2}{9}; m = \frac{4}{3}` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2004 (July), Geology Faculty, 2.; answer block on page 238) |
-| `dgn.parameters.p238.7` | wrong answer |  | `page 238, item 7 of the 'Ответы' block: 7. { (5π/6 + 2πm; 2π/3 + 2πk); (−5π/6 + 2πm; π/3 + 2πk); k, m ∈ Z }  (printed problem, page 238: 'Задача 7. Решите систему уравнений  sin x / cos(x−y) = 1/√3,  cos x / cos(x−y) = −1/√3.')` | `\varnothing \text{ (the system has no solutions)}` | With c = cos(x−y) the two printed equations give sin x = c/√3 and cos x = −c/√3, so sin²x + cos²x = 2c²/3 = 1 forces c² = 3/2 > 1 — impossible for a cosine, hence the printed system has no solutions at all. The printed family indeed fails the second equation: for (5π/6, 2π/3) it gives cos x/cos(x−y) = −1 and for (−5π/6, π/3) it gives +1, never −1/√3 (that family would be correct if the second numerator were 'cos y'). The engine's complaint is legitimate. |
+| `dgn.parameters.p238.7` | wrong answer |  | `page 238, item 7 of the 'Ответы' block: 7. { (5π/6 + 2πm; 2π/3 + 2πk); (−5π/6 + 2πm; π/3 + 2πk); k, m ∈ Z } (printed problem, page 238: 'Задача 7. Решите систему уравнений sin x / cos(x−y) = 1/√3, cos x / cos(x−y) = −1/√3.')` | `\varnothing \text{ (the system has no solutions)}` | With c = cos(x−y) the two printed equations give sin x = c/√3 and cos x = −c/√3, so sin²x + cos²x = 2c²/3 = 1 forces c² = 3/2 > 1 — impossible for a cosine, hence the printed system has no solutions at all. The printed family indeed fails the second equation: for (5π/6, 2π/3) it gives cos x/cos(x−y) = −1 and for (−5π/6, π/3) it gives +1, never −1/√3 (that family would be correct if the second numerator were 'cos y'). The engine's complaint is legitimate. |
 | `dgn.parameters.p238.8` | answer recovered |  | `` | `2` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2004 (July), Geology Faculty, 2.; answer block on page 238) |
 | `dgn.parameters.p239.Task 1` | answer recovered |  | `` | `\left(-\frac{1}{2}; \frac{11}{6}\right)` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2004 (July), Philology Faculty, 2.; answer block on page 240) |
 | `dgn.parameters.p239.Task 2` | answer recovered |  | `` | `\pm \frac{\pi}{12} + \frac{\pi n}{2}, n \in \mathbb{Z}` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2004 (July), Philology Faculty, 2.; answer block on page 240) |
@@ -1307,7 +1292,7 @@ Answers. 1. $0, \pm 1, \pm 2, \pm 3$. 2. $240$. 3. $(-3/4; 0) \cup (3; +\infty)$
 | `dgn.parameters.Variant 2004 (iyul), psychology Faculty, 2. Task 3` | answer recovered |  | `` | `\pi/2 + \pi m, m \in \mathbb{Z}; x = \pi/6 + 2\pi n, n \in \mathbb{Z}; \text{the sum of roots belonging to the segment } [-\pi; \pi] \text{ equals } \pi/6` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2004 (July), Psychology Faculty, 2.; answer block on page 248) |
 | `dgn.parameters.Variant 2004 (iyul), psychology Faculty, 2. Task 4` | answer recovered |  | `` | `\angle KMN = \arcsin(7/8), KL = 784/(33\sqrt{15})` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2004 (July), Psychology Faculty, 2.; answer block on page 248) |
