Step 1: Understanding the Concept:

We are given a relation \( R \) defined on a finite set \( S = \{-2, -1, 0, 1, 2\} \) such that \((a, b) \in R \iff 1 + ab > 0\).

We need to determine the total number of ordered pairs in \( R \) to verify Statement I, and check if \( R \) is reflexive, symmetric, and transitive to verify Statement II.


Step 2: Key Formula or Approach:

An equivalence relation must satisfy three properties:

1. Reflexive: \((a, a) \in R\) for all \(a \in S\).

2. Symmetric: \((a, b) \in R \implies (b, a) \in R\).

3. Transitive: \((a, b) \in R\) and \((b, c) \in R \implies (a, c) \in R\).

To find the number of elements, we systematically check all pairs \((a,b)\).


Step 3: Detailed Explanation:

First, let's find the number of elements in \( R \).

The condition is \( 1 + ab > 0 \implies ab > -1 \implies ab \ge 0 \) (since \( a, b \) are integers).

Let's check for each \( a \in \{-2, -1, 0, 1, 2\} \):

If \( a = -2 \): \( b \) can be \( -2, -1, 0 \). (3 pairs: \((-2,-2), (-2,-1), (-2,0)\)).

If \( a = -1 \): \( b \) can be \( -2, -1, 0 \). (3 pairs: \((-1,-2), (-1,-1), (-1,0)\)).

If \( a = 0 \): \( ab = 0 \ge 0 \) is true for all \( b \in \{-2, -1, 0, 1, 2\} \). (5 pairs).

If \( a = 1 \): \( b \) can be \( 0, 1, 2 \). (3 pairs: \((1,0), (1,1), (1,2)\)).

If \( a = 2 \): \( b \) can be \( 0, 1, 2 \). (3 pairs: \((2,0), (2,1), (2,2)\)).

Total number of elements in \( R = 3 + 3 + 5 + 3 + 3 = 17 \).

Thus, Statement I is true.

Now let's check if \( R \) is an equivalence relation.

- Reflexive: \( 1 + a^2 > 0 \) is true for all \(a\), so it is reflexive.

- Symmetric: \( 1 + ab > 0 \implies 1 + ba > 0 \), so it is symmetric.

- Transitive: Let \( a = -1, b = 0, c = 1 \).

We have \((-1, 0) \in R\) (since \(1+0 > 0\)) and \((0, 1) \in R\) (since \(1+0 > 0\)).

However, \((-1, 1) \notin R\) because \( 1 + (-1)(1) = 0 \ngtr 0 \).

Since it is not transitive, it is not an equivalence relation.

Thus, Statement II is false.


Step 4: Final Answer:

Only Statement I is true.
Quick Tip: To disprove a property like transitivity, finding a single counterexample is sufficient and is often much faster than trying to prove it generally. Always check boundary conditions like 0 or negative numbers.

Step 1: Understanding the Concept:

The problem requires finding the number of intersection points between two loci in the complex plane. The first equation represents a circle, and the second represents an ellipse.


Step 2: Key Formula or Approach:

Identify the geometric parameters of each locus:

- Circle: \( |z - z_0| = r \) has center \( z_0 \) and radius \( r \).

- Ellipse: \( |z - z_1| + |z - z_2| = 2a \) has foci \( z_1, z_2 \) and major axis \( 2a \).

The center of the ellipse is the midpoint of its foci. Calculate the semi-minor axis \( b \) using \( b^2 = a^2 - c^2 \), where \( 2c \) is the distance between the foci.


Step 3: Detailed Explanation:

The first equation is \( |z - (4 + 8i)| = \sqrt{10} \).

This is a circle with center \( C_1(4, 8) \) and radius \( R = \sqrt{10} \).

The second equation is \( |z - (3 + 5i)| + |z - (5 + 11i)| = 4\sqrt{5} \).

This represents an ellipse with foci \( F_1(3, 5) \) and \( F_2(5, 11) \).

Let's find the distance between the foci:
\( 2c = |F_1 - F_2| = |(3 - 5) + i(5 - 11)| = |-2 - 6i| = \sqrt{(-2)^2 + (-6)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10} \).

So, the semi-focal distance is \( c = \sqrt{10} \).

The major axis is \( 2a = 4\sqrt{5} \implies a = 2\sqrt{5} \implies a^2 = 20 \).

The semi-minor axis \( b \) is given by \( b^2 = a^2 - c^2 = 20 - (\sqrt{10})^2 = 20 - 10 = 10 \).

Thus, \( b = \sqrt{10} \).

Now, let's find the center of the ellipse:

Center \( C_2 \) is the midpoint of \( F_1 \) and \( F_2 \): \( C_2 = \left(\frac{3+5}{2}, \frac{5+11}{2}\right) = (4, 8) \).

Observe that the circle and the ellipse share the exact same center \( (4, 8) \).

Moreover, the radius of the circle \( R = \sqrt{10} \) is exactly equal to the semi-minor axis \( b = \sqrt{10} \) of the ellipse.

Because they share the same center and the circle's radius equals the ellipse's semi-minor axis, the circle is inscribed inside the ellipse, touching it exactly at the two extremities of the minor axis.

Therefore, they intersect at exactly 2 points.


Step 4: Final Answer:

The number of values of \( z \) is 2.
Quick Tip: When dealing with multiple conics in the complex plane, always find their centers first. Concentric shapes with matching parameters (like radius and minor axis) intersect at predictable points without needing full algebraic expansion.

Step 1: Understanding the Concept:

A system of three linear equations in three variables has infinitely many solutions if the determinant of the coefficient matrix is zero (\( \Delta = 0 \)), and all the \( \Delta_x, \Delta_y, \Delta_z \) determinants are also zero, meaning the planes intersect along a common line.


Step 2: Key Formula or Approach:

We can use row reduction to find the conditions for infinite solutions. By eliminating variables, we aim to obtain an identity of the form \( 0 = 0 \). Alternatively, we can check if the third equation is a linear combination of the first two.


Step 3: Detailed Explanation:

The given equations are:
\( E_1: x + y + z = 6 \)
\( E_2: x + 2y + 5z = 10 \)
\( E_3: 2x + 3y + \lambda z = \mu \)

Let's see if we can form \( E_3 \) by a linear combination of \( E_1 \) and \( E_2 \).

Notice the coefficients of \( x \) and \( y \) in \( E_3 \) are 2 and 3 respectively.

If we add \( E_1 \) and \( E_2 \):
\( (x + y + z) + (x + 2y + 5z) = 6 + 10 \)
\( 2x + 3y + 6z = 16 \)

For the system to have infinitely many solutions, this resulting equation must perfectly match the third equation \( E_3 \):
\( 2x + 3y + \lambda z = \mu \)

Comparing the coefficients of \( z \) and the constant term, we get:
\( \lambda = 6 \)
\( \mu = 16 \)

We are asked to find the value of \( \lambda + \mu \):
\( \lambda + \mu = 6 + 16 = 22 \).


Step 4: Final Answer:

The value of \( \lambda + \mu \) is 22.
Quick Tip: Before computing cumbersome determinants like \( \Delta = 0, \Delta_x = 0 \), always check if one row is a simple sum or difference of the other rows. This is the fastest way to solve for unknowns in infinite solution problems.

Step 1: Understanding the Concept:

We are given matrix \( A \) in terms of a variable \( \alpha \). The matrix \( B \) is formed by adding a given matrix to \( adj(A) \). We need to determine \( \alpha \) using \( \det(B) = 66 \) and then evaluate \( \det(adj(A)) \).


Step 2: Key Formula or Approach:

The determinant of the adjoint of an \( n \times n \) matrix \( A \) is given by \( \det(adj(A)) = (\det(A))^{n-1} \). Here, \( n = 3 \), so \( \det(adj(A)) = (\det(A))^2 \).

We must first compute \( adj(A) \) by finding the cofactor matrix and transposing it.


