JEE Main 2024 question paper pdf with solutions- Download April 9 Shift 1 Mathematics Question Paper pdf

Step 1: Simplifying the integrand.

We begin by considering the given integral: \[ I = \int \frac{2 - \tan x}{3 + \tan x} \, dx \]
To solve this, let's use a substitution method. Let: \[ t = \tan x \]
Then, the differential \( dt = \sec^2 x \, dx \). Now, we can rewrite the integral in terms of \( t \). First, express \( \sec^2 x \) in terms of \( t \): \[ \sec^2 x = 1 + \tan^2 x = 1 + t^2 \]

Step 2: Substituting into the integral.

Substitute into the integral: \[ I = \int \frac{2 - t}{3 + t} \cdot \frac{dt}{1 + t^2} \]
Now simplify the fraction: \[ I = \int \frac{(2 - t)}{(3 + t)(1 + t^2)} \, dt \]

Step 3: Breaking the fraction into partial fractions.

Next, we break the fraction into partial fractions to make it easier to integrate. The denominator factors as: \[ (3 + t)(1 + t^2) \]
We can express the integrand as: \[ \frac{2 - t}{(3 + t)(1 + t^2)} = \frac{A}{3 + t} + \frac{Bt + C}{1 + t^2} \]
Multiplying both sides by \( (3 + t)(1 + t^2) \) and solving for \( A \), \( B \), and \( C \), we get: \[ 2 - t = A(1 + t^2) + (Bt + C)(3 + t) \]
Expanding both sides and equating coefficients gives the system of equations for \( A \), \( B \), and \( C \). After solving, we find that: \[ A = 1, \quad B = -1, \quad C = 0 \]

Step 4: Integrating.

Now, the integral becomes: \[ I = \int \frac{1}{3 + t} \, dt - \int \frac{t}{1 + t^2} \, dt \]
The first integral is straightforward: \[ \int \frac{1}{3 + t} \, dt = \ln |3 + t| \]
The second integral is a standard arctangent form: \[ \int \frac{t}{1 + t^2} \, dt = \frac{1}{2} \ln |1 + t^2| \]
So, the integral becomes: \[ I = \ln |3 + t| - \frac{1}{2} \ln |1 + t^2| + C \]
Finally, substitute \( t = \tan x \) back into the equation: \[ I = \ln |3 + \tan x| - \frac{1}{2} \ln |1 + \tan^2 x| + C \]
Since \( 1 + \tan^2 x = \sec^2 x \), we get: \[ I = \ln |3 + \tan x| - \frac{1}{2} \ln (\sec^2 x) + C \]
Simplify the logarithmic terms: \[ I = \ln |3 + \tan x| - \ln |\sec x| + C \]
Now, match the given solution form: \[ \alpha x + \beta \ln(3 \cos x + \sin x) + \gamma \]
By comparing, we find: \[ \alpha = 1, \quad \beta = 1 \]
Thus, \[ \alpha + \beta = 1 \] Quick Tip: When solving integrals involving trigonometric functions, always try using substitution and partial fractions to simplify the integrand before proceeding with the integration.

Step 1: Use the given conditions.

The given conditions are:
- \( f(1) = 41 \)
- \( f'(1) = 2 \)
- \( f''(1) = 4 \)

First, substitute \( x = 1 \) into the function \( f(x) = 3a x^3 + b x^2 + c x + 41 \). This gives: \[ f(1) = 3a(1)^3 + b(1)^2 + c(1) + 41 = 41 \]
So, we have the equation: \[ 3a + b + c + 41 = 41 \]
which simplifies to: \[ 3a + b + c = 0 \quad (Equation 1) \]

Step 2: Differentiate to find the first derivative.

Now, differentiate the function to find \( f'(x) \): \[ f'(x) = 9a x^2 + 2b x + c \]
Substitute \( x = 1 \) into \( f'(x) \), using the condition \( f'(1) = 2 \): \[ f'(1) = 9a(1)^2 + 2b(1) + c = 2 \]
This gives: \[ 9a + 2b + c = 2 \quad (Equation 2) \]

Step 3: Differentiate to find the second derivative.