 | `dgn.parameters.Variant 2004 (iyul), psychology Faculty, 2. Task 5` | mispaired answer |  | `1. $x = \log_{5}(5 + \sqrt{28})$. 2. $x \in [-5; 1) \cup (4; 5] \cup (7; +\infty)$. 3. $\pi/2 + \pi m$, $m \in \mathbb{Z}$; $x = \pi/6 + 2\pi n$, $n \in \mathbb{Z}$; the sum of the roots belonging to $[-\pi; \pi]$ equals $\pi/6$. 4. $\angle KMN = \arcsin(7/8)$, $KL = 784/(33\sqrt{15})$. 5. $a = 0$; $a = -4$.` | `a = 0; a = -4` | the answer was taken from the answer block of the neighbouring exam variant printed on the same page; the page was read again and the printed answer of this problem is published instead (variant: Variant 2004 (July), Psychology Faculty, 2.; answer block on page 248) |
-| `dgn.parameters.Variant 2004 (iyul), psychology Faculty, 3. Task 1` | wrong answer |  | `page 249, item 1 of the 'Ответы' block for ВАРИАНТ 2004 (июль), психологический факультет, 3.: x = log₂(6 + √37)  (printed problem, page 248: 'Задача 1. Решите уравнение 16^x − 8 · 4^x − 2 = 0.')` | `x = \log_{4}(4 + 3\sqrt{2})` | With t = 4^x > 0 the equation becomes t^2 - 8t - 2 = 0, whose only positive root is t = 4 + 3√2, hence x = log_4(4 + 3√2) = 1.52155 (residual 16^x - 8·4^x - 2 = 0 exactly). The printed value x = log_2(6 + √37) = 3.59488 leaves a residual of about +20144 in the printed equation, so it is not a root; it would be the root of 4^x - 12·2^x - 1 = 0. The engine's complaint is legitimate. |
+| `dgn.parameters.Variant 2004 (iyul), psychology Faculty, 3. Task 1` | wrong answer |  | `page 249, item 1 of the 'Ответы' block for ВАРИАНТ 2004 (июль), психологический факультет, 3.: x = log₂(6 + √37) (printed problem, page 248: 'Задача 1. Решите уравнение 16^x − 8 · 4^x − 2 = 0.')` | `x = \log_{4}(4 + 3\sqrt{2})` | With t = 4^x > 0 the equation becomes t^2 - 8t - 2 = 0, whose only positive root is t = 4 + 3√2, hence x = log_4(4 + 3√2) = 1.52155 (residual 16^x - 8·4^x - 2 = 0 exactly). The printed value x = log_2(6 + √37) = 3.59488 leaves a residual of about +20144 in the printed equation, so it is not a root; it would be the root of 4^x - 12·2^x - 1 = 0. The engine's complaint is legitimate. |
 | `dgn.parameters.Variant 2004 (iyul), psychology Faculty, 3. Task 2` | wrong answer |  | `Item 2 of the «Ответы» block on p. 249: «2. x ∈ [6; 8).»` | `[-5;\, -1) \cup (3;\, 4] \cup (7;\, +\infty)` | For log_{(x+1)/(2x−6)}((x+6)/4) ⩽ 1 the domain splits into (−6;−1) (base<1), (3;7) (base>1) and (7;+∞) (base<1), and the comparison reduces to (x+5)(x−4) ≤ 0 or ≥ 0 respectively, giving [−5;−1) ∪ (3;4] ∪ (7;+∞) — not [6;8). The engine's check is right (at x=6: log_{7/6}(3) − 1 = 6.1269 > 0); the printed key is stale — the very same key block is printed verbatim for the psychology-faculty-4 variant at the bottom of p. 249, while the faculty-2 key on p. 248 does match its own problem. |
 | `dgn.parameters.Variant 2004 (iyul), psychology Faculty, 3. Task 3` | answer recovered |  | `` | `\pi/2 + 2\pi m, m \in \mathbb{Z}; x = -\pi/3 + (-1)^{n+1}\pi/6 + \pi n, n \in \mathbb{Z}; \text{the sum of roots belonging to the segment } [-\pi; \pi] \text{ equals } 5\pi/6` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2004 (July), Psychology Faculty, 3.; answer block on page 249) |
 | `dgn.parameters.2004 (July)` | mispaired answer |  | `4. $\angle MKN = \arcsin(6/7)$, $ML = 1764/(95\sqrt{13})$` | `1. x = \log_{2}(6 + \sqrt{37}); 2. x \in [6; 8); 3. \pi/2 + 2\pi m, m \in \mathbb{Z}; x = -\pi/3 + (-1)^{n+1}\pi/6 + \pi n, n \in \mathbb{Z}; \text{the sum of the roots belonging to the segment } [-\pi; \pi] \text{ equals } 5\pi/6; 4. \angle MKN = \arcsin(6/7), ML = 1764/(95\sqrt{13}); 5. a = 0; a = 1` | the answer was taken from the answer block of the neighbouring exam variant printed on the same page; the page was read again and the printed answer of this problem is published instead (variant: Variant 2004 (July), Psychology Faculty, 4.; answer block on page 249) |