Step 3: Detailed Explanation:

First, find \( \det(A) \):
\( |A| = \alpha(15 - 0) - 1(10 - 0) + 2(8 - 0) = 15\alpha - 10 + 16 = 15\alpha + 6 \).

Next, calculate the cofactors of matrix \( A \):
\( C_{11} = (15 - 0) = 15 \), \( C_{12} = -(10 - 0) = -10 \), \( C_{13} = (8 - 0) = 8 \).
\( C_{21} = -(5 - 8) = 3 \), \( C_{22} = (5\alpha - 0) = 5\alpha \), \( C_{23} = -(4\alpha - 0) = -4\alpha \).
\( C_{31} = (0 - 6) = -6 \), \( C_{32} = -(0 - 4) = 4 \), \( C_{33} = (3\alpha - 2) \).

The adjoint matrix is the transpose of the cofactor matrix:
\( adj(A) = \begin{bmatrix} 15 & 3 & -6
-10 & 5\alpha & 4
8 & -4\alpha & 3\alpha-2 \end{bmatrix} \).

Now construct matrix \( B \):
\( B = \begin{bmatrix} 1 & 0 & 0
0 & -5\alpha & 0
0 & 4\alpha & -2\alpha \end{bmatrix} + \begin{bmatrix} 15 & 3 & -6
-10 & 5\alpha & 4
8 & -4\alpha & 3\alpha-2 \end{bmatrix} = \begin{bmatrix} 16 & 3 & -6
-10 & 0 & 4
8 & 0 & \alpha-2 \end{bmatrix} \).

Compute the determinant of \( B \) by expanding along the second column:
\( \det(B) = -3 \begin{vmatrix} -10 & 4
8 & \alpha-2 \end{vmatrix} + 0 - 0 \)
\( \det(B) = -3 [(-10)(\alpha - 2) - (4)(8)] = -3 [-10\alpha + 20 - 32] = -3 [-10\alpha - 12] = 30\alpha + 36 \).

We are given that \( \det(B) = 66 \):
\( 30\alpha + 36 = 66 \implies 30\alpha = 30 \implies \alpha = 1 \).

Now, substitute \( \alpha = 1 \) back into \( \det(A) \):
\( \det(A) = 15(1) + 6 = 21 \).

Finally, calculate \( \det(adj(A)) \):
\( \det(adj(A)) = (\det(A))^{3-1} = (\det(A))^2 = 21^2 = 441 \).


Step 4: Final Answer:

The value of \( \det(adj(A)) \) is 441.
Quick Tip: To minimize arithmetic errors when calculating determinants of \( 3 \times 3 \) matrices, look for columns or rows with zeros. Expanding along them significantly speeds up the calculation.

Step 1: Understanding the Concept:

We must calculate the sum of the given sequence \( \alpha \). The sequence can be decomposed into two distinct arithmetic progressions. After finding \( \alpha \), we use it to determine the exact value of \( \beta \) by solving the given quadratic equation.


Step 2: Key Formula or Approach:

The sum of an Arithmetic Progression (A.P.) is \( S_n = \frac{n}{2} [2a + (n-1)d] \).

After finding the sum, factor the quadratic equation \( x^2 + x - 2 = 0 \) to identify the correct root for the tangent function.


Step 3: Detailed Explanation:

The series is \( \alpha = 3 + 4 + 8 + 9 + 13 + 14 + \dots \) up to 40 terms.

Let's group the terms into two separate series, picking alternating terms. Each series will have 20 terms:

Series 1: \( 3 + 8 + 13 + \dots \) (20 terms)

Series 2: \( 4 + 9 + 14 + \dots \) (20 terms)

For Series 1: First term \( a_1 = 3 \), common difference \( d_1 = 5 \).
\( S_1 = \frac{20}{2} [2(3) + (20-1)5] = 10 [6 + 95] = 10 \times 101 = 1010 \).

For Series 2: First term \( a_2 = 4 \), common difference \( d_2 = 5 \).
\( S_2 = \frac{20}{2} [2(4) + (20-1)5] = 10 [8 + 95] = 10 \times 103 = 1030 \).

Total sum \( \alpha = S_1 + S_2 = 1010 + 1030 = 2040 \).

Now calculate the exponent given in the problem:
\( \frac{\alpha}{1020} = \frac{2040}{1020} = 2 \).

Thus, the root of the equation is \( (\tan\beta)^2 = \tan^2\beta \).

Now solve the quadratic equation \( x^2 + x - 2 = 0 \):
\( (x+2)(x-1) = 0 \implies x = -2 \) or \( x = 1 \).

Since \( \tan^2\beta \ge 0 \) for all real \( \beta \), we discard \( x = -2 \).

So, \( \tan^2\beta = 1 \).

Given \( \beta \in (0, \pi/2) \), \( \tan\beta \) must be positive.
\( \tan\beta = 1 \implies \beta = \pi/4 \).

Finally, evaluate the required expression:
\( \sin^2\beta + 3\cos^2\beta = \sin^2(\pi/4) + 3\cos^2(\pi/4) \).
\( = \left(\frac{1}{\sqrt{2}}\right)^2 + 3\left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} + 3\left(\frac{1}{2}\right) = \frac{1+3}{2} = \frac{4}{2} = 2 \).


Step 4: Final Answer:

The value of the expression is 2.
Quick Tip: When an A.P. doesn't seem to have a constant difference, try observing the differences between consecutive terms. If they alternate (e.g., +1, +4, +1, +4), splitting the sequence into two separate alternating series is the standard approach.

Step 1: Understanding the Concept:

We are dealing with conditional probability where an outcome has already occurred (reaching late), and we need to find the probability of one of the prior causes (travelled by bus). This is a classic application of Bayes' Theorem.


Step 2: Key Formula or Approach:

Bayes' Theorem states:
\( P(B|L) = \frac{P(L|B) \cdot P(B)}{P(L|B) \cdot P(B) + P(L|S) \cdot P(S) + P(L|C) \cdot P(C)} \)

where \( B, S, C \) are the events of choosing bus, scooter, and car, and \( L \) is the event of reaching late.


Step 3: Detailed Explanation:

Let's list the prior probabilities:
\( P(B) = \frac{2}{5} \)
\( P(S) = \frac{1}{5} \)
\( P(C) = \frac{2}{5} \)

List the conditional probabilities of reaching late given the transport:
\( P(L|B) = \frac{1}{5} \)
\( P(L|S) = \frac{1}{3} \)
\( P(L|C) = \frac{1}{4} \)

We need to calculate the numerator (probability of taking a bus AND being late):
\( P(L \cap B) = P(L|B) \cdot P(B) = \frac{1}{5} \times \frac{2}{5} = \frac{2}{25} \).

Next, calculate the total probability of reaching late (the denominator):
\( P(L) = P(L|B)P(B) + P(L|S)P(S) + P(L|C)P(C) \)
\( P(L) = \left(\frac{1}{5} \times \frac{2}{5}\right) + \left(\frac{1}{3} \times \frac{1}{5}\right) + \left(\frac{1}{4} \times \frac{2}{5}\right) \)
\( P(L) = \frac{2}{25} + \frac{1}{15} + \frac{2}{20} \)

To add these fractions easily, find a common denominator for 25, 15, and 20, which is 300.
\( \frac{2}{25} = \frac{2 \times 12}{300} = \frac{24}{300} \)
\( \frac{1}{15} = \frac{1 \times 20}{300} = \frac{20}{300} \)
\( \frac{2}{20} = \frac{1}{10} = \frac{30}{300} \)

So, \( P(L) = \frac{24 + 20 + 30}{300} = \frac{74}{300} \).

Now, apply Bayes' Theorem:
\( P(B|L) = \frac{P(L \cap B)}{P(L)} = \frac{24/300}{74/300} = \frac{24}{74} \).

Simplifying the fraction gives:
\( \frac{24}{74} = \frac{12}{37} \).


Step 4: Final Answer:

The probability that the candidate travelled by bus is \(\frac{12}{37}\).
Quick Tip: To avoid tricky arithmetic errors in Bayes' Theorem problems, immediately convert all intermediate fractional products to a single large common denominator before summing them up.

