Step 1: Understanding the Concept:

The domain of a composite function is the intersection of the domains of its individual parts.

For \(\log_e(g(x))\), \(g(x) > 0\).

For \(\sin^{-1}(h(x))\) and \(\cos^{-1}(k(x))\), \(-1 \leq h(x) \leq 1\) and \(-1 \leq k(x) \leq 1\).


Step 2: Key Formula or Approach:

We need to solve three simultaneous inequalities for \(x\):

1) \(4x^2 + 11x + 6 > 0\)

2) \(-1 \leq 4x + 3 \leq 1\)

3) \(-1 \leq \frac{10x+6}{3} \leq 1\)


Step 3: Detailed Explanation:

Condition 1: \(4x^2 + 8x + 3x + 6 > 0 \Rightarrow (4x+3)(x+2) > 0\).

The solution is \(x \in (-\infty, -2) \cup (-0.75, \infty)\).

Condition 2: \(-4 \leq 4x \leq -2 \Rightarrow -1 \leq x \leq -0.5\).

Condition 3: \(-3 \leq 10x + 6 \leq 3 \Rightarrow -9 \leq 10x \leq -3 \Rightarrow -0.9 \leq x \leq -0.3\).

Taking the intersection of (2) and (3): \(x \in [-0.9, -0.5]\).

Now intersect with (1): \(x \in (-0.75, -0.5]\).

Comparing with \((\alpha, \beta]\), we get \(\alpha = -0.75\) and \(\beta = -0.5\).


Step 4: Final Answer:

We need to calculate \(36|\alpha + \beta|\):
\[ 36|-0.75 - 0.5| = 36|-1.25| = 36 \times \frac{5}{4} = 9 \times 5 = 45. \]
Quick Tip: When finding the domain of the sum of inverse trigonometric functions, always check the overlap of intervals first, as the intersection often simplifies significantly.

Step 1: Understanding the Concept:

Equations involving absolute values must be solved by dividing the real number line into intervals based on the critical points of the modulus functions (\(x=0\) and \(x=-2\)).


Step 2: Detailed Explanation:

Case 1: \(x \geq 0\)
\(x^2 - 5(x+2) + 6 = 0 \Rightarrow x^2 - 5x - 4 = 0\).
\(x = \frac{5 \pm \sqrt{25+16}}{2} = \frac{5 \pm \sqrt{41}}{2}\).

Since \(x \geq 0\), only \(x = \frac{5 + \sqrt{41}}{2}\) is a valid root.


Case 2: \(-2 \leq x < 0\)
\(-x^2 - 5(x+2) + 6 = 0 \Rightarrow -x^2 - 5x - 4 = 0 \Rightarrow x^2 + 5x + 4 = 0\).
\((x+4)(x+1) = 0 \Rightarrow x = -1, -4\).

Only \(x = -1\) lies in \([-2, 0)\).


Case 3: \(x < -2\)
\(-x^2 - 5(-(x+2)) + 6 = 0 \Rightarrow -x^2 + 5x + 10 + 6 = 0 \Rightarrow x^2 - 5x - 16 = 0\).
\(x = \frac{5 \pm \sqrt{25+64}}{2} = \frac{5 \pm \sqrt{89}}{2}\).
\(x = \frac{5 - \sqrt{89}}{2} \approx \frac{5 - 9.43}{2} = -2.215\), which is valid for \(x < -2\).


Step 3: Final Answer:

The real roots are \(\frac{5+\sqrt{41}}{2}, -1,\) and \(\frac{5-\sqrt{89}}{2}\).

The total number of real roots is 3.
Quick Tip: Always check if the calculated value of \(x\) actually falls within the domain of the case you are solving. This prevents counting "false" roots.

Step 1: Understanding the Concept:

Let \(z = 3 + iy\) where \(y \in \mathbb{R}\). Then \(\bar{z} = 3 - iy\).

We substitute these into the expression and find the range of its real part.


Step 2: Detailed Explanation:
\(z - \bar{z} = 2iy\) and \(z\bar{z} = 9 + y^2\).

Numerator \(N = 9 + y^2 + 2iy\).

Denominator \(D = 2 - 3(3+iy) + 5(3-iy) = 2 - 9 - 3iy + 15 - 5iy = 8 - 8iy = 8(1-iy)\).

Expression \(E = \frac{(9+y^2) + 2iy}{8(1-iy)} \times \frac{1+iy}{1+iy} = \frac{(9+y^2) + i(9y+y^3) + 2iy - 2y^2}{8(1+y^2)}\).

Real part \(f(y) = \frac{9-y^2}{8(1+y^2)}\).

Let \(t = y^2\) where \(t \in [0, \infty)\).
\(f(t) = \frac{1}{8} \left( \frac{9-t}{1+t} \right) = \frac{1}{8} \left( \frac{10 - (1+t)}{1+t} \right) = \frac{1}{8} \left( \frac{10}{1+t} - 1 \right)\).

As \(t \to \infty, f(t) \to -1/8\).

At \(t = 0, f(t) = 9/8\).

The range is \((-1/8, 9/8]\), so \(\alpha = -1/8\) and \(\beta = 9/8\).


Step 3: Final Answer:
\(24(\beta - \alpha) = 24 \left( \frac{9}{8} - \left(-\frac{1}{8}\right) \right) = 24 \left( \frac{10}{8} \right) = 30\).
Quick Tip: For rational functions of \(y^2\), substitute \(t = y^2\) and use the properties of basic hyperbolic or reciprocal curves to find the range quickly.

Step 1: Understanding the Concept:

First, find \(m\) and \(n\) using the linear equations. Then use the determinant property: \(\det(k \cdot adj(adj(B))) = k^m (\det B)^{(m-1)^2}\).


Step 2: Detailed Explanation:

Solving linear equations: \(4m+n=22 \Rightarrow n = 22-4m\).
\(17m + 4(22-4m) = 93 \Rightarrow 17m + 88 - 16m = 93 \Rightarrow m = 5\).

Then \(n = 22-20 = 2\).

Determinant of \(A\): \(|A| = m-n = 5-2 = 3\).

Let \(B = mA = 5A\). Since \(A\) is \(5 \times 5\), \(|B| = 5^5 |A| = 5^5 \cdot 3\).

Now, \(\det(n \cdot adj(adj(B))) = 2^5 (|B|)^{(5-1)^2} = 2^5 |B|^{16}\).

Substituting \(|B|\): \(2^5 \cdot (5^5 \cdot 3)^{16} = 2^5 \cdot 5^{80} \cdot 3^{16}\).

Given this is \(3^a 5^b 6^c = 3^a 5^b (2 \cdot 3)^c = 2^c 3^{a+c} 5^b\).

Comparing powers: \(c = 5, b = 80, a+c = 16 \Rightarrow a = 11\).


Step 3: Final Answer:
\(a+b+c = 11 + 80 + 5 = 96\).
Quick Tip: Remember that for a matrix \(A\) of order \(n\), \(\det(k A) = k^n \det(A)\). Forgetting the power \(n\) is a common source of error.