@@ -1374,7 +1359,7 @@ Answers. 1. $0, \pm 1, \pm 2, \pm 3$. 2. $240$. 3. $(-3/4; 0) \cup (3; +\infty)$
 | `dgn.parameters.p266.Task 3` | mispaired answer |  | `$S_{100} = 15050,\ b_{40} = 81$` | `S_{90} = 16470, a_{35} = 71` | the answer was taken from the answer block of the neighbouring exam variant printed on the same page; the page was read again and the printed answer of this problem is published instead (variant: Variant 2005 (July), Faculty of Computational Mathematics and Cybernetics, 2.; answer block on page 267) |
 | `dgn.parameters.p266.Task 4` | mispaired answer |  | `$4\sqrt{10} - 2\sqrt{15}$` | `R = 2\sqrt{15}` | the answer was taken from the answer block of the neighbouring exam variant printed on the same page; the page was read again and the printed answer of this problem is published instead (variant: Variant 2005 (July), Faculty of Computational Mathematics and Cybernetics, 2.; answer block on page 267) |
 | `dgn.parameters.p267.1` | answer recovered |  | `` | `-150` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2005 (July), Faculty of Computational Mathematics and Cybernetics (Bachelor's Division), 1.; answer block on page 268) |
-| `dgn.parameters.p267.2` | wrong answer |  | `page 268, item 2 of the 'Ответы' block for ВАРИАНТ 2005 (июль), факультет вычислительной математики и кибернетики (отделение бакалавров), 1.: (0; 1)  (printed problem, page 267: 'Задача 2. Решите неравенство log₃((x² + \|2x−5\| + 5)/(x+2)) − \|log₃x − 1\| > log₃x + 1.')` | `\varnothing` | Domain x > 0. Put A(x) = (x² + \|2x−5\| + 5)/(x+2); for 0 < x ⩽ 2.5 this is (x²−2x+10)/(x+2) ∈ (3, 5), and for x > 2.5 it is exactly x. If \|log₃x − 1\| = 1 − log₃x (x < 3), the inequality collapses to log₃ A > 2, i.e. A > 9 — impossible since A < 5. If x > 3, \|log₃x − 1\| = log₃x − 1 and the inequality becomes A > x², which for x > 2.5 means x > x², impossible. Hence the solution set is EMPTY and the printed '(0; 1)' is wrong (it looks like the result of mishandling \|log₃x − 1\| on (0, 1)). A dense scan over (0, 50) with step 1e−3 finds no point with F > 0. The engine's probe is legitimate: x = 0.05 gives F = −0.566421 (exactly the engine's value) < 0. |
+| `dgn.parameters.p267.2` | wrong answer |  | `page 268, item 2 of the 'Ответы' block for ВАРИАНТ 2005 (июль), факультет вычислительной математики и кибернетики (отделение бакалавров), 1.: (0; 1) (printed problem, page 267: 'Задача 2. Решите неравенство log₃((x² + \|2x−5\| + 5)/(x+2)) − \|log₃x − 1\| > log₃x + 1.')` | `\varnothing` | Domain x > 0. Put A(x) = (x² + \|2x−5\| + 5)/(x+2); for 0 < x ⩽ 2.5 this is (x²−2x+10)/(x+2) ∈ (3, 5), and for x > 2.5 it is exactly x. If \|log₃x − 1\| = 1 − log₃x (x < 3), the inequality collapses to log₃ A > 2, i.e. A > 9 — impossible since A < 5. If x > 3, \|log₃x − 1\| = log₃x − 1 and the inequality becomes A > x², which for x > 2.5 means x > x², impossible. Hence the solution set is EMPTY and the printed '(0; 1)' is wrong (it looks like the result of mishandling \|log₃x − 1\| on (0, 1)). A dense sweep over (0, 50) with step 1e−3 finds no point with F > 0. The engine's probe is legitimate: x = 0.05 gives F = −0.566421 (exactly the engine's value) < 0. |
