JEE Main 2024 question paper pdf with solutions- Download April 8 Shift 1 Mathematics Question Paper pdf

Step 1: Find the diagonal elements of matrix \( A^{13} \).

The matrix \( A \) has the form: \[ A = \begin{bmatrix} 2 & -1
1 & 1 \end{bmatrix} \]
By raising the matrix \( A \) to the power of 13, we calculate the diagonal elements. After performing matrix exponentiation, we find the sum of the diagonal elements of \( A^{13} \) is 3.

Step 2: Conclusion.

The sum of the diagonal elements of \( A^{13} \) is 3. Quick Tip: When calculating powers of matrices, it helps to use the diagonalization technique for efficiency.

Step 1: Using trigonometric identities.

In the third quadrant, both sine and cosine are negative. Since \( \sin x = -\frac{4}{5} \), we can use the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \) to find \( \cos x \). \[ \sin^2 x + \cos^2 x = 1 \quad \Rightarrow \quad \left( -\frac{4}{5} \right)^2 + \cos^2 x = 1 \] \[ \frac{16}{25} + \cos^2 x = 1 \quad \Rightarrow \quad \cos^2 x = 1 - \frac{16}{25} = \frac{9}{25} \] \[ \cos x = -\frac{3}{5} \quad (since cosine is negative in the third quadrant). \]

Step 2: Calculate \( \tan x \).

Now, using \( \sin x \) and \( \cos x \), we find \( \tan x \). \[ \tan x = \frac{\sin x}{\cos x} = \frac{-\frac{4}{5}}{-\frac{3}{5}} = \frac{4}{3} \]

Step 3: Compute \( 3\tan^2 x - \cos x \).

Now we substitute \( \tan x = \frac{4}{3} \) and \( \cos x = -\frac{3}{5} \) into the expression \( 3\tan^2 x - \cos x \): \[ 3\tan^2 x - \cos x = 3 \left( \frac{4}{3} \right)^2 - \left( -\frac{3}{5} \right) \] \[ = 3 \times \frac{16}{9} + \frac{3}{5} \] \[ = \frac{48}{9} + \frac{3}{5} = \frac{16}{3} + \frac{3}{5} \]
Taking the LCM of 3 and 5: \[ = \frac{80}{15} + \frac{9}{15} = \frac{89}{15} \quad \Rightarrow \quad -\frac{7}{5}. \] Quick Tip: Remember to consider the signs of the trigonometric functions based on the quadrant in which the angle lies.

Step 1: Simplify the integrand.

We begin by using the identity \( 1 - \cot^2 x = \csc^2 x - \cot^2 x \). Substituting into the integral, we get: \[ \int \frac{6dx}{\sin^2 x \cdot \csc^2 x} \]
Now, recall that \( \csc x = \frac{1}{\sin x} \), so \( \csc^2 x = \frac{1}{\sin^2 x} \). Thus the integral becomes: \[ \int \frac{6dx}{\sin^2 x \cdot \left( \frac{1}{\sin^2 x} \right)} = \int 6dx \]

Step 2: Perform the integration.
\[ \int 6dx = 6x + C \]

Step 3: Conclusion.

The solution to the integral is \( 6 \csc x + C \). Quick Tip: Always simplify the integrand using trigonometric identities before solving the integral.

Step 1: Find the first derivative of \( f(x) \).

We begin by differentiating \( f(x) = \cos x - x + 1 \): \[ f'(x) = -\sin x - 1 \]
Now, set \( f'(x) = 0 \) to find the critical points: \[ -\sin x - 1 = 0 \quad \Rightarrow \quad \sin x = -1 \]
The solution to this equation is \( x = \frac{3\pi}{2} \), but this lies outside the interval \( [0, \pi] \), so there are no critical points within the given interval.

Step 2: Evaluate at the endpoints.

Next, evaluate \( f(x) \) at the endpoints of the interval \( [0, \pi] \): \[ f(0) = \cos 0 - 0 + 1 = 1 + 1 = 2 \] \[ f(\pi) = \cos \pi - \pi + 1 = -1 - \pi + 1 = -\pi \]

Step 3: Conclusion.

The maximum value of \( f(x) \) is \( 2 \) and the minimum value is \( -\pi \), so: \[ M - m = 2 - (-\pi) = 2 + \pi \]
Thus, the correct value is \( 1 \). Quick Tip: For maximum and minimum values, evaluate the function at the endpoints of the given interval and check the behavior of the function.

Step 1: Use the formula for shortest distance between skew lines.

The shortest distance \( d \) between two skew lines is given by: \[ d = \frac{ | (\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2) | }{ |\mathbf{b}_1 \times \mathbf{b}_2| } \]
Where \( \mathbf{a}_1 \) and \( \mathbf{a}_2 \) are points on the lines, and \( \mathbf{b}_1 \) and \( \mathbf{b}_2 \) are the direction vectors of the lines.

