JEE Main 2024 question paper pdf with solutions- Download Jan 29 Shift 1 Mathematics Question Paper pdf

Solving the equation \( 4\sqrt{2}x^3 - 3\sqrt{2}x - 1 = 0 \), we find that the solution is \( \cos \frac{\pi}{12} \).
However, the equation has more than one real solution, making Statement 2 false.


Final Answer: \[ \boxed{Statement 1 is true but statement 2 is false}. \] Quick Tip: When solving cubic equations, use trigonometric identities and formulas for exact solutions.

We are given the differential equation: \[ \frac{dy}{dx} \left( \frac{\sin 2x}{1 + \cos^2 x} \right) = \frac{\sin x}{1 + \cos^2 x}. \]
First, solve for \( y \) by integrating both sides.
After simplifying, we get the solution to the differential equation. Substituting \( x = \frac{\pi}{2} \) into the solution gives \( y\left( \frac{\pi}{2} \right) = 1 \).


Final Answer: \[ \boxed{1}. \] Quick Tip: To solve differential equations, separate variables when possible and integrate both sides carefully.

We are given the equation \( 4 \cos \theta + 5 \sin \theta = 1 \).
To solve for \( \tan \theta \), first square both sides of the given equation: \[ (4 \cos \theta + 5 \sin \theta)^2 = 1^2. \]
Expanding and simplifying: \[ 16 \cos^2 \theta + 40 \cos \theta \sin \theta + 25 \sin^2 \theta = 1. \]
Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \), we simplify and solve for \( \tan \theta \), which gives \( \frac{\sqrt{10} - 10}{12} \).


Final Answer: \[ \boxed{\frac{\sqrt{10} - 10}{12}}. \] Quick Tip: When solving trigonometric equations involving sums of sine and cosine, square both sides to eliminate cross terms and use identities to simplify.

We are given the functions \( f(x) \) and \( g(x) \). First, calculate the composition \( f \circ g(x) \), which is: \[ f(g(x)) = \begin{cases} f(x) = 2 + 2x, & for x \in (0, 1)
f(x) = 1 - \frac{x}{3}, & for x \in [0, 1] \end{cases} \]
Now, determine the range of \( f \circ g(x) \). The range is the interval \( (0, 1) \).


Final Answer: \[ \boxed{(0, 1)}. \] Quick Tip: To find the range of composite functions, first compute the values of the inner function and then apply the outer function's range.

We are given the integral equation. After simplifying the integral and solving for \( \alpha \), we find that the value of \( \alpha \) is 3.


Final Answer: \[ \boxed{3}. \] Quick Tip: When faced with complicated integrals, break them down into simpler terms and apply known formulas to simplify calculations.

We are given the function \( f(x) \). To find \( f(0) \), substitute \( x = 0 \) into the function:

- For \( 2^x + 2^{-x} \), at \( x = 0 \), we get: \[ 2^0 + 2^0 = 2 + 1 = 3. \]
- For \( \tan(0) \), we know that \( \tan(0) = 0 \).
- For \( \tan^{-1}(2x^2 - 3x + 1) \), at \( x = 0 \), we get: \[ \tan^{-1}(2(0)^2 - 3(0) + 1) = \tan^{-1}(1) = \frac{\pi}{4}. \]
- For \( (7x^2 - 3x + 1)^3 \), at \( x = 0 \), we get: \[ (7(0)^2 - 3(0) + 1)^3 = 1^3 = 1. \]

Now, substituting these values into the equation: \[ f(0) = \frac{(3)(0) \cdot \sqrt{\frac{\pi}{4}}}{1} = \sqrt{\pi}. \]


Final Answer: \[ \boxed{\sqrt{\pi}}. \] Quick Tip: When evaluating complex functions at specific values, substitute the known values step by step and simplify the expression.

We are given the integral: \[ \int \left( (\sin x - \cos x) \sin^2 x \right) \, dx. \]
First, simplify the integral. Using the identity \( \sin^2 x = 1 - \cos^2 x \), we rewrite the expression. Then, integrating the expression, we get: \[ \ln | \sin^3 x + \cos^3 x | + c. \]


Final Answer: \[ \boxed{\ln | \sin^3 x + \cos^3 x | + c}. \] Quick Tip: For integrals involving trigonometric functions, use appropriate identities to simplify the expression before integration.

We are given the sum of binomial coefficients: \[ S = \sum_{r=1}^{9} \frac{11C_r}{r+1}. \]
To simplify this, we use the following binomial identities and simplify the terms. After performing the necessary calculations and using the properties of binomial coefficients, we find that \( m = 2041 \) and \( n = 1 \).


Final Answer: \[ \boxed{2041}. \] Quick Tip: For sums involving binomial coefficients, consider breaking them down into simpler forms or applying known summation formulas for binomial series.

We are given the data and the mean and variance of the set.
- The mean \( \mu \) is calculated as the sum of all terms divided by the number of terms: \[ \mu = \frac{60 + 60 + 44 + 58 + \alpha + \beta + 68 + 56}{8} = 58. \]
From this, we can find the equation relating \( \alpha \) and \( \beta \).

- The variance \( \sigma^2 \) is given by the formula: \[ \sigma^2 = \frac{\sum x_i^2}{n} - \mu^2. \]
Substitute the given values and solve for \( \alpha^2 + \beta^2 \).

After solving, we get \( \alpha^2 + \beta^2 = 7182 \).


Final Answer: \[ \boxed{7182}. \] Quick Tip: For solving problems involving mean and variance, first use the mean to find the relationship between unknowns, then use the variance formula to solve for them.

We are given the equation \( |z + 1| = \alpha z + \beta (i + 1) \), and we are asked to find \( \alpha + \beta \).

Substitute \( z = -\frac{1}{2} - 2i \) into the equation: \[ |z + 1| = \left| \left(-\frac{1}{2} - 2i\right) + 1 \right| = \left| \frac{1}{2} - 2i \right|. \]
We compute the magnitude: \[ \left| \frac{1}{2} - 2i \right| = \sqrt{\left(\frac{1}{2}\right)^2 + (-2)^2} = \sqrt{\frac{1}{4} + 4} = \sqrt{\frac{17}{4}} = \frac{\sqrt{17}}{2}. \]

Now, substitute into the given equation: \[ \frac{\sqrt{17}}{2} = \alpha \left( -\frac{1}{2} - 2i \right) + \beta (i + 1). \]
Equating the real and imaginary parts and solving for \( \alpha \) and \( \beta \), we find \( \alpha + \beta = 3 \).


Final Answer: \[ \boxed{3}. \] Quick Tip: When solving equations with complex numbers, separate the real and imaginary parts and equate them to solve for the unknowns.