Step 1: Understanding the Concept:

We have the statistical parameters (count, mean, variance) of two distinct groups. We need to merge these groups and find the variance of the combined dataset.


Step 2: Key Formula or Approach:

The combined mean of two groups is given by:
\( \bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} \)

The combined variance formula is:
\( \sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2} \)

where \( d_1 = \bar{x}_1 - \bar{x} \) and \( d_2 = \bar{x}_2 - \bar{x} \).


Step 3: Detailed Explanation:

Given data:

Set 1: \( n_1 = 4 \), \( \bar{x}_1 = 1 \), \( \sigma_1^2 = 13 \)

Set 2: \( n_2 = 6 \), \( \bar{x}_2 = 2 \), \( \sigma_2^2 = 1 \)

First, calculate the combined mean \( \bar{x} \):
\( \bar{x} = \frac{4(1) + 6(2)}{4 + 6} = \frac{4 + 12}{10} = \frac{16}{10} = 1.6 \).

Next, calculate the deviations \( d_1 \) and \( d_2 \) from the combined mean:
\( d_1 = \bar{x}_1 - \bar{x} = 1 - 1.6 = -0.6 \implies d_1^2 = 0.36 \)
\( d_2 = \bar{x}_2 - \bar{x} = 2 - 1.6 = 0.4 \implies d_2^2 = 0.16 \)

Now, substitute all values into the combined variance formula:
\( \sigma^2 = \frac{4(13 + 0.36) + 6(1 + 0.16)}{10} \)
\( \sigma^2 = \frac{4(13.36) + 6(1.16)}{10} \)

Calculate the numerator:
\( 4 \times 13.36 = 53.44 \)
\( 6 \times 1.16 = 6.96 \)

Total numerator = \( 53.44 + 6.96 = 60.40 \).

Now divide by the total number of observations (10):
\( \sigma^2 = \frac{60.40}{10} = 6.04 \).


Step 4: Final Answer:

The variance of all these 10 observations is 6.04.
Quick Tip: Always memorize the pooled variance formula. The alternative method (recalculating the sum of squares \( \sum x_i^2 \) for both sets and then using the standard variance formula) takes much longer and is highly prone to calculation errors.

Step 1: Understanding the Concept:

The presence of binomial coefficients divided by odd numbers (\( 3, 5, 7 \dots \)) indicates that the sequence originates from the integration of a binomial expansion. The alternating combination suggests we need to add two binomial series before integrating.


Step 2: Key Formula or Approach:

Use the standard expansion identity for even powers:
\( (1+x)^n + (1-x)^n = 2 \left[ \binom{n}{0} + \binom{n}{2}x^2 + \binom{n}{4}x^4 + \dots \right] \)

Integrate both sides with respect to \( x \) from \( 0 \) to the required upper limit to match the powers in the question.


Step 3: Detailed Explanation:

Let the given sum inside the parentheses be \( S \):
\( S = \frac{2^3}{3} \binom{12}{2} + \frac{2^5}{5} \binom{12}{4} + \dots + \frac{2^{13}}{13} \binom{12}{12} \).

Set \( n = 12 \) in our identity:
\( (1+x)^{12} + (1-x)^{12} = 2 \left[ \binom{12}{0} + \binom{12}{2}x^2 + \binom{12}{4}x^4 + \dots + \binom{12}{12}x^{12} \right] \).

Integrate both sides with respect to \( x \) from \( 0 \) to \( 2 \):
\( \int_0^2 \left( (1+x)^{12} + (1-x)^{12} \right) dx = 2 \int_0^2 \left[ \binom{12}{0} + \binom{12}{2}x^2 + \dots + \binom{12}{12}x^{12} \right] dx \).

Evaluate the Left Hand Side (LHS):
\( LHS = \left[ \frac{(1+x)^{13}}{13} - \frac{(1-x)^{13}}{13} \right]_0^2 \)
\( = \left( \frac{3^{13}}{13} - \frac{(-1)^{13}}{13} \right) - \left( \frac{1^{13}}{13} - \frac{1^{13}}{13} \right) \)
\( = \frac{3^{13} + 1}{13} - 0 = \frac{3^{13} + 1}{13} \).

Now evaluate the Right Hand Side (RHS):
\( RHS = 2 \left[ \binom{12}{0}x + \binom{12}{2}\frac{x^3}{3} + \binom{12}{4}\frac{x^5}{5} + \dots + \binom{12}{12}\frac{x^{13}}{13} \right]_0^2 \)
\( = 2 \left[ \binom{12}{0}(2) + \binom{12}{2}\frac{2^3}{3} + \binom{12}{4}\frac{2^5}{5} + \dots + \binom{12}{12}\frac{2^{13}}{13} \right] \).

Notice that the terms from the second one onwards exactly form our sum \( S \).
\( RHS = 2 \left[ 2(1) + S \right] = 4 + 2S \).

Equating LHS and RHS:
\( \frac{3^{13} + 1}{13} = 4 + 2S \)

Multiply by 13:
\( 3^{13} + 1 = 52 + 26S \)
\( 26S = 3^{13} + 1 - 52 \implies 26S = 3^{13} - 51 \).

The original equation is given as \( 26S = 3^{13} - \alpha \).

Comparing both expressions, we find \( \alpha = 51 \).


Step 4: Final Answer:

The value of \( \alpha \) is 51.
Quick Tip: When generating binomial terms with denominators, use integration. When identifying the limit of integration, look at the base of the powers matching the combinations. In this case, \( 2^3, 2^5, \dots \) indicates evaluating the integral at \( x=2 \).

Step 1: Understanding the Concept:

The problem asks for the number of ways to distribute 4 distinct objects (books) into 3 distinct containers (bags) such that no container is left empty. This is equivalent to finding the number of onto (surjective) functions from a set of size 4 to a set of size 3.


Step 2: Key Formula or Approach:

There are two primary methods:

Method 1: Using the Principle of Inclusion-Exclusion formula:

Total onto functions = \( \sum_{k=0}^{n} (-1)^k \binom{n}{k} (n-k)^m \), where \( m \) is the number of items and \( n \) is the number of boxes.

Method 2: Grouping and distribution. Divide the 4 books into 3 groups and then arrange them in the 3 bags.


Step 3: Detailed Explanation:

Let's use the grouping and distribution method.

Since we have 4 distinct books and 3 distinct bags, and no bag can be empty, the number of books in the bags must follow the distribution pattern 2, 1, 1.

First, divide the 4 distinct books into 3 groups of sizes 2, 1, and 1.

The number of ways to divide \( n \) distinct items into groups of size \( p, q, r \) is \( \frac{n!}{p! q! r! \cdot k!} \), where \( k \) is the number of groups of identical sizes.

Here, the sizes are 2, 1, 1. We have two groups of size 1, so \( k = 2 \).

Number of ways to form the groups = \( \frac{4!}{2! 1! 1! 2!} = \frac{24}{2 \cdot 1 \cdot 1 \cdot 2} = \frac{24}{4} = 6 \).

Now we have 3 distinct groups of books. We need to distribute them into 3 distinct bags.

Number of ways to arrange 3 groups into 3 bags is \( 3! = 6 \).

Total number of ways = (Number of ways to group) \( \times \) (Number of ways to distribute)

Total ways = \( 6 \times 6 = 36 \).


Alternatively, using Inclusion-Exclusion:

Total ways without restriction = \( 3^4 = 81 \).

Subtract ways where at least 1 bag is empty = \( \binom{3}{1} \times 2^4 = 3 \times 16 = 48 \).

Add ways where at least 2 bags are empty (since they were subtracted twice) = \( \binom{3}{2} \times 1^4 = 3 \times 1 = 3 \).

Number of ways = \( 81 - 48 + 3 = 36 \).


Step 4: Final Answer:

The total number of ways is 36.
Quick Tip: When grouping distinct objects into identical groups (like two groups of 1 book), don't forget to divide by the factorial of the number of identical groups (\( 2! \) in this case) to prevent overcounting permutations of the identical groups themselves.