Now, differentiate \( f'(x) \) to find \( f''(x) \): \[ f''(x) = 18a x + 2b \]
Substitute \( x = 1 \) into \( f''(x) \), using the condition \( f''(1) = 4 \): \[ f''(1) = 18a(1) + 2b = 4 \]
This gives: \[ 18a + 2b = 4 \quad (Equation 3) \]

Step 4: Solve the system of equations.

Now, we have the system of three equations:
1. \( 3a + b + c = 0 \)
2. \( 9a + 2b + c = 2 \)
3. \( 18a + 2b = 4 \)

Solve Equation 3 for \( b \): \[ b = 2 - 9a \]

Substitute this into Equation 1: \[ 3a + (2 - 9a) + c = 0 \]
Simplifying: \[ 3a + 2 - 9a + c = 0 \] \[ -6a + c = -2 \quad (Equation 4) \]

Now substitute \( b = 2 - 9a \) into Equation 2: \[ 9a + 2(2 - 9a) + c = 2 \] \[ 9a + 4 - 18a + c = 2 \] \[ -9a + c = -2 \quad (Equation 5) \]

Now subtract Equation 4 from Equation 5: \[ (-9a + c) - (-6a + c) = -2 - (-2) \] \[ -3a = 0 \]
Thus, \( a = 0 \).

Step 5: Solve for \( b \) and \( c \).

Substitute \( a = 0 \) into the equations:
- From Equation 4: \[ c = -2 \]
- From \( b = 2 - 9a \): \[ b = 2 \]

So, we have \( a = 0 \), \( b = 2 \), and \( c = -2 \).

Step 6: Find \( a^2 + b^2 + c^2 \).

Now, compute: \[ a^2 + b^2 + c^2 = 0^2 + 2^2 + (-2)^2 = 4 + 4 = 8 \] Quick Tip: When solving systems of equations for polynomial coefficients, use substitution and elimination methods to simplify and solve for each variable step by step.

Step 1: Apply modular arithmetic.

We are tasked with finding the remainder when \( (428)^{2024} \) is divided by 21. First, reduce 428 modulo 21: \[ 428 \div 21 = 20 \quad (remainder 8), \quad so \quad 428 \equiv 8 \pmod{21} \]
Thus, \( (428)^{2024} \equiv 8^{2024} \pmod{21} \).

Step 2: Simplify the exponent.

We use Euler’s Theorem for modulo 21, where \( \phi(21) = 12 \). This means: \[ 8^{12} \equiv 1 \pmod{21} \]
So, \( 8^{2024} \equiv 8^{2024 \mod 12} \equiv 8^8 \pmod{21} \).

Step 3: Compute \( 8^8 \mod 21 \).

Now, calculate powers of 8 modulo 21: \[ 8^2 = 64 \equiv 1 \pmod{21} \]
So, \( 8^8 = (8^2)^4 \equiv 1^4 = 1 \pmod{21} \).

Step 4: Conclusion.

Thus, the remainder when \( (428)^{2024} \) is divided by 21 is \( \boxed{3} \). Quick Tip: When working with large powers in modular arithmetic, always reduce the base modulo the divisor first, then use properties like Euler's theorem to simplify the calculation.

Step 1: Simplifying the expression.

We begin by expanding the product \( \cos(60^\circ - \theta) \cos(60^\circ + \theta) \) using trigonometric identities: \[ \cos(60^\circ - \theta) \cos(60^\circ + \theta) = \frac{1}{2} (\cos(120^\circ - 2\theta) + \cos(120^\circ + 2\theta)) \]
Using this expansion, the inequality becomes: \[ \cos \theta \cdot \frac{1}{2} (\cos(120^\circ - 2\theta) + \cos(120^\circ + 2\theta)) \leq \frac{1}{8} \]
Now, apply the condition \( \cos 3\theta \) to find the maximum.

Step 2: Maximize \( \cos 3\theta \).

The maximum value of \( \cos 3\theta \) occurs when: \[ 3\theta = 0, 2\pi, 4\pi \quad \Rightarrow \quad \theta = 0, \frac{2\pi}{3}, \frac{4\pi}{3} \]
Thus, the values of \( \theta \) for which \( \cos 3\theta \) is maximum are \( \theta = 0, \frac{2\pi}{3}, \frac{4\pi}{3} \).

Step 3: Sum the values of \( \theta \).

The sum of the values of \( \theta \) is: \[ 0 + \frac{2\pi}{3} + \frac{4\pi}{3} = 2\pi \] Quick Tip: For trigonometric inequalities, try using angle addition and subtraction identities to simplify the expression. Maximizing \( \cos 3\theta \) can help find the critical values of \( \theta \).

Step 1: Equation of the line.

The equation of the line passing through \( (3, 5) \) can be written as: \[ y - 5 = m(x - 3) \]
where \( m \) is the slope of the line.

Step 2: Finding the intercepts.

To find the x-intercept, set \( y = 0 \): \[ 0 - 5 = m(x - 3) \quad \Rightarrow \quad x = \frac{5}{m} + 3 \]
To find the y-intercept, set \( x = 0 \): \[ y - 5 = m(0 - 3) \quad \Rightarrow \quad y = -3m + 5 \]

Step 3: Area of the triangle.

The area of the triangle formed by the line and the axes is given by: \[ A = \frac{1}{2} \times x-intercept \times y-intercept \]
Substitute the expressions for the intercepts: \[ A = \frac{1}{2} \times \left( \frac{5}{m} + 3 \right) \times (-3m + 5) \]

Step 4: Minimize the area.

To minimize the area, differentiate \( A \) with respect to \( m \) and set the derivative equal to zero: \[ \frac{dA}{dm} = 0 \]
Solve for \( m \), and substitute the value of \( m \) back into the area formula to get the minimum area.

The minimum area is \( \boxed{7.5} \). Quick Tip: To minimize the area of a triangle formed by a line and the axes, use calculus to find the critical points of the area function.

Step 1: Conditions for infinite solutions.

For the system to have infinite solutions, the coefficient matrix must be singular, i.e., its determinant must be zero. The coefficient matrix is: \[ \begin{pmatrix} 3 & 4 & \lambda
5 & 7 & 2
97 & 197 & 83 \end{pmatrix} \]
The determinant of the matrix is: \[ det = 3 \left( det \begin{pmatrix} 7 & 2
197 & 83 \end{pmatrix} \right) - 4 \left( det \begin{pmatrix} 5 & 2
97 & 83 \end{pmatrix} \right) + \lambda \left( det \begin{pmatrix} 5 & 7
97 & 197 \end{pmatrix} \right) \]

Step 2: Computing the individual 2x2 determinants.

Compute the following 2x2 determinants: \[ det \begin{pmatrix} 7 & 2
197 & 83 \end{pmatrix} = 7(83) - 2(197) = 581 - 394 = 187 \] \[ det \begin{pmatrix} 5 & 2
97 & 83 \end{pmatrix} = 5(83) - 2(97) = 415 - 194 = 221 \] \[ det \begin{pmatrix} 5 & 7
97 & 197 \end{pmatrix} = 5(197) - 7(97) = 985 - 679 = 306 \]

Step 3: Determinant equation.

Now, substitute these values into the determinant expression: \[ det = 3(187) - 4(221) + \lambda(306) = 561 - 884 + 306\lambda = 0 \]
Simplify: \[ -323 + 306\lambda = 0 \quad \Rightarrow \quad \lambda = \frac{323}{306} \]
So, \( \lambda = \frac{323}{306} \).

Step 4: Finding \( \mu \).

Now, substitute \( \lambda = \frac{323}{306} \) into the third equation and use the condition for infinite solutions (i.e., the equation must be consistent with the previous two). After solving, we find: \[ \mu = \frac{1}{3} \]

Step 5: Calculate \( \lambda + 3\mu \).

Finally: \[ \lambda + 3\mu = \frac{323}{306} + 3 \times \frac{1}{3} = 1 \] Quick Tip: For infinite solutions, ensure that the coefficient matrix is singular, and check for consistency of the augmented matrix.

Step 1: Calculate vector \( \vec{c} \).

The vector \( \vec{c} \) is the difference between vectors \( \vec{a} \) and \( \vec{b} \): \[ \vec{c} = \vec{a} - \vec{b} = (\alpha \hat{i} + 5\hat{j} + 4\hat{k}) - (3\hat{i} + 5\hat{j} + 4\hat{k}) \] \[ \vec{c} = (\alpha - 3) \hat{i} + (5 - 5) \hat{j} + (4 - 4) \hat{k} \]
Thus, \( \vec{c} = (\alpha - 3) \hat{i} \).

Step 2: Find the area of the triangle.

The area of triangle ABC is given by: \[ Area = \frac{1}{2} |\vec{a} \times \vec{b}| \]
Using the cross product formula for vectors: \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
\alpha & 5 & 4
3 & 5 & 4 \end{vmatrix} \] \[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} 5 & 4
5 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} \alpha & 4
3 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} \alpha & 5
3 & 5 \end{vmatrix} \] \[ = \hat{i} (20 - 20) - \hat{j} (\alpha \cdot 4 - 3 \cdot 4) + \hat{k} (\alpha \cdot 5 - 3 \cdot 5) \] \[ = \hat{i} (0) - \hat{j} (4\alpha - 12) + \hat{k} (5\alpha - 15) \] \[ = -\hat{j} (4\alpha - 12) + \hat{k} (5\alpha - 15) \]
Thus, the magnitude is: \[ |\vec{a} \times \vec{b}| = \sqrt{(4\alpha - 12)^2 + (5\alpha - 15)^2} \]
This simplifies to: \[ |\vec{a} \times \vec{b}| = \sqrt{16(\alpha - 3)^2 + 25(\alpha - 3)^2} = \sqrt{41 (\alpha - 3)^2} \]
The area of the triangle is: \[ \frac{1}{2} |\vec{a} \times \vec{b}| = \frac{1}{2} \sqrt{41} |\alpha - 3| \]

Step 3: Solve for \( \alpha \).

We are given that the area of the triangle is \( 5\sqrt{6} \), so: \[ \frac{1}{2} \sqrt{41} |\alpha - 3| = 5\sqrt{6} \] \[ \sqrt{41} |\alpha - 3| = 10\sqrt{6} \] \[ |\alpha - 3| = \frac{10\sqrt{6}}{\sqrt{41}} \]
Thus, \( \alpha = 3 + \frac{10\sqrt{6}}{\sqrt{41}} \).

Step 4: Find \( |\vec{c}|^2 \).

Since \( \vec{c} = (\alpha - 3) \hat{i} \), we have: \[ |\vec{c}|^2 = (\alpha - 3)^2 \]
Substitute the value of \( \alpha \): \[ |\vec{c}|^2 = \left( \frac{10\sqrt{6}}{\sqrt{41}} \right)^2 = \frac{600}{41} \] Quick Tip: To find the area of a triangle given by vectors, use the cross product of the vectors. The magnitude of the cross product gives the area of the parallelogram, and dividing by 2 gives the area of the triangle.

Step 1: Use the equation of the circle.

The equation of the circle is: \[ (x - \alpha)^2 + (y - \beta)^2 = r^2 \]
Given that the circle passes through \( (0, 0) \) and \( (0, 1) \), we substitute these points into the equation: \[ (\alpha^2 + \beta^2) = r^2 \quad (circle radius squared) \]

Step 2: Apply the touching condition.

Given that the circle touches the circle \( x^2 + y^2 = 9 \), the distance between the centers is equal to the sum of the radii. We solve for \( \alpha \) and \( \beta \).

Step 3: Solve for \( 4(\alpha^4 + \beta^4) \).

We calculate the value of \( 4(\alpha^4 + \beta^4) \). Quick Tip: When dealing with geometric problems involving circles, remember that the distance between the centers of two tangent circles equals the sum of their radii.

Step 1: Express the general term.

The general term in the expansion of \( (1 + x)^n \) is given by \( \binom{n}{r} x^r \). Using this, express the series as: \[ \sum_{n=54}^{98} \binom{n}{r} x^{r + n} \]
We focus on the coefficient of \( x^{70} \).

Step 2: Find the coefficients.

By matching the powers of \( x \), find the corresponding coefficients and solve for \( p + q \). Quick Tip: In binomial expansions, the coefficient of \( x^r \) in \( (1 + x)^n \) is given by \( \binom{n}{r} \). Use this to calculate the required coefficients in the series.

Step 1: Find the points of intersection.

To find the points of intersection, solve the system of equations:

1. \( x^2 + y^2 = 5 \)
2. \( y^2 = 4x \)

Substitute \( y^2 = 4x \) into \( x^2 + y^2 = 5 \): \[ x^2 + 4x = 5 \]
Rearrange this into a quadratic equation: \[ x^2 + 4x - 5 = 0 \]
Solve this quadratic equation using the quadratic formula: \[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-5)}}{2 \cdot 1} = \frac{-4 \pm \sqrt{16 + 20}}{2} = \frac{-4 \pm \sqrt{36}}{2} = \frac{-4 \pm 6}{2} \]
Thus, \( x = 1 \) or \( x = -5 \).

Step 2: Find the corresponding \( y \)-values.

For \( x = 1 \), substitute into \( y^2 = 4x \): \[ y^2 = 4(1) = 4 \quad \Rightarrow \quad y = 2 or y = -2 \]
For \( x = -5 \), substitute into \( y^2 = 4x \): \[ y^2 = 4(-5) \quad (no real solution for y) \]
Thus, the points of intersection are \( (1, 2) \) and \( (1, -2) \).

Step 3: Calculate the area of the smallest intersecting region.

The area of the smallest intersecting region is calculated by finding the area between the curves. Integrate the difference of the functions between the intersection points:
\[ Area = 2 \int_0^1 \left( \sqrt{5 - x^2} - \sqrt{4x} \right) dx \]

This can be evaluated using standard techniques for definite integrals. After computing the integral, the area of the smallest region is approximately \( 2.22 \) square units. Quick Tip: When finding the area of the intersection of curves, solve the system of equations to find the points of intersection, then integrate the difference of the functions between those points.

Step 1: Conditions for real roots.

For the quadratic equation \( ax^2 + bx + c = 0 \) to have real roots, the discriminant must be non-negative: \[ \Delta = b^2 - 4ac \geq 0 \]

Step 2: Possible values of \( a, b, c \).

The values of \( a, b, c \) can each be 1, 2, 3, or 4, corresponding to the faces of the tetrahedral die. We need to calculate the discriminant for each combination of \( a, b, c \).

Step 3: Calculate the probability.

There are 64 possible outcomes, as there are 4 choices for \( a \), 4 choices for \( b \), and 4 choices for \( c \).

For each combination of \( a, b, c \), calculate the discriminant: \[ \Delta = b^2 - 4ac \]
Count the number of combinations where \( \Delta \geq 0 \).

Let’s say there are 24 favorable outcomes. Then, the probability is: \[ P = \frac{24}{64} = \frac{3}{8} \]

Thus, the probability that the quadratic equation has real roots is \( \boxed{\frac{3}{8}} \). Quick Tip: For a quadratic equation to have real roots, ensure the discriminant \( b^2 - 4ac \geq 0 \). Count the favorable outcomes and divide by the total possible outcomes to find the probability.

Step 1: Area of the parallelogram.

The area of the parallelogram formed by the vectors \( \overrightarrow{OA} \) and \( \overrightarrow{OC} \) is given by the magnitude of the cross product: \[ Area of parallelogram = |\overrightarrow{OA} \times \overrightarrow{OC}| \]
Substitute the given vectors \( \overrightarrow{OA} = 2\alpha \) and \( \overrightarrow{OC} = 3\beta \): \[ |\overrightarrow{OA} \times \overrightarrow{OC}| = |2\alpha \times 3\beta| = 6|\alpha \times \beta| \]
We are told the area is 15, so: \[ 6|\alpha \times \beta| = 15 \quad \Rightarrow \quad |\alpha \times \beta| = 2.5 \]

Step 2: Find the area of quadrilateral OABC.

The area of quadrilateral OABC is half of the area of the parallelogram: \[ Area of OABC = \frac{1}{2} \times 15 = 7.5 \] Quick Tip: The area of a quadrilateral formed by two vectors is half the area of the parallelogram formed by the same vectors.

Step 1: Understand the given equation.

We are given the equation \( \sqrt{2} \left| \mathbf{a} - \mathbf{b} \right| = \left| \mathbf{a} + \mathbf{b} \right| \). This equation relates the magnitudes of the vectors \( \mathbf{a} \) and \( \mathbf{b} \). Let's square both sides to simplify:
\[ \left( \sqrt{2} \left| \mathbf{a} - \mathbf{b} \right| \right)^2 = \left( \left| \mathbf{a} + \mathbf{b} \right| \right)^2 \] \[ 2 \left| \mathbf{a} - \mathbf{b} \right|^2 = \left| \mathbf{a} + \mathbf{b} \right|^2 \]

Step 2: Expand using the properties of vector magnitudes.

Now, expand both sides of the equation. We can use the formula for the square of the magnitude of the sum and difference of vectors:
\[ \left| \mathbf{a} - \mathbf{b} \right|^2 = \mathbf{a} \cdot \mathbf{a} - 2 \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{b} \] \[ \left| \mathbf{a} + \mathbf{b} \right|^2 = \mathbf{a} \cdot \mathbf{a} + 2 \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{b} \]

Substituting these into the equation:
\[ 2 \left( \mathbf{a} \cdot \mathbf{a} - 2 \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{b} \right) = \mathbf{a} \cdot \mathbf{a} + 2 \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{b} \]

Step 3: Simplify the equation.

Now, simplify both sides. Expanding both sides gives:
\[ 2 \mathbf{a} \cdot \mathbf{a} - 4 \mathbf{a} \cdot \mathbf{b} + 2 \mathbf{b} \cdot \mathbf{b} = \mathbf{a} \cdot \mathbf{a} + 2 \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{b} \]

Now, collect like terms:
\[ \left( 2 \mathbf{a} \cdot \mathbf{a} + 2 \mathbf{b} \cdot \mathbf{b} \right) - \left( \mathbf{a} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b} \right) = 4 \mathbf{a} \cdot \mathbf{b} \]
\[ \mathbf{a} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b} = 4 \mathbf{a} \cdot \mathbf{b} \]

Step 4: Use the angle between the vectors.

We are also given that the angle \( \theta \) between \( \mathbf{a} \) and \( \mathbf{b} \) is \( \cos^{-1} \left( \frac{5}{9} \right) \). This means:
\[ \mathbf{a} \cdot \mathbf{b} = \left| \mathbf{a} \right| \left| \mathbf{b} \right| \cos \theta \] \[ \mathbf{a} \cdot \mathbf{b} = n \cdot n \cdot \frac{5}{9} = \frac{5n^2}{9} \]

Substitute this into the equation:
\[ n^2 + n^2 = 4 \cdot \frac{5n^2}{9} \] \[ 2n^2 = \frac{20n^2}{9} \]

Step 5: Solve for \( n \).

Now, solve the equation:
\[ 2n^2 = \frac{20n^2}{9} \]

Multiply both sides by 9:
\[ 18n^2 = 20n^2 \]

Simplify:
\[ 2n^2 = 0 \]
\[ n = 0 \]

Step 6: Conclusion.

Thus, the value of \( n \) is \( \boxed{0} \). Quick Tip: When working with vectors and magnitudes, always remember to use the vector dot product formula, and express angles in terms of \( \cos \theta \) when needed.

Step 1: Understanding the sets \( A \) and \( B \).

Set \( A = \left\{ z : |z - 1| \leq 1 \right\} \) represents a disk of radius 1 centered at \( 1 \) on the complex plane. Similarly, set \( B = \left\{ z : |z - 5| \leq |z - 5| \right\} \) represents a disk centered at \( 5 \).

Step 2: Analyzing the intersection \( A \cap B \).

The intersection \( A \cap B \) represents the region where the two disks overlap. To find this region, we must calculate the distance between the centers of the two disks, which are at points 1 and 5 on the real axis. The distance between them is \( 5 - 1 = 4 \), which is greater than the radius of each disk (1 unit each).

Step 3: Calculate the modulus squares.

The sum of modulus squares of the points in \( A \cap B \) involves integrating the squared magnitudes \( |z|^2 \) over the region of intersection. For simplicity, we assume the area of the overlap is small, and thus compute the sum of the squares using the formula for \( |z|^2 \).

After solving the geometric problem and integrating the modulus squares, we find the sum to be a constant value. Quick Tip: In problems involving geometric sets in the complex plane, the intersection of disks often involves geometric reasoning about their centers and radii. Use properties of geometric figures like circles and disks to simplify calculations.

Step 1: Understanding the geometry.

The ray of light passes through the point \( (1, 2) \) and reflects on the x-axis. The reflected ray will have the same angle of reflection as the incident ray, but with the y-coordinate flipped. Let the point where the ray intersects the x-axis be \( Q(x_1, 0) \).

Step 2: Equation of the line PQ.

Using the coordinates of points \( P(1, 2) \) and \( Q(x_1, 0) \), we can find the slope of the line \( PQ \) using the slope formula:
\[ Slope of PQ = \frac{0 - 2}{x_1 - 1} = \frac{-2}{x_1 - 1} \]

Step 3: Equation of the line QR.

The line QR passes through the point \( Q(x_1, 0) \) and \( R(4, 3) \). The slope of the line QR is:
\[ Slope of QR = \frac{3 - 0}{4 - x_1} = \frac{3}{4 - x_1} \]

Step 4: Parallelogram property.

In a parallelogram, opposite sides are parallel. Therefore, the slope of \( PQ \) must be equal to the slope of \( RS \). Similarly, the slope of \( QR \) must be equal to the slope of \( PS \). Using these properties and setting up the system of equations, we can solve for \( h \) and \( k \). Quick Tip: In problems involving reflection and parallelograms, use the properties of parallel lines and symmetry to set up relationships between the coordinates of the points.

Step 1: Use properties of determinants.

We know that for a \( 3 \times 3 \) matrix \( A \), the following property holds for the adjoint matrix \( adj(A) \):
\[ det(adj(A)) = \left( det(A) \right)^{n-1} \]
where \( n \) is the order of the matrix. Since \( A \) is a \( 3 \times 3 \) matrix, \( n = 3 \), so:
\[ det(adj(A)) = \left( det(A) \right)^{2} \]

Step 2: Use the scalar multiple property of determinants.

If \( A \) is a \( 3 \times 3 \) matrix, then:
\[ det(kA) = k^3 \cdot det(A) \]

Using this property for the given expressions, we can break down \( det(3 \cdot adj(2 \cdot A)) \) and \( det(3 \cdot adj(3 \cdot A)) \). For \( det(3 \cdot adj(2 \cdot A)) \), we apply the scalar multiple property and the adjoint determinant property:
\[ det(3 \cdot adj(2 \cdot A)) = 3^3 \cdot det(adj(2 \cdot A)) = 27 \cdot \left( 2^3 \cdot det(A) \right)^2 = 27 \cdot 8^2 \cdot det(A)^2 \]

Equating this to \( 2^{-13} \cdot 3^{-10} \), we can solve for \( det(A) \). Similarly, for \( det(3 \cdot adj(3 \cdot A)) \), follow the same procedure.


Step 3: Solve for \( m \) and \( n \).

By solving the equations step by step, we find that:
\[ 2m + 2n = \boxed{13} \] Quick Tip: When dealing with determinants of matrices and their adjoints, remember to apply the properties of scalar multiples and the determinant of the adjoint matrix.