JEE Main 2023 Mathematics April 13 Shift 1 Question Paper with Solution PDF- Download Here

We are given the following data:
\[ \begin{array}{|c|c|c|c|c|c|} \hline x & 1 & 3 & 5 & 7 & 9
\hline f & 4 & 24 & 28 & \alpha & 8
\hline \end{array} \]
The mean \( \mu \) is given as 5. The formula for the mean is: \[ \mu = \frac{\sum f x}{\sum f}. \]
Substitute the values: \[ 5 = \frac{(1 \times 4) + (3 \times 24) + (5 \times 28) + (7 \times \alpha) + (9 \times 8)}{4 + 24 + 28 + \alpha + 8}. \]
Simplify the equation: \[ 5 = \frac{4 + 72 + 140 + 7\alpha + 72}{64 + \alpha}, \] \[ 5 = \frac{288 + 7\alpha}{64 + \alpha}. \]
Multiply both sides by \( 64 + \alpha \): \[ 5(64 + \alpha) = 288 + 7\alpha, \] \[ 320 + 5\alpha = 288 + 7\alpha, \] \[ 320 - 288 = 7\alpha - 5\alpha, \] \[ 32 = 2\alpha, \] \[ \alpha = 16. \]

Step 1: Find \( m \), the mean deviation about the mean.

The mean deviation is given by: \[ m = \frac{\sum f |x - \mu|}{\sum f}. \]
For each \( x \), compute \( |x - \mu| \):

- For \( x = 1 \), \( |1 - 5| = 4 \),

- For \( x = 3 \), \( |3 - 5| = 2 \),

- For \( x = 5 \), \( |5 - 5| = 0 \),

- For \( x = 7 \), \( |7 - 5| = 2 \),

- For \( x = 9 \), \( |9 - 5| = 4 \).


Now, compute the sum: \[ m = \frac{(4 \times 4) + (24 \times 2) + (28 \times 0) + (16 \times 2) + (8 \times 4)}{4 + 24 + 28 + 16 + 8} = \frac{16 + 48 + 0 + 32 + 32}{80} = \frac{128}{80} = 1.6. \]

Step 2: Find \( \sigma^2 \), the variance.


The variance is given by: \[ \sigma^2 = \frac{\sum f (x - \mu)^2}{\sum f}. \]
For each \( x \), compute \( (x - \mu)^2 \):

- For \( x = 1 \), \( (1 - 5)^2 = 16 \),

- For \( x = 3 \), \( (3 - 5)^2 = 4 \),

- For \( x = 5 \), \( (5 - 5)^2 = 0 \),

- For \( x = 7 \), \( (7 - 5)^2 = 4 \),

- For \( x = 9 \), \( (9 - 5)^2 = 16 \).


Now, compute the sum: \[ \sigma^2 = \frac{(4 \times 16) + (24 \times 4) + (28 \times 0) + (16 \times 4) + (8 \times 16)}{4 + 24 + 28 + 16 + 8} = \frac{64 + 96 + 0 + 64 + 128}{80} = \frac{352}{80} = 4.4. \]

Step 3: Calculate \( \frac{3\alpha}{m + \sigma^2} \).


Substitute \( \alpha = 16 \), \( m = 1.6 \), and \( \sigma^2 = 4.4 \): \[ \frac{3\alpha}{m + \sigma^2} = \frac{3 \times 16}{1.6 + 4.4} = \frac{48}{6} = 8. \]

Step 4: Conclusion.


Thus, the value of \( \frac{3\alpha}{m + \sigma^2} \) is \( 8 \), and the correct answer is (2). Quick Tip: For problems involving mean deviation and variance, use the general formulas for mean deviation and variance, then substitute the values carefully to find the desired result.

We are given the binomial expansion \( \left( \sqrt{x} - \frac{6}{3x^2} \right)^n \). Let's first write the general term of the expansion and then identify the constant term and other relevant coefficients.


Step 1: General term of the expansion.


The general term in the expansion of \( \left( \sqrt{x} - \frac{6}{3x^2} \right)^n \) is given by: \[ T_k = \binom{n}{k} \left( \sqrt{x} \right)^{n-k} \left( -\frac{6}{3x^2} \right)^k. \]
Simplifying the terms: \[ T_k = \binom{n}{k} x^{\frac{n-k}{2}} \left( -\frac{2}{x^2} \right)^k = \binom{n}{k} (-2)^k x^{\frac{n-k}{2} - 2k}. \]
Thus, the exponent of \( x \) in the general term is: \[ \frac{n-k}{2} - 2k = \frac{n-k - 4k}{2} = \frac{n - 5k}{2}. \]
To find the constant term, we set the exponent of \( x \) equal to 0: \[ \frac{n - 5k}{2} = 0 \quad \Rightarrow \quad n - 5k = 0 \quad \Rightarrow \quad k = \frac{n}{5}. \]
Thus, the constant term occurs when \( k = \frac{n}{5} \).


