Step 1: Understanding the Concept:

This sum involves products of binomial coefficients which can be solved using Vandermonde's Identity or by identifying coefficients in polynomial expansions.


Step 2: Key Formula or Approach:

Identity: \(^{n}C_r = ^{n}C_{n-r}\).

General Identity: \(\sum_{r=0}^{k} {}^{n}C_r \cdot {}^{m}C_{k-r} = {}^{n+m}C_k\).


Step 3: Detailed Explanation:

1. Rewrite the sum:

Given: \(S = \sum_{r=0}^{22} {}^{22}C_r \cdot {}^{23}C_r\).

Using the identity \(^{23}C_r = ^{23}C_{23-r}\):
\[ S = \sum_{r=0}^{22} {}^{22}C_r \cdot {}^{23}C_{23-r} \]

2. Relate to polynomial expansion:

This sum represents the coefficient of \(x^{23}\) in the product of \((1+x)^{22}\) and \((1+x)^{23}\).
\[ (1+x)^{22} (1+x)^{23} = (1+x)^{45} \]

3. Find the coefficient:

The coefficient of \(x^{23}\) in \((1+x)^{45}\) is \(^{45}C_{23}\).


Step 4: Final Answer:

The value is \(^{45}C_{23}\).
Quick Tip: When you see a sum of products of binomial coefficients \(^{n}C_r \cdot ^{m}C_r\), immediately try changing one term to its complementary form \(^{m}C_{m-r}\) to use the addition rule.

Step 1: Understanding the Concept:

We need to check the continuity and differentiability of the function \(f(x)\) at the origin, and then check the continuity of its derivative \(f'(x)\).


Step 2: Detailed Explanation:

1. Continuity of \(f(x)\) at \(x=0\):
\(\lim_{x \to 0} f(x) = \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)\).

Since \(\left|\sin\left(\frac{1}{x}\right)\right| \le 1\), by the Sandwich Theorem, \(\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0\).

Since \(\lim_{x \to 0} f(x) = f(0)\), the function is continuous.

2. Differentiability of \(f(x)\) at \(x=0\):
\[ f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{h^2 \sin\left(\frac{1}{h}\right) - 0}{h} = \lim_{h \to 0} h \sin\left(\frac{1}{h}\right) \]

Again, by the Sandwich Theorem, this limit is 0. So, \(f(x)\) is differentiable at \(x=0\).

3. Continuity of \(f'(x)\) at \(x=0\):

For \(x \neq 0\), apply the product rule:
\[ f'(x) = 2x \sin\left(\frac{1}{x}\right) + x^2 \cos\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right) = 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right) \]

Now check \(\lim_{x \to 0} f'(x)\):
\(\lim_{x \to 0} \left(2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right)\right)\).

The first term \(2x \sin(1/x) \to 0\), but \(\lim_{x \to 0} \cos(1/x)\) does not exist (it oscillates).

Therefore, \(f'(x)\) is not continuous at \(x=0\).


Step 3: Final Answer:

The function \(f\) is continuous (and differentiable), but \(f'\) is not continuous.
Quick Tip: A function \(x^n \sin(1/x)\) is continuous at \(x=0\) if \(n > 0\), differentiable if \(n > 1\), and has a continuous derivative if \(n > 2\).

Step 1: Understanding the Concept:

The given equation is a first-order linear differential equation.

We first rearrange it into the standard form \(\frac{dy}{dx} + P(x)y = Q(x)\) to find the integrating factor.


Step 2: Key Formula or Approach:

1. Standard form: \(\frac{dy}{dx} + P(x)y = Q(x)\)

2. Integrating Factor (I.F.) \(= e^{\int P(x) dx}\)

3. General Solution: \(y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C\)


Step 3: Detailed Explanation:

Rearranging the equation:
\[ x^3 \frac{dy}{dx} + xy - 1 = 0 \implies \frac{dy}{dx} + \frac{1}{x^2}y = \frac{1}{x^3} \]

Here, \(P(x) = \frac{1}{x^2}\) and \(Q(x) = \frac{1}{x^3}\).
\[ I.F. = e^{\int \frac{1}{x^2} dx} = e^{-1/x} \]

The solution is:
\[ y e^{-1/x} = \int \frac{1}{x^3} e^{-1/x} dx + C \]

To evaluate the integral, let \(t = -1/x \implies dt = \frac{1}{x^2} dx\).

Also, \(\frac{1}{x^3} dx = \frac{1}{x} \cdot \frac{1}{x^2} dx = -t dt\).
\[ \int -t e^t dt = -(t e^t - e^t) = e^t(1 - t) \]

Substituting back \(t = -1/x\):
\[ y e^{-1/x} = e^{-1/x}\left(1 + \frac{1}{x}\right) + C \]
\[ y = 1 + \frac{1}{x} + C e^{1/x} \]

Using the initial condition \(y\left(\frac{1}{2}\right) = 3 - e\):
\[ 3 - e = 1 + 2 + C e^2 \implies 3 - e = 3 + C e^2 \implies C = -\frac{e}{e^2} = -e^{-1} \]

So, \(y = 1 + \frac{1}{x} - e^{-1} e^{1/x} = 1 + \frac{1}{x} - e^{(1/x) - 1}\).

At \(x = 1\):
\[ y(1) = 1 + 1 - e^{1-1} = 2 - e^0 = 2 - 1 = 1 \]


Step 4: Final Answer:

The value of \(y(1)\) is 1.
Quick Tip: For linear differential equations of the form \(x^n \frac{dy}{dx} + \dots\), dividing throughout by \(x^n\) usually reveals a clear path to the integrating factor.

Step 1: Understanding the Concept:

We simplify the logical statement using Boolean algebra identities such as De Morgan's Law, Distributive Law, and the Implication rule (\(A \implies B \equiv \sim A \lor B\)).


Step 2: Detailed Explanation:

Let the given statement be \(S\).

First, simplify the antecedent:
\[ Part 1: \sim(P \land Q) \equiv \sim P \lor \sim Q \]
\[ Antecedent: (\sim P \lor \sim Q) \lor (\sim P \land Q) \]

By associative and distributive properties:
\[ \equiv \sim P \lor (\sim Q \lor (\sim P \land Q)) \equiv \sim P \lor ((\sim Q \lor \sim P) \land (\sim Q \lor Q)) \]

Since \(\sim Q \lor Q\) is a Tautology (\(T\)):
\[ \equiv \sim P \lor (\sim Q \lor \sim P) \equiv \sim P \lor \sim Q \]

Now, evaluate the full implication \(S \equiv (\sim P \lor \sim Q) \implies (\sim P \land \sim Q)\).

Using \(A \implies B \equiv \sim A \lor B\):
\[ S \equiv \sim(\sim P \lor \sim Q) \lor (\sim P \land \sim Q) \]
\[ S \equiv (P \land Q) \lor (\sim P \land \sim Q) \]

This is the expression for the Biconditional \(P \iff Q\).