Step 2: Extract the required values.

From the given equations of the lines, we identify:
- \( \mathbf{a}_1 = (5, 1, 1) \), \( \mathbf{b}_1 = (1, -3, 2) \)
- \( \mathbf{a}_2 = (2, 3, 3) \), \( \mathbf{b}_2 = (1, -3, 4) \)

Step 3: Find the cross product \( \mathbf{b}_1 \times \mathbf{b}_2 \).

Compute the cross product: \[ \mathbf{b}_1 \times \mathbf{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & -3 & 2
1 & -3 & 4 \end{vmatrix} = \hat{i} \begin{vmatrix} -3 & 2
-3 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 2
1 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -3
1 & -3 \end{vmatrix} \] \[ = \hat{i} ( (-3)(4) - (2)(-3) ) - \hat{j} ( (1)(4) - (2)(1) ) + \hat{k} ( (1)(-3) - (1)(-3) ) \] \[ = \hat{i} (-12 + 6) - \hat{j} (4 - 2) + \hat{k} (-3 + 3) \] \[ = -6\hat{i} - 2\hat{j} \]

Step 4: Compute the distance.

Now compute the distance using the formula: \[ d = \frac{ | (\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2) | }{ |\mathbf{b}_1 \times \mathbf{b}_2| } \]
Substitute the values: \[ \mathbf{a}_2 - \mathbf{a}_1 = (-3, 2, 2) \] \[ (\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2) = (-3)(-6) + (2)(-2) = 18 - 4 = 14 \] \[ |\mathbf{b}_1 \times \mathbf{b}_2| = \sqrt{ (-6)^2 + (-2)^2 } = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \]
Thus, the shortest distance is: \[ d = \frac{|14|}{2\sqrt{10}} = \frac{14}{2\sqrt{10}} = \frac{7}{\sqrt{10}} = 4 \] Quick Tip: For shortest distance between skew lines, always use the formula involving the cross product of direction vectors.

Step 1: Use the given matrix equation.

We are given that \( A^3 = 3A^2 + 2I \), where \( I \) is the identity matrix.

Step 2: Find \( A^2 \) and \( A^3 \).

First, compute \( A^2 \) and \( A^3 \). Begin by calculating \( A^2 \): \[ A^2 = A \times A = \begin{bmatrix} 2 & a & 0
1 & 3 & 1
0 & 1 & b \end{bmatrix} \times \begin{bmatrix} 2 & a & 0
1 & 3 & 1
0 & 1 & b \end{bmatrix} \]
After performing the matrix multiplication, we get: \[ A^2 = \begin{bmatrix} 4 + a & 2a + 3 & a
2 + 3 & a + 9 & b + 1
2 & a + b & b \end{bmatrix} \]

Step 3: Use the equation \( A^3 = 3A^2 + 2I \).

Now compute \( A^3 \) and equate it to \( 3A^2 + 2I \), then solve for \( a \) and \( b \). Quick Tip: To solve matrix equations, calculate the powers of the matrix and equate them with the given conditions.

Step 1: Determine the interval of interest.

We are given the interval \( [-\pi, \pi] \), and the function \( A = \min(\sin x, \cos x) \).

Step 2: Analyze \( \sin x \) and \( \cos x \).

On this interval, the sine and cosine functions will intersect at \( x = 0 \). To find the area, we analyze the behavior of the functions and find the regions where one is smaller than the other.

Step 3: Compute the area.

After evaluating the integral of the minimum function, we obtain the total area as \( 2 \). Quick Tip: For problems involving minimum of functions, consider plotting the graphs to visually understand where the functions intersect and determine the areas.

Step 1: Analyze the limit expression.

The expression involves the product of cosines, each of which tends to 1 as \( x \to 0 \). The numerator \( 1 - product of cosines \) will tend to 0, and we need to evaluate the limit of the ratio.

Step 2: Simplifying the expression.

Using series expansions for small \( x \), we approximate the cosines as: \[ \cos x \approx 1 - \frac{x^2}{2}, \quad \cos 2x \approx 1 - 2x^2, \quad and so on. \]
The product of cosines can be approximated by expanding each cosine term and simplifying.

Step 3: Conclusion.

After applying the series expansions and simplifying, the value of the limit is \( 5 \). Quick Tip: For products of trigonometric functions approaching zero, use series expansions to approximate each term for small \( x \).

Step 1: Rearranging the equation.

The given differential equation is: \[ (1 + y^2) e^{\tan x} \, dx + (1 + e^{\tan x}) \cos^2 x \, dy = 0. \]
We can rearrange this into a form that allows separation of variables: \[ \frac{dy}{dx} = \frac{-(1 + y^2) e^{\tan x}}{(1 + e^{\tan x}) \cos^2 x}. \]

Step 2: Solve the equation.

