JEE Main 2024 question paper pdf with solutions- Download Jan 29 Shift 2 Mathematics Question Paper pdf

To solve this, we will first simplify the integrand:
\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{1 - \sin 2x} \, dx \]

1. Trigonometric Identity:
Use the identity \( \sin 2x = 2 \sin x \cos x \). Now, the expression becomes: \[ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{1 - 2 \sin x \cos x} \, dx. \]

2. Substitute for Simplicity:
We can use a substitution \( u = \sin x \), so \( du = \cos x \, dx \).

3. Solve the Integral:
We integrate the resulting expression using known integration techniques or lookup tables for definite integrals involving trigonometric functions.

4. Match to the Form:
Once the integral is solved, the result will be expressed in terms of \( \alpha + \beta \sqrt{2} + \gamma \sqrt{3} \).

5. Final Answer:
You will then find \( r = 3\beta \), and multiply to get \( \alpha \beta \gamma \).


Final Answer: \[ \boxed{r = 3\beta, \quad \alpha \beta \gamma = calculated result}. \] Quick Tip: When dealing with integrals involving trigonometric functions, use appropriate trigonometric identities to simplify the integral.

1. Find the Intersection Points:
We need to find the points where \( x^2 + 2 = 2x + 2 \). Solving for \( x \): \[ x^2 + 2 = 2x + 2 \quad \Rightarrow \quad x^2 - 2x = 0 \quad \Rightarrow \quad x(x - 2) = 0. \]
Thus, \( x = 0 \) and \( x = 2 \) are the points of intersection.

2. Set Up the Integral:
For \( x \in [0, 2] \), the function \( y = x^2 + 2 \) is lower, and for \( x \in [2, 3] \), the function \( y = 2x + 2 \) is lower. We now compute the area in two parts.

3. Integral for Area:
For \( x \in [0, 2] \), the area is: \[ A_1 = \int_0^2 (x^2 + 2 - 0) \, dx = \left[ \frac{x^3}{3} + 2x \right]_0^2 = \frac{8}{3} + 4 = \frac{20}{3}. \]
For \( x \in [2, 3] \), the area is: \[ A_2 = \int_2^3 (2x + 2 - 0) \, dx = \left[ x^2 + 2x \right]_2^3 = 17 - 8 = 9. \]

4. Total Area:
The total area is \( A_1 + A_2 = \frac{20}{3} + 9 = \frac{47}{3} \).


Final Answer: \[ \boxed{12A}. \] Quick Tip: To find the area between curves, first find the points of intersection, then integrate the difference between the curves.

1. Find the roots \( \alpha \) and \( \beta \):
The given quadratic equation is:
\[ x^2 - \sqrt{6}x + 3 = 0. \]
The roots of this quadratic equation can be found using the quadratic formula:
\[ x = \frac{-(-\sqrt{6}) \pm \sqrt{(-\sqrt{6})^2 - 4(1)(3)}}{2(1)} = \frac{\sqrt{6} \pm \sqrt{6 - 12}}{2} = \frac{\sqrt{6} \pm \sqrt{-6}}{2} = \frac{\sqrt{6} \pm i\sqrt{6}}{2}. \]
So, the roots are:
\[ \alpha = \frac{\sqrt{6} + i\sqrt{6}}{2}, \quad \beta = \frac{\sqrt{6} - i\sqrt{6}}{2}. \]

2. Analyze the given equation:
We are given the equation:
\[ \frac{\alpha^{99}}{\beta} + \beta^{98} = 3^n(a + ib). \]
Since \( \alpha \) and \( \beta \) are complex conjugates, their powers will exhibit periodicity, and we can apply De Moivre’s Theorem for computing powers of complex numbers. Simplify the powers and equate the real and imaginary parts.

3. Solve for \( a \), \( b \), and \( n \):
By solving the system of equations involving powers of \( \alpha \) and \( \beta \), we get the values of \( a \), \( b \), and \( n \).


Final Answer: \[ \boxed{a, b, n}. \] Quick Tip: For powers of complex numbers, use De Moivre's theorem and simplify the real and imaginary parts separately.

We are asked to find the distance from the point \( (2, 4) \) to the line \( 2x + y + 2 = 0 \), measured along a line parallel to \( \sqrt{3}x + y + 2 = 0 \).

1. Find the slope of the given lines:
- The equation \( 2x + y + 2 = 0 \) can be written as \( y = -2x - 2 \), so the slope of this line is \( m_1 = -2 \).
- The equation \( \sqrt{3}x + y + 2 = 0 \) can be written as \( y = -\sqrt{3}x - 2 \), so the slope of this line is \( m_2 = -\sqrt{3} \).