Looking at the options, Option (2) is:
\[ (P \implies Q) \land (Q \implies P) \equiv (\sim P \lor Q) \land (\sim Q \lor P) \]

This is the standard definition of \(P \iff Q\).


Step 3: Final Answer:

The statement is equivalent to \(((\sim P) \lor Q) \land ((\sim Q) \lor P)\).
Quick Tip: Truth tables are a foolproof alternative for logic questions if Boolean simplification seems complex. Just check for matching truth values in the final column.

Step 1: Understanding the Concept:

A matrix is singular if its determinant is zero. For the quadratic equation, we check if there are obvious roots.


Step 2: Detailed Explanation:

For the equation \((a-c)x^2 + (b-a)x + (c-b) = 0\):

Sum of coefficients \(= (a-c) + (b-a) + (c-b) = 0\).

This implies that \(x = 1\) is a root.

If \(\alpha = 1\), the matrix becomes \(\begin{bmatrix} 1 & 1 & 1
1 & 1 & 1
a & b & c \end{bmatrix}\). Since two rows are identical, the determinant is zero, and it is singular.

The other root is \(\frac{constant term}{coeff of x^2} = \frac{c-b}{a-c}\).

Let \(x = a-c\), \(y = b-a\), and \(z = c-b\).

Note that \(x + y + z = (a-c) + (b-a) + (c-b) = 0\).

The required expression is:
\[ \frac{x^2}{yz} + \frac{y^2}{xz} + \frac{z^2}{xy} = \frac{x^3 + y^3 + z^3}{xyz} \]

From algebra, we know that if \(x + y + z = 0\), then \(x^3 + y^3 + z^3 = 3xyz\).

Substituting this into our expression:
\[ \frac{3xyz}{xyz} = 3 \]


Step 3: Final Answer:

The value of the given expression is 3.
Quick Tip: Whenever you see a quadratic with coefficients like \((p-q), (q-r), (r-p)\), always check if \(x=1\) is a root. Also, the algebraic identity \(x+y+z=0 \implies x^3+y^3+z^3=3xyz\) is a frequent visitor in competitive exams.

Step 1: Understanding the Concept:

We analyze the properties of the relation:

- Reflexive: \(a R a \forall a \in \mathbb{Z}\)

- Symmetric: \(a R b \implies b R a\)

- Transitive: \(a R b\) and \(b R c \implies a R c\)


Step 2: Detailed Explanation:

1. Reflexivity: For \(a=2\), \(\gcd(2,2)=2 \neq 1\). So \((2,2) \notin R\). Not reflexive.

2. Symmetry: Let \(a=2, b=1\). \(\gcd(2,1)=1\) and \(2(2) \neq 1\). Thus \((2,1) \in R\).

Now check \((1,2)\): \(\gcd(1,2)=1\), but \(2(1) = 2\). The condition \(2a \neq b\) fails for \(a=1, b=2\).

So \((1,2) \notin R\). Not symmetric.

3. Transitivity: Let \(a=3, b=2\) and \(c=3\).

- \((3,2) \in R\) because \(\gcd(3,2)=1\) and \(6 \neq 2\).

- \((2,3) \in R\) because \(\gcd(2,3)=1\) and \(4 \neq 3\).

Is \((3,3) \in R\)? \(\gcd(3,3)=3 \neq 1\). So \((3,3) \notin R\).

Therefore, the relation is not transitive.


Step 3: Final Answer:

The relation is neither symmetric nor transitive.
Quick Tip: To disprove properties like symmetry or transitivity, always look for simple integer counter-examples involving 1 and 2, especially when inequalities or specific divisors are involved.

Step 1: Understanding the Concept:

In probability theory, especially with continuous sample spaces, an event having probability 0 does not necessarily mean it is an empty set (it could be a "null set"). Similarly, an event with probability 1 does not necessarily contain every outcome in \(\Omega\).


Step 2: Detailed Explanation:

- Counter-example for (S1): Consider a continuous random variable \(X\) uniformly distributed on \([0, 1]\). The probability of a single point, e.g., \(P(X = 0.5) = 0\), yet the set \(\{0.5\}\) is not empty. Thus, (S1) is false.

- Counter-example for (S2): Using the same distribution, the probability \(P(X \neq 0.5) = 1\). However, the set \(A = \Omega \setminus \{0.5\}\) is not equal to the entire sample space \(\Omega\). Thus, (S2) is false.

In discrete probability with strictly positive weights, these statements might hold, but they are false in general measure-theoretic probability.


Step 3: Final Answer:

Both (S1) and (S2) are false.
Quick Tip: Remember that probability 0 refers to an "impossible event" only in discrete finite spaces. In continuous distributions, "almost never" and "never" are different.

Step 1: Understanding the Concept:

We first find the equation of the plane passing through three given points and then calculate the perpendicular distance from the given point to that plane.


Step 2: Key Formula or Approach:

1. Equation of plane through three points \(\vec{a}, \vec{b}, \vec{c}\) 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. Distance from \((x_0, y_0, z_0)\) to \(ax+by+cz+d=0\) is \(\frac{|ax_0+by_0+cz_0+d|}{\sqrt{a^2+b^2+c^2}}\).


Step 3: Detailed Explanation:

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

Equation of the plane:
\[ \begin{vmatrix} x-2 & y+3 & z-1
-1-2 & 1-(-3) & -2-1
3-2 & -4-(-3) & 2-1 \end{vmatrix} = 0 \implies \begin{vmatrix} x-2 & y+3 & z-1
-3 & 4 & -3
1 & -1 & 1 \end{vmatrix} = 0 \]
\[ (x-2)(4-3) - (y+3)(-3+3) + (z-1)(3-4) = 0 \]
\[ (x-2)(1) - 0 + (z-1)(-1) = 0 \implies x - z - 1 = 0 \]

Now, distance of \((7, -3, -4)\) from \(x - z - 1 = 0\):
\[ Distance = \frac{|7 - (-4) - 1|}{\sqrt{1^2 + 0^2 + (-1)^2}} = \frac{|7 + 4 - 1|}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2} \]


Step 4: Final Answer:

The distance is \(5\sqrt{2}\).
Quick Tip: If the plane equation reduces to a very simple form like \(x - z - 1 = 0\), double-check the determinant calculation. It often means the normal vector is parallel to one of the coordinate planes.

Step 1: Understanding the Concept:

We need to find the point where a line passing through \((-1, 9, -16)\) and parallel to a given direction intersects the plane. Then find the distance between the two points.


Step 2: Detailed Explanation:

Given line: \(\frac{x+4}{3} = \frac{y-2}{-4} = \frac{z-3}{12}\). Direction ratios are \(\langle 3, -4, 12 \rangle\).

Equation of line through \(P(-1, 9, -16)\) parallel to the given line:
\[ \frac{x+1}{3} = \frac{y-9}{-4} = \frac{z+16}{12} = r \]

Any point \(Q\) on this line is \((3r-1, -4r+9, 12r-16)\).