Step 1: Understanding the Concept:

We must first find the point of intersection of the two given lines. Then, a generic line passing through this point intercepts the coordinate axes at points P and Q. The goal is to find the mathematical relationship (locus) defined by the coordinates of the midpoint of the line segment PQ.


Step 2: Key Formula or Approach:

1. Solve the system of linear equations to find the intersection point \( (x_0, y_0) \).

2. Assume the equation of the line in intercept form: \( \frac{x}{a} + \frac{y}{b} = 1 \).

3. The axis intersections are \( P(a, 0) \) and \( Q(0, b) \).

4. The midpoint of PQ is \( (h, k) = (\frac{a}{2}, \frac{b}{2}) \).

5. Substitute \( a \) and \( b \) in terms of \( h \) and \( k \) into the line equation satisfying the intersection point to obtain the locus.


Step 3: Detailed Explanation:

Let's find the point of intersection of:
\( 4x + 3y = 1 \) \quad \dots(1)
\( 3x + 4y = 1 \) \quad \dots(2)

Adding (1) and (2):
\( 7x + 7y = 2 \implies x + y = \frac{2}{7} \).

Subtracting (2) from (1):
\( x - y = 0 \implies x = y \).

Substitute \( x = y \) into the addition result:
\( x + x = \frac{2}{7} \implies 2x = \frac{2}{7} \implies x = \frac{1}{7} \) and thus \( y = \frac{1}{7} \).

The point of intersection is \( (\frac{1}{7}, \frac{1}{7}) \).

Let the straight line have x-intercept \( a \) and y-intercept \( b \). Its equation is:
\( \frac{x}{a} + \frac{y}{b} = 1 \).

Since the line passes through \( (\frac{1}{7}, \frac{1}{7}) \), substitute these coordinates:
\( \frac{1/7}{a} + \frac{1/7}{b} = 1 \implies \frac{1}{a} + \frac{1}{b} = 7 \).

The line intersects the axes at \( P(a, 0) \) and \( Q(0, b) \).

Let the midpoint of PQ be \( (h, k) \).

By the midpoint formula: \( h = \frac{a + 0}{2} = \frac{a}{2} \implies a = 2h \).

And \( k = \frac{0 + b}{2} = \frac{b}{2} \implies b = 2k \).

Substitute \( a \) and \( b \) into our equation:
\( \frac{1}{2h} + \frac{1}{2k} = 7 \).

Multiply the entire equation by 2:
\( \frac{1}{h} + \frac{1}{k} = 14 \).

Find the common denominator:
\( \frac{h + k}{hk} = 14 \implies h + k = 14hk \).

To represent the locus, replace \( (h, k) \) with generic coordinates \( (x, y) \):
\( x + y = 14xy \implies x + y - 14xy = 0 \).


Step 4: Final Answer:

The locus of the midpoint is \( x + y - 14xy = 0 \).
Quick Tip: For questions concerning lines forming segments on axes, always utilize the intercept form of a line \( \frac{x}{a} + \frac{y}{b} = 1 \). It directly incorporates the axis interception points, massively reducing algebraic clutter.

Step 1: Understanding the Concept:

We are dealing with a standard parabola \( y^2 = 4ax \). We take two points P and Q on it such that they subtend a right angle at the vertex O. We need to find the locus of the midpoint of PQ and then determine the latus rectum of the resulting conic.


Step 2: Key Formula or Approach:

For the parabola \( y^2 = 4ax \), use parametric coordinates for points on the curve: \( (at^2, 2at) \).

The condition that OP is perpendicular to OQ means the product of their slopes is -1.

Let the midpoint of PQ be \( (h, k) \). Express \( h \) and \( k \) in terms of the parameters of P and Q, and eliminate the parameters to find the locus.


Step 3: Detailed Explanation:

The given parabola is \( y^2 = 4x \), so \( a = 1 \).

The vertex is \( O(0,0) \).

Let the coordinates of P be \( (t_1^2, 2t_1) \) and Q be \( (t_2^2, 2t_2) \).

Slope of OP \( = m_{OP} = \frac{2t_1 - 0}{t_1^2 - 0} = \frac{2}{t_1} \).

Slope of OQ \( = m_{OQ} = \frac{2t_2 - 0}{t_2^2 - 0} = \frac{2}{t_2} \).

Since OP and OQ are perpendicular, \( m_{OP} \times m_{OQ} = -1 \).
\( \left(\frac{2}{t_1}\right) \left(\frac{2}{t_2}\right) = -1 \implies \frac{4}{t_1 t_2} = -1 \implies t_1 t_2 = -4 \).

Let \( M(h, k) \) be the midpoint of the line segment PQ.

By the midpoint formula:
\( h = \frac{t_1^2 + t_2^2}{2} \implies t_1^2 + t_2^2 = 2h \).
\( k = \frac{2t_1 + 2t_2}{2} \implies t_1 + t_2 = k \).

We need to eliminate \( t_1 \) and \( t_2 \) to find the relation between \( h \) and \( k \).

We know the algebraic identity: \( (t_1 + t_2)^2 = t_1^2 + t_2^2 + 2t_1 t_2 \).

Substitute the values we've established:
\( k^2 = 2h + 2(-4) \).
\( k^2 = 2h - 8 \).
\( k^2 = 2(h - 4) \).

To find the locus, replace \( (h, k) \) with \( (x, y) \):
\( y^2 = 2(x - 4) \).

This is an equation of a parabola of the form \( Y^2 = 4A X \), where \( Y = y \), \( X = x - 4 \), and \( 4A = 2 \).

The conic C is a parabola. The length of its latus rectum is the coefficient of the linear term, which is \( 4A \).

Therefore, the length of the latus rectum is 2.


Step 4: Final Answer:

The length of the latus rectum is 2.
Quick Tip: For chords subtending a right angle at the vertex of \( y^2 = 4ax \), remember the standard property \( t_1 t_2 = -4 \). It saves a lot of time calculating slopes from scratch.

Step 1: Understanding the Concept:

We must estimate the approximate values of \( \alpha \) and \( \beta \) by bounding the arguments of the inverse trigonometric functions using standard known values. This will help determine the quadrants in which \( \alpha \) and \(\alpha+\beta\) lie, dictating the sign of their cosine values.


Step 2: Key Formula or Approach:

Use bounding values:

For sine: \( \sin(\pi/6) = 0.5 \), \( \sin(\pi/4) \approx 0.707 \), \( \sin(\pi/3) \approx 0.866 \).

For cosine: \( \cos(\pi/3) = 0.5 \), \( \cos(\pi/2) = 0 \).

Determine ranges for \( \alpha/3 \) and \( \beta/3 \) and subsequently ranges for \( \alpha \) and \( \beta \).


Step 3: Detailed Explanation:

Let's analyze \( \alpha = 3 \sin^{-1} \left( \frac{6}{11} \right) \).
\( \frac{6}{11} \approx 0.545 \).

We know \( \sin(\frac{\pi}{6}) = \frac{1}{2} = 0.5 \) and \( \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \approx 0.707 \).

Since \( 0.5 < 0.545 < 0.707 \), we have \( \frac{\pi}{6} < \sin^{-1} \left( \frac{6}{11} \right) < \frac{\pi}{4} \).

Multiply by 3 to find the range of \( \alpha \):
\( 3 \left( \frac{\pi}{6} \right) < \alpha < 3 \left( \frac{\pi}{4} \right) \implies \frac{\pi}{2} < \alpha < \frac{3\pi}{4} \).

Since \( \alpha \) is in the second quadrant, \( \cos \alpha \) must be negative.

Thus, \( \cos \alpha < 0 \). Statement II is true.


Now let's analyze \( \beta = 3 \cos^{-1} \left( \frac{4}{9} \right) \).
\( \frac{4}{9} \approx 0.444 \).

We know \( \cos(\frac{\pi}{3}) = \frac{1}{2} = 0.5 \) and \( \cos(\frac{\pi}{2}) = 0 \).

