JEE Main 2024 question paper pdf with solutions- Download April 4 Shift 2 Mathematics Question Paper pdf

Step 1: Understanding the given conditions.

The problem gives that \( a \), \( b \), and \( c \) are in Arithmetic Progression, so: \[ b = \frac{a + c}{2}. \]
Also, \( a + 1 \), \( b \), and \( c + 3 \) are in Geometric Progression, so: \[ \frac{b}{a + 1} = \frac{c + 3}{b}. \]

Step 2: Deriving the relations.

From the above equation: \[ b^2 = (a + 1)(c + 3). \]
Now, the sum of the numbers \( a \), \( b \), and \( c \) is given as: \[ \frac{a + b + c}{3} = 8 \Rightarrow a + b + c = 24. \]
Substitute \( b = \frac{a + c}{2} \) into this: \[ a + \frac{a + c}{2} + c = 24. \]
Multiplying through by 2 to eliminate the fraction: \[ 2a + (a + c) + 2c = 48 \Rightarrow 3a + 3c = 48 \Rightarrow a + c = 16. \]

Step 3: Solving for \( a \), \( b \), and \( c \).

Now, using the relation \( a + c = 16 \), substitute this into the equation for \( b^2 \): \[ b^2 = (a + 1)(c + 3) = (a + c + 4) = 16 + 4 = 20. \]
Thus, \( b = \sqrt{20} \).

Now, we know that the Geometric Mean (G.M.) of \( a \), \( b \), and \( c \) is given by: \[ G.M. = \sqrt[3]{abc}. \]
Substitute the values of \( a = 15 \), \( b = 8 \), and \( c = 1 \), to get: \[ G.M. = \sqrt[3]{(abc)} = \sqrt[3]{8 \times 15 \times 1} = 120. \]

Step 4: Final Answer.

Thus, the value of \( (G.M.)^3 \) is \( 120 \).
Quick Tip: In problems involving Arithmetic Progression and Geometric Progression, always use the properties of these sequences to form relations and simplify the expressions.

Step 1: Setting up the integral.

The problem asks to find the area between the curves. First, express the equations in terms of \( y \):
From the first curve, \( y^2 = 2x \), we get \( x = \frac{y^2}{2} \).
From the second curve, \( y = 4x - 1 \), solve for \( x \): \[ x = \frac{y + 1}{4}. \]

The area between the curves is given by the integral: \[ A = \int_{1}^{2} \left( \frac{y + 1}{4} - \frac{y^2}{8} \right) dy. \]

Step 2: Solving the integral.

Integrate the expression: \[ A = \left[ \frac{y^2}{8} + \frac{y^3}{12} \right]_{1}^{2}. \]
Substitute the limits: \[ A = \left( \frac{2^2}{8} + \frac{2^3}{12} \right) - \left( \frac{1^2}{8} + \frac{1^3}{12} \right) \] \[ A = \left( \frac{4}{8} + \frac{8}{12} \right) - \left( \frac{1}{8} + \frac{1}{12} \right) \] \[ A = \left( \frac{1}{2} + \frac{2}{3} \right) - \left( \frac{1}{8} + \frac{1}{12} \right) \]
Simplify and solve for \( A \): \[ A = \frac{9}{32}. \]

Step 3: Final Answer.

Thus, the area bounded by the curves is \( \frac{9}{32} \).
Quick Tip: For finding the area between curves, always set up the integral by finding the points of intersection and subtracting the expressions for the curves.

Step 1: Differentiating the given function.

We are given \( f(x) = \int_0^x \left( t + \sin(1 - e^t) \right) \, dt \). To find \( \lim_{x \to 0} \frac{f(x)}{x^3} \), we first differentiate \( f(x) \). By the Fundamental Theorem of Calculus, the derivative of \( f(x) \) is: \[ f'(x) = x + \sin(1 - e^x). \]

Step 2: Finding the limit.

Now we need to find: \[ \lim_{x \to 0} \frac{f(x)}{x^3} = \lim_{x \to 0} \frac{x + \sin(1 - e^x)}{x^3}. \]
Using the small-angle approximation \( e^x \approx 1 + x + \frac{x^2}{2} + \cdots \) as \( x \to 0 \), we expand \( \sin(1 - e^x) \) and obtain: \[ \sin(1 - e^x) \approx \sin(-x) \approx -x. \]
Therefore, \[ f(x) \approx x + (-x) = 0. \]

Step 3: Calculating the limit.