Step 1: Understanding the Concept:

Solve the first three equations to find \(x, y, z\). Since the solution is unique, it must satisfy the fourth equation, allowing us to find \(\lambda\).


Step 2: Detailed Explanation:

From (1) and (2): Subtracting gives \(y + 16z = 2\).

From (1) and (3): Multiply (1) by 2 and subtract (3) to eliminate \(x\):
\((-2x + 4y - 18z) - (-2x + y + 5z) = 14 - 8 \Rightarrow 3y - 23z = 6\).

Solving \(y+16z=2\) and \(3y-23z=6\):
\(3(2-16z) - 23z = 6 \Rightarrow 6 - 48z - 23z = 6 \Rightarrow -71z = 0 \Rightarrow z = 0\).

Then \(y = 2\) and \(x = -3\).

Substitute \((-3, 2, 0)\) in the fourth equation: \(-3(-3) + 2 + 13(0) = \lambda \Rightarrow \lambda = 11\).


Step 3: Calculating Distance:

Distance of \((-3, 2, 0)\) from \(2x - 2y + z - 11 = 0\):
\[ d = \frac{|2(-3) - 2(2) + 0 - 11|}{\sqrt{2^2 + (-2)^2 + 1^2}} = \frac{|-6-4-11|}{3} = \frac{21}{3} = 7. \]


Step 4: Final Answer:

The distance is 7.
Quick Tip: When four equations are given with three variables, the "unique solution" exists only if the determinant of the augmented matrix of any three equations is non-zero and the fourth is consistent.

Step 1: Understanding the Concept:

A number is divisible by 3 if the sum of its digits is divisible by 3.

The available digits and their remainders mod 3 are: \(\{1 (rem 1), 3 (rem 0), 5 (rem 2), 8 (rem 2)\}\).


Step 2: Detailed Explanation:

Combinations of remainders that sum to a multiple of 3:

1) \((0, 0, 0) \Rightarrow \{3, 3, 3\} \rightarrow 1\) number.

2) \((1, 1, 1) \Rightarrow \{1, 1, 1\} \rightarrow 1\) number.

3) \((2, 2, 2) \Rightarrow \{5, 8\} \rightarrow\) each place has 2 choices. \(2^3 = 8\) numbers.

4) \((0, 1, 2) \Rightarrow\) Set \(A = \{3\}\), Set \(B = \{1\}\), Set \(C = \{5, 8\}\).

Using \(\{3, 1, 5\}\): \(3! = 6\) numbers.

Using \(\{3, 1, 8\}\): \(3! = 6\) numbers.


Step 3: Final Answer:

Total numbers = \(1 + 1 + 8 + 6 + 6 = 22\).
Quick Tip: For divisibility by 3, group digits by their remainders (0, 1, and 2). The sum of three remainders must be 0, 3, or 6.

Step 1: Understanding the Concept:

Using the multinomial expansion, the general term is \(\frac{10!}{n_1! n_2! n_3!} a^{n_1} (bx)^{n_2} (cx^2)^{n_3}\) where \(n_1+n_2+n_3 = 10\).


Step 2: Detailed Explanation:

For \(p_1\) (coeff of \(x\)): \(n_2 = 1, n_3 = 0, n_1 = 9\).
\(p_1 = \frac{10!}{9!1!0!} a^9 b^1 = 10 a^9 b = 20 \Rightarrow a^9 b = 2\).

Since \(a, b \in \mathbb{N}\), only \(a = 1, b = 2\) is possible.

For \(p_2\) (coeff of \(x^2\)):

Case 1: \(n_2 = 2, n_3 = 0, n_1 = 8 \Rightarrow \frac{10!}{8!2!0!} a^8 b^2 = 45(1)^8 (2)^2 = 180\).

Case 2: \(n_2 = 0, n_3 = 1, n_1 = 9 \Rightarrow \frac{10!}{9!0!1!} a^9 c^1 = 10(1)^9 c = 10c\).
\(p_2 = 180 + 10c = 210 \Rightarrow 10c = 30 \Rightarrow c = 3\).


Step 3: Final Answer:
\(2(a+b+c) = 2(1+2+3) = 12\).
Quick Tip: In multinomial expansions, terms of low powers like \(x\) or \(x^2\) can often be solved by manually checking the distribution of powers among the variables.

Step 1: Understanding the Concept:

Let the numbers be \(a\) and \(b\). Arithmetic means satisfy \(A_1+A_2 = a+b\).

Geometric means satisfy \(G_1 = a(b/a)^{1/4}, G_2 = \sqrt{ab}, G_3 = b(a/b)^{1/4}\).


Step 2: Detailed Explanation:
\(G_1 G_3 = ab\) and \(G_2 = \sqrt{ab} \Rightarrow G_2^4 = (ab)^2\).
\(G_1^4 = a^3b\) and \(G_3^4 = ab^3\).

Expression \(E = a^3b + (ab)^2 + ab^3 + (ab)^2 = a^3b + 2a^2b^2 + ab^3\).
\(E = ab(a^2 + 2ab + b^2) = ab(a+b)^2\).

Substitute \(ab = G_1 G_3\) and \(a+b = A_1 + A_2\):
\(E = G_1 G_3 (A_1 + A_2)^2\).


Step 3: Final Answer:

The expression equals \((A_1 + A_2)^2 G_1 G_3\).
Quick Tip: Properties of means are powerful: \(\sum AM = n(\frac{a+b}{2})\) and \(\prod GM = (ab)^{n/2}\). Use these to eliminate intermediate variables.

Step 1: Understanding the Concept:

Break the function into intervals based on the greatest integer function.


Step 2: Detailed Explanation:

Case 1: \(x \in [0, 1) \Rightarrow [x]=0\).
\(f(x) = \max\{1+x, 2+x, x\} = 2+x\).

Case 2: \(x \in [1, 2) \Rightarrow [x]=1\).
\(f(x) = \max\{2+x, 2+x, x+2\} = 2+x\).

Case 3: \(x=2 \Rightarrow [x]=2\).
\(f(2) = \max\{5, 4, 6\} = 6\).

Function is \(2+x\) for \(x \in [0, 2)\) and \(f(2)=6\).

Continuity at \(x=1\): \(\lim_{x \to 1} f(x) = 3 = f(1)\). Continuous.

Continuity at \(x=2\): \(\lim_{x \to 2^-} f(x) = 4 \neq f(2) = 6\). Discontinuous at \(x=2\).

So, \(m=1\).

In \((0, 2)\), \(f(x)=2+x\), which is differentiable everywhere. So, \(n=0\).


Step 3: Final Answer:
\((m+n)^2 + 2 = (1+0)^2 + 2 = 3\).
Quick Tip: Always simplify the "max" or "min" function expressions in intervals where \([x]\) is constant. Often, one branch dominates the others entirely.

Step 1: Understanding the Concept:

Use the "King's property": \(\int_a^b f(x) dx = \int_a^b f(a+b-x) dx\).


Step 2: Detailed Explanation:

Replace \(x\) with \(1-x\). The quadratic \(5+2x-2x^2\) remains unchanged.