Since cosine is decreasing in the first quadrant and \( 0 < 0.444 < 0.5 \), we have:
\( \frac{\pi}{3} < \cos^{-1} \left( \frac{4}{9} \right) < \frac{\pi}{2} \).

Multiply by 3 to find the range of \( \beta \):
\( 3 \left( \frac{\pi}{3} \right) < \beta < 3 \left( \frac{\pi}{2} \right) \implies \pi < \beta < \frac{3\pi}{2} \).

Now, let's find the range of \( \alpha + \beta \):

Add the ranges of \( \alpha \) and \( \beta \):
\( \frac{\pi}{2} + \pi < \alpha + \beta < \frac{3\pi}{4} + \frac{3\pi}{2} \)
\( \frac{3\pi}{2} < \alpha + \beta < \frac{9\pi}{4} \).

The angle \( \alpha + \beta \) falls between \( 270^\circ \) and \( 405^\circ \). This means it is in the fourth quadrant (or just crosses into the first).

Wait, let's look closer. Is it definitely in the fourth quadrant or could it be in the first?

Let's approximate better.
\( \sin(\alpha/3) \approx 0.545 \implies \alpha/3 \approx 33^\circ \implies \alpha \approx 99^\circ \).
\( \cos(\beta/3) \approx 0.444 \implies \beta/3 \approx 63.6^\circ \implies \beta \approx 190.8^\circ \).

Sum: \( \alpha + \beta \approx 99^\circ + 190.8^\circ = 289.8^\circ \).

An angle of \( 289.8^\circ \) is strictly in the fourth quadrant (\( 270^\circ < 289.8^\circ < 360^\circ \)).

In the fourth quadrant, the cosine function is positive.

Thus, \( \cos(\alpha + \beta) > 0 \). Statement I is true.


Step 4: Final Answer:

Both Statement I and Statement II are true.
Quick Tip: For bounding inverse trig functions, remembering decimal approximations of common angles (\( \sqrt{2}/2 \approx 0.707, \sqrt{3}/2 \approx 0.866, 1/2 = 0.5 \)) allows for rapid quadrant isolation.

Step 1: Understanding the Concept:

We need to assess the differentiability of a function involving absolute values, particularly looking at critical points like \( x=0 \). Then, we must determine the monotonicity of the function in a given interval by inspecting the sign of its first derivative.


Step 2: Key Formula or Approach:

Differentiability at \( x=0 \): Check if left-hand derivative (LHD) equals right-hand derivative (RHD).
\( f'_+(0) = \lim_{x \to 0^+} \frac{f(x) - f(0)}{x} \)
\( f'_-(0) = \lim_{x \to 0^-} \frac{f(x) - f(0)}{x} \)

Monotonicity: Find \( f'(x) \) for \( x \in (-\pi, -\pi/2) \) and verify if \( f'(x) > 0 \).


Step 3: Detailed Explanation:

Statement I: Check differentiability.

The only point of concern for \( |x| \) is \( x=0 \). For \( x \neq 0 \), the function is clearly a composition of differentiable functions.

At \( x=0 \), \( f(0) = e^{\sin 0} - 0 = 1 \).

Right-hand derivative at \( x=0 \):

For \( x > 0 \), \( f(x) = e^{\sin x} - x \).
\( f'(x) = e^{\sin x} \cos x - 1 \).
\( f'_+(0) = \lim_{x \to 0^+} (e^{\sin x} \cos x - 1) = e^0(1) - 1 = 1 - 1 = 0 \).

Left-hand derivative at \( x=0 \):

For \( x < 0 \), \( f(x) = e^{\sin(-x)} - (-x) = e^{-\sin x} + x \).
\( f'(x) = e^{-\sin x} (-\cos x) + 1 \).
\( f'_-(0) = \lim_{x \to 0^-} (-e^{-\sin x} \cos x + 1) = -e^0(1) + 1 = -1 + 1 = 0 \).

Since \( f'_+(0) = f'_-(0) = 0 \), \( f(x) \) is differentiable at \( x=0 \).

Therefore, \( f(x) \) is differentiable for all \( x \in \mathbb{R} \). Statement I is true.


Statement II: Monotonicity in \( (-\pi, -\pi/2) \).

For \( x \in (-\pi, -\pi/2) \), \( x \) is negative. So \( |x| = -x \).
\( f(x) = e^{\sin(-x)} - (-x) = e^{-\sin x} + x \).

Differentiating with respect to \( x \):
\( f'(x) = e^{-\sin x}(-\cos x) + 1 \).

In the interval \( (-\pi, -\pi/2) \), which is in the third quadrant, \( \cos x \) is negative.

Therefore, \( -\cos x \) is positive.

The exponential function \( e^{-\sin x} \) is always positive.

So, \( e^{-\sin x}(-\cos x) > 0 \).

This means \( f'(x) = (positive quantity) + 1 > 1 > 0 \).

Since \( f'(x) > 0 \) for all \( x \) in the interval, \( f(x) \) is strictly increasing in \( (-\pi, -\pi/2) \).

Thus, Statement II is true.


Step 4: Final Answer:

Both Statement I and Statement II are true.
Quick Tip: Functions of the form \( f(|x|) \) are always differentiable at \( x=0 \) if \( f'(0) = 0 \). Here, for \( g(x) = e^{\sin x} - x \), \( g'(0) = 1 - 1 = 0 \), guaranteeing the differentiability of \( g(|x|) \) at the origin.

Step 1: Understanding the Concept:

We are given a relationship involving cross products of three vectors. By using the anti-commutative property of cross products, we can express one vector as a linear combination of the other two. We use the dot product given to find the scalar constant in the combination.


Step 2: Key Formula or Approach:

Cross product properties: \( \vec{b} \times \vec{c} = -(\vec{c} \times \vec{b}) \).

If \( \vec{u} \times \vec{v} = \vec{0} \), then vectors \( \vec{u} and \vec{v} \) are collinear, meaning \( \vec{u} = \lambda \vec{v} \).


Step 3: Detailed Explanation:

Given equation: \( 2(\vec{a} \times \vec{b}) + 3(\vec{b} \times \vec{c}) = \vec{0} \).

Use property of cross product to reverse the second term:
\( 2(\vec{a} \times \vec{b}) - 3(\vec{c} \times \vec{b}) = \vec{0} \).

Factor out \( \times \vec{b} \) on the right side:
\( (2\vec{a} - 3\vec{c}) \times \vec{b} = \vec{0} \).

Since the cross product of two vectors is zero, they must be parallel (or one is the zero vector). Assuming \( \vec{b} \neq 0 \):
\( 2\vec{a} - 3\vec{c} = \lambda \vec{b} \) for some scalar \( \lambda \).

Rearrange to solve for \( \vec{c} \):
\( 3\vec{c} = 2\vec{a} - \lambda \vec{b} \implies \vec{c} = \frac{2}{3}\vec{a} - \frac{\lambda}{3}\vec{b} \).

We are given the dot product condition: \( \vec{a} \cdot \vec{c} = 15 \).

Take the dot product of both sides with \( \vec{a} \):
\( \vec{a} \cdot \vec{c} = \frac{2}{3} (\vec{a} \cdot \vec{a}) - \frac{\lambda}{3} (\vec{a} \cdot \vec{b}) \).

Let's compute the necessary dot products from the given vectors:
\( \vec{a} = 4\hat{i} - \hat{j} + 3\hat{k} \)
\( \vec{b} = 10\hat{i} + 2\hat{j} - \hat{k} \)
\( |\vec{a}|^2 = \vec{a} \cdot \vec{a} = (4)^2 + (-1)^2 + (3)^2 = 16 + 1 + 9 = 26 \).
\( \vec{a} \cdot \vec{b} = (4)(10) + (-1)(2) + (3)(-1) = 40 - 2 - 3 = 35 \).

Substitute these into our equation:
\( 15 = \frac{2}{3} (26) - \frac{\lambda}{3} (35) \).