 | `dgn.parameters.p267.3` | answer recovered |  | `` | `\operatorname{arcctg} 3 + 2\pi n, n \in \mathbb{Z}` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2005 (July), Faculty of Computational Mathematics and Cybernetics (Bachelor's Division), 1.; answer block on page 268) |
 | `dgn.parameters.p267.4` | answer recovered |  | `` | `(x; y) = ((-13 - \sqrt{481})/10; (-13 + \sqrt{481})/10), ((-13 + \sqrt{481})/10; (-13 - \sqrt{481})/10)` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2005 (July), Faculty of Computational Mathematics and Cybernetics (Bachelor's Division), 1.; answer block on page 268) |
 | `dgn.parameters.p267.5` | answer recovered |  | `` | `4\sqrt{14} - 2\sqrt{7}` | the page-image extraction did not attach the printed answer to this problem; the answer block of the printed variant was read again and the printed answer of this problem is published instead (variant: Variant 2005 (July), Faculty of Computational Mathematics and Cybernetics (Bachelor's Division), 1.; answer block on page 268) |
@@ -1487,8 +1472,8 @@ Answers. 1. $0, \pm 1, \pm 2, \pm 3$. 2. $240$. 3. $(-3/4; 0) \cup (3; +\infty)$
 | `dgn.parameters.p296.10` | mispaired answer |  | `$x\in[\pi/4+\pi k;\,3\pi/4+\pi k],\ k\in\mathbb{Z}$` | `x \in [\pi/4 + \pi k; 3\pi/4 + \pi k], k \in \mathbb{Z}` | the answer was taken from the answer block of the neighbouring exam variant printed on the same page; the page was read again and the printed answer of this problem is published instead (variant: Variant 2006 (May), "Lomonosov-2006", 2.; answer block on page 296) |
 | `dgn.syllabus.p668.8` | answer recovered |  | `` | `a \in (-\infty; -7/3] \cup [15 + 4\sqrt{17}; +\infty)` | the exercise carried no answer in the extraction; the answer was recomputed with the symbolic engine and confirmed by an independent computation of the range of the parameter |
 | `dgn.parameters.p280.4` | transcription error |  | `Item 4 of the «Ответы» block on p. 280: «4. (−∞; 1 − √14] ∪ [−√(17/3); −2) ∪ (1; √(17/3)].» (the radical bar covers 17/3)` | `(-\infty;\, 1-\sqrt{14}] \cup [-\sqrt{\frac{17}{3}};\, -2) \cup (1;\, \sqrt{\frac{17}{3}}]` | the printed answer was mis-transcribed during extraction (the radical covers 17/3, not 17/3 with the root outside the fraction) - Our published answer writes \sqrt{17}/3, but the 600 dpi crop of p. 280 shows the radical bar over «17/3», i.e. √(17/3) — our transcription moved the «/3» out of the radical. The printed answer is mathematically right: putting c = √2−1, 3+√8 = c⁻², 3−√8 = c², 17−6√8 = c⁴, the inequality is equivalent to x²+x−2 ≤ (3−x)\|2x+5\| on x<3, x≠−5/2, x<−2 or x>1, i.e. x ≤ 1−√14 (for x<−5/2) or 3x² ≤ 17 (for x>−5/2), giving exactly (−∞;1−√14] ∪ [−√(17/3);−2) ∪ (1;√(17/3)]. The engine's complaint (x = −2.19 missing) is therefore an artefact of our mis-transcription: −2.19 does lie in [−√(17/3);−2). |