The exponential becomes \(e^{2-4(1-x)} = e^{4x-2} = e^{-(2-4x)}\).
\(2I = \int_0^1 \frac{1}{5+2x-2x^2} \left( \frac{1}{1+e^{2-4x}} + \frac{1}{1+e^{-(2-4x)}} \right) dx = \int_0^1 \frac{1}{5+2x-2x^2} dx\).
\(2I = \frac{1}{2} \int_0^1 \frac{1}{2.5 + x - x^2} dx = \frac{1}{2} \int_0^1 \frac{1}{11/4 - (x-1/2)^2} dx\).
\(2I = \frac{1}{2} \left[ \frac{1}{2 \cdot \frac{\sqrt{11}}{2}} \ln \left| \frac{\sqrt{11}/2 + x - 1/2}{\sqrt{11}/2 - x + 1/2} \right| \right]_0^1 = \frac{1}{2\sqrt{11}} \cdot 2 \ln \left( \frac{\sqrt{11}+1}{\sqrt{11}-1} \right)\).
\(I = \frac{1}{2\sqrt{11}} \ln \left( \frac{(\sqrt{11}+1)^2}{10} \right) = \frac{1}{\sqrt{11}} \ln \left( \frac{\sqrt{11}+1}{\sqrt{10}} \right)\).
\(\alpha = \sqrt{11}, \beta = \sqrt{10}\).


Step 3: Final Answer:
\(\alpha^4 - \beta^4 = 11^2 - 10^2 = 121 - 100 = 21\).
Quick Tip: The expression \(\frac{1}{1+e^{g(x)}}\) combined with a symmetric function often reduces to \(\frac{1}{2}\) when the King's property is applied if \(g(x) = -g(a+b-x)\).

Step 1: Understanding the Concept:

Rearrange to linear form: \(\frac{dx}{dy} + P(y)x = Q(y)\).


Step 2: Detailed Explanation:
\(\frac{dx}{dy} + \frac{x}{2(y+2)\ln(y+2)} = \frac{2\ln(y+2)-4}{2(y+2)\ln(y+2)}\).

Integrating factor \(IF = e^{\int \frac{1}{2(y+2)\ln(y+2)} dy} = \sqrt{\ln(y+2)}\).
\(x \cdot \sqrt{\ln(y+2)} = \int \frac{\ln(y+2)-2}{(y+2)\ln(y+2)} \cdot \sqrt{\ln(y+2)} dy\).

Let \(t = \ln(y+2) \Rightarrow x\sqrt{t} = \int \frac{t-2}{t} \sqrt{t} dt = \int (t^{1/2} - 2t^{-1/2}) dt\).
\(x\sqrt{t} = \frac{2}{3}t^{3/2} - 4t^{1/2} + C \Rightarrow x = \frac{2}{3}t - 4 + \frac{C}{\sqrt{t}}\).

Using \(x(e^4-2)=1 \Rightarrow t=4\): \(1 = \frac{8}{3} - 4 + \frac{C}{2} \Rightarrow C = 14/3\).

For \(y = e^9-2 \Rightarrow t=9\): \(x = \frac{2}{3}(9) - 4 + \frac{14/3}{3} = 6 - 4 + 14/9 = 32/9\).


Step 3: Final Answer:

The value is 32/9.
Quick Tip: When a differential equation is hard to solve in terms of \(y=f(x)\), try \(x=f(y)\). It often converts the equation into a standard linear form.

Step 1: Understanding the Concept:

Find the intersection of altitudes.


Step 2: Detailed Explanation:

Slope of \(BC = \frac{3}{5} \Rightarrow\) Altitude from \(A\) is \(y+7 = -\frac{5}{3}(x-3) \Rightarrow 5x+3y = -6\).

Slope of \(AC = \frac{12}{1} = 12 \Rightarrow\) Altitude from \(B\) is \(y-2 = -\frac{1}{12}(x+1) \Rightarrow x+12y = 23\).

Solving gives \(y = \frac{121}{57}\) and \(x = -\frac{141}{57}\).

Expression \(E = 9(-\frac{141}{57}) - 6(\frac{121}{57}) + 60 = \frac{-1269 - 726}{57} + 60\).
\(E = -35 + 60 = 25\).


Step 3: Final Answer:

The value is 25.
Quick Tip: Orthocenter \((\alpha, \beta)\) satisfies \(\sum \cos A \cdot x_A\). Alternatively, if slopes are clean, find two altitude lines and intersect.

Step 1: Understanding the Concept:

Common tangents depend on the distance \(d\) between centers and radii \(r_1, r_2\).


Step 2: Detailed Explanation:

Circle 1: \(C_1(9, 7.5), r_1 = \sqrt{81+56.25-131} = 2.5\).

Circle 2: \(C_2(3, 3), r_2 = \sqrt{9+9+7} = 5\).
\(d = \sqrt{6^2 + 4.5^2} = \sqrt{36+20.25} = \sqrt{56.25} = 7.5\).
\(r_1 + r_2 = 2.5 + 5 = 7.5\).

Since \(d = r_1+r_2\), the circles touch externally.


Step 3: Final Answer:

The number of common tangents is 3.
Quick Tip: \(d = r_1+r_2 \rightarrow 3\) tangents. \(d > r_1+r_2 \rightarrow 4\). \(|r_1-r_2| < d < r_1+r_2 \rightarrow 2\). \(d = |r_1-r_2| \rightarrow 1\).

Step 1: Understanding the Concept:

The problem requires finding the equation of a plane passing through three given points. Once the plane equation is established, we find the foot of the perpendicular from point \(P\) to this plane. Finally, we calculate the Euclidean distance of the foot \(Q\) from the origin \((0, 0, 0)\).


Step 2: Key Formula or Approach:

1. The equation of a plane passing through \((x_1, y_1, z_1)\), \((x_2, y_2, z_2)\), and \((x_3, y_3, z_3)\) is: \[ \begin{vmatrix} x - x_1 & y - y_1 & z - z_1
x_2 - x_1 & y_2 - y_1 & z_2 - z_1
x_3 - x_1 & y_3 - y_1 & z_3 - z_1 \end{vmatrix} = 0 \]
2. The foot of the perpendicular \((\alpha, \beta, \gamma)\) from point \((x_1, y_1, z_1)\) to the plane \(ax + by + cz + d = 0\) is given by: \[ \frac{\alpha - x_1}{a} = \frac{\beta - y_1}{b} = \frac{\gamma - z_1}{c} = -\frac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2} \]

Step 3: Detailed Explanation:

1. Finding the Plane Equation:

Let the points be \(A(-1, -2, -3)\), \(B(9, 3, 4)\), and \(C(9, -2, 1)\).

Vector \(\vec{AB} = (9 - (-1), 3 - (-2), 4 - (-3)) = (10, 5, 7)\).

Vector \(\vec{AC} = (9 - (-1), -2 - (-2), 1 - (-3)) = (10, 0, 4)\).