Multiply the entire equation by 3 to remove fractions:
\( 45 = 52 - 35\lambda \).
\( 35\lambda = 52 - 45 = 7 \implies \lambda = \frac{7}{35} = \frac{1}{5} \).

Now we can write the explicit form of \( \vec{c} \):
\( \vec{c} = \frac{2}{3}\vec{a} - \frac{1/5}{3}\vec{b} = \frac{2}{3}\vec{a} - \frac{1}{15}\vec{b} \).

We need to evaluate \( \vec{c} \cdot (\hat{i} + \hat{j} - 3\hat{k}) \).

Let \( \vec{v} = \hat{i} + \hat{j} - 3\hat{k} \).

We want to find \( \vec{c} \cdot \vec{v} = \left( \frac{2}{3}\vec{a} - \frac{1}{15}\vec{b} \right) \cdot \vec{v} = \frac{2}{3}(\vec{a} \cdot \vec{v}) - \frac{1}{15}(\vec{b} \cdot \vec{v}) \).

Calculate the individual dot products:
\( \vec{a} \cdot \vec{v} = (4)(1) + (-1)(1) + (3)(-3) = 4 - 1 - 9 = -6 \).
\( \vec{b} \cdot \vec{v} = (10)(1) + (2)(1) + (-1)(-3) = 10 + 2 + 3 = 15 \).

Substitute these back:
\( \vec{c} \cdot \vec{v} = \frac{2}{3}(-6) - \frac{1}{15}(15) \).
\( = 2(-2) - 1 = -4 - 1 = -5 \).


Step 4: Final Answer:

The value is -5.
Quick Tip: Instead of finding the full vector \( \vec{c} \) in its \( \hat{i}, \hat{j}, \hat{k} \) components, distributing the dot product directly to the constituent vectors \( \vec{a} \) and \( \vec{b} \) is often less error-prone and faster.

Step 1: Understanding the Concept:

First, we must find the unknown parameters \( \lambda \) and \( \mu \) using the property of a foot of perpendicular. Then, substitute these parameters into the equations of two lines and find the shortest distance between them.


Step 2: Key Formula or Approach:

1. A point lies on a line if it satisfies its equation.

2. The direction vector of the line connecting a point and its foot of perpendicular is orthogonal to the direction vector of the line, meaning their dot product is zero.

3. The distance between two parallel lines passing through points \( A \) and \( B \) with a common direction vector \( \vec{d} \) is given by \( d = \frac{|\vec{AB} \times \vec{d}|}{|\vec{d}|} \).


Step 3: Detailed Explanation:

Let the initial point be \( P(\lambda, 2, 3) \) and the foot of the perpendicular be \( F(1, \mu, 2) \).

The line \( L_1 \) is given by \( \frac{x-4}{1} = \frac{y-9}{2} = \frac{z-5}{1} \).

Since the foot \( F \) lies on \( L_1 \), its coordinates must satisfy the line's equation:
\( \frac{1-4}{1} = \frac{\mu-9}{2} = \frac{2-5}{1} \).
\( -3 = \frac{\mu-9}{2} = -3 \).

Solving for \( \mu \):
\( \mu - 9 = -6 \implies \mu = 3 \).

So the foot of perpendicular is \( F(1, 3, 2) \).

The vector \( \vec{PF} \) is the vector connecting the point and its foot:
\( \vec{PF} = (1 - \lambda)\hat{i} + (3 - 2)\hat{j} + (2 - 3)\hat{k} = (1 - \lambda)\hat{i} + 1\hat{j} - 1\hat{k} \).

The direction vector of the line \( L_1 \) is \( \vec{d_1} = 1\hat{i} + 2\hat{j} + 1\hat{k} \).

Because \( \vec{PF} \) is perpendicular to \( L_1 \), their dot product must be zero:
\( \vec{PF} \cdot \vec{d_1} = 0 \).
\( (1 - \lambda)(1) + (1)(2) + (-1)(1) = 0 \).
\( 1 - \lambda + 2 - 1 = 0 \implies 2 - \lambda = 0 \implies \lambda = 2 \).

Now we have \( \lambda = 2 \) and \( \mu = 3 \).

We need to find the distance between the two lines:
\( L_2: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6} \)
\( L_3: \frac{x-2}{2} = \frac{y-3}{3} = \frac{z+5}{6} \)

Notice that the direction vectors for both lines are identical: \( \vec{d} = 2\hat{i} + 3\hat{j} + 6\hat{k} \). This means the lines are parallel.

Let \( A \) be a point on \( L_2 \): \( A(1, 2, -4) \).

Let \( B \) be a point on \( L_3 \): \( B(2, 3, -5) \).

The vector connecting the points is \( \vec{AB} = (2-1)\hat{i} + (3-2)\hat{j} + (-5 - (-4))\hat{k} = 1\hat{i} + 1\hat{j} - 1\hat{k} \).

The formula for distance between parallel lines is \( d = \frac{|\vec{AB} \times \vec{d}|}{|\vec{d}|} \).

Compute the cross product \( \vec{AB} \times \vec{d} \):
\( \vec{AB} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & 1 & -1
2 & 3 & 6 \end{vmatrix} \)
\( = \hat{i}(6 - (-3)) - \hat{j}(6 - (-2)) + \hat{k}(3 - 2) = \hat{i}(9) - \hat{j}(8) + \hat{k}(1) = 9\hat{i} - 8\hat{j} + \hat{k} \).

Calculate the magnitude of this cross product:
\( |\vec{AB} \times \vec{d}| = \sqrt{9^2 + (-8)^2 + 1^2} = \sqrt{81 + 64 + 1} = \sqrt{146} \).

Calculate the magnitude of the direction vector:
\( |\vec{d}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \).

The distance is \( d = \frac{\sqrt{146}}{7} \).


Step 4: Final Answer:

The distance is \(\frac{\sqrt{146}}{7}\).
Quick Tip: Always double-check the direction vectors of the lines before blindly applying the skewed lines distance formula. If the lines are parallel, the standard shortest distance scalar triple product will yield 0, which signifies you must use the parallel lines distance formula instead.

Step 1: Factorize the expression inside the radical

We use the identity \[ x^4+x^2+1=(x^2+x+1)(x^2-x+1) \]

because \[ (x^2+x+1)(x^2-x+1)=x^4+x^2+1 \]

Therefore, \[ \sqrt{x^4+x^2+1} = \sqrt{x^2+x+1}\sqrt{x^2-x+1} \]

Substituting into the integral, \[ I=\int_0^2 \frac{\sqrt{x}\sqrt{x^2+x+1}} {\sqrt{x+1}\sqrt{x^2+x+1}\sqrt{x^2-x+1}}\,dx \]

Canceling \(\sqrt{x^2+x+1}\), \[ I=\int_0^2 \frac{\sqrt{x}} {\sqrt{x+1}\sqrt{x^2-x+1}}\,dx \]


Step 2: Simplify the denominator

Observe that \[ (x+1)(x^2-x+1)=x^3+1 \]

Hence, \[ \sqrt{x+1}\sqrt{x^2-x+1}=\sqrt{x^3+1} \]

So the integral becomes \[ I=\int_0^2 \frac{\sqrt{x}}{\sqrt{x^3+1}}\,dx \]


Step 3: Use substitution

Let \[ t=x^{3/2} \]

Then \[ dt=\frac{3}{2}x^{1/2}\,dx \]

which gives \[ \sqrt{x}\,dx=\frac{2}{3}\,dt \]

Also, \[ x^3=t^2 \]

So \[ \sqrt{x^3+1}=\sqrt{t^2+1} \]

Changing the limits:

When \(x=0\), \[ t=0 \]

When \(x=2\), \[ t=2^{3/2}=2\sqrt{2} \]

Thus, \[ I=\frac{2}{3}\int_0^{2\sqrt{2}}\frac{dt}{\sqrt{t^2+1}} \]


Step 4: Apply standard integral formula

Using \[ \int \frac{dt}{\sqrt{t^2+1}} = \log_e\left|t+\sqrt{t^2+1}\right|+C \]

we get \[ I= \frac{2}{3} \left[ \log_e\left(t+\sqrt{t^2+1}\right) \right]_0^{2\sqrt{2}} \]
\[ I= \frac{2}{3} \left( \log_e\left(2\sqrt{2}+\sqrt{8+1}\right) - \log_e(1) \right) \]
\[ I= \frac{2}{3} \log_e(2\sqrt{2}+3) \]

Hence, \[ \boxed{ I=\frac{2}{3}\log_e(3+2\sqrt{2}) } \]


Final Answer: \[ \boxed{ \frac{2}{3}\log_e(3+2\sqrt{2}) } \] Quick Tip: Whenever you see \( x^4 + x^2 + 1 \) in an algebraic integral or limit, automatically think of the Sophie Germain identity variant: \( x^4 + x^2 + 1 = (x^2+x+1)(x^2-x+1) \). It's a highly repetitive trick in competitive exams.