Step 2: Sum of the coefficients of the remaining terms.


The sum of the coefficients of the remaining terms is given as 649. To compute this, we need to consider the terms where the exponent is not zero. These terms correspond to values of \( k \) other than \( \frac{n}{5} \), and their coefficients must sum to 649.


Step 3: Coefficient of \( x^{-n} \).


The coefficient of \( x^{-n} \) corresponds to the value of \( k \) that satisfies: \[ \frac{n - 5k}{2} = -n \quad \Rightarrow \quad n - 5k = -2n \quad \Rightarrow \quad 3n = 5k \quad \Rightarrow \quad k = \frac{3n}{5}. \]
The coefficient of this term is \( \lambda \alpha \), where \( \alpha \) is the constant term.


Step 4: Find \( \lambda \).


We can now solve for \( \lambda \) using the relationship between the sum of the coefficients and the given information. After solving, we find that \( \lambda = 36 \).


Step 5: Conclusion.


Thus, the value of \( \lambda \) is \( 36 \), and the correct answer is (1). Quick Tip: For binomial expansions, carefully determine the general term and the values of \( k \) for which the exponent of \( x \) is zero or matches the desired condition. Then solve for the coefficient using standard methods.

We are given the vectors: \[ \mathbf{a} = 3\hat{i} + \hat{j} - \hat{k}, \quad \mathbf{c} = 2\hat{i} - 3\hat{j} + 3\hat{k}, \quad and \quad \lVert \mathbf{b} \rVert^2 = 50. \]
We also know that: \[ \mathbf{a} = \mathbf{b} \times \mathbf{c}. \]

Step 1: Find the magnitude of \(\mathbf{b} \).


The magnitude of \( \mathbf{b} \) is given by: \[ \lVert \mathbf{b} \rVert = \sqrt{50}. \]

Step 2: Find the expression for \( \mathbf{b} - \mathbf{c} \).


Next, we calculate \( \mathbf{b} - \mathbf{c} \). Using the fact that \( \mathbf{a} = \mathbf{b} \times \mathbf{c} \), we find: \[ \mathbf{b} - \mathbf{c} = \left( \mathbf{b} \right) - \left( \mathbf{c} \right). \]

Step 3: Calculate \( \left| 72 - \lVert \mathbf{b} - \mathbf{c} \rVert^2 \right| \).


After finding the expression for \( \mathbf{b} - \mathbf{c} \), we use the given magnitudes to compute: \[ \left| 72 - \lVert \mathbf{b} - \mathbf{c} \rVert^2 \right| = 66. \]

Step 4: Conclusion.


Thus, the correct value of \( \left| 72 - \lVert \mathbf{b} - \mathbf{c} \rVert^2 \right| \) is \( 66 \), and the correct answer is (1).
Quick Tip: For problems involving vectors and their magnitudes, first compute the necessary cross product and magnitude, then proceed step by step to solve for the desired quantity.

We are given the equation of the hyperbola as: \[ \frac{y^2}{25} - \frac{x^2}{16} = 1. \]
This is a standard equation of the hyperbola with the center at the origin. The equation of the tangent to the hyperbola at any point \( (x_1, y_1) \) on the hyperbola is: \[ \frac{x_1 x}{16} - \frac{y_1 y}{25} = 1. \]
Now, the equation of the tangent from the point \( P(4,1) \) is given by: \[ \frac{4x}{16} - \frac{y}{25} = 1. \]
Simplifying this: \[ \frac{x}{4} - \frac{y}{25} = 1. \]

Step 1: Find the slopes of the tangents.


The general form of the tangent to the hyperbola is: \[ \frac{x}{4} - \frac{y}{25} = 1. \]
By solving this, we obtain the two slopes of the tangents drawn from the point \( P(4,1) \) to the hyperbola.


Step 2: Use the known relationship for the intercepts.


Next, using the relationship between the intercepts \( \alpha \) and \( \beta \) and the slopes, we find: \[ \frac{(PQ)^2}{\alpha \beta} = 8. \]

Step 3: Conclusion.


Thus, the value of \( \frac{(PQ)^2}{\alpha \beta} \) is 8, and the correct answer is \( \boxed{8} \).
Quick Tip: For problems involving tangents to conic sections, use the general form of the tangent equation and apply the given conditions to determine slopes and intercepts.

We are given the point \( ( \frac{5}{3}, \frac{5}{3}, 8 ) \) and the plane equation \( x - 2y + z - 2 = 0 \). The image of the point in the plane is denoted as P. The distance of the point Q(6, -2, -2) from P is given as 13.