The normal vector \(\vec{n}\) is the cross product of \(\vec{AB}\) and \(\vec{AC}\): \[ \vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
10 & 5 & 7
10 & 0 & 4 \end{vmatrix} = \hat{i}(20 - 0) - \hat{j}(40 - 70) + \hat{k}(0 - 50) = 20\hat{i} + 30\hat{j} - 50\hat{k} \]
Dividing by 10, we get the simplified normal vector \(\vec{n} = (2, 3, -5)\).

Equation of the plane: \(2(x + 1) + 3(y + 2) - 5(z + 3) = 0 \Rightarrow 2x + 3y - 5z - 7 = 0\).


2. Finding the Foot of Perpendicular \(Q(\alpha, \beta, \gamma)\):

From \(P(3, -2, -9)\) to the plane \(2x + 3y - 5z - 7 = 0\): \[ \frac{\alpha - 3}{2} = \frac{\beta + 2}{3} = \frac{\gamma + 9}{-5} = -\frac{2(3) + 3(-2) - 5(-9) - 7}{2^2 + 3^2 + (-5)^2} \] \[ = -\frac{6 - 6 + 45 - 7}{4 + 9 + 25} = -\frac{38}{38} = -1 \]
Calculating coordinates: \(\alpha - 3 = -2 \Rightarrow \alpha = 1\).
\(\beta + 2 = -3 \Rightarrow \beta = -5\).
\(\gamma + 9 = 5 \Rightarrow \gamma = -4\).

Thus, \(Q = (1, -5, -4)\).


3. Distance from Origin:
\[ OQ = \sqrt{1^2 + (-5)^2 + (-4)^2} = \sqrt{1 + 25 + 16} = \sqrt{42} \]


Step 4: Final Answer:

The distance of point \(Q\) from the origin is \(\sqrt{42}\).
Quick Tip: To find the foot of the perpendicular, always use the formula approach instead of finding the intersection of a line and a plane manually; it significantly reduces the margin for error in calculations.

Step 1: Understanding the Concept:

The shortest distance (SD) between two skew lines \(\vec{r} = \vec{a_1} + t\vec{b_1}\) and \(\vec{r} = \vec{a_2} + s\vec{b_2}\) is the projection of the vector joining the points on the lines onto the vector perpendicular to both lines.


Step 2: Key Formula or Approach:
The shortest distance \(d\) is given by: \[ d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} \]

Step 3: Detailed Explanation:

Line 1: Passes through \(A(\lambda, 3, -6)\) with direction \(\vec{b_1} = (0, 4, 1)\).

Line 2: Passes through \(B(-\lambda, 0, 6)\) with direction \(\vec{b_2} = (3, -4, 0)\).

1. Compute \(\vec{b_1} \times \vec{b_2}\):
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
0 & 4 & 1
3 & -4 & 0 \end{vmatrix} = \hat{i}(0 - (-4)) - \hat{j}(0 - 3) + \hat{k}(0 - 12) = 4\hat{i} + 3\hat{j} - 12\hat{k} \]
The magnitude \(|\vec{b_1} \times \vec{b_2}| = \sqrt{4^2 + 3^2 + (-12)^2} = \sqrt{16 + 9 + 144} = \sqrt{169} = 13\).


2. Compute \((\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})\):
\(\vec{a_2} - \vec{a_1} = (-\lambda - \lambda, 0 - 3, 6 - (-6)) = (-2\lambda, -3, 12)\).

Dot product: \((-2\lambda)(4) + (-3)(3) + (12)(-12) = -8\lambda - 9 - 144 = -8\lambda - 153\).


3. Solve for \(\lambda\):

Shortest distance \(d = \frac{|-8\lambda - 153|}{13} = 13\).
\(|8\lambda + 153| = 169\).

Case 1: \(8\lambda + 153 = 169 \Rightarrow 8\lambda = 16 \Rightarrow \lambda_1 = 2\).

Case 2: \(8\lambda + 153 = -169 \Rightarrow 8\lambda = -322 \Rightarrow \lambda_2 = -\frac{322}{8}\).


4. Finding the required value:

The set \(S = \{2, -\frac{322}{8}\}\).

Required value = \(8 \left| \lambda_1 + \lambda_2 \right| = 8 \left| 2 - \frac{322}{8} \right| = \left| 16 - 322 \right| = 306\).


Step 4: Final Answer:

The result is 306.
Quick Tip: For lines in symmetric form \(\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}\), always write out the point vector \(\vec{a}\) and direction vector \(\vec{b}\) explicitly to avoid sign errors when calculating \(\vec{a_2} - \vec{a_1}\).

Step 1: Understanding the Concept:

Three vectors are coplanar if their scalar triple product is zero, which is equivalent to the determinant of their components being zero.

Step 2: Key Formula or Approach:

For vectors \(\vec{v_1}, \vec{v_2}, \vec{v_3}\) to be coplanar: \[ \begin{vmatrix} a_1 & b_1 & c_1
a_2 & b_2 & c_2
a_3 & b_3 & c_3 \end{vmatrix} = 0 \]

Step 3: Detailed Explanation:

Given vectors: \(\vec{v_1} = (\lambda, -1, 1)\), \(\vec{v_2} = (1, 2, \mu)\), and \(\vec{v_3} = (3, -4, 5)\).

Determinant condition: \[ \begin{vmatrix} \lambda & -1 & 1
1 & 2 & \mu
3 & -4 & 5 \end{vmatrix} = 0 \] \[ \lambda(10 + 4\mu) + 1(5 - 3\mu) + 1(-4 - 6) = 0 \] \[ 10\lambda + 4\lambda\mu + 5 - 3\mu - 10 = 0 \] \[ 10\lambda + 4\lambda\mu - 3\mu - 5 = 0 \quad --- (Eq. 1) \]
Substitute \(\lambda = \mu + 5\) (from \(\lambda - \mu = 5\)) into (Eq. 1): \[ 10(\mu + 5) + 4(\mu + 5)\mu - 3\mu - 5 = 0 \] \[ 10\mu + 50 + 4\mu^2 + 20\mu - 3\mu - 5 = 0 \] \[ 4\mu^2 + 27\mu + 45 = 0 \]
Factoring the quadratic: \[ 4\mu^2 + 12\mu + 15\mu + 45 = 0 \] \[ 4\mu(\mu + 3) + 15(\mu + 3) = 0 \Rightarrow (\mu + 3)(4\mu + 15) = 0 \]
The values for \(\mu\) are \(\mu_1 = -3\) and \(\mu_2 = -\frac{15}{4}\).

Corresponding values for \(\lambda\):
If \(\mu_1 = -3 \Rightarrow \lambda_1 = -3 + 5 = 2\).
If \(\mu_2 = -\frac{15}{4} \Rightarrow \lambda_2 = -\frac{15}{4} + \frac{20}{4} = \frac{5}{4}\).


Now, compute \(\sum 80(\lambda^2 + \mu^2)\):
For \((2, -3)\): \(80(2^2 + (-3)^2) = 80(4 + 9) = 80(13) = 1040\).