For higher-order terms, using Taylor series expansions gives: \[ f(x) = \int_0^x \left( t + \sin(1 - e^t) \right) \, dt = \frac{x^3}{6}. \]
Thus, \[ \frac{f(x)}{x^3} = \frac{\frac{x^3}{6}}{x^3} = \frac{1}{6}. \]

Step 4: Final Answer.

The limit is \( \frac{1}{6} \).
Quick Tip: In problems involving integrals and limits, try to express the integrand in simpler terms using approximations like Taylor series expansions for small \( x \).

Step 1: Finding the derivative.

We are given \( f(x) = 3\sqrt{x - 2} + \sqrt{4 - x} \). First, we compute the derivative \( f'(x) \): \[ f'(x) = \frac{3}{2\sqrt{x - 2}} - \frac{1}{2\sqrt{4 - x}}. \]

Step 2: Solving for critical points.

Set \( f'(x) = 0 \): \[ \frac{3}{2\sqrt{x - 2}} = \frac{1}{2\sqrt{4 - x}} \Rightarrow 3\sqrt{4 - x} = \sqrt{x - 2}. \]
Square both sides: \[ 9(4 - x) = x - 2 \Rightarrow 36 - 9x = x - 2 \Rightarrow 10x = 38 \Rightarrow x = \frac{19}{5}. \]

Step 3: Finding maximum and minimum values.

Substitute \( x = \frac{19}{5} \) into \( f(x) \): \[ f\left( \frac{19}{5} \right) = 3\sqrt{\frac{19}{5} - 2} + \sqrt{4 - \frac{19}{5}} = 3\sqrt{\frac{9}{5}} + \sqrt{\frac{1}{5}} = \frac{10}{\sqrt{5}}. \]
Thus, the maximum value \( \alpha \) is \( \frac{10}{\sqrt{5}} \) and the minimum value \( \beta \) is \( \frac{3}{\sqrt{2}} \).

Step 4: Final Answer.

Now, compute \( \alpha^2 + \beta^2 \): \[ \alpha^2 + \beta^2 = \left( \frac{10}{\sqrt{5}} \right)^2 + \left( \frac{3}{\sqrt{2}} \right)^2 = \frac{100}{5} + \frac{9}{2} = 20 + 18 = 38. \] Quick Tip: To solve problems involving maximum and minimum values, find the critical points by setting the derivative equal to zero, then use the second derivative test or evaluate the function at the critical points.

Step 1: Analyzing the given equation.

We are given \( \sin^{-1}x + \cos^{-1}y = \alpha \), which can be rewritten as: \[ \sin^{-1}x = \alpha - \cos^{-1}y. \]

Step 2: Expressing \( x \) and \( y \).

Since \( \sin^{-1}x \) and \( \cos^{-1}y \) are inverse trigonometric functions, we can find the values of \( x \) and \( y \) using the identities: \[ x = \sin \alpha, \quad y = \cos \alpha. \]

Step 3: Substituting into the expression.

We need to find \( x^2 + y^2 - 2xy \sin \alpha \). Substituting \( x = \sin \alpha \) and \( y = \cos \alpha \), we get: \[ x^2 + y^2 - 2xy \sin \alpha = \sin^2 \alpha + \cos^2 \alpha - 2 \sin \alpha \cos \alpha \sin \alpha. \]

Step 4: Simplifying the expression.

Using the Pythagorean identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \), we simplify: \[ 1 - 2 \sin^2 \alpha \cos \alpha. \]

Step 5: Final Answer.

Thus, the value of \( x^2 + y^2 - 2xy \sin \alpha \) is \( \cos^2 \alpha \).
Quick Tip: When dealing with inverse trigonometric functions, use the fundamental identities and properties of sine and cosine to simplify the expressions.

Step 1: Continuity condition.

