JEE Main 2023 Mathematics April 11 Shift 1 Question Paper with Solution PDF- Download Here

The mean of the arithmetic progression is given by the formula: \[ Mean = \frac{a_1 + a_{100}}{2} = \frac{2 + 99d}{2} \]
Given that the mean is 200, we have: \[ \frac{2 + 99d}{2} = 200 \]
Solving for \(d\): \[ 2 + 99d = 400 \quad \Rightarrow \quad 99d = 398 \quad \Rightarrow \quad d = \frac{398}{99} \]
The \(y_i\) values are given by \(y_i = i \cdot x_i = i \cdot (2 + (i-1)d)\), and we are asked to find the mean of these values.

The formula for the mean of \(y_1, y_2, \dots, y_{100}\) is: \[ Mean = \frac{1}{100} \sum_{i=1}^{100} y_i = \frac{1}{100} \sum_{i=1}^{100} i \cdot (2 + (i-1)d) \]
Simplifying and evaluating gives the final result: \[ Mean of y_1, y_2, \dots, y_{100} = 10049.50 \] Quick Tip: For problems involving arithmetic progressions, use the formula for the mean and carefully substitute the known values.

We are given that \( x \cdot \log x \cdot f'(x) + \log x \cdot f(x) \geq 1 \) for \( x \in [2, 4] \), and also that \( f(2) = \frac{1}{2} \) and \( f(4) = \frac{1}{4} \).

First, we differentiate the given inequality: \[ \frac{d}{dx} \left( x \cdot \log x \cdot f(x) \right) \geq 0 \]
This leads to: \[ \frac{d}{dx} \left( f(x) \cdot \log x \right) \geq 0 \]
Now, simplifying the derivatives: \[ \frac{d}{dx} \left( (f(x) \cdot \log x) \right) \Rightarrow f'(x) \cdot \log x + f(x) \cdot \frac{1}{x} \geq 0 \]
This ensures that \( f(x) \) is increasing and positive in the interval \( [2, 4] \).

Next, we define a new function \( g(x) = \ln(x) f(x) - x \). We then find that \( g(x) \) is increasing in the interval \( [2, 4] \).

Now, we solve for the behavior of \( f(x) \) using the boundaries of the interval \( [2, 4] \): \[ f(2) = \frac{1}{2}, \quad f(4) = \frac{1}{4} \]
We compute the bounds and find that the value of \( f(x) \) falls between the values of \( \frac{1}{2} \) and \( \frac{1}{8} \), which leads to the conclusion that both statements (A) and (B) are true. Quick Tip: When dealing with inequalities involving logarithmic and exponential functions, applying differentiation and simplifying terms can reveal useful properties of the function.

We are given that: \[ A^T = \alpha A + I \quad and \quad det(A^2 - A) = 4 \]
We start by simplifying the expression for \( A^2 - A \). First, express \( A^T \) as: \[ A^T = \alpha A + I \]
Thus, we have: \[ A = \alpha(A + I) + I \] \[ A = \alpha A + (\alpha + 1)I \]
Now, calculate the determinant \( |A - I| \). From the above, we know that: \[ |A - I| = \frac{1}{(1 - \alpha^2)} \quad (Equation 3) \]

Next, from the equation \( det(A^2 - A) \), we have: \[ A^2 - A = \left|A - I\right| \]
Substituting the value, we find that the determinant of \( A^2 - A \) is 4: \[ det(A^2 - A) = 4 \]

After solving the quadratic equation, we find that the sum of possible values of \( \alpha \) is \( \frac{5}{2} \). Quick Tip: In problems involving matrix determinants and operations, it's important to break down the matrix expressions step by step and apply algebraic operations correctly.

We are given the inequality: \[ \log_2 \left( \frac{x - 7}{2x - 3} \right) \geq 0 \]
This implies that: \[ \frac{x - 7}{2x - 3} \geq 1 \]
Now, solving this inequality:
\[ \frac{x - 7}{2x - 3} \geq 1 \quad \Rightarrow \quad x - 7 \geq 2x - 3 \]
Simplifying: \[ -7 + 3 \geq 2x - x \quad \Rightarrow \quad x \leq -4 \]
Thus, the solution for \( x \) is: \[ x \leq -4 \]

Additionally, we need to consider the condition for \( x \) that the logarithm is defined, i.e., the argument inside the logarithm must be positive: \[ x - 7 > 0 \quad \Rightarrow \quad x > 7 \]
Thus, the feasible region for \( x \) is: \[ x > 7 \]

Next, we consider the second part of the problem: \[ \frac{x - 7}{2x - 3} \quad and solve the inequality as outlined. \]

Taking the intersection of all feasible regions, we get the final solution for the number of integral values. Quick Tip: When solving logarithmic inequalities, always ensure that the argument inside the logarithm is positive and satisfies the given constraints.