-| `dgn.parameters.Task 1 (Variant 2006, aprel)` | wrong answer |  | `page 293, item 1 of the 'Ответы' block for ВАРИАНТ 2006 (апрель): [−2; (7+√21)/21)  (printed problem, page 293: 'Задача 1. Решите неравенство √(x+2) > x−3.')` | `[-2; (7+\sqrt{21})/2)` | √(x+2) > x−3 has domain x ≥ −2. For x ≤ 3 the right side is ≤ 0 while the left side is ≥ 0, so every x ∈ [−2, 3] satisfies it; for x > 3 both sides are positive and squaring gives x+2 > x²−6x+9 ⟺ x²−7x+7 < 0 ⟺ x ∈ ((7−√21)/2, (7+√21)/2), so the whole solution is [−2, (7+√21)/2). The book printed the upper endpoint as (7+√21)/21 = 0.5515512 instead of (7+√21)/2 = 5.7912878 (a typesetting slip in the denominator — the companion variant on the same page prints (5+√21)/2 for its own variant). The engine's complaint is valid: 0.5515512 satisfies the inequality (√(2.5515512) − 0.5515512 = 4.0458 > 0) yet is excluded from the printed set. |
-| `dgn.parameters.p289.2` | wrong answer |  | `page 289, item 2 of the 'Ответы' block for ВАРИАНТ 2005 (июль), московская школа экономики, 1.: (−∞; 2]  (printed problem, page 289: 'Задача 2. Решите неравенство √(x²+2x) − x > 1.')` | `(-\infty; -2]` | Domain x²+2x ≥ 0, i.e. x ⩽ −2 or x ≥ 0. On x ⩽ −2 the left side is √(x²+2x) − x ⩾ −x ⩾ 2 > 1, so the whole branch x ⩽ −2 satisfies the inequality (at x = −2 the value is 1 + 2 = 3). On x ≥ 0 both sides are non-negative and squaring gives x²+2x > x²+2x+1, i.e. 0 > 1 — impossible, so that branch is empty. The solution is (−∞; −2]; the printed answer (−∞; 2] has the sign of the endpoint wrong. The engine's probe is legitimate and reproduces exactly: F(2) = √8 − 2 − 1 = −0.17157288 (the engine's value) < 0, so the printed endpoint 2 does not satisfy the printed inequality. |
-| `dgn.parameters.p263.3` | wrong answer |  | `page 263, item 3 of the 'Ответы' block for ВАРИАНТ 2005 (июль), механико-математический факультет, 1.: x ∈ [1; 2) ∪ (2; √5)  (printed problem, page 263: 'Задача 3. Решите неравенство (3 − x − √(5−x²))/(cos((2x−7)/4) − cos((x−5)/4)) ⩾ 0.')` | `x \in [1; 2) \cup (2; \sqrt{5}]` | The domain is x ∈ [−√5, √5] with the denominator non-zero. The numerator N = 3 − x − √(5−x²) has N > 0 for x < 1 and for x > 2, N < 0 for 1 < x < 2 (its zeros are x = 1 and x = 2). The denominator D = cos((2x−7)/4) − cos((x−5)/4) = −2·sin(3(x−4)/8)·sin((x−2)/8) is negative for x < 2, zero at x = 2 and positive for 2 < x ⩽ √5 (D(√5) = 0.0362478 ≠ 0). The quotient is therefore ⩾ 0 exactly on [1; 2) ∪ (2; √5], and √5 must be INCLUDED (N(√5) = 3 − √5 = 0.7639 > 0, D(√5) = 0.0362 > 0, quotient = 21.0753). The printed answer leaves the upper endpoint √5 open, so it omits a genuine solution — the engine's complaint is valid. |
-| `dgn.parameters.p271.2` | wrong answer |  | `page 271, item 2 of the SECOND 'Ответы' block (the one following ВАРИАНТ 2005 (июль), химический факультет, 2.): 2. (−∞; 1/4]  (printed problem, page 271: 'Задача 2. Решите неравенство √5 · 9^x ⩽ √3 · 25^x.')` | `[1/4; +\infty)` | √5·9^x ⩽ √3·25^x is equivalent to 3^(2x−1/2) ≤ 5^(2x−1/2); since 3 < 5 this holds iff 2x − 1/2 ≥ 0, i.e. x ≥ 1/4 (equality at x = 1/4, both sides = √15). The printed answer (−∞; 1/4] is the solution set of the reversed inequality '⩾'. The engine's probe x = −0.75 lies in the printed interval and gives √5·9^(−0.75) − √3·25^(−0.75) = +0.275412 (matching its F) > 0, so the complaint is valid. |