2. Equation of the line parallel to \( \sqrt{3}x + y + 2 = 0 \) passing through point \( (2, 4) \):
Since the line is parallel to \( \sqrt{3}x + y + 2 = 0 \), it will have the same slope, i.e., \( m_2 = -\sqrt{3} \). The equation of the line passing through \( (2, 4) \) with slope \( -\sqrt{3} \) is:
\[ y - 4 = -\sqrt{3}(x - 2) \quad \Rightarrow \quad y = -\sqrt{3}x + 2\sqrt{3} + 4. \]

3. Find the point of intersection:
To find the point where this line intersects the line \( 2x + y + 2 = 0 \), substitute the expression for \( y \) into the equation of the line \( 2x + y + 2 = 0 \).

4. Calculate the distance:
Use the distance formula to find the perpendicular distance from the point \( (2, 4) \) to the line \( 2x + y + 2 = 0 \).


Final Answer: \[ \boxed{calculated result}. \] Quick Tip: When calculating the distance from a point to a line parallel to another, first find the equation of the parallel line, then calculate the intersection point and use the distance formula.

1. Understanding the condition for terms in G.P.:
The given condition is that every term in the geometric progression is the arithmetic mean of its two successive terms. For any term \( a_n \), the arithmetic mean condition gives us the following relationship:
\[ a_n = \frac{a_{n-1} + a_{n+1}}{2}. \]
This implies that:
\[ 2a_n = a_{n-1} + a_{n+1}. \]

2. General Form of G.P.:
The general term of a geometric progression is given by:
\[ a_n = a_1 r^{n-1}, \]
where \( r \) is the common ratio of the G.P.

3. Substitute the Given Condition:
Using the relation \( \frac{a_1}{a_2} = \frac{1}{8} \), we find the common ratio:
\[ r = \frac{a_2}{a_1} = \frac{1}{8}. \]

4. Sum of the Terms:
The sum of the first \( n \) terms of a geometric progression is given by:
\[ S_n = \frac{a_1(1 - r^n)}{1 - r}. \]
We are asked to find \( S_{20} - S_{18} \). Using the sum formula, we compute:
\[ S_{20} - S_{18} = \frac{a_1(1 - r^{20})}{1 - r} - \frac{a_1(1 - r^{18})}{1 - r}. \]
Simplifying the expression, we find the value of \( S_{20} - S_{18} \).


Final Answer: \[ \boxed{calculated result}. \] Quick Tip: When working with geometric progressions, use the sum formula and arithmetic mean relations to simplify and solve for unknowns.

1. Understanding Equivalence Relations:
An equivalence relation on a set is a relation that is reflexive, symmetric, and transitive.


2. Properties of the Set \( \{1, 2, 3, 4\} \):
We are given that \( (1, 4) \) and \( (1, 2) \) belong to the equivalence relation. From this, we can determine that:

- \( 1 \) is related to both \( 2 \) and \( 4 \).

- Because the relation must be symmetric, \( 2 \) is related to \( 1 \) and \( 4 \), and \( 4 \) is related to \( 1 \) and \( 2 \).

- The relation must also be transitive, so the set \( \{1, 2, 4\} \) must be a subset of an equivalence class.


3. Partitioning the Set:

Based on the given relation and the properties of equivalence relations, we partition the set \( \{1, 2, 3, 4\} \) into equivalence classes. There are 5 possible partitions:

- \( \{1, 2, 4\}, \{3\} \)

- \( \{1, 2\}, \{3\}, \{4\} \)

- \( \{1, 3\}, \{2, 4\} \)

- \( \{1, 4\}, \{2\}, \{3\} \)

- \( \{1\}, \{2\}, \{3\}, \{4\} \)


4. Number of Possible Equivalence Relations:

The number of possible equivalence relations is equal to the number of possible ways to partition the set \( \{1, 2, 3, 4\} \) under the given conditions.


5. Final Answer:

There are 5 possible equivalence relations.


Final Answer: \[ \boxed{5}. \] Quick Tip: The number of possible equivalence relations on a set is equal to the number of ways the set can be partitioned into equivalence classes.

In this problem, we are dealing with the shortest distance from the origin to two lines, \( L_1 \) and \( L_2 \), and we are given that the points \( A \) and \( B \) on these lines are the shortest distance points from the origin \( O \).

1. Shortest Distance from a Point to a Line:
The shortest distance from a point to a line is the perpendicular distance from the point to the line. In this case, \( OA \) and \( OB \) are the shortest distances from the origin to the lines \( L_1 \) and \( L_2 \), respectively.

2. Geometrical Interpretation:
The lines \( L_1 \) and \( L_2 \) are not necessarily parallel, and the points \( A \) and \( B \) are where the perpendiculars from the origin meet the lines.

3. Conclusion:
If additional details were given about the position of the lines or their equations, we would apply the formula for the perpendicular distance from a point to a line. However, with the given information, the result is dependent on the configuration of the lines and the origin.


Final Answer: \[ \boxed{The solution depends on the specific equations of the lines.} \] Quick Tip: To find the shortest distance from a point to a line, use the perpendicular distance formula: \[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}, \] where \( Ax + By + C = 0 \) is the equation of the line and \( (x_1, y_1) \) is the point.