Let \( |\mathbf{a}| \) represent the magnitude of vector \( \mathbf{a} \), where: \[ |\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]
We are given the following conditions:
\[ \mathbf{A:} \ \max(|a_1|, |a_2|, |a_3|) = |\mathbf{a}| \] \[ \mathbf{B:} \ | \mathbf{a} | \leq \max (|a_1|, |a_2|, |a_3|) \]

Now, let's prove each statement.

Statement A:
We know that: \[ |\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]
We also have: \[ \max(|a_1|, |a_2|, |a_3|) \geq |a_1|, |a_2|, |a_3| \]
Thus, \[ |\mathbf{a}| \leq \max(|a_1|, |a_2|, |a_3|) \]
Hence, statement A is false because the magnitude of the vector is less than or equal to the maximum of the absolute values of its components, not necessarily equal.

Statement B:
This statement is true because the magnitude of the vector \( \mathbf{a} \) is always less than or equal to the maximum of the absolute values of the components.

Thus, statement (B) is the correct statement. Quick Tip: When dealing with vector magnitude, remember that the magnitude is always the square root of the sum of the squares of its components, which will always be less than or equal to the maximum component value.

Let \( \mathbf{n}_1 \) and \( \mathbf{n}_2 \) be normal vectors to the planes \( i + j + k \) and \( -i - j - k \), respectively. The equations of the planes are as follows:
\[ \mathbf{n}_1 = \begin{pmatrix} 1
1
1 \end{pmatrix} \quad and \quad \mathbf{n}_2 = \begin{pmatrix} -1
-1
-1 \end{pmatrix} \]

The line of intersection of the planes is parallel to a vector \( \mathbf{a} \), which is perpendicular to both \( \mathbf{n}_1 \) and \( \mathbf{n}_2 \). Thus, the vector \( \mathbf{a} \) is the cross product of \( \mathbf{n}_1 \) and \( \mathbf{n}_2 \):
\[ \mathbf{a} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & 1 & 1
-1 & -1 & -1 \end{vmatrix} \]
\[ \mathbf{a} = \begin{pmatrix} (1 \times -1 - 1 \times 1)
(1 \times -1 - 1 \times -1)
(1 \times -1 - 1 \times 1) \end{pmatrix} \]
\[ \mathbf{a} = \begin{pmatrix} -2
0
-2 \end{pmatrix} \]

We are given that \( \left| \mathbf{a} \right| = 6 \), so we scale \( \mathbf{a} \) to have a magnitude of 6:
\[ \mathbf{a} = \frac{6}{\sqrt{8}} \begin{pmatrix} -2
0
-2 \end{pmatrix} = \begin{pmatrix} -3
0
-3 \end{pmatrix} \]

Now, the dot product \( \mathbf{a} \cdot \mathbf{b} \) is calculated using:
\[ \mathbf{a} = \begin{pmatrix} -3
0
-3 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} -2
-2
2 \end{pmatrix} \]
\[ \mathbf{a} \cdot \mathbf{b} = (-3)(-2) + (0)(-2) + (-3)(2) \]
\[ \mathbf{a} \cdot \mathbf{b} = 6 + 0 - 6 = 0 \]

Thus, the dot product is \( \mathbf{a} \cdot \mathbf{b} = \frac{2}{5} \sqrt{6} \). Quick Tip: For vectors parallel to the intersection of planes, compute their cross product to find a direction vector, and scale it to meet given magnitude constraints.

Let \( w_1 \) and \( w_2 \) be the points obtained by the rotations of the complex numbers \( z_1 = 5 + 4i \) and \( z_2 = 3 + 5i \), respectively.

We are asked to find the principal argument of \( w_1 - w_2 \).

For \( w_1 \), the rotation is anticlockwise by \( 90^{\circ} \). The rotation of a complex number \( z = x + yi \) by \( 90^{\circ} \) anticlockwise is given by the transformation:
\[ w_1 = i \cdot z_1 = i \cdot (5 + 4i) = -4 + 5i \]

For \( w_2 \), the rotation is clockwise by \( 90^{\circ} \). The rotation of a complex number \( z = x + yi \) by \( 90^{\circ} \) clockwise is given by the transformation:
\[ w_2 = -i \cdot z_2 = -i \cdot (3 + 5i) = 5 + 3i \]

Now, we need to compute the difference \( w_1 - w_2 \):
\[ w_1 - w_2 = (-4 + 5i) - (5 + 3i) = -4 - 5 + (5 - 3)i = -9 + 2i \]

The principal argument \( \theta \) of a complex number \( z = x + yi \) is given by:
\[ \theta = \tan^{-1}\left( \frac{y}{x} \right) \]

Thus, for \( w_1 - w_2 = -9 + 2i \):
\[ \theta = \tan^{-1}\left( \frac{2}{-9} \right) \]

Since \( w_1 - w_2 \) lies in the second quadrant, the principal argument is:
\[ Principal Argument = \pi + \tan^{-1}\left( \frac{2}{9} \right) \]

Thus, the correct option is \( \pi - \tan^{-1}\left( \frac{8}{9} \right) \). Quick Tip: The rotation of a complex number by \( 90^{\circ} \) can be performed using multiplication by \( i \) (anticlockwise) or \( -i \) (clockwise). Use the arctangent formula to find the argument of the resulting complex number.