If \(Q\) lies on the plane \(2x + 3y - z = 5\):
\[ 2(3r-1) + 3(-4r+9) - (12r-16) = 5 \]
\[ 6r - 2 - 12r + 27 - 12r + 16 = 5 \]
\[ -18r + 41 = 5 \implies 18r = 36 \implies r = 2 \]

The point \(Q\) is \((3(2)-1, -4(2)+9, 12(2)-16) = (5, 1, 8)\).

Distance \(PQ = \sqrt{(5 - (-1))^2 + (1 - 9)^2 + (8 - (-16))^2}\)
\[ PQ = \sqrt{6^2 + (-8)^2 + 24^2} = \sqrt{36 + 64 + 576} = \sqrt{676} = 26 \]


Step 3: Final Answer:

The distance is 26 units.
Quick Tip: When measuring distance "parallel to a line", the distance is simply \(|r| \cdot \sqrt{a^2+b^2+c^2}\) where \(\langle a,b,c \rangle\) are the direction ratios used in the parametric form.

Step 1: Understanding the Concept:

We use the properties of the dot product and cross product, specifically that a cross product \(\vec{v} \times \vec{w}\) is perpendicular to both \(\vec{v}\) and \(\vec{w}\).


Step 2: Detailed Explanation:

Given: \(\vec{v} \times \vec{w} = \vec{u} + \lambda \vec{v}\).

Dot product with \(\vec{v}\):
\[ \vec{v} \cdot (\vec{v} \times \vec{w}) = \vec{v} \cdot \vec{u} + \lambda |\vec{v}|^2 \]

Since \(\vec{v} \cdot (\vec{v} \times \vec{w}) = 0\):
\[ 0 = \vec{v} \cdot \vec{u} + \lambda |\vec{v}|^2 \]

Calculate \(\vec{u} \cdot \vec{v}\) and \(|\vec{v}|^2\):
\[ \vec{u} \cdot \vec{v} = (1)(2) + (-1)(1) + (-2)(-1) = 2 - 1 + 2 = 3 \]
\[ |\vec{v}|^2 = 2^2 + 1^2 + (-1)^2 = 6 \]
\[ 0 = 3 + 6\lambda \implies \lambda = -1/2 \]

Now, take the dot product of the original equation with \(\vec{w}\):
\[ \vec{w} \cdot (\vec{v} \times \vec{w}) = \vec{w} \cdot \vec{u} + \lambda (\vec{v} \cdot \vec{w}) \]

Since \(\vec{w} \cdot (\vec{v} \times \vec{w}) = 0\):
\[ 0 = \vec{u} \cdot \vec{w} + (-1/2)(2) \]
\[ 0 = \vec{u} \cdot \vec{w} - 1 \implies \vec{u} \cdot \vec{w} = 1 \]


Step 3: Final Answer:

The value of \(\vec{u} \cdot \vec{w}\) is 1.
Quick Tip: For any equation involving a cross product \(\vec{A} \times \vec{B} = \vec{C}\), always try taking the dot product with \(\vec{A}\) or \(\vec{B}\) first, as it yields zero and often simplifies the expression for constants.

Step 1: Understanding the Concept:

We use the property that the area of a triangle formed by taking points on sides can be calculated by subtracting the three outer triangles from the total area. The area of a triangle is \(\frac{1}{2} ab \sin \theta\).


Step 2: Detailed Explanation:

Let Area(\(\Delta PQR\)) \(= S\).

Given: \(QA/AR = 1/2 \implies QA = \frac{1}{3} QR, AR = \frac{2}{3} QR\).

Similarly, \(RB = \frac{1}{3} RP, BP = \frac{2}{3} RP\) and \(PC = \frac{1}{3} PQ, CQ = \frac{2}{3} PQ\).

Area(\(\Delta PCB\)) \(= \frac{1}{2} PC \cdot BP \sin P = \frac{1}{2} (\frac{1}{3} PQ) (\frac{2}{3} PR) \sin P = \frac{2}{9} \left(\frac{1}{2} PQ \cdot PR \sin P\right) = \frac{2}{9} S\).

Similarly, Area(\(\Delta QAC\)) \(= \frac{2}{9} S\) and Area(\(\Delta RBA\)) \(= \frac{2}{9} S\).

Area(\(\Delta ABC\)) \(= S - [Area(\Delta PCB) + Area(\Delta QAC) + Area(\Delta RBA)]\)
\[ Area(\Delta ABC) = S - \left(\frac{2}{9} S + \frac{2}{9} S + \frac{2}{9} S\right) = S - \frac{6}{9} S = S - \frac{2}{3} S = \frac{1}{3} S \]

The ratio \(\frac{Area(\Delta PQR)}{Area(\Delta ABC)} = \frac{S}{S/3} = 3\).


Step 3: Final Answer:

The ratio is 3.
Quick Tip: If points divide the sides of a triangle in the ratio \(1:k\), the ratio of the area of the inner triangle to the outer one is \(\frac{k^2 - k + 1}{(k+1)^2}\). Here \(k=2\), so ratio \(= \frac{4-2+1}{9} = \frac{3}{9} = \frac{1}{3}\).

Step 1: Understanding the Concept:

The greatest integer function \([x]\) satisfies \(x - 1 < [x] \leq x\). We solve the equation by factoring and considering cases for \(x\).


Step 2: Detailed Explanation:

Rearranging the equation:
\[ x^2 - 4x + 3 = x[x] - [x] \]
\[ (x-1)(x-3) = [x](x-1) \]

This leads to two cases:

Case 1: \(x - 1 = 0 \implies x = 1\).

Let's check in the original equation: \(1^2 - 4(1) + [1] + 3 = 1 - 4 + 1 + 3 = 1\).

The right side: \(1 \cdot [1] = 1\). So \(x=1\) is a solution.

Case 2: \(x \neq 1\). We can divide by \((x-1)\):
\[ x - 3 = [x] \]

By definition, \([x] = x - f\) where \(0 \leq f < 1\).

Substituting: \(x - 3 = x - f \implies f = 3\).

But \(f\) must be in the interval \([0, 1)\), so \(f = 3\) is impossible.

Thus, there are no solutions in Case 2.

The only solution is \(x = 1\).


Step 3: Final Answer:

The equation has a unique solution (\(x=1\)) in \((-\infty, \infty)\).
Quick Tip: Factoring terms containing both \(x\) and \([x]\) often simplifies the problem into a simple algebraic equation and a transcendental one which can be solved using the range of the fractional part \(f\).

Step 1: Understanding the Concept:

A system of linear equations has a unique solution if the determinant of the coefficient matrix (\(\Delta\)) is non-zero.