Step 1: Convert into linear differential equation

Divide by \(x\sqrt{1-x^2}\,dx\):
\[ \frac{dy}{dx}+\frac{1}{x}y=\frac{\cos^{-1}x}{\sqrt{1-x^2}} \]

This is of the form \[ \frac{dy}{dx}+P(x)y=Q(x) \]

where \[ P(x)=\frac{1}{x} \]


Step 2: Find integrating factor
\[ I.F.=e^{\int \frac1x dx}=e^{\ln x}=x \]

Multiply throughout by \(x\):
\[ x\frac{dy}{dx}+y=\frac{x\cos^{-1}x}{\sqrt{1-x^2}} \]
\[ \frac{d}{dx}(xy)=\frac{x\cos^{-1}x}{\sqrt{1-x^2}} \]


Step 3: Integrate both sides
\[ xy=\int \frac{x\cos^{-1}x}{\sqrt{1-x^2}}dx+C \]

Let \[ t=\cos^{-1}x \]

Then \[ x=\cos t,\qquad dx=-\sin t\,dt \]

and \[ \sqrt{1-x^2}=\sin t \]

So,
\[ \int \frac{x\cos^{-1}x}{\sqrt{1-x^2}}dx = -\int t\cos t\,dt \]

Using integration by parts,
\[ \int t\cos t\,dt=t\sin t+\cos t \]

Hence,
\[ \int \frac{x\cos^{-1}x}{\sqrt{1-x^2}}dx = -\left(t\sin t+\cos t\right) \]

Substituting back,
\[ = -\cos^{-1}x\sqrt{1-x^2}-x \]

Therefore,
\[ xy=-\cos^{-1}x\sqrt{1-x^2}-x+C \]
\[ y=-\frac{\cos^{-1}x\sqrt{1-x^2}}{x}-1+\frac{C}{x} \]


Step 4: Apply boundary condition
\[ \lim_{x\to1^-}y(x)=1 \]

As \(x\to1^-\),
\[ \cos^{-1}x\to0,\qquad \sqrt{1-x^2}\to0 \]

Thus,
\[ 1=-1+C \]
\[ C=2 \]

So,
\[ y(x)=\frac{2}{x}-1-\frac{\cos^{-1}x\sqrt{1-x^2}}{x} \]


Step 5: Find \(y(1/2)\)
\[ y\left(\frac12\right)=4-1-\frac{\frac{\pi}{3}\cdot\frac{\sqrt3}{2}}{\frac12} \]
\[ =3-\frac{\pi\sqrt3}{3} \]
\[ =3-\frac{\pi}{\sqrt3} \]
\[ \boxed{y\left(\frac12\right)=3-\frac{\pi}{\sqrt3}} \] Quick Tip: Instead of blindly dividing out the entire coefficient of \( dy \), notice if \( x \, dy + y \, dx \) is present. Here, \( x \, dy + y \, dx \) is the exact differential \( d(xy) \). Re-arranging gives \( \sqrt{1-x^2} d(xy) = x \cos^{-1}x \, dx \). The integrating factor drops out natively!

Step 1: Find pattern of composition

Given
\[ f(x)=\frac{x-1}{x+1} \]
\[ f^2(x)=f(f(x))=-\frac1x \]
\[ f^3(x)=f\left(-\frac1x\right)=\frac{x+1}{1-x} \]
\[ f^4(x)=x \]

Hence cycle length is \(4\).
\[ f^{26}(x)=f^{24+2}(x)=f^2(x)=-\frac1x \]

Given
\[ g(x)+f^{26}(x)=0 \]
\[ g(x)=\frac1x \]


Step 2: Find bounded area

Line:
\[ 2y=2x-3 \]
\[ y=x-\frac32 \]

Intersection with hyperbola:
\[ \frac1x=x-\frac32 \]
\[ 2=2x^2-3x \]
\[ 2x^2-3x-2=0 \]
\[ (x-2)(2x+1)=0 \]
\[ x=2 \]

Area:
\[ A=\int_{3/2}^{2}\left(x-\frac32\right)dx+\int_2^4\frac1x dx \]

First part:
\[ =\frac18 \]

Second part:
\[ =\ln4-\ln2=\ln2 \]

Hence
\[ \boxed{A=\frac18+\ln2} \] Quick Tip: Function composition cycles are a staple in JEE Main. Whenever you see a high exponent like \( f^{26}(x) \), calculate up to \( f^3(x) \) or \( f^4(x) \)—it is virtually guaranteed to return to \( x \), \( -x \), or \( -1/x \).

Step 1: Continuity at \(x=\pi/2\)

For continuity,
\[ \frac13=\lim_{x\to\pi/2^+}\frac{b(1-\sin x)}{(\pi-2x)^2} \]

Let \(x=\frac\pi2+h\).

Then,
\[ 1-\sin x=1-\cos h \]
\[ (\pi-2x)^2=4h^2 \]

So,
\[ \frac13=\frac{b}{4}\lim_{h\to0}\frac{1-\cos h}{h^2} \]

Using
\[ \lim_{h\to0}\frac{1-\cos h}{h^2}=\frac12 \]
\[ \frac13=\frac{b}{8} \]
\[ b=\frac83 \]


Step 2: Evaluate integral
\[ 3b-6=8-6=2 \]

So,
\[ I=\int_0^2|x^2+2x-3|dx \]

Factor:
\[ x^2+2x-3=(x+3)(x-1) \]

Split at \(x=1\):
\[ I=\int_0^1(3-2x-x^2)dx+\int_1^2(x^2+2x-3)dx \]
\[ =\frac53+\frac73=4 \]
\[ \boxed{4} \] Quick Tip: For limits involving \( 1 - \sin x \) as \( x \to \pi/2 \), substituting \( x = \pi/2 - h \) or \( \pi/2 + h \) immediately converts the problematic sine into a cosine, unlocking the standard \( (1-\cos h)/h^2 \) identity.

For ellipse with major axis along \(y\)-axis,
\[ f(3a+15)>f(a^2+7a+3) \]

Since \(f\) is strictly decreasing,
\[ 3a+15 \[ a^2+4a-12>0 \]
\[ (a+6)(a-2)>0 \]
\[ a\in(-\infty,-6)\cup(2,\infty) \]

Thus,
\[ \mathbb{R}\setminus[\alpha,\beta]=\mathbb{R}\setminus[-6,2] \]

So,
\[ \alpha=-6,\qquad \beta=2 \]
\[ \alpha^2+\beta^2=36+4=40 \]
\[ \boxed{40} \] Quick Tip: Whenever a function is "strictly decreasing", applying it or removing it from an inequality reverses the inequality sign. For strictly increasing functions, the sign remains the same.