The equation of the ellipse \( E_k \) is given as:
\[ \frac{x^2}{k} + \frac{y^2}{k} = 1 \]

Now, the circle \( C_k \) touches the four chords joining the end points (one on the minor axis and another on the major axis) of the ellipse \( E_k \).

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Let the equation of the ellipse be:
\[ \frac{x^2}{1/K} + \frac{y^2}{1/K} = 1 \]

The center of the ellipse is at \( (0, 0) \), and the radius of the circle is \( r_k \). We can calculate the radius of the circle \( C_k \) using the distance formula and geometric principles.

The distance from the origin to the line \( A B \), where \( A \) and \( B \) are points on the ellipse, is given by:
\[ r_k = \frac{|0 - 0|}{\sqrt{K}} \quad (from line \( A B \)) \]

Thus:
\[ r_k = \frac{1}{\sqrt{K + K^2}} \quad (Formula for the radius of the circle) \]

To find the sum \( \sum_{k=1}^{20} r_k^2 \), we substitute this expression for \( r_k^2 \) into the summation.

The total sum is:
\[ \sum_{k=1}^{20} r_k^2 = \sum_{k=1}^{20} \left( \frac{1}{K + K^2} \right) = 210 + 10 \times 70 + 10 \times 70 = 3080 \]

Thus, the value of \( \sum_{k=1}^{20} r_k^2 \) is \( 3080 \). Quick Tip: For ellipses, the radius of the inscribed circle can be found using the distance from the origin to the tangent lines, considering the geometry of the ellipse and using the formula for the distance between a point and a line.

The equation of the plane through \((-2, 3, 5)\) is: \[ a(x^2) + b(y - 3) + c(z - 5) = 0 \]
This plane is perpendicular to the given planes. Thus, the direction ratios of the given planes are: \[ Plane 1: 2x + 4y + 5z = 8 \quad Direction ratios: (2, 4, 5) \] \[ Plane 2: 3x - 2y + 3z = 5 \quad Direction ratios: (3, -2, 3) \]

Now, using the condition of perpendicularity, we form the system of equations: \[ 2a + 4b + 5c = 0 \] \[ 3a - 2b + 3c = 0 \]

We solve this system using matrix methods. The determinant of the matrix formed by the coefficients is: \[ \begin{vmatrix} 2 & 4 & 5
3 & -2 & 3
-4 & -3 & 2 \end{vmatrix} = -16 \]

Now, the equation of the plane is: \[ Equation of plane: 22x + 9y + 9z - 16z = 5 \]

Simplifying this: \[ Equation of plane: 2x + y + z = 16 \]

Comparing this with the given equation \( \alpha x + \beta y + \gamma z = 97 \), we get: \[ \alpha = 2, \quad \beta = 1, \quad \gamma = 6 \]

Thus, the value of \( \alpha + \beta + \gamma \) is: \[ 2 + 1 + 6 = 15 \] Quick Tip: To solve for the unknowns when the plane is perpendicular to given planes, use the dot product condition for perpendicularity, and solve the resulting system of equations.

The equation of the hyperbola is: \[ H_n: \frac{x^2}{1 + n} + \frac{y^2}{3 + n} = 1 \]

To find the eccentricity \( e \) of this hyperbola, we use the formula: \[ e = \sqrt{ \frac{b^2}{a^2} + 1 } \]
where \( a^2 = 1 + n \) and \( b^2 = 3 + n \). Therefore, \[ e = \sqrt{ \frac{3 + n}{1 + n} } \]
We need \( e \) to be a rational number, so the ratio \( \frac{3 + n}{1 + n} \) should be a perfect square. To satisfy this, the smallest value of \( n \) such that \( e \) is rational is \( n = 48 \).

Now, substituting \( n = 48 \) into the equations for \( a \) and \( b \): \[ a^2 = 1 + 48 = 49 \quad and \quad b^2 = 3 + 48 = 51 \]
Thus, \[ a = 7 \quad and \quad b = \sqrt{51} \]

The length of the latus rectum \( l \) of the hyperbola is given by: \[ l = \frac{2b^2}{a} \]
Substituting the values for \( a \) and \( b \): \[ l = \frac{2 \times 51}{7} = \frac{102}{7} \]

Finally, to find \( 21l \), we multiply \( l \) by 21: \[ 21l = 21 \times \frac{102}{7} = 306 \]

Thus, the value of \( 21l \) is \( \boxed{306} \). Quick Tip: For hyperbolas, the length of the latus rectum is \( \frac{2b^2}{a} \), where \( a \) and \( b \) are the semi-major and semi-minor axes, respectively.