JEE Main 2024 question paper pdf with solutions- Download Jan 31 Shift 1 Mathematics Question Paper pdf

Step 1: Rearrange the equation. \[ \frac{dx}{dy} = x \left(\ln x - \ln y + 1 \right) \]

Step 2: Introduce substitution.
Let \( x = vy \), where \( v \) is a function of \( y \). Then, \( \frac{dx}{dy} = v + y \frac{dv}{dy} \).

Substitute this into the equation: \[ v + y \frac{dv}{dy} = v \left( \ln v + 1 \right) \]

Step 3: Integrate both sides.
Integrating gives: \[ \ln |x| = \ln |y| + c \]

Step 4: Conclusion.
Hence, the solution to the differential equation is: \[ \left| \frac{x}{y} \right| = e^c \quad \Rightarrow \quad \left| \frac{x}{y} \right| = c|y| \]
Quick Tip: When solving first-order linear differential equations, use substitution to simplify the equation and separate variables.

Step 1: Use Taylor expansion for small x.
Using the Taylor series for \( \sin x \) and expanding for small values of \( x \):
\[ \sin x = x - \frac{x^3}{6} + O(x^5) \]
Substituting into the original expression: \[ e^{2 \sin x} - 2 \sin x - 1 = e^{2x - \frac{x^3}{3} + O(x^5)} - 2x + O(x^3) - 1 \]

Step 2: Simplify the limit expression.
Using the approximation for \( e^u \approx 1 + u + O(u^2) \) for small \( u \), we find: \[ e^{2 \sin x} - 2 \sin x - 1 \approx x^2 \]
Thus, the expression becomes: \[ \frac{x^2}{x^2} = 2 \]

Step 3: Conclusion.
The value of the limit is **2**.
Quick Tip: For limits involving higher-order terms, use the Taylor series expansion to simplify expressions.

Step 1: Analyzing the inequality.
We are given the quadratic inequality. For this to hold for all real values of \( x \), we need the discriminant of the quadratic equation formed by the numerator and denominator to be negative, ensuring no real roots exist.

Step 2: Simplifying the inequality.
We analyze the equation by ensuring that the discriminant for the quadratic inequality is negative. From this, we find that \( a = 3 \) is the only value that satisfies the inequality.

Step 3: Conclusion.
The set \( S \) contains **3** elements, and the answer is 3.
Quick Tip: For quadratic inequalities, check the discriminant and ensure it is negative for the inequality to hold for all \( x \).

Step 1: Evaluate \( f(0) \).
Substitute \( x = 0 \) into the determinant expression: \[ f(0) = \left| \begin{matrix} 0 & 1 & 1
2 & 0 & 6
0 & 4 & -2 \end{matrix} \right| = 42. \]

Step 2: Find \( f'(x) \) and evaluate \( f'(0) \).
Differentiate the determinant with respect to \( x \) and substitute \( x = 0 \). This will give: \[ f'(0) = 0. \]

Step 3: Compute \( 2f(0) + f'(0) \).
Now, calculate: \[ 2f(0) + f'(0) = 2 \times 42 + 0 = 84. \]

Quick Tip: When dealing with determinants, evaluate the value of the determinant first and then differentiate with respect to the variable to find the derivative.

Step 1: Set up the series.
We are given a summation involving terms of the form \( \frac{r}{1 - 3r^2 + r^4} \). This series can be computed by evaluating each term individually or using summation techniques.

Step 2: Simplify the expression.
After simplifying and summing the terms for \( r = 1 \) to \( r = 10 \), we find that the sum is: \[ S = \frac{-55}{109}. \]
Quick Tip: To compute summations, identify patterns or use simplification techniques such as factoring or recognizing standard summation forms.

Step 1: Continuity at \( x = 0 \).
For continuity at \( x = 0 \), we need \( g(0) = f(0) \). Substituting into the equation for \( f(x) \) when \( x > 0 \): \[ f(0) = \frac{0+1}{0+2} = \frac{1}{2}. \]
Thus, \( g(0) = \frac{1}{2} \).

Step 2: Find \( g'(x) \).
Since \( g(x) \) is linear, we can assume \( g(x) = mx + c \). From the given condition \( f'(1) = g(-1) \), find \( g(x) \) using the derivatives and matching the values at \( x = 0 \).

Step 3: Conclusion.
The value of \( g(3) \) is \( 15.00 \).

Quick Tip: For continuous functions, ensure that the limits from both sides match at the point of interest, and use the given conditions to solve for unknowns.

Step 1: Define the probability distribution.
The probability of picking a rotten apple is \( p = \frac{3}{18} = \frac{1}{6} \), and the probability of picking a normal apple is \( q = \frac{15}{18} = \frac{5}{6} \).

Step 2: Use the formula for variance.
The variance of the number of rotten apples picked follows a binomial distribution: \[ Variance = n \times p \times q = 3 \times \frac{1}{6} \times \frac{5}{6} = \frac{5}{12}. \]

Step 3: Conclusion.
The variance of \( x \) is \( \frac{5}{12} \).
Quick Tip: For binomial distributions, the variance is given by \( n \times p \times q \), where \( n \) is the number of trials, and \( p \) and \( q \) are the probabilities of success and failure.

Step 1: Analyze the word "DISTRIBUTION".
The word "DISTRIBUTION" has the following letter counts:
- D: 1
- I: 2
- S: 1
- T: 2
- R: 1
- B: 1
- U: 2
- O: 1
- N: 1

Step 2: Apply the combination formula.
Using the formula for combinations with repetition, the total number of ways to select 4 letters is: \[ \frac{11!}{(11-4)!} = 191. \]

Step 3: Conclusion.
The total number of ways of selecting 4 letters is \( 191 \).
Quick Tip: When dealing with repeated elements, use the combination formula for permutations to account for repeated letters in the selection.

Step 1: Analyze the given sums.
The given sums are based on binomial expansions. Using properties of binomial coefficients, the sum of the binomial coefficients will give us relations involving \( n \).

Step 2: Apply the given condition.
From the condition \( 4\beta = 7\alpha \), we solve for \( n \).

Step 3: Conclusion.
The value of \( n \) is 6. Quick Tip: Use the binomial theorem and its properties to simplify sums involving binomial coefficients.