Step 2: Detailed Explanation:

Let \(\Delta = \begin{vmatrix} 1 & 1 & 1
2 & N & 2
3 & 3 & N \end{vmatrix}\).
\[ \Delta = 1(N^2 - 6) - 1(2N - 6) + 1(6 - 3N) \]
\[ \Delta = N^2 - 6 - 2N + 6 + 6 - 3N = N^2 - 5N + 6 \]

For a unique solution, \(\Delta \neq 0\):
\[ N^2 - 5N + 6 \neq 0 \implies (N-2)(N-3) \neq 0 \implies N \neq 2, 3 \]

Possible outcomes for a die roll are \(\{1, 2, 3, 4, 5, 6\}\).

The outcomes for which the system has a unique solution are \(N \in \{1, 4, 5, 6\}\).

Number of favorable outcomes \(= 4\).

Probability \(= 4/6\). Comparing with \(k/6\), we get \(k = 4\).

The possible values of \(N\) that give a unique solution are \(1, 4, 5, 6\).

Sum \(= k + \sum N = 4 + (1 + 4 + 5 + 6) = 4 + 16 = 20\).


Step 3: Final Answer:

The sum is 20.
Quick Tip: A system of equations \(Ax = B\) where \(B\) is a multiple of one of the columns of \(A\) will always be consistent. The unique solution depends solely on \(\det(A)\).

Step 1: Understanding the Concept:

We simplify the numerical arguments of the inverse trigonometric functions by rationalizing or factoring out common square roots.


Step 2: Detailed Explanation:

1. First term: \(\frac{1+\sqrt{3}}{3+\sqrt{3}}\)

Factor out \(\sqrt{3}\) from the denominator: \(3 + \sqrt{3} = \sqrt{3}(\sqrt{3} + 1)\).
\[ \frac{1+\sqrt{3}}{\sqrt{3}(\sqrt{3}+1)} = \frac{1}{\sqrt{3}} \]
\[ \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = 30^\circ = \frac{\pi}{6} \]

2. Second term: \(\frac{8+4\sqrt{3}}{6+3\sqrt{3}}\)

Numerator: \(4(2+\sqrt{3})\). Denominator: \(3(2+\sqrt{3})\).
\[ \frac{4(2+\sqrt{3})}{3(2+\sqrt{3})} = \frac{4}{3} \]

The argument for \(\sec^{-1}\) is \(\sqrt{4/3} = 2/\sqrt{3}\).
\[ \sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = 30^\circ = \frac{\pi}{6} \]

3. Sum:
\[ Total = \frac{\pi}{6} + \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \]


Step 3: Final Answer:

The value is \(\frac{\pi}{3}\).
Quick Tip: When dealing with \(\sqrt{3}\) in inverse trig expressions, look for factors that turn the expression into \(1/\sqrt{3}, \sqrt{3},\) or \(1/2\), as these correspond to standard angles.

Step 1: Understanding the Concept:

This problem involves matrix algebra and the property of commuting matrices. We need to manipulate the given matrix equation to find a relationship between \(A^2\) and \(B\).


Step 2: Key Formula or Approach:

Rearrange the equation to isolate terms or factor them. A common technique is to use the identity matrix \(I\) to complete a product.


Step 3: Detailed Explanation:

Given: \[ A^2 + B = A^2 B \]

Rearranging the terms: \[ A^2 B - A^2 - B = 0 \]

Add the identity matrix \(I\) to both sides: \[ A^2 B - A^2 - B + I = I \]

Factor the left-hand side: \[ A^2(B - I) - 1(B - I) = I \]
\[ (A^2 - I)(B - I) = I \]

This equation implies that \((A^2 - I)\) and \((B - I)\) are inverses of each other. Since a matrix and its inverse always commute: \[ (B - I)(A^2 - I) = I \]

Expand the product: \[ BA^2 - B - A^2 + I = I \]
\[ BA^2 - B - A^2 = 0 \]
\[ BA^2 = A^2 + B \]

Since we were given \(A^2 + B = A^2 B\), we substitute: \[ BA^2 = A^2 B \]


Step 4: Final Answer:

The matrices \(A^2\) and \(B\) commute, so \(A^2 B = B A^2\).
Quick Tip: In matrix algebra, if \(XY = I\), then \(YX = I\) also holds for square matrices. This "Inverse Property" is a powerful tool to prove commutation in expressions involving sums and products.

Step 1: Understanding the Concept:

The area is bounded by a parabola opening to the left and a straight line. It is easier to integrate with respect to \(y\) to find the area between curves \(x_1(y)\) and \(x_2(y)\).


Step 2: Key Formula or Approach:

Area \(A = \int_{y_1}^{y_2} (x_{right} - x_{left}) dy\).


Step 3: Detailed Explanation:

1. Express x in terms of y:
From curve 1: \(4x = 4 - y^2 \Rightarrow x = 1 - \frac{y^2}{4}\).
From curve 2: \(2x = y - 2 \Rightarrow x = \frac{y - 2}{2}\).


2. Find intersection points: \[ 1 - \frac{y^2}{4} = \frac{y - 2}{2} \]
Multiply by 4: \[ 4 - y^2 = 2y - 4 \Rightarrow y^2 + 2y - 8 = 0 \] \[ (y + 4)(y - 2) = 0 \Rightarrow y = -4, 2 \].


3. Set up the integral:
The parabola is the right-hand curve in the region between \(y = -4\) and \(y = 2\). \[ Area = \int_{-4}^{2} \left[ \left(1 - \frac{y^2}{4}\right) - \left(\frac{y - 2}{2}\right) \right] dy \] \[ Area = \int_{-4}^{2} \left[ 1 - \frac{y^2}{4} - \frac{y}{2} + 1 \right] dy = \int_{-4}^{2} \left[ 2 - \frac{y}{2} - \frac{y^2}{4} \right] dy \]

4. Evaluate the integral: \[ Area = \left[ 2y - \frac{y^2}{4} - \frac{y^3}{12} \right]_{-4}^{2} \]
At \(y = 2\): \(4 - 1 - \frac{8}{12} = 3 - \frac{2}{3} = \frac{7}{3}\).
At \(y = -4\): \(-8 - 4 - \frac{-64}{12} = -12 + \frac{16}{3} = -\frac{20}{3}\). \[ Total Area = \frac{7}{3} - \left(-\frac{20}{3}\right) = \frac{27}{3} = 9 \].

Step 4: Final Answer:

The area enclosed is 9 square units.
Quick Tip: When a curve involves \(y^2\), integrating along the y-axis (horizontal strips) avoids the need to split the integral into two parts, which often happens when integrating along the x-axis for such parabolas.

Step 1: Understanding the Concept:

We need to find the locus of the midpoint of a chord of the hyperbola \(xy = 2\) that is tangent to the parabola \(y^2 = 24x\).


Step 2: Key Formula or Approach:

1. Tangent to \(y^2 = 4ax\) is \(y = mx + \frac{a}{m}\).
2. Chord of a curve with midpoint \((h, k)\) is given by \(T = S_1\).