For \((\frac{5}{4}, -\frac{15}{4})\): \(80((\frac{5}{4})^2 + (-\frac{15}{4})^2) = 80(\frac{25}{16} + \frac{225}{16}) = 80(\frac{250}{16}) = 5 \times 250 = 1250\).

Total sum: \(1040 + 1250 = 2290\).


Step 4: Final Answer:

The value is 2290.
Quick Tip: When a quadratic equation appears in a vector problem, factor it immediately to avoid working with fractional roots as much as possible until the final summation step.

Step 1: Understanding the Concept:

This problem is solved using position vectors. Let the position vectors of vertices \(A, B, C, D\) be \(\vec{a}, \vec{b}, \vec{c}, \vec{d}\) respectively. We express all given vectors in terms of these position vectors.


Step 2: Key Formula or Approach:

Position vector of the midpoint of a segment with endpoints \(\vec{u}\) and \(\vec{v}\) is \(\frac{\vec{u} + \vec{v}}{2}\).
Vector \(\vec{XY}\) in terms of position vectors is \(\vec{y} - \vec{x}\).


Step 3: Detailed Explanation:

Midpoint of \(AC\) is \(E \Rightarrow \vec{e} = \frac{\vec{a} + \vec{c}}{2}\).

Midpoint of \(BD\) is \(F \Rightarrow \vec{f} = \frac{\vec{b} + \vec{d}}{2}\).

Then, \(\vec{FE} = \vec{e} - \vec{f} = \frac{\vec{a} + \vec{c} - \vec{b} - \vec{d}}{2}\).


Expanding the LHS of the given equation: \((\vec{AB} - \vec{BC}) + (\vec{AD} - \vec{DC}) = [(\vec{b} - \vec{a}) - (\vec{c} - \vec{b})] + [(\vec{d} - \vec{a}) - (\vec{c} - \vec{d})]\)
\(= (\vec{b} - \vec{a} - \vec{c} + \vec{b}) + (\vec{d} - \vec{a} - \vec{c} + \vec{d})\)
\(= 2\vec{b} - \vec{a} - \vec{c} + 2\vec{d} - \vec{a} - \vec{c}\)
\(= 2\vec{b} + 2\vec{d} - 2\vec{a} - 2\vec{c}\)
\(= -2 (2\vec{a} + 2\vec{c} - 2\vec{b} - 2\vec{d}) / 2\)

Actually, regrouping: \(2\vec{b} + 2\vec{d} - 2\vec{a} - 2\vec{c} = -2 (\vec{a} + \vec{c} - \vec{b} - \vec{d})\).


Comparing with \(k \vec{FE} = k \left( \frac{\vec{a} + \vec{c} - \vec{b} - \vec{d}}{2} \right)\): \(-2 (\vec{a} + \vec{c} - \vec{b} - \vec{d}) = \frac{k}{2} (\vec{a} + \vec{c} - \vec{b} - \vec{d})\).

Thus, \(\frac{k}{2} = -2 \Rightarrow k = -4\).


Step 4: Final Answer:

The value of \(k\) is -4.
Quick Tip: In vector geometry problems involving midpoints, setting the origin at one of the vertices or using position vectors directly is almost always the most efficient route.

Step 1: Understanding the Concept:

This is a problem of total probability. Let \(D\) be the outcome of the die roll, where \(D \in \{1, 2, 3, 4, 5, 6\}\). Each outcome has probability \(P(D=k) = \frac{1}{6}\). If \(k\) balls are drawn, we need them all to be white.


Step 2: Key Formula or Approach:

Total Probability \(P(W) = \sum_{k=1}^{6} P(D=k) \times P(all k balls are white)\).
The probability of drawing \(k\) white balls out of a total of \(N\) balls (where \(W\) are white) is \(\frac{\binom{W}{k}}{\binom{N}{k}}\).

Step 3: Detailed Explanation:

Total balls = 10, White balls = 6. \(P(W) = \frac{1}{6} \left[ \frac{\binom{6}{1}}{\binom{10}{1}} + \frac{\binom{6}{2}}{\binom{10}{2}} + \frac{\binom{6}{3}}{\binom{10}{3}} + \frac{\binom{6}{4}}{\binom{10}{4}} + \frac{\binom{6}{5}}{\binom{10}{5}} + \frac{\binom{6}{6}}{\binom{10}{6}} \right]\)
Calculating each term:
1. \(\frac{6}{10}\)
2. \(\frac{15}{45} = \frac{1}{3}\)
3. \(\frac{20}{120} = \frac{1}{6}\)
4. \(\frac{15}{210} = \frac{1}{14}\)
5. \(\frac{6}{252} = \frac{1}{42}\)
6. \(\frac{1}{210}\)

Summing inside the bracket (common denominator 210): \(\frac{126}{210} + \frac{70}{210} + \frac{35}{210} + \frac{15}{210} + \frac{5}{210} + \frac{1}{210} = \frac{252}{210}\)

Simplifying \(\frac{252}{210} = \frac{126}{105} = \frac{42}{35} = \frac{6}{5}\).

Final probability = \(\frac{1}{6} \times \frac{6}{5} = \frac{1}{5}\).


Step 4: Final Answer:

The probability is \(\frac{1}{5}\).
Quick Tip: When dealing with multiple cases in total probability, check if a common denominator simplifies the summation of fractions before reducing each individual term.

Step 1: Understanding the Concept:

Variance is defined as \(\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2\). To find the correct variance, we must correct both the sum of observations (\(\sum x_i\)) to get the correct mean, and the sum of squares of observations (\(\sum x_i^2\)).


Step 2: Key Formula or Approach:

1. \(\sum x_{new} = \sum x_{old} - x_{wrong} + x_{correct}\)

2. \(\sum x_{new}^2 = \sum x_{old}^2 - x_{wrong}^2 + x_{correct}^2\)


Step 3: Detailed Explanation:

Given \(n = 10, \bar{x}_{old} = 20, \sigma_{old} = 8\).
Wrong value = 50, Correct value = 40.

1. Correcting the Mean:
\(\sum x_{old} = n \bar{x}_{old} = 10 \times 20 = 200\).

Correct \(\sum x = 200 - 50 + 40 = 190\).

Correct mean \(\bar{x}_{new} = \frac{190}{10} = 19\).


2. Correcting the Sum of Squares:

Variance \(\sigma_{old}^2 = 8^2 = 64\).
\(64 = \frac{\sum x_{old}^2}{10} - (20)^2 \Rightarrow 64 = \frac{\sum x_{old}^2}{10} - 400 \Rightarrow \frac{\sum x_{old}^2}{10} = 464 \Rightarrow \sum x_{old}^2 = 4640\).

Correct \(\sum x^2 = 4640 - 50^2 + 40^2 = 4640 - 2500 + 1600 = 3740\).


3. Calculating Correct Variance:

Correct variance \(\sigma_{new}^2 = \frac{3740}{10} - (19)^2 = 374 - 361 = 13\).


Step 4: Final Answer:

The correct variance is 13.
Quick Tip: Always find the sum of squares first before correcting the mean in such problems to maintain a clear workflow. A small error in correcting the mean can propagate through the whole calculation.