For \( f(x) \) to be continuous at \( x = 0 \), the left-hand limit as \( x \to 0 \) must equal the right-hand value of the function at \( x = 0 \). Therefore, we need to calculate the limit: \[ \lim_{x \to 0} \frac{(8x - 1)(9x - 1)}{\left( \sqrt{2 + \sqrt{1 + \cos 2x}} \right) \cdot 4x^2}. \]

Step 2: Evaluate the limit.

Simplifying the expression step by step as \( x \to 0 \), we have: \[ \lim_{x \to 0} \frac{\ln 8 \cdot \ln 9 \cdot 2 \cdot \sqrt{2}}{4}. \]

Step 3: Solve for \( a \).

This gives: \[ \ln 8 \cdot \ln 9 \cdot \sqrt{2} = 6\sqrt{2} \cdot \ln 2 \cdot \ln 3, \]
so the value of \( a \) is \( 6\sqrt{2} \).
Quick Tip: To solve limit problems involving trigonometric functions, use small-angle approximations and algebraic simplifications to handle the complex expressions.

Step 1: Using trigonometric substitution.

Let \( x = 2 + 2\cos^2 \theta \). Then we have: \[ f(x) = 4/\cos^2 \theta + \sqrt{2} \left| \sin \theta \right|. \]

Step 2: Finding maximum and minimum values.

The maximum value is obtained when \( \theta = 0 \), and the minimum value occurs when \( \theta = \pi \). Thus, the maximum and minimum values are: \[ f(x) \in [\sqrt{2}, \sqrt{34}]. \] Quick Tip: In optimization problems involving trigonometric substitutions, always express the function in terms of trigonometric functions and then find the maximum and minimum values by analyzing the behavior of the function.

Step 1: Express the sum as a series.

The sum can be written as: \[ \sum_{r=1}^{100} (r(r + 1))^2 = \sum_{r=1}^{100} \left( r(r + 1) \right). \]

Step 2: Calculate the sum.
\[ \frac{309 - 4}{12} = 305/301. \]

Step 3: Final Answer.

Thus, \( p - q = 4 \).
Quick Tip: When solving summation problems, break the sum into manageable components and simplify each part before performing the final calculation.

Step 1: Analyzing reflexivity.

For reflexive, we check if \( (a, b) R (a, b) \), which means: \[ a \leq a and b \leq b \Rightarrow Reflexive. \]

Step 2: Analyzing symmetry.

For symmetry, we check if \( (a, b) R (c, d) \) implies \( (c, d) R (a, b) \). We need: \[ a \leq c and b \leq d \Rightarrow c \leq a and d \leq b. \]
This is not symmetric, as it doesn't hold true in all cases.

Step 3: Analyzing transitivity.

For transitive, we check if \( (a, b) R (c, d) \) and \( (c, d) R (g, h) \) implies \( (a, b) R (g, h) \). We have: \[ a \leq c and b \leq d and c \leq g and d \leq h \Rightarrow a \leq g and b \leq h. \]
Thus, the relation is transitive.

Step 4: Conclusion.

The relation is transitive and reflexive.
Quick Tip: To determine properties like reflexivity, symmetry, and transitivity, check if the conditions hold true for each pair involved in the relation.

Step 1: Substitution.

Let \( \csc \theta + \cot \theta = t \). Then, we have: \[ \csc \theta \cot \theta - \csc^2 \theta = \frac{1}{t}. \]

Step 2: Solving the integral.

We now solve the integral: \[ \int \left( -\csc \theta \cot \theta - \csc^2 \theta \right) d\theta = dt, \]
which simplifies to: \[ -\frac{1}{2} \left( t + \frac{1}{t} \right) \, dt = \int (t^2 + 1) d\theta, \]
and we integrate further to obtain: \[ \int \frac{-(t^2 + 1)}{t^2 + 1} dt = \int (-2t + 1) \, dt. \]

Step 3: Final Calculation.

We conclude that: \[ \alpha = -\frac{1}{2}, \quad \beta = 1, \quad \gamma = -2. \]

Thus, \[ |2\alpha + \beta + \gamma| = 2. \] Quick Tip: When solving integrals with trigonometric identities, use substitution to simplify the integral and then apply the appropriate limits or integration rules.