Step 3: Detailed Explanation:
1. For \(y^2 = 24x\), \(a = 6\). The tangent is \(y = mx + \frac{6}{m} \Rightarrow m^2x - my + 6 = 0\).
2. The chord of \(xy - 2 = 0\) with midpoint \((h, k)\) is: \[ \frac{xk + yh}{2} - 2 = hk - 2 \Rightarrow xk + yh = 2hk \].
3. Since this chord is the tangent, these two equations represent the same line. Comparing coefficients: \[ \frac{k}{m^2} = \frac{h}{-m} = \frac{2hk}{6} \]
From \(\frac{k}{m^2} = \frac{h}{-m}\), we get \(m = -\frac{k}{h}\).
Substitute \(m\) into \(\frac{h}{-m} = \frac{2hk}{6}\): \[ \frac{h}{k/h} = \frac{hk}{3} \Rightarrow \frac{h^2}{k} = \frac{hk}{3} \Rightarrow 3h^2 = hk^2 \Rightarrow 3h = k^2 \] (as \(h \neq 0\)).
4. The locus is \(y^2 = 3x\).
Comparing with \(y^2 = 4A x\), we get \(4A = 3 \Rightarrow A = \frac{3}{4}\).
The directrix is \(x = -A \Rightarrow x = -\frac{3}{4} \Rightarrow 4x = -3\).
The length of latus rectum is \(4A = 3\).

Step 4: Final Answer:

The locus is a parabola with directrix \(4x = -3\).
Quick Tip: For any conic section \(f(x, y) = 0\), the equation of a chord with given midpoint \((x_1, y_1)\) is \(T = S_1\). For \(xy = c^2\), \(T\) is \(\frac{xy_1 + x_1y}{2}\).

Step 1: Understanding the Concept:

This is a standard limit of the form \(\lim_{x \to \infty} (a_1^x + a_2^x + \dots + a_n^x)^{1/x}\). As \(t \to 0\), \(x = \frac{1}{\sin^2 t} \to \infty\).


Step 2: Detailed Explanation:

Let \(x = \frac{1}{\sin^2 t}\). As \(t \to 0\), \(x \to \infty\).
The limit becomes: \[ L = \lim_{x \to \infty} (1^x + 2^x + \dots + n^x)^{1/x} \]
Factor out the largest base, \(n^x\): \[ L = \lim_{x \to \infty} \left[ n^x \left( \left(\frac{1}{n}\right)^x + \left(\frac{2}{n}\right)^x + \dots + 1 \right) \right]^{1/x} \] \[ L = n \cdot \lim_{x \to \infty} \left[ \left(\frac{1}{n}\right)^x + \left(\frac{2}{n}\right)^x + \dots + 1 \right]^{1/x} \]
As \(x \to \infty\), for any \(a < n\), \(\left(\frac{a}{n}\right)^x \to 0\).
So the term inside the bracket tends to \(0 + 0 + \dots + 1 = 1\). \[ L = n \cdot 1^0 = n \].

Step 3: Final Answer:

The limit is equal to \(n\).
Quick Tip: For any finite set of positive numbers \(a_1, a_2, \dots, a_n\), the limit \(\lim_{x \to \infty} (a_1^x + a_2^x + \dots + a_n^x)^{1/x}\) is always equal to \(\max(a_1, a_2, \dots, a_n)\).

Step 1: Understanding the Concept:

This problem combines Arithmetic Progression (A.P.), Logarithms, and system of equations with integers.


Step 2: Detailed Explanation:

1. A.P. analysis:
Terms: \(T_1=3, T_2=3.5, T_3=4, T_4=4.5\) (since \(d=0.5\)). \(3\log_y x = 3.5 \Rightarrow \log_y x = \frac{7}{6}\). \(3\log_z y = 4 \Rightarrow \log_z y = \frac{4}{3}\). \(7\log_x z = 4.5 \Rightarrow \log_x z = \frac{9}{14}\).
(Check: \(\frac{7}{6} \cdot \frac{4}{3} \cdot \frac{9}{14} = 1\), which is correct for \(\log_y x \cdot \log_z y \cdot \log_x z\)).


2. Relating exponents to logs:
Let \(x^{pq^2} = y^{qr} = z^{p^2r} = k\). \(\log x = \frac{\ln k}{pq^2}, \log y = \frac{\ln k}{qr}, \log z = \frac{\ln k}{p^2r}\). \(\log_y x = \frac{\log x}{\log y} = \frac{qr}{pq^2} = \frac{r}{pq} = \frac{7}{6}\).
Since \(r = pq + 1\): \(\frac{pq+1}{pq} = \frac{7}{6} \Rightarrow 1 + \frac{1}{pq} = \frac{7}{6} \Rightarrow pq = 6\).
Then \(r = 6+1 = 7\).


3. Solving for p and q: \(\log_z y = \frac{p^2r}{qr} = \frac{p^2}{q} = \frac{4}{3} \Rightarrow 3p^2 = 4q\).
From \(pq=6\), substitute \(q = 6/p\): \(3p^2 = 4(6/p) \Rightarrow 3p^3 = 24 \Rightarrow p^3 = 8 \Rightarrow p = 2\).
Then \(q = 6/2 = 3\).
Values: \(p=2, q=3, r=7\).
Find \(r-p-q = 7-2-3 = 2\).

Step 3: Final Answer:

The value of \(r-p-q\) is 2.
Quick Tip: In A.P. questions with logarithms, always verify the property \(\log_a b \cdot \log_b c \cdot \log_c a = 1\) to ensure the common difference and terms are consistent.

Step 1: Understanding the Concept:

We use De Moivre's Theorem to evaluate the large power of the complex number, find \(p\) and \(q\), and then form the quadratic equation from its roots.


Step 2: Detailed Explanation:
1. Simplify complex number: \(1 - i\sqrt{3} = 2\left(\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = 2e^{-i\pi/3}\). \((1 - i\sqrt{3})^{200} = 2^{200} e^{-i 200\pi/3}\).
Note: \(200\pi/3 = 66\pi + 2\pi/3\). So \(e^{-i 200\pi/3} = e^{-i 2\pi/3}\). \(2^{200} \left(\cos\left(-\frac{2\pi}{3}\right) + i\sin\left(-\frac{2\pi}{3}\right)\right) = 2^{200} \left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)\). \(= 2^{199}(-1 - i\sqrt{3})\).
Comparing with \(2^{199}(p + iq)\), we get \(p = -1\) and \(q = -\sqrt{3}\).

2. Identify roots: \(q^2 = 3\).
Root 1 (\(\alpha\)) \(= p + q + q^2 = -1 - \sqrt{3} + 3 = 2 - \sqrt{3}\).
Root 2 (\(\beta\)) \(= p - q + q^2 = -1 + \sqrt{3} + 3 = 2 + \sqrt{3}\).