Step 1: Understanding the Concept:

This problem requires simplifying a logical expression using Boolean algebra laws (like De Morgan's Law, Distributive Law, etc.) and then finding its negation.


Step 2: Key Formula or Approach:

1. De Morgan's Law: \(\sim(a \wedge b) = \sim a \vee \sim b\).

2. Negation of a contradiction is a tautology.


Step 3: Detailed Explanation:

Let the expression be \(E = p \wedge (q \wedge \sim(p \wedge q))\).

Simplify inside:
\(q \wedge \sim(p \wedge q) = q \wedge (\sim p \vee \sim q)\) (by De Morgan's Law).

Using Distributive Law: \((q \wedge \sim p) \vee (q \wedge \sim q)\).

Since \((q \wedge \sim q)\) is False (\(F\)): \((q \wedge \sim p) \vee F = q \wedge \sim p\).


Now substitute back into \(E\):
\(E = p \wedge (q \wedge \sim p) = (p \wedge \sim p) \wedge q\) (by Associative and Commutative Laws).

Since \((p \wedge \sim p)\) is False (\(F\)): \(F \wedge q = F\).

The entire original expression is a contradiction (always false).


Step 4: Finding Negation:

Negation of \(E\) is \(\sim F = T\) (a tautology).

Now, check which option is a tautology:

(A) \(\sim(p \vee q)\) is not a tautology.

(B) \(p \vee q\) is not a tautology.

(C) \((\sim(p \wedge q)) \wedge q = (\sim p \vee \sim q) \wedge q = \sim p \wedge q\) (not a tautology).

(D) \((\sim(p \wedge q)) \vee p = (\sim p \vee \sim q) \vee p = (\sim p \vee p) \vee \sim q = T \vee \sim q = T\).

Option (D) is a tautology.

Step 5: Final Answer:

The negation is \((\sim(p \wedge q)) \vee p\).
Quick Tip: If an expression simplifies to \(F\) (False), its negation must be \(T\) (True). Evaluating which option is a tautology is often faster than performing long symbolic manipulations on the options.

Step 1: Understanding the Concept:

A relation \(R\) on \(A \times A\) consists of ordered pairs of ordered pairs.

We need to find the number of quadruplets \((a, b, c, d)\) where each variable is in \(\{1, 2, 3, 4\}\) and satisfy the linear equation \(2a + 3b = 4c + 5d\).


Step 2: Key Formula or Approach:

Since the set \(A\) is small, we can list all possible values for the left-hand side (LHS) \(2a + 3b\) and the right-hand side (RHS) \(4c + 5d\) and identify matches.


Step 3: Detailed Explanation:

Let \(X = 2a + 3b\) for \(a, b \in \{1, 2, 3, 4\}\).

Possible values for \(X\):

For \(a=1: \{5, 8, 11, 14\}\)

For \(a=2: \{7, 10, 13, 16\}\)

For \(a=3: \{9, 12, 15, 18\}\)

For \(a=4: \{11, 14, 17, 20\}\)

Combined set of LHS values with counts: \(\{5(1), 7(1), 8(1), 9(1), 10(1), 11(2), 12(1), 13(1), 14(2), 15(1), 16(1), 17(1), 18(1), 20(1)\}\).


Let \(Y = 4c + 5d\) for \(c, d \in \{1, 2, 3, 4\}\).

Possible values for \(Y\):

For \(c=1: \{9, 14, 19, 24\}\)

For \(c=2: \{13, 18, 23, 28\}\)

For \(c=3: \{17, 22, 27, 32\}\)

For \(c=4: \{21, 26, 31, 36\}\)


Identifying common values where \(X = Y\):

1. \(X=Y=9\): LHS has \((3,1)\), RHS has \((1,1)\). (1 pair)

2. \(X=Y=13\): LHS has \((2,3)\), RHS has \((2,1)\). (1 pair)

3. \(X=Y=14\): LHS has \((1,4), (4,2)\), RHS has \((1,2)\). (2 pairs)

4. \(X=Y=17\): LHS has \((4,3)\), RHS has \((3,1)\). (1 pair)

5. \(X=Y=18\): LHS has \((3,4)\), RHS has \((2,2)\). (1 pair)


Step 4: Final Answer:

Total number of elements in \(R = 1 + 1 + 2 + 1 + 1 = 6\).
Quick Tip: For finite set relation problems, constructing a small frequency table for the values of the expressions on both sides is the most reliable way to ensure no cases are missed.

Step 1: Understanding the Concept:

Let the pin be \((x, y, z, w)\).

Constraints: \(x, y, z, w \in \{0, 1, 2, 3, 4, 5, 6, 7\}\) (distinct digits), \(\max(x, y, z, w) = 7\), and \(x+y = z+w\).


Step 2: Key Formula or Approach:

We need to find sets of four distinct digits containing 7 that can be partitioned into two pairs with equal sums.


Step 3: Detailed Explanation:

One pair must contain 7. Let the sum be \(S\).

Since digits are distinct and the max is 7, the possible other digit in the pair with 7 is \(k \in \{0, 1, 2, 3, 4\}\).

1. \(S = 7+0 = 7\): Other pairs from \(\{1,2,3,4,5,6\}\) are \(\{1,6\}, \{2,5\}, \{3,4\}\). (3 sets)

2. \(S = 7+1 = 8\): Other pairs from \(\{0,2,3,4,5,6\}\) are \(\{2,6\}, \{3,5\}\). (2 sets)

3. \(S = 7+2 = 9\): Other pairs from \(\{0,1,3,4,5,6\}\) are \(\{3,6\}, \{4,5\}\). (2 sets)

4. \(S = 7+3 = 10\): Other pairs from \(\{0,1,2,4,5,6\}\) is \(\{4,6\}\). (1 set)

5. \(S = 7+4 = 11\): Other pairs from \(\{0,1,2,3,5,6\}\) is \(\{5,6\}\). (1 set)


Total unique sets of digits \(= 3 + 2 + 2 + 1 + 1 = 9\).

For each set \(\{a, b, c, d\}\) where \(a+b = c+d\):

- The pair \(\{a, b\}\) can be at positions \((1,2)\) or \((3,4)\) (\(2\) ways).

- Digits within the first pair can be swapped (\(2\) ways).

- Digits within the second pair can be swapped (\(2\) ways).

Total arrangements per set \(= 2 \times 2 \times 2 = 8\).


Step 4: Final Answer:

Total number of trials \(= 9 \times 8 = 72\).
Quick Tip: Partitioning the problem by the specific "sum" value ensures you don't double count sets. Remember that once the set of 4 digits is chosen, the "equal sum" constraint limits permutations compared to a standard distinct digit problem.

Step 1: Understanding the Concept:

The condition \(3^n - 3\) is a multiple of 7 means \(3^n \equiv 3 \pmod 7\).

We need to find the pattern of powers of 3 modulo 7.


Step 2: Key Formula or Approach:

Calculate the powers of 3 mod 7 until the cycle repeats.


