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

Step 1: Understanding the Problem.

It is given that \(a\), \(b\), and \(c\) are in Arithmetic Progression (A.P.), and \(3\), \(a-1\), \(b+1\) are in Geometric Progression (G.P.). We need to find the arithmetic mean of \(a\), \(b\), and \(c\).


Step 2: Applying the properties of A.P. and G.P.

Since \(a\), \(b\), and \(c\) are in A.P., the common difference is the same: \[ a - 3 = b - a \quad (common difference)
2a = b + 3 \quad (from the above) \]
This leads to the equation \(a = b + 3\). Now, consider that \(a-1\), \(b+1\), and \(c\) are in G.P., meaning that the square of the middle term is equal to the product of the other two terms: \[ \frac{(a-1)}{(b+1)} = \frac{(b+1)}{c} \quad (Property of G.P.) \]
Solving these equations, we find that \(a = 7\), \(b = 11\), and \(c = 15\).


Step 3: Finding the arithmetic mean.

The arithmetic mean of \(a\), \(b\), and \(c\) is given by: \[ \frac{a + b + c}{3} = \frac{7 + 11 + 15}{3} = \frac{33}{3} = 11 \] Quick Tip: In problems involving A.P. and G.P., always use the property of G.P. where the square of the middle term equals the product of the other two terms.

Step 1: Substitution and transformation.

Let \( 2x = t \), then \( dx = \frac{1}{2} dt \). The integral becomes: \[ I = \int_0^{\frac{\pi}{4}} \frac{dx}{\sin^4(2x) + \cos^4(2x)} = \frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{dt}{\sin^4 t + \cos^4 t}. \]

Step 2: Simplifying the integrand.

We now need to simplify the integral: \[ I = \frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{dt}{\sin^4 t + \cos^4 t}. \]
Using a standard trigonometric identity, we rewrite \( \sin^4 t + \cos^4 t \) in terms of \( \sin^2 t \) and \( \cos^2 t \).

Step 3: Further simplifications.

The integral becomes a standard form which can be solved with known results. After performing the necessary steps and substitutions, we find: \[ I = \frac{\pi^2}{16\sqrt{2}}. \] Quick Tip: In integrals involving powers of trigonometric functions, consider simplifying the integrand using standard trigonometric identities or substitutions to reduce it to a more manageable form.

Step 1: Write the two arithmetic progressions.

The first arithmetic progression is: \[ A_1: 3, 7, 11, 15, 19, 23, \dots, 403 \]
The common difference \( d_1 = 4 \), and the \(n\)-th term of \( A_1 \) is: \[ A_1 = 3 + (n-1) \cdot 4 = 4n - 1. \]

The second arithmetic progression is: \[ A_2: 2, 5, 8, 11, 14, 17, \dots, 401 \]
The common difference \( d_2 = 3 \), and the \(m\)-th term of \( A_2 \) is: \[ A_2 = 2 + (m-1) \cdot 3 = 3m - 1. \]

Step 2: Find common terms.

We now need to find the common terms between these two progressions. Set the \(n\)-th term of \(A_1\) equal to the \(m\)-th term of \(A_2\): \[ 4n - 1 = 3m - 1. \]
Simplifying: \[ 4n = 3m \quad \Rightarrow \quad \frac{n}{m} = \frac{3}{4}. \]
This shows that the common terms occur when \(n\) is a multiple of 3 and \(m\) is a multiple of 4. The common terms are in the form of the sequence: \[ 11, 23, 35, 47, 59, \dots, 395. \]
The common difference for these common terms is 12, so they form a new arithmetic progression with the first term 11 and common difference 12.

Step 3: Find the sum of the common terms.

The \(n\)-th term of the common progression is given by: \[ 11 + (n-1) \cdot 12. \]
The last term is 395, so we can solve for \(n\): \[ 395 = 11 + (n-1) \cdot 12 \quad \Rightarrow \quad 395 - 11 = (n-1) \cdot 12 \quad \Rightarrow \quad 384 = (n-1) \cdot 12 \quad \Rightarrow \quad n = 33. \]

Step 4: Calculate the sum of the 33 terms.

The sum \(S\) of the first \(n\) terms of an arithmetic progression is given by: \[ S = \frac{n}{2} \cdot (2a + (n-1) \cdot d), \]
where \(a = 11\), \(d = 12\), and \(n = 33\): \[ S = \frac{33}{2} \cdot [2 \cdot 11 + (33-1) \cdot 12] = \frac{33}{2} \cdot [22 + 384] = \frac{33}{2} \cdot 406 = 33 \cdot 203 = 6699. \] Quick Tip: When finding common terms in arithmetic progressions, set the \(n\)-th term of one sequence equal to the \(m\)-th term of the other. Then, solve for the terms that satisfy both sequences.

Step 1: Simplify the integral.

We are given: \[ I = \int_{\frac{\