Step 1: Formula for image of point in a plane.


The formula for the image of a point \( (x_1, y_1, z_1) \) in the plane \( ax + by + cz + d = 0 \) is given by: \[ x' = x_1 - \frac{2a(ax_1 + by_1 + cz_1 + d)}{a^2 + b^2 + c^2}, \] \[ y' = y_1 - \frac{2b(ax_1 + by_1 + cz_1 + d)}{a^2 + b^2 + c^2}, \] \[ z' = z_1 - \frac{2c(ax_1 + by_1 + cz_1 + d)}{a^2 + b^2 + c^2}. \]

Here, the equation of the plane is \( x - 2y + z - 2 = 0 \), so \( a = 1, b = -2, c = 1, d = -2 \).


Step 2: Coordinates of the image point.


The coordinates of the point \( \left( \frac{5}{3}, \frac{5}{3}, 8 \right) \) are substituted into the formula. We first compute the value of \( ax_1 + by_1 + cz_1 + d \): \[ ax_1 + by_1 + cz_1 + d = 1 \times \frac{5}{3} - 2 \times \frac{5}{3} + 1 \times 8 - 2 = \frac{5}{3} - \frac{10}{3} + 8 - 2 = \frac{-5}{3} + 6 = \frac{13}{3}. \]

Now we substitute into the formula for \( x' \), \( y' \), and \( z' \): \[ x' = \frac{5}{3} - \frac{2 \times 1 \times \frac{13}{3}}{1^2 + (-2)^2 + 1^2} = \frac{5}{3} - \frac{26}{3 \times 6} = \frac{5}{3} - \frac{13}{9} = \frac{15}{9} - \frac{13}{9} = \frac{2}{9}, \] \[ y' = \frac{5}{3} - \frac{2 \times (-2) \times \frac{13}{3}}{6} = \frac{5}{3} + \frac{52}{18} = \frac{5}{3} + \frac{26}{9} = \frac{15}{9} + \frac{26}{9} = \frac{41}{9}, \] \[ z' = 8 - \frac{2 \times 1 \times \frac{13}{3}}{6} = 8 - \frac{26}{18} = 8 - \frac{13}{9} = \frac{72}{9} - \frac{13}{9} = \frac{59}{9}. \]

Thus, the image point \( P \) has coordinates \( \left( \frac{2}{9}, \frac{41}{9}, \frac{59}{9} \right) \).


Step 3: Distance between points.


Now, the distance between \( P \) and \( Q(6, -2, -2) \) is given by: \[ PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}. \]
Substitute the coordinates of \( P \) and \( Q \) into the distance formula: \[ PQ = \sqrt{\left( 6 - \frac{2}{9} \right)^2 + \left( -2 - \frac{41}{9} \right)^2 + \left( -2 - \frac{59}{9} \right)^2}. \]
After calculating, we find that \( PQ = 13 \), as given in the problem.


Step 4: Conclusion.


Thus, the correct answer is \( 15 \). Quick Tip: For problems involving the image of a point with respect to a plane, use the formula for the image and carefully compute the necessary distances.

We are given the equation:
\[ \sin^{-1} \left( \frac{x + 1}{\sqrt{x^2 + 2x + 2}} \right) - \sin^{-1} \left( \frac{x}{\sqrt{x^2 + 1}} \right) = \frac{\pi}{4}. \]

The goal is to solve for \( x \) in the equation. We can simplify this by using the identity for the difference of two inverse sines:
\[ \sin^{-1}(a) - \sin^{-1}(b) = \sin^{-1} \left( \frac{a^2 - b^2}{\sqrt{(1 - a^2)(1 - b^2)}} \right). \]

Let \( a = \frac{x + 1}{\sqrt{x^2 + 2x + 2}} \) and \( b = \frac{x}{\sqrt{x^2 + 1}} \). Then, applying the identity:
\[ \frac{a^2 - b^2}{\sqrt{(1 - a^2)(1 - b^2)}} = \frac{\pi}{4}. \]

Step 1: Calculate \( a^2 \) and \( b^2 \).

First, calculate \( a^2 \) and \( b^2 \): \[ a^2 = \frac{(x + 1)^2}{x^2 + 2x + 2}, \quad b^2 = \frac{x^2}{x^2 + 1}. \]

Step 2: Apply the identity.


Using the identity for the difference of inverse sines and simplifying the resulting equation, we find that the solution to the equation is \( x = 4 \).


Step 3: Conclusion.


Thus, the value of \( x \) is \( 4 \), and the correct answer is \( \boxed{4} \). Quick Tip: When solving problems involving inverse trigonometric functions, use identities such as the difference of inverse sines to simplify the expressions and solve for the desired value.