Step 3: Detailed Explanation:
\(3^1 \equiv 3 \pmod 7\)
\(3^2 \equiv 9 \equiv 2 \pmod 7\)
\(3^3 \equiv 6 \pmod 7\)
\(3^4 \equiv 18 \equiv 4 \pmod 7\)
\(3^5 \equiv 12 \equiv 5 \pmod 7\)
\(3^6 \equiv 15 \equiv 1 \pmod 7\)

The cycle length is 6.

The condition \(3^n \equiv 3 \pmod 7\) holds if \(n \equiv 1 \pmod 6\).

We need \(10 \leq n \leq 100\) and \(n = 6k + 1\).
\(10 \leq 6k + 1 \leq 100\)
\(9 \leq 6k \leq 99\)
\(1.5 \leq k \leq 16.5\).

Since \(k\) must be an integer, \(k \in \{2, 3, 4, \dots, 16\}\).


Step 4: Final Answer:

Number of values for \(k = 16 - 2 + 1 = 15\).
Quick Tip: Fermat's Little Theorem states \(a^{p-1} \equiv 1 \pmod p\). For \(p=7\), \(3^6 \equiv 1 \pmod 7\), so the cycle will always be a divisor of 6.

Step 1: Understanding the Concept:

The \(n\)-th bracket is a geometric series sum of the form \(a^n - a^{n-1}b + a^{n-2}b^2 \dots \pm b^n\).

Specifically, the \(n\)-th term \(T_n = \sum_{r=0}^n \left(\frac{1}{2}\right)^{n-r} \left(-\frac{1}{3}\right)^r\).


Step 2: Key Formula or Approach:

Use the formula for the sum of a finite geometric progression: \(\sum_{r=0}^n x^{n-r}y^r = \frac{x^{n+1} - y^{n+1}}{x - y}\).


Step 3: Detailed Explanation:

Here \(x = \frac{1}{2}\) and \(y = -\frac{1}{3}\).
\(T_n = \frac{(1/2)^{n+1} - (-1/3)^{n+1}}{1/2 - (-1/3)} = \frac{6}{5} \left[ \left(\frac{1}{2}\right)^{n+1} - \left(-\frac{1}{3}\right)^{n+1} \right]\).

Total sum \(S = \sum_{n=1}^\infty T_n = \frac{6}{5} \left[ \sum_{n=1}^\infty \left(\frac{1}{2}\right)^{n+1} - \sum_{n=1}^\infty \left(-\frac{1}{3}\right)^{n+1} \right]\).

Sum of first infinite GP: \(\frac{1/4}{1 - 1/2} = \frac{1}{2}\).

Sum of second infinite GP: \(\frac{1/9}{1 - (-1/3)} = \frac{1/9}{4/3} = \frac{1}{12}\).
\(S = \frac{6}{5} \left[ \frac{1}{2} - \frac{1}{12} \right] = \frac{6}{5} \left[ \frac{5}{12} \right] = \frac{1}{2}\).
\(\alpha = 1, \beta = 2\).


Step 4: Final Answer:
\(\alpha + 3\beta = 1 + 3(2) = 7\).
Quick Tip: The identity \(x^n + x^{n-1}y + \dots + y^n = \frac{x^{n+1}-y^{n+1}}{x-y}\) is a common shortcut in series problems to convert a complex sum into two separate geometric progressions.

Step 1: Understanding the Concept:

Perimeter \(P = AB + BC + AC\). Since \(A\) and \(B\) are fixed, we need to find values of \(t\) that minimize and maximize \(f(t) = BC + AC\).
\(A = (2, 1), B = (0, 0), C = (t, 4)\).


Step 2: Key Formula or Approach:
\(f(t) = \sqrt{t^2 + 16} + \sqrt{(t-2)^2 + 9}\).

For minimum, use the reflection principle: find the shortest path from \(B\) to \(A\) via the line \(y=4\).


Step 3: Detailed Explanation:

Minimization (\(\beta\)):

Reflect \(B(0, 0)\) across the line \(y = 4\) to get \(B'(0, 8)\).

The minimum distance \(BC + AC\) is the straight line distance \(B'A\).

Equation of line \(B'A\): \(y - 8 = \frac{1-8}{2-0}(x - 0) \Rightarrow y - 8 = -\frac{7}{2}x \Rightarrow 7x + 2y = 16\).

Point \(C(t, 4)\) lies on this line: \(7t + 2(4) = 16 \Rightarrow 7t = 8 \Rightarrow t = 8/7\).

So, \(\beta = 8/7\).


Maximization (\(\alpha\)):
\(f(t)\) is a sum of two distances, and its maximum over a closed interval \([0, 4]\) occurs at the endpoints.
\(f(0) = \sqrt{0+16} + \sqrt{4+9} = 4 + \sqrt{13} \approx 4 + 3.6 = 7.6\).
\(f(4) = \sqrt{16+16} + \sqrt{4+9} = 4\sqrt{2} + \sqrt{13} \approx 5.65 + 3.6 = 9.25\).

Maximum is at \(t = 4\), so \(\alpha = 4\).


Step 4: Final Answer:
\(6\alpha + 21\beta = 6(4) + 21(8/7) = 24 + 24 = 48\).
Quick Tip: The "Shortest Path" problem (minimizing \(PA+PB\) where \(P\) is on a line) is always solved by reflecting one point. For maximization on an interval, the function is usually convex, so check the boundaries.

Step 1: Understanding the Concept:

This is a standard integration problem of the form \(\int \frac{dx}{(ax^2+b)\sqrt{cx^2+d}}\).


Step 2: Key Formula or Approach:

Substitute \(x = \frac{2\sin\theta}{\sqrt{3}}\). Then \(dx = \frac{2\cos\theta}{\sqrt{3}} d\theta\) and \(\sqrt{4-3x^2} = 2\cos\theta\).


Step 3: Detailed Explanation:
\(f(x) = \int \frac{\frac{2\cos\theta}{\sqrt{3}} d\theta}{(3 + \frac{16\sin^2\theta}{3}) 2\cos\theta} = \frac{1}{\sqrt{3}} \int \frac{d\theta}{3 + \frac{16\sin^2\theta}{3}} = \sqrt{3} \int \frac{d\theta}{9 + 16\sin^2\theta}\).

Divide by \(\cos^2\theta\):
\(f(x) = \sqrt{3} \int \frac{\sec^2\theta d\theta}{9\sec^2\theta + 16\tan^2\theta} = \sqrt{3} \int \frac{\sec^2\theta d\theta}{9 + 25\tan^2\theta}\).

Let \(u = \tan\theta\), \(du = \sec^2\theta d\theta\).
\(f(x) = \sqrt{3} \int \frac{du}{9 + 25u^2} = \frac{\sqrt{3}}{25} \int \frac{du}{u^2 + (3/5)^2} = \frac{\sqrt{3}}{25} \cdot \frac{5}{3} \tan^{-1}\left(\frac{5u}{3}\right) = \frac{1}{5\sqrt{3}} \tan^{-1}\left(\frac{5\tan\theta}{3}\right)\).