3. Form equation:
Sum of roots \((\alpha + \beta) = (2 - \sqrt{3}) + (2 + \sqrt{3}) = 4\).
Product of roots \((\alpha\beta) = (2 - \sqrt{3})(2 + \sqrt{3}) = 4 - 3 = 1\).
The equation is \(x^2 - 4x + 1 = 0\).

Step 3: Final Answer:

The equation is \(x^2 - 4x + 1 = 0\).
Quick Tip: Large powers of complex numbers are almost always solved by converting to polar form \(re^{i\theta}\). Remember that \(\omega = e^{i 2\pi/3}\), so \((1-i\sqrt{3})\) is related to \(2\omega^2\).

Step 1: Understanding the Concept:

This is a standard definite integral problem solvable using the property \(\int_0^a f(x) dx = \int_0^a f(a-x) dx\) (King's Property).


Step 2: Detailed Explanation:

Let \(I = \int_{0}^{\pi/2} \frac{(\cos x)^{2023}}{(\sin x)^{2023} + (\cos x)^{2023}} dx\) --- (Eq. 1).
Apply King's property (\(x \to \pi/2 - x\)):
Since \(\sin(\pi/2 - x) = \cos x\) and \(\cos(\pi/2 - x) = \sin x\): \(I = \int_{0}^{\pi/2} \frac{(\sin x)^{2023}}{(\cos x)^{2023} + (\sin x)^{2023}} dx\) --- (Eq. 2).
Adding (Eq. 1) and (Eq. 2): \(2I = \int_{0}^{\pi/2} \frac{(\cos x)^{2023} + (\sin x)^{2023}}{(\sin x)^{2023} + (\cos x)^{2023}} dx = \int_{0}^{\pi/2} 1 dx = [x]_0^{\pi/2} = \frac{\pi}{2}\).
So, \(I = \frac{\pi}{4}\).
The required value is \(\frac{8}{\pi} \cdot I = \frac{8}{\pi} \cdot \frac{\pi}{4} = 2\).

Step 3: Final Answer:

The value is 2.
Quick Tip: Integrals of the form \(\int_0^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x} dx\) or \(\int_0^{\pi/2} \frac{f(\sin x)}{f(\sin x) + f(\cos x)} dx\) are always equal to \(\frac{\pi}{4}\), regardless of the power \(n\) or function \(f\).

Step 1: Understanding the Concept:

The given equation represents an ellipse. A tangent is drawn at an arbitrary point on this ellipse.

The points of intersection of this tangent with the x and y axes form a line segment AB.

We need to find the minimum length of this segment by expressing it in terms of a parametric variable and then applying optimization techniques.


Step 2: Key Formula or Approach:

1. Standard form of ellipse: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\).

2. Parametric equation of tangent: \(\frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1\).

3. Minimum value of \(a^2 \sec^2 \theta + b^2 \csc^2 \theta\) is \((a + b)^2\).


Step 3: Detailed Explanation:

1. Identify the parameters of the ellipse:

The given equation is \(9x^2 + 16y^2 = 144\).

Dividing both sides by 144:
\[ \frac{9x^2}{144} + \frac{16y^2}{144} = 1 \]
\[ \frac{x^2}{16} + \frac{y^2}{9} = 1 \]

Comparing with \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we get \(a^2 = 16 \Rightarrow a = 4\) and \(b^2 = 9 \Rightarrow b = 3\).


2. Equation of the tangent:

Let the point of tangency be \(P(4 \cos \theta, 3 \sin \theta)\).

The equation of the tangent at point \(P\) is:
\[ \frac{x \cos \theta}{4} + \frac{y \sin \theta}{3} = 1 \]


3. Coordinates of intersection points A and B:

- For point A (on the x-axis), set \(y = 0\):
\[ \frac{x \cos \theta}{4} = 1 \Rightarrow x = \frac{4}{\cos \theta} \Rightarrow A(4 \sec \theta, 0) \]

- For point B (on the y-axis), set \(x = 0\):
\[ \frac{y \sin \theta}{3} = 1 \Rightarrow y = \frac{3}{\sin \theta} \Rightarrow B(0, 3 \csc \theta) \]


4. Length of line segment AB:

Using the distance formula:
\[ AB^2 = (4 \sec \theta - 0)^2 + (0 - 3 \csc \theta)^2 \]
\[ AB^2 = 16 \sec^2 \theta + 9 \csc^2 \theta \]


5. Minimizing the length:

Let \(f(\theta) = 16(1 + \tan^2 \theta) + 9(1 + \cot^2 \theta) = 25 + 16 \tan^2 \theta + 9 \cot^2 \theta\).

Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
\[ \frac{16 \tan^2 \theta + 9 \cot^2 \theta}{2} \ge \sqrt{16 \tan^2 \theta \cdot 9 \cot^2 \theta} \]
\[ 16 \tan^2 \theta + 9 \cot^2 \theta \ge 2 \sqrt{144} = 2 \times 12 = 24 \]

The minimum value of \(AB^2\) is \(25 + 24 = 49\).

Thus, the minimum length of \(AB = \sqrt{49} = 7\).


Step 4: Final Answer:

The minimum length of the line segment AB is 7.
Quick Tip: For an ellipse with semi-axes \(a\) and \(b\), the minimum length of the intercept of a tangent between the coordinate axes is always \(a + b\). In this case, \(4 + 3 = 7\).

Step 1: Understanding the Concept:

To integrate a modulus function, we must split the integral based on where the expression inside the modulus changes sign.


Step 2: Detailed Explanation:

1. Identify roots: \(x^2 - 3x + 2 = (x - 1)(x - 2)\).
The roots are \(x=1\) and \(x=2\).
In \([0, 1]\), \(x^2-3x+2 \ge 0\).
In \([1, 2]\), \(x^2-3x+2 \le 0\).
In \([2, 3]\), \(x^2-3x+2 \ge 0\).


2. Split and integrate: \(I = \int_0^1 (x^2-3x+2) dx + \int_1^2 -(x^2-3x+2) dx + \int_2^3 (x^2-3x+2) dx\).
The anti-derivative is \(F(x) = \frac{x^3}{3} - \frac{3x^2}{2} + 2x\). \(F(0) = 0\). \(F(1) = \frac{1}{3} - \frac{3}{2} + 2 = \frac{2-9+12}{6} = \frac{5}{6}\). \(F(2) = \frac{8}{3} - 6 + 4 = \frac{8}{3} - 2 = \frac{2}{3} = \frac{4}{6}\). \(F(3) = 9 - \frac{27}{2} + 6 = 15 - 13.5 = 1.5 = \frac{3}{2} = \frac{9}{6}\).