+| `dgn.parameters.Task 1 (Variant 2006, aprel)` | wrong answer |  | `page 293, item 1 of the 'Ответы' block for ВАРИАНТ 2006 (апрель): [−2; (7+√21)/21) (printed problem, page 293: 'Задача 1. Решите неравенство √(x+2) > x−3.')` | `[-2; (7+\sqrt{21})/2)` | √(x+2) > x−3 has domain x ≥ −2. For x ≤ 3 the right side is ≤ 0 while the left side is ≥ 0, so every x ∈ [−2, 3] satisfies it; for x > 3 both sides are positive and squaring gives x+2 > x²−6x+9 ⟺ x²−7x+7 < 0 ⟺ x ∈ ((7−√21)/2, (7+√21)/2), so the whole solution is [−2, (7+√21)/2). The source printed the upper endpoint as (7+√21)/21 = 0.5515512 instead of (7+√21)/2 = 5.7912878 (a typesetting slip in the denominator — the companion variant on the same page prints (5+√21)/2 for its own variant). The engine's complaint is valid: 0.5515512 satisfies the inequality (√(2.5515512) − 0.5515512 = 4.0458 > 0) yet is excluded from the printed set. |
+| `dgn.parameters.p289.2` | wrong answer |  | `page 289, item 2 of the 'Ответы' block for ВАРИАНТ 2005 (июль), московская школа экономики, 1.: (−∞; 2] (printed problem, page 289: 'Задача 2. Решите неравенство √(x²+2x) − x > 1.')` | `(-\infty; -2]` | Domain x²+2x ≥ 0, i.e. x ⩽ −2 or x ≥ 0. On x ⩽ −2 the left side is √(x²+2x) − x ⩾ −x ⩾ 2 > 1, so the whole branch x ⩽ −2 satisfies the inequality (at x = −2 the value is 1 + 2 = 3). On x ≥ 0 both sides are non-negative and squaring gives x²+2x > x²+2x+1, i.e. 0 > 1 — impossible, so that branch is empty. The solution is (−∞; −2]; the printed answer (−∞; 2] has the sign of the endpoint wrong. The engine's probe is legitimate and reproduces exactly: F(2) = √8 − 2 − 1 = −0.17157288 (the engine's value) < 0, so the printed endpoint 2 does not satisfy the printed inequality. |
+| `dgn.parameters.p263.3` | wrong answer |  | `page 263, item 3 of the 'Ответы' block for ВАРИАНТ 2005 (июль), механико-математический факультет, 1.: x ∈ [1; 2) ∪ (2; √5) (printed problem, page 263: 'Задача 3. Решите неравенство (3 − x − √(5−x²))/(cos((2x−7)/4) − cos((x−5)/4)) ⩾ 0.')` | `x \in [1; 2) \cup (2; \sqrt{5}]` | The domain is x ∈ [−√5, √5] with the denominator non-zero. The numerator N = 3 − x − √(5−x²) has N > 0 for x < 1 and for x > 2, N < 0 for 1 < x < 2 (its zeros are x = 1 and x = 2). The denominator D = cos((2x−7)/4) − cos((x−5)/4) = −2·sin(3(x−4)/8)·sin((x−2)/8) is negative for x < 2, zero at x = 2 and positive for 2 < x ⩽ √5 (D(√5) = 0.0362478 ≠ 0). The quotient is therefore ⩾ 0 exactly on [1; 2) ∪ (2; √5], and √5 must be INCLUDED (N(√5) = 3 − √5 = 0.7639 > 0, D(√5) = 0.0362 > 0, quotient = 21.0753). The printed answer leaves the upper endpoint √5 open, so it omits a genuine solution — the engine's complaint is valid. |
+| `dgn.parameters.p271.2` | wrong answer |  | `page 271, item 2 of the SECOND 'Ответы' block (the one following ВАРИАНТ 2005 (июль), химический факультет, 2.): 2. (−∞; 1/4] (printed problem, page 271: 'Задача 2. Решите неравенство √5 · 9^x ⩽ √3 · 25^x.')` | `[1/4; +\infty)` | √5·9^x ⩽ √3·25^x is equivalent to 3^(2x−1/2) ≤ 5^(2x−1/2); since 3 < 5 this holds iff 2x − 1/2 ≥ 0, i.e. x ≥ 1/4 (equality at x = 1/4, both sides = √15). The printed answer (−∞; 1/4] is the solution set of the reversed inequality '⩾'. The engine's probe x = −0.75 lies in the printed interval and gives √5·9^(−0.75) − √3·25^(−0.75) = +0.275412 (matching its F) > 0, so the complaint is valid. |
 | `dgn.parameters.p274.6` | wrong answer |  | `3` | `1 : 3` | the question asks in what ratio the point K divides the diagonal AC, and the printed key gives the single number 3; the diagonals AC and BD meet at K = A + (1/4)(C - A), so AK : KC = 1 : 3, and the same problem of the neighbouring variant is keyed 1 : 3 on the printed page 273 |