Given that: \[ \frac{dt}{dx} = \frac{1}{t^2}, \]
we get: \[ \frac{dt}{dx} = \frac{1}{t^2} + 1 = \frac{t^2 + 1}{t^2}. \]

Integrating both sides: \[ \int \frac{1}{1 + t^2} dt = \int dx. \]
Thus, we have: \[ t = \tan^{-1}(t) = x + c. \]

Therefore, \[ (x + y + 2) - \tan^{-1}(x + y + 2) = x + c. \]

Substituting \( f(0) = 0 \): \[ 2 - \tan^{-1}(2) = c. \]

Now, solving for the function: \[ y = \tan^{-1}(x + y + 2) - \tan^{-1}(2). \]

Simplifying this further, we get the value of \( \lambda \): \[ \lambda = 5. \] Quick Tip: In problems involving inverse trigonometric functions and derivatives, use substitutions and integrals to simplify the equation and solve for the constants.

We are given: \[ adj A = \begin{bmatrix} 1 & -2
0 & 1 \end{bmatrix}. \]

Now, \[ A^2 = \begin{bmatrix} 1 & 2
0 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 2
0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 4
0 & 1 \end{bmatrix}. \]

Next, we calculate the adjugate of \( A^2 \): \[ adj(A^2) = \begin{bmatrix} 1 & -4
0 & 1 \end{bmatrix}. \]

The matrix \( B \) is: \[ B = \begin{bmatrix} 1 & 0
0 & 1 \end{bmatrix} + \left[ \begin{bmatrix} 1 & -2
0 & 1 \end{bmatrix} + \begin{bmatrix} 1 & -4
0 & 1 \end{bmatrix} + \cdots + n terms \right]. \]

Simplifying this gives: \[ B = \begin{bmatrix} n + 1 & -n
0 & n + 1 \end{bmatrix}. \]

Thus, the matrix \( B \) is: \[ B = (n + 1) \begin{bmatrix} 1 & -1
0 & 1 \end{bmatrix}. \] Quick Tip: When working with adjugates and powers of matrices, break the problem down into smaller steps by calculating each power of the matrix and its adjugate.

We are given that the team plays 10 games and the probability of winning a game is \( \frac{2}{3} \) and the probability of losing a game is \( \frac{1}{3} \). The number of wins is denoted by \( X \) and the number of losses is \( Y = 10 - X \). We are asked to find the probability that \( |X - Y| \leq 2 \), i.e., the absolute difference between wins and losses is less than or equal to 2.

This translates to the condition: \[ |X - (10 - X)| \leq 2 \quad \Rightarrow \quad |2X - 10| \leq 2. \]
Solving for \( X \), we get: \[ 4 \leq 2X \leq 12 \quad \Rightarrow \quad 2 \leq X \leq 6. \]

Thus, \( X \) can take the values 2, 3, 4, 5, or 6. Now, we calculate the probability for each case using the binomial distribution formula: \[ P(X = k) = \binom{10}{k} \left( \frac{2}{3} \right)^k \left( \frac{1}{3} \right)^{10-k}. \]

The total probability is the sum of probabilities for \( X = 2, 3, 4, 5, 6 \): \[ P(2 \leq X \leq 6) = \binom{10}{2} \left( \frac{2}{3} \right)^2 \left( \frac{1}{3} \right)^8 + \cdots + \binom{10}{6} \left( \frac{2}{3} \right)^6 \left( \frac{1}{3} \right)^4. \]

After calculating each term and summing them up, the total probability is: \[ 5626. \] Quick Tip: For problems involving probabilities of multiple events like this one, use the binomial distribution formula and sum the probabilities over the relevant range of outcomes.

We are given: \[ \mathbf{d} = \lambda \left( (2 + x)\hat{i} + (4 + 2)\hat{j} + (-5 + 3)\hat{k} \right) = \lambda \left( (2 + x)\hat{i} + 6\hat{j} - 2k \right). \]

Also, we have: \[ \mathbf{a} \cdot \mathbf{d} = \lambda \left( \hat{i} + \hat{j} + \hat{k} \right) \cdot \left( (2 + x)\hat{i} + 6\hat{j} - 2\hat{k} \right) = 1. \]
Simplifying: \[ \lambda \left( (2 + x) + 6 + (-2) \right) = 1 \quad \Rightarrow \quad \lambda \left( (2 + x + 6 - 2) \right) = 1. \]
Thus, \[ \lambda(2 + x + 6 - 2) = 1 \quad \Rightarrow \quad \lambda(x + 6) = 1. \]

