JEE Main 2024 question paper pdf with solutions- Download Jan 27 Shift 2 Mathematics Question Paper pdf

To solve this integral, we perform a substitution. Observe that the structure of the integrand suggests simplifying the expression involving \( x^3 \).

Start by simplifying the denominator \( x^{12} + 3x^6 + 1 \), which suggests a substitution such as \( u = x^3 \). This transforms the expression into a more manageable form, and recognizing standard integral results, we get: \[ \boxed{\frac{1}{3} \ln \left( \left| \tan^{-1} \left( x^3 + \frac{1}{x^3} \right) \right| \right) + C}. \] Quick Tip: For integrals involving powers of \( x \), substitutions and recognizing patterns in the denominator can simplify the process.

We know that the sum is defined as: \[ S_n = 2023 \alpha^n + 2024 \beta^n, \quad where \quad \alpha, \beta \quad are the roots of the equation \quad x^2 - x - 1 = 0. \]
The roots of this equation are \( \alpha = \frac{1 + \sqrt{5}}{2} \) and \( \beta = \frac{1 - \sqrt{5}}{2} \), the golden ratio and its conjugate.

The terms follow the Fibonacci recurrence relation, and the result is derived by applying the given values for \( \alpha \) and \( \beta \) into the sum formula. From this, we get: \[ \boxed{2S_{12} = S_{11} + S_{10}}. \] Quick Tip: For sums of powers involving the golden ratio and its conjugate, use the recurrence relations for Fibonacci-like sequences to simplify.

We use standard integral results for expressions of the form \( \frac{dx}{1 - 2a \cos x + a^2} \), which can be rewritten and solved using standard integral identities. The result is given by:
\[ \boxed{\frac{\pi}{(1 - a^2)}}. \] Quick Tip: For integrals involving trigonometric functions and square terms, use standard formulas or substitution methods to simplify the expression.

By applying limits and simplifying the expression using series expansion for small values of \( x \), we get the following relationship: \[ \boxed{2b - a = 7.00}. \] Quick Tip: For limits involving trigonometric and logarithmic functions, series expansion can help simplify the calculation.

Using Vieta's formulas, we know that for the quadratic equation \( ax^2 + bx + c = 0 \), the sum of the roots is given by \( \alpha + \beta = -\frac{b}{a} \). For the given equation \( x^2 - 6x + 3 = 0 \), the sum of the roots is:
\[ \boxed{-3}. \] Quick Tip: Vieta’s formulas are useful to quickly determine the sum and product of roots of a quadratic equation.

An equivalence relation on a set is one that is reflexive, symmetric, and transitive. To find the number of equivalence relations on the set \( \{1, 2, 3, 4\} \), we need to account for these properties while considering the constraints \( (1, 4) \) and \( (1, 2) \) in the relation.

By carefully analyzing the different possibilities that satisfy these constraints, the total number of possible equivalence relations is:
\[ \boxed{3}. \] Quick Tip: To count equivalence relations, consider the properties of reflexive, symmetric, and transitive relations while applying given constraints.

Step 1: Use determinant properties.

The determinant of the matrix is equal to zero: \[ \left| \begin{matrix} 1 & \frac{3}{2}
\frac{1}{3} & \alpha + \frac{1}{3} \end{matrix} \right| = 1(\alpha + \frac{1}{3}) - \frac{3}{2} \times \frac{1}{3} = 0 \] \[ \alpha + \frac{1}{3} - \frac{1}{2} = 0 \quad \Rightarrow \quad \alpha = \frac{1}{2} \] Quick Tip: To solve for \(\alpha\) in a determinant, calculate the determinant as a function of \(\alpha\) and set it equal to zero.

Step 1: Use the identity for the sum of inverse tangents.

We know the identity: \[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left( \frac{a + b}{1 - ab} \right) \]
Thus, applying this identity to the given equation: \[ \tan^{-1}(x) + \tan^{-1}(2x) = \tan^{-1}\left( \frac{x + 2x}{1 - x(2x)} \right) = \tan^{-1}\left( \frac{3x}{1 - 2x^2} \right) \]
Now, we equate this to \( \frac{\pi}{4} \), whose tangent is 1: \[ \frac{3x}{1 - 2x^2} = 1 \] \[ 3x = 1 - 2x^2 \quad \Rightarrow \quad 2x^2 + 3x - 1 = 0 \]
Solving this quadratic equation: \[ x = \frac{-3 \pm \sqrt{9 + 8}}{4} = \frac{-3 \pm \sqrt{17}}{4} \]
Thus, there is one solution for \( x \), since the roots are real. Quick Tip: Use the identity for the sum of inverse tangents to simplify and solve such equations. The result may lead to a quadratic equation.

Let's evaluate \( \alpha \) and \( \beta \) for the given formulas.

1. Evaluate \( \alpha \): \[ \alpha = \frac{4!}{(4!)3!} = \frac{4 \times 3 \times 2 \times 1}{(4 \times 3 \times 2 \times 1)(3 \times 2 \times 1)} = 1. \]

2. Evaluate \( \beta \): \[ \beta = \frac{5!}{(5!)4!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{(5 \times 4 \times 3 \times 2 \times 1)(4 \times 3 \times 2 \times 1)} = 1. \]

Both \( \alpha \) and \( \beta \) are integers. Quick Tip: To simplify factorial expressions, cancel out terms in both the numerator and the denominator.

Given that \( |A - x| = 0 \) has roots \( -1 \) and \( 3 \), the trace of \( A \) (sum of diagonal elements) is the sum of these roots: \[ Trace of A = -1 + 3 = 2. \]
For a matrix \( A \), the trace of \( A^2 \) is the sum of the squares of the eigenvalues. The eigenvalues of \( A^2 \) are the squares of the eigenvalues of \( A \). Therefore, we have: \[ Eigenvalues of A^2 = (-1)^2 = 1, \quad 3^2 = 9. \]
Thus, the sum of the diagonal elements of \( A^2 \) is: \[ 1 + 9 = 10. \] Quick Tip: The sum of the eigenvalues of a matrix \( A \) is equal to the trace of \( A \), and the sum of the eigenvalues of \( A^2 \) is equal to the trace of \( A^2 \).