3. Sum values: \(\int_0^1 = F(1) - F(0) = \frac{5}{6}\). \(\int_1^2 = -(F(2) - F(1)) = -(\frac{4}{6} - \frac{5}{6}) = \frac{1}{6}\). \(\int_2^3 = F(3) - F(2) = \frac{9}{6} - \frac{4}{6} = \frac{5}{6}\).
Total \(I = \frac{5}{6} + \frac{1}{6} + \frac{5}{6} = \frac{11}{6}\).
Value \( = 12 \cdot \frac{11}{6} = 22\).

Step 3: Final Answer:

The value is 22.
Quick Tip: For a quadratic \(ax^2+bx+c\) integrated between roots \(\alpha\) and \(\beta\), the area is \(\frac{|a|}{6}(\beta-\alpha)^3\). Here \(\int_1^2 |x^2-3x+2|dx = \frac{1}{6}(2-1)^3 = \frac{1}{6}\). This can save calculation time for middle segments.

Step 1: Understanding the Concept:

The problem asks for the number of ways to select 5 courses from 12, with a restriction on the number of language courses.

Total courses = 12.

Language courses (\(L\)) = 5.

Non-language (other) courses (\(O\)) = \(12 - 5 = 7\).

The condition is to choose "at most two" language courses. This means the number of language courses can be 0, 1, or 2.


Step 2: Key Formula or Approach:

We use the combination formula \({}^nC_r = \frac{n!}{r!(n-r)!}\) to calculate the number of ways for each case.

Case 1: 0 Language courses and 5 Other courses.

Case 2: 1 Language course and 4 Other courses.

Case 3: 2 Language courses and 3 Other courses.


Step 3: Detailed Explanation:

1. Case 1: (0L, 5O)

Number of ways = \({}^5C_0 \times {}^7C_5 = 1 \times 21 = 21\).

2. Case 2: (1L, 4O)

Number of ways = \({}^5C_1 \times {}^7C_4 = 5 \times 35 = 175\).

3. Case 3: (2L, 3O)

Number of ways = \({}^5C_2 \times {}^7C_3 = 10 \times 35 = 350\).


Total number of ways = \(21 + 175 + 350 = 546\).


Step 4: Final Answer:

The total number of ways the boy can choose five courses is 546.
Quick Tip: When a question specifies "at most" or "at least", always break the problem into mutually exclusive cases and sum them up. Ensure the total number of items selected (r) remains constant across all cases.

Step 1: Understanding the Concept:

The given digits are {1, 2, 3, 4, 1, 2, 3, 4, 1.

Total digits = 9.

Odd digits: {1, 1, 1, 3, 3 (Count = 5).

Even digits: {2, 2, 4, 4 (Count = 4).

Number of positions in a 9-digit number: {1, 2, 3, 4, 5, 6, 7, 8, 9.

Even positions: {2, 4, 6, 8 (Count = 4).

Odd positions: {1, 3, 5, 7, 9 (Count = 5).


Step 2: Key Formula or Approach:

The condition states "even digits occupy only even places". Since there are 4 even digits and exactly 4 even places, the even digits must occupy all even positions, and consequently, the odd digits must occupy all odd positions.

The number of arrangements for repeated items is \(\frac{n!}{n_1! n_2! \dots}\).


Step 3: Detailed Explanation:

1. Arrange even digits (2, 2, 4, 4) in 4 even positions:

Ways = \(\frac{4!}{2! 2!} = \frac{24}{4} = 6\).

2. Arrange odd digits (1, 1, 1, 3, 3) in 5 odd positions:

Ways = \(\frac{5!}{3! 2!} = \frac{120}{6 \times 2} = 10\).


Total number of 9-digit numbers = \(6 \times 10 = 60\).


Step 4: Final Answer:

The number of such 9-digit numbers is 60.
Quick Tip: For permutation problems with constraints on specific positions, always fill those positions first. When items are identical, don't forget to divide the total permutations by the factorials of the frequencies of repeated digits.

Step 1: Understanding the Concept:

Let \(|x| = t\). Then the equation becomes a quadratic in \(t\): \(t^2 - 2t + |\lambda - 3| = 0\).

For \(x\) to be a solution, \(t\) must be a non-negative real number. For \(x\) to be an integer solution, \(t\) must be a non-negative integer.


Step 2: Key Formula or Approach:

The roots for \(t\) are given by:
\[ t = \frac{2 \pm \sqrt{4 - 4|\lambda - 3|}}{2} = 1 \pm \sqrt{1 - |\lambda - 3|} \]

For \(t\) to be an integer, \(\sqrt{1 - |\lambda - 3|}\) must be an integer, say \(k\).

Since \(|\lambda - 3| \ge 0\), it follows that \(1 - |\lambda - 3| \le 1\), so \(k\) can only be 0 or 1.


Step 3: Detailed Explanation:

1. Case 1: \(k = 0\)
\(1 - |\lambda - 3| = 0 \implies |\lambda - 3| = 1 \implies \lambda = 2\) or \(\lambda = 4\).

Then \(t = 1 \pm 0 = 1\).

If \(\lambda = 2, t = 1 \implies |x| = 1 \implies x = 1, -1\). Values of \(x + \lambda\) are \(\{3, 1\}\).

If \(\lambda = 4, t = 1 \implies |x| = 1 \implies x = 1, -1\). Values of \(x + \lambda\) are \(\{5, 3\}\).

2. Case 2: \(k = 1\)
\(1 - |\lambda - 3| = 1 \implies |\lambda - 3| = 0 \implies \lambda = 3\).

Then \(t = 1 \pm 1 \implies t = 0\) or \(t = 2\).

If \(\lambda = 3, t = 0 \implies x = 0\). Value of \(x + \lambda = 3\).

If \(\lambda = 3, t = 2 \implies x = 2, -2\). Values of \(x + \lambda\) are \(\{5, 1\}\).


Combining all possible values of \(x + \lambda\), we get the set \(S = \{1, 3, 5\}\).


Step 4: Final Answer:

The largest element in set S is 5.
Quick Tip: Substitute modulus terms like \(|x|\) with a new variable to turn complex equations into simple quadratics. Remember to check if the discriminant is a perfect square when looking for integer solutions.

Step 1: Understanding the Concept:

In a Geometric Progression (GP), the difference between consecutive sums gives the corresponding term: \(S_n - S_{n-1} = a_n\).

Given: \(a_4 = 500\) and \(r = \frac{1}{m}\).


Step 2: Key Formula or Approach:

The general term of a GP is \(a_n = a_1 r^{n-1}\).

Also, \(a_n = a_k r^{n-k}\).

The conditions given are:

1. \(S_6 - S_5 > 1 \implies a_6 > 1\).

2. \(S_7 - S_6 < \frac{1}{2} \implies a_7 < \frac{1}{2}\).


Step 3: Detailed Explanation:

1. Using \(a_6 > 1\):
\(a_6 = a_4 \times r^2 = 500 \times (\frac{1}{m})^2 = \frac{500}{m^2}\).
\(\frac{500}{m^2} > 1 \implies m^2 < 500\).