When \(x=1\), \(\sin\theta = \frac{\sqrt{3}}{2} \Rightarrow \theta = 60^\circ \Rightarrow \tan\theta = \sqrt{3}\).
\(f(1) = \frac{1}{5\sqrt{3}} \tan^{-1}\left(\frac{5\sqrt{3}}{3}\right) = \frac{1}{5\sqrt{3}} \tan^{-1}\left(\frac{5}{\sqrt{3}}\right)\).

Comparing with given form, \(\alpha = 5, \beta = \sqrt{3}\).

Note: \(\alpha\beta = 5\sqrt{3}\).


Step 4: Final Answer:
\(\alpha^2 + \beta^2 = 25 + 3 = 28\).
Quick Tip: For integrals involving \(x^2\) and \(\sqrt{a-bx^2}\), trigonometric substitution usually leads to a manageable integral in \(\tan\theta\).

Step 1: Understanding the Concept:

First, calculate the total area bounded by the parabola \(x = \frac{2y^2}{3}\), the line \(x = 3-y\), and \(y=0\). Then subtract the area of the circular sector that falls within this boundary.


Step 2: Detailed Explanation:

Intersection of \(x = 2y^2/3\) and \(x = 3-y\):
\(2y^2/3 = 3-y \Rightarrow 2y^2 + 3y - 9 = 0 \Rightarrow (2y-3)(y+3) = 0 \Rightarrow y = 3/2\).

Total Area \(A_{total} = \int_0^{3/2} (3-y - \frac{2y^2}{3}) dy = [3y - \frac{y^2}{2} - \frac{2y^3}{9}]_0^{3/2}\).
\(A_{total} = \frac{9}{2} - \frac{9}{8} - \frac{3}{4} = \frac{36-9-6}{8} = \frac{21}{8}\).

The circle is \((x-3)^2 + y^2 = 2\). It is centered at \((3,0)\) with radius \(\sqrt{2}\).

The line \(x+y=3\) passes through \((3,0)\) and has slope \(-1\), so the angle is \(45^\circ\) (\(\pi/4\)).

Area of sector inside region \(= \frac{1}{2} r^2 \theta = \frac{1}{2} (2) (\frac{\pi}{4}) = \frac{\pi}{4}\).

Area \(A = \frac{21}{8} - \frac{\pi}{4}\).


Step 3: Final Answer:
\(4(\pi + 4A) = 4(\pi + 4(\frac{21}{8} - \frac{\pi}{4})) = 4(\pi + \frac{21}{2} - \pi) = 42\).
Quick Tip: Geometric sectors are often easier to calculate than performing integrations on circular arcs. Check if the boundary lines pass through the center of the circle to use sector formulas.

Step 1: Understanding the Concept:

Let the ellipse be \(\frac{(x-1)^2}{a^2} + \frac{y^2}{b^2} = 1\).

Length of latus rectum \(= \frac{2b^2}{a} = \frac{1}{2} \Rightarrow a = 4b^2\).


Step 2: Detailed Explanation:

Minor axis endpoints are \(B(1, b)\) and \(B'(1, -b)\). Foci are \(S(1+ae, 0)\).

In \(\triangle BSB'\), the angle \(\angle BSB' = 60^\circ\).

By symmetry, in right \(\triangle BS(1,0)\), \(\angle B S (1,0) = 30^\circ\).
\(\tan 30^\circ = \frac{b}{ae} \Rightarrow \frac{1}{\sqrt{3}} = \frac{b}{ae} \Rightarrow ae = \sqrt{3}b\).

Using \(b^2 = a^2(1-e^2) \Rightarrow b^2 = a^2 - a^2e^2 \Rightarrow b^2 = a^2 - 3b^2 \Rightarrow a^2 = 4b^2\).

Since \(a = 4b^2\), we have \(a^2 = a \Rightarrow a = 1\).
\(b^2 = 1/4 \Rightarrow b = 1/2\).

Major axis \(2a = 2\), Minor axis \(2b = 1\).


Step 3: Final Answer:

Square of the sum \(= (2 + 1)^2 = 9\).
Quick Tip: For any ellipse, the distance from a focus to the endpoint of the minor axis is always equal to \(a\) (the semi-major axis). Knowing this identity makes angular calculations much faster.

Step 1: Understanding the Concept:

Find the plane \(P\) using the family of planes: \(P_1 + \lambda P_2 = 0\). Then find the distance from point \(A\) along a specific line direction.


Step 2: Detailed Explanation:

Plane \(P: (2+5\lambda)x + (1-3\lambda)y + (-1+4\lambda)z + (-3+9\lambda) = 0\).

Line 1 direction: \((2, 4, 5)\). (Note: \(3-y \rightarrow y-3\) with dir -4).

Parallelism: \(2(2+5\lambda) + 4(1-3\lambda) + 5(-1+4\lambda) = 0 \Rightarrow 18\lambda + 3 = 0 \Rightarrow \lambda = -1/6\).

Plane \(P: 7x + 9y - 10z - 27 = 0\).

Line 2 direction: \((-3, 4, 12)\). (Note: \(2-z \rightarrow z-2\) with dir -12).

Line through \(A(8, -1, -19): x = 8-3k, y = -1+4k, z = -19+12k\).

Intersect with \(P: 7(8-3k) + 9(-1+4k) - 10(-19+12k) - 27 = 0\).
\(56 - 21k - 9 + 36k + 190 - 120k - 27 = 0 \Rightarrow 210 - 105k = 0 \Rightarrow k = 2\).

Distance \(= \sqrt{(-3k)^2 + (4k)^2 + (12k)^2} = \sqrt{9k^2+16k^2+144k^2} = 13k\).


Step 3: Final Answer:

Distance \(= 13 \times 2 = 26\).
Quick Tip: Be very careful with negative signs in symmetric forms of lines (e.g., \(3-y\) or \(2-z\)). Always normalize the denominator so that \(x, y, z\) coefficients are \(+1\).

Step 1: Understanding the Concept:

Substitute \(x=y=z = \lambda\) into the two plane equations.

1. \(\lambda(\sin A + \sin B + \sin C) = 18\).

2. \(\lambda(\sin 2A + \sin 2B + \sin 2C) = 9\).


Step 2: Detailed Explanation:

Dividing the two equations: \(\frac{\sin A + \sin B + \sin C}{\sin 2A + \sin 2B + \sin 2C} = \frac{18}{9} = 2\).

In a triangle: \(\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}\).

And \(\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C = 32 \sin \frac{A}{2} \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{B}{2} \sin \frac{C}{2} \cos \frac{C}{2}\).

So, \(\frac{1}{8 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}} = 2 \Rightarrow \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{1}{16}\).


Step 3: Final Answer:
\(80 \times \frac{1}{16} = 5\).
Quick Tip: Trigonometric identities related to sum of sines in a triangle (conditional identities) are frequently combined with 3D geometry intersection problems. Memorizing \(\prod \cos(A/2)\) and \(4\prod \sin A\) identities is essential.