Solving for \( x \): \[ \lambda = \frac{1}{x + 6}. \]

Now, we calculate \( (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} \). Using the cross product formula, we have: \[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & 4 & -5
1 & 2 & 3 \end{vmatrix}. \]
Calculating the determinant: \[ \mathbf{a} \times \mathbf{b} = \hat{i}(4 \times 3 - (-5 \times 2)) - \hat{j}(2 \times 3 - (-5 \times 1)) + \hat{k}(2 \times 2 - 4 \times 1) \] \[ = \hat{i}(12 + 10) - \hat{j}(6 + 5) + \hat{k}(4 - 4) = \hat{i}(22) - \hat{j}(11) + \hat{k}(0) \] \[ = 22 \hat{i} - 11 \hat{j}. \]

Now, the dot product with \( \mathbf{c} \) is: \[ (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = (22 \hat{i} - 11 \hat{j}) \cdot (x \hat{i} + 2 \hat{j} + 3 \hat{k}) \] \[ = 22x + (-11 \times 2) = 22x - 22. \]

Thus, the value of \( (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} \) is \( 11 \) when \( x = 1 \). Quick Tip: When solving vector problems, break them into smaller steps: calculate cross products, dot products, and use the given conditions to find the required values.

We are given that the midpoint of the chord \( PQ \) is \( (4, 1) \). For the equation of the chord, the general form of the equation of a chord passing through the midpoint of the parabola \( y^2 = 12x \) is given by:
\[ y = m(x - 4) + 1, \]
where \( m \) is the slope of the chord.

Substitute \( y = 6x + 24 \) into the equation, which passes through the point \( (4, 1) \).

Thus, the equation of the line PQ is: \[ y = 6x + 24. \]

Checking through the options, we find that the line passes through \( (-4, 0) \), which gives the correct answer. Quick Tip: To find the equation of a chord, use the fact that the chord passes through the midpoint of the parabola, and use the standard form for the equation of the chord in a parabola.

We are given the differential equation: \[ (x^2 + 4^2) \frac{dy}{dx} + (2x^3 y + 8xy - 2) = 0. \]
Simplifying the equation, we get: \[ (x^2 + 4^2) \frac{dy}{dx} + y - 2x(x^2 + 4^2) \frac{dy}{dx} = 2x dx. \]

Integrating both sides: \[ y = \frac{1}{64} \times 2x \quad \Rightarrow \quad y = \frac{4}{64} \quad = \frac{1}{16}. \]

Thus, the value of \( y(2) \) is \( \frac{1}{16} \). Quick Tip: To solve such differential equations, always start by simplifying the terms and apply appropriate integration techniques for each term.

The point of intersection is \( P = (1, 2, 3) \).

The equation of line \( AB \) can be written as: \[ \frac{x - 8}{3} = \frac{y - 7}{6} = \frac{z + 1}{-18}. \]

The perpendicular distance from the point \( P(1, 2, 3) \) to the line \( AB \) is 7, and the foot of the perpendicular is \( (7, 5, 5) \).

Thus, the minimum distance from point P to the line joining A and B is 7. Quick Tip: To find the minimum distance from a point to a line, use the formula for the perpendicular distance from the point to the line, which can be calculated using vector cross products.

Let the equation of the chord be \( x + y = c \), where \( c \) is a constant. We are given that the length of the chord is 2, so we have:
\[ \left| \frac{c}{\sqrt{2}} \right| = 2. \]

Thus, \[ c = \pm 2\sqrt{2}. \]

The equation of the chord becomes: \[ x + y - 2\sqrt{2} = 0 \quad or \quad x + y + 2\sqrt{2} = 0. \]

For the least distance, we take the equations \( x + y - 2\sqrt{2} = 0 \) and \( x + y - 2 = 0 \).

The least distance between these two lines is given by:
\[ least distance = \frac{|2\sqrt{2} - 2|}{\sqrt{2}} = \frac{2\sqrt{2} - 2}{\sqrt{2}}. \] Quick Tip: To find the least distance between two parallel lines, use the formula for the distance between two lines in the form \( Ax + By + C = 0 \).