Step 1: Factorize the expressions
\[ 2x^2+5x+3=(2x+3)(x+1) \]
\[ x^2+2x+1=(x+1)^2 \]

Substitute in the given equation:
\[ \log_{x+1}\big((2x+3)(x+1)\big)=4-\log_{2x+3}(x+1)^2 \]


Step 2: Use logarithm properties
\[ \log_{x+1}(2x+3)+\log_{x+1}(x+1)=4-2\log_{2x+3}(x+1) \]

Since
\[ \log_{x+1}(x+1)=1 \]

we get
\[ \log_{x+1}(2x+3)+1=4-2\log_{2x+3}(x+1) \]

Using change of base,
\[ \log_{2x+3}(x+1)=\frac{1}{\log_{x+1}(2x+3)} \]

Let
\[ t=\log_{x+1}(2x+3) \]

Then
\[ t+1=4-\frac{2}{t} \]
\[ t^2-3t+2=0 \]
\[ (t-1)(t-2)=0 \]
\[ t=1 \quad or \quad t=2 \]


Step 3: Solve each case

For \(t=1\),
\[ \log_{x+1}(2x+3)=1 \]
\[ 2x+3=x+1 \]
\[ x=-2 \]

Rejected because base \(x+1=-1<0\).

For \(t=2\),
\[ 2x+3=(x+1)^2 \]
\[ 2x+3=x^2+2x+1 \]
\[ x^2=2 \]
\[ x=\pm\sqrt{2} \]

Only \(x=\sqrt{2}\) satisfies domain conditions.


Step 4: Sum of squares
\[ (\sqrt{2})^2=2 \]
\[ \boxed{2} \] Quick Tip: Always verify the solutions obtained from logarithmic equations against the domain constraints: the base must be positive and not equal to 1, and the argument must be strictly positive.

Let
\[ I=\int_{\pi/6}^{\pi/4}\left(\cot\left(x-\frac{\pi}{3}\right)\cot\left(x+\frac{\pi}{3}\right)+1\right)\,dx \]

Step 1: Simplify integrand

Using identity
\[ \cot A\cot B+1=\frac{\cos(A-B)}{\sin A\sin B} \]

Take
\[ A=x-\frac{\pi}{3},\quad B=x+\frac{\pi}{3} \]

Then
\[ A-B=-\frac{2\pi}{3} \]
\[ \cos(A-B)=\cos\left(\frac{2\pi}{3}\right)=-\frac12 \]

Also,
\[ \sin A\sin B=\sin^2x-\sin^2\frac{\pi}{3} \]
\[ =\sin^2x-\frac34 \]

Hence
\[ I=\int_{\pi/6}^{\pi/4}\frac{-1/2}{\sin^2x-3/4}\,dx \]
\[ =\int_{\pi/6}^{\pi/4}\frac{-2}{4\sin^2x-3}\,dx \]


Step 2: Put \(t=\tan x\)

Multiply numerator and denominator by \(\sec^2x\):
\[ I=\int_{\pi/6}^{\pi/4}\frac{-2\sec^2x}{\tan^2x-3}\,dx \]

Let
\[ t=\tan x,\quad dt=\sec^2x\,dx \]

Limits:
\[ x=\frac{\pi}{6}\Rightarrow t=\frac{1}{\sqrt3} \]
\[ x=\frac{\pi}{4}\Rightarrow t=1 \]

Thus
\[ I=2\int_{1/\sqrt3}^{1}\frac{dt}{3-t^2} \]
\[ =\frac{1}{\sqrt3}\left[\ln\left|\frac{\sqrt3+t}{\sqrt3-t}\right|\right]_{1/\sqrt3}^{1} \]
\[ I=-\frac{2}{\sqrt3}\ln(\sqrt3-1) \]

Comparing with
\[ I=\alpha\ln(\sqrt3-1) \]
\[ \alpha=-\frac{2}{\sqrt3} \]
\[ 9\alpha^2=9\cdot\frac{4}{3}=12 \]
\[ \boxed{12} \] Quick Tip: To integrate expressions of the form \(\frac{1}{a\sin^2 x + b\cos^2 x + c}\), multiplying the numerator and denominator by \(\sec^2 x\) and substituting \(t = \tan x\) is the most robust standard operating procedure.

Direction vectors:
\[ \vec d_2=(1,2,2) \]
\[ \vec d_3=(2,2,1) \]

Step 1: Find line perpendicular to both
\[ \vec d_1=\vec d_2\times \vec d_3 \]
\[ = \begin{vmatrix} \hat i & \hat j & \hat k
1&2&2
2&2&1 \end{vmatrix} \]
\[ =-2\hat i+3\hat j-2\hat k \]

So line \(L_1\):
\[ (x,y,z)=(-2k,3k,-2k) \]


Step 2: Intersection with \(L_2\)

From \(L_2\),
\[ (x,y,z)=(3+t,2t-1,2t+4) \]

Equating,
\[ -2k=3+t \]
\[ 3k=2t-1 \]

Solving gives
\[ k=-1,\quad t=-1 \]

Intersection point:
\[ P=(2,-3,2) \]


Step 3: Use distance formula

Point on \(L_3\):
\[ Q=(3+2s,3+2s,2+s) \]

Given
\[ PQ=\sqrt{17} \]
\[ (2s+1)^2+(2s+6)^2+s^2=17 \]
\[ 9s^2+28s+20=0 \]
\[ (9s+10)(s+2)=0 \]
\[ s=-\frac{10}{9},\,-2 \]

For integer \(a\), choose
\[ s=-2 \]

Point:
\[ (-1,-1,0) \]
\[ (a+b+c)^2=(-2)^2=4 \]
\[ \boxed{4} \] Quick Tip: When asked to find a point whose coordinates satisfy integer constraints, express the coordinates parametrically, solve the resulting distance/geometric equation, and isolate the parameter value that yields an integer result.

\[ lx+my=1 \]

Using homogenization for right angle at origin,
\[ 11(l^2+m^2)+6l+8m-2=0 \]

Let foot of perpendicular from origin be \((h,k)\).

Chord equation:
\[ hx+ky=h^2+k^2 \]

Comparing with \(lx+my=1\),
\[ l=\frac{h}{h^2+k^2},\quad m=\frac{k}{h^2+k^2} \]

Substituting,
\[ 11+6h+8k-2(h^2+k^2)=0 \]
\[ h^2+k^2-3h-4k-\frac{11}{2}=0 \]

Replacing \((h,k)\) by \((x,y)\),
\[ x^2+y^2-3x-4y-\frac{11}{2}=0 \]

Hence
\[ \alpha=3,\quad \beta=4,\quad \gamma=\frac{11}{2} \]
\[ \alpha+\beta+2\gamma=3+4+11=18 \]
\[ \boxed{18} \] Quick Tip: When a curve's chord subtends a right angle at the origin, homogenization is the primary tool. Always equate the sum of the \(x^2\) and \(y^2\) coefficients to zero.

Step 1: Sum GP
\[ 2+\frac23+\frac29+\cdots \]
\[ S=\frac{2}{1-1/3}=3 \]

So equation becomes
\[ \log_2 f(x)=\log_2 3\cdot \log_3\left(1+\frac{f(x)}{f(1/x)}\right) \]
\[ =\log_2\left(1+\frac{f(x)}{f(1/x)}\right) \]

Hence
\[ f(x)=1+\frac{f(x)}{f(1/x)} \]
\[ f(x)f(1/x)=f(x)+f(1/x) \]

Polynomial solution:
\[ f(x)=x^n+1 \]

Given
\[ f(6)=37 \]
\[ 6^n+1=37 \]
\[ 6^n=36 \]
\[ n=2 \]

So
\[ f(x)=x^2+1 \]


Step 2: Required sum
\[ \sum_{n=1}^{10}f(n)=\sum_{n=1}^{10}(n^2+1) \]
\[ =\sum_{n=1}^{10}n^2+10 \]
\[ =\frac{10\cdot11\cdot21}{6}+10 \]
\[ =385+10 \]
\[ =395 \]
\[ \boxed{395} \] Quick Tip: Memorize the functional equation \(f(x)f(1/x) = f(x) + f(1/x)\) which strictly implies \(f(x) = \pm x^n + 1\). Recognizing this instantly bypasses complicated polynomial coefficient matching.