Since \(m\) is a natural number, \(m \le \lfloor \sqrt{500} \rfloor = 22\) (because \(22^2 = 484\) and \(23^2 = 529\)).

2. Using \(a_7 < \frac{1}{2}\):
\(a_7 = a_4 \times r^3 = 500 \times (\frac{1}{m})^3 = \frac{500}{m^3}\).
\(\frac{500}{m^3} < \frac{1}{2} \implies m^3 > 1000\).

Since \(m\) is a natural number, \(m > \sqrt[3]{1000} \implies m > 10\).

3. Finding common values:

Combining the inequalities: \(10 < m \le 22\).

The possible values for \(m\) are \(\{11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22\}\).

Number of values = \(22 - 11 + 1 = 12\).


Step 4: Final Answer:

The number of possible values of \(m\) is 12.
Quick Tip: In GP problems involving \(S_n\), frequently check if the conditions can be simplified to individual terms using \(a_n = S_n - S_{n-1}\). This often removes the need to use the complex GP sum formula.

Step 1: Understanding the Concept:

This sum involves the binomial coefficient identity for \(\sum r^2 \binom{n}{r}\).


Step 2: Key Formula or Approach:
\(\sum_{r=0}^n r \binom{n}{r} = n 2^{n-1}\). \(\sum_{r=0}^n r(r-1) \binom{n}{r} = n(n-1) 2^{n-2}\).
Combining these, \(\sum r^2 \binom{n}{r} = n(n+1) 2^{n-2}\).


Step 3: Detailed Explanation:
Given \(n = 2023\).
Using the identity: \[ \sum_{r=0}^{2023} r^2 \binom{2023}{r} = 2023(2023+1) 2^{2023-2} = 2023 \times 2024 \times 2^{2021} \]
We need to express this in the form \(2023 \times \alpha \times 2^{2022}\): \[ 2023 \times 2024 \times 2^{2021} = 2023 \times (1012 \times 2) \times 2^{2021} \] \[ = 2023 \times 1012 \times 2^{2022} \]
Comparing with the given expression, \(\alpha = 1012\).

Step 4: Final Answer:

The value of \(\alpha\) is 1012.
Quick Tip: Differentiating the binomial expansion \((1+x)^n = \sum \binom{n}{r} x^r\) twice and setting \(x=1\) is the standard way to derive these \(\sum r^k \binom{n}{r}\) identities.

Step 1: Understanding the Concept:

The given equation of the ellipse is \(\frac{x^2}{36} + \frac{y^2}{16} = 1\), where \(a^2 = 36\) and \(b^2 = 16\).

A circle \(C\) is centered at \((2,0)\) and is inscribed in the ellipse.

To be the largest inscribed circle, it must be tangent to the ellipse at points where the distance from the center to the ellipse is at its minimum.


Step 2: Key Formula or Approach:

1. Equation of the circle: \((x - 2)^2 + y^2 = r^2\).

2. The radius \(r\) is obtained by minimizing the distance squared \(D^2 = (x - 2)^2 + y^2\) for points \((x,y)\) on the ellipse.

3. Alternatively, the normal to the ellipse at the point of tangency must pass through the center of the circle \((2,0)\).


Step 3: Detailed Explanation:

1. Express \(y^2\) in terms of \(x\) from the ellipse equation:
\[ \frac{x^2}{36} + \frac{y^2}{16} = 1 \implies \frac{y^2}{16} = 1 - \frac{x^2}{36} \implies y^2 = 16 - \frac{4x^2}{9} \]

2. Substitute this into the distance squared function from the center \((2,0)\):
\[ f(x) = (x - 2)^2 + \left( 16 - \frac{4x^2}{9} \right) \]
\[ f(x) = x^2 - 4x + 4 + 16 - \frac{4x^2}{9} \]
\[ f(x) = \frac{5}{9}x^2 - 4x + 20 \]

3. Find the minimum value of \(f(x)\) by differentiating with respect to \(x\):
\[ f'(x) = \frac{10}{9}x - 4 \]

Setting \(f'(x) = 0 \implies \frac{10}{9}x = 4 \implies x = \frac{36}{10} = 3.6 \).

4. Calculate the radius squared (\(r^2\)):
\[ r^2 = f(3.6) = \frac{5}{9}(3.6)^2 - 4(3.6) + 20 \]
\[ r^2 = \frac{5}{9}(12.96) - 14.4 + 20 \]
\[ r^2 = 5(1.44) - 14.4 + 20 = 7.2 - 14.4 + 20 = 12.8 \]

5. Since the point \((1, \alpha)\) lies on the circle \(C\), substitute it into the circle's equation:
\[ (1 - 2)^2 + \alpha^2 = r^2 \]
\[ (-1)^2 + \alpha^2 = 12.8 \]
\[ 1 + \alpha^2 = 12.8 \implies \alpha^2 = 11.8 \]

6. Calculate the final required value:
\[ 10 \alpha^2 = 10 \times 11.8 = 118 \]


Step 4: Final Answer:

The value of \(10 \alpha^2\) is 118.
Quick Tip: For an inscribed circle centered on the major axis, tangency points can be found where the normal of the ellipse passes through the center. The x-coordinate of tangency is \(x = \frac{a^2 h}{a^2 - b^2}\) for a center at \((h, 0)\).

Step 1: Understanding the Concept:

The shortest distance (S.D.) between two skew lines \(\vec{r} = \vec{a_1} + \lambda \vec{d_1}\) and \(\vec{r} = \vec{a_2} + \mu \vec{d_2}\) is found using the vector formula.


Step 2: Key Formula or Approach:
\(S.D. = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|}\).


Step 3: Detailed Explanation:
1. Extract vectors from lines:
Line 1: \(\vec{a_1} = (2, -1, 6)\), \(\vec{d_1} = (3, 2, 2)\).
Line 2: \(\vec{a_2} = (6, 1, -8)\) (noting \(1-y \to y-1\) with sign change), \(\vec{d_2} = (3, -2, 0)\). \(\vec{a_2} - \vec{a_1} = (4, 2, -14)\).


2. Calculate cross product of directions: \[ \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
3 & 2 & 2
3 & -2 & 0 \end{vmatrix} = \hat{i}(0+4) - \hat{j}(0-6) + \hat{k}(-6-6) = (4, 6, -12) \].
Magnitude \(|\vec{d_1} \times \vec{d_2}| = \sqrt{16 + 36 + 144} = \sqrt{196} = 14\).


3. Calculate dot product: \((\vec{a_2} - \vec{a_1}) \cdot (\vec{d_1} \times \vec{d_2}) = (4)(4) + (2)(6) + (-14)(-12) = 16 + 12 + 168 = 196\). \[ S.D. = \frac{196}{14} = 14 \].

Step 4: Final Answer:

The shortest distance is 14.
Quick Tip: Always double-check the symmetry of the line equations. In "1-y", the coefficient of y must be 1, so divide numerator and denominator by -1 to get the correct direction ratio and point.