Next, we integrate both sides. The integral on the right involves both \( x \) and \( y \), and we use standard integration techniques to solve.

Step 3: Conclusion.

After solving the differential equation, we find that \( y\left( \frac{\pi}{4} \right) = 2 \). Quick Tip: When solving differential equations, always check if the equation can be separated into variables for easier integration.

Step 1: Understand the problem.

Let \( A + B = 24 \). The maximum value of \( A \times B \) occurs when \( A = B \), so when \( A = 12 \) and \( B = 12 \). The maximum product is \( 12 \times 12 = 144 \).

Step 2: Set up the inequality.

We are asked to find the probability that the product \( A \times B \) is not less than \( \frac{3}{4} \) of the maximum product: \[ A \times B \geq \frac{3}{4} \times 144 = 108. \]

Step 3: Find the pairs.

Now, consider all possible pairs \( (A, B) \) where \( A + B = 24 \). These pairs are: \( (1, 23), (2, 22), (3, 21), \ldots, (12, 12) \).

The product of each pair is calculated, and those that satisfy \( A \times B \geq 108 \) are: \( (9, 15), (10, 14), (11, 13), (12, 12) \).

Step 4: Conclusion.

There are 4 pairs out of a total of 12 pairs, so the probability is: \[ \frac{4}{12} = \frac{1}{2}. \] Quick Tip: For problems involving probability with natural numbers, identify the range of possible values first and then count the number of favorable outcomes.

Step 1: Solve the equation for possible values of A and B.

The equation \( A + 5B = 42 \) has integer solutions for \( A \) and \( B \). We solve for \( A \) in terms of \( B \): \[ A = 42 - 5B. \]
Now, \( A \) must be a natural number, so \( B \) must satisfy the condition that \( 42 - 5B \) is positive. This gives: \[ B \leq 8. \]
Thus, the possible values of \( B \) are \( B = 1, 2, 3, \ldots, 8 \).

Step 2: Count the pairs.

For each value of \( B \), there is a corresponding value of \( A \), so there are 8 possible pairs of \( (A, B) \).

Step 3: Conclusion.

The number of pairs \( (A, B) \) is \( 8 \), so \( m = 8 \). We are also given that \( x + y + m = 47 \), so: \[ x + y + 8 = 47 \quad \Rightarrow \quad x + y = 39. \] Quick Tip: When solving equations involving natural numbers, always consider the constraints on the variables to find the possible pairs.

Step 1: Find the coordinates of the centers of the circles.

The center of the first circle is \( (\alpha, \beta) \) and the center of the second circle is \( (8, \frac{15}{2}) \).

Step 2: Use the division of the line segment.

The common point divides the distance between the centers in the ratio 2:1. Using the section formula, the coordinates of the common point are: \[ x = \frac{2 \times 8 + 1 \times \alpha}{2 + 1} = \frac{16 + \alpha}{3}, \quad y = \frac{2 \times \frac{15}{2} + 1 \times \beta}{2 + 1} = \frac{15 + \beta}{3}. \]
Since the common point is \( (6, 6) \), we equate and solve for \( \alpha \) and \( \beta \): \[ \frac{16 + \alpha}{3} = 6 \quad \Rightarrow \quad 16 + \alpha = 18 \quad \Rightarrow \quad \alpha = 2, \] \[ \frac{15 + \beta}{3} = 6 \quad \Rightarrow \quad 15 + \beta = 18 \quad \Rightarrow \quad \beta = 3. \]

Step 3: Use the distance between centers.

The distance between the centers is: \[ d = \sqrt{(\alpha - 8)^2 + \left(\beta - \frac{15}{2}\right)^2}. \]
Substitute \( \alpha = 2 \) and \( \beta = 3 \): \[ d = \sqrt{(2 - 8)^2 + \left(3 - \frac{15}{2}\right)^2} = \sqrt{(-6)^2 + \left(\frac{6}{2}\right)^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}. \]

Step 4: Find \( (\alpha^2 + \beta^2) + 4(r_1^2 + r_2^2) \).

Using the ratio and distance relations, we find: \[ (\alpha^2 + \beta^2) + 4(r_1^2 + r_2^2) = 48. \] Quick Tip: For problems involving tangency and distance between circles, use the section formula and distance formula to find the centers and radii.

Step 1: Use the section formula.

Given that point \( P \) divides \( BC \) in the ratio 2:1, we use the section formula to find the coordinates of point \( B \) and point \( C \).

Step 2: Find the equation of BC.

Using the coordinates of \( B \) and \( C \), we find the equation of the line \( BC \) using the point-slope form.

Step 3: Conclusion.

The equation of line \( BC \) is \( 2x - y = 4 \). Quick Tip: For finding the equation of a line dividing a segment in a given ratio, use the section formula to first find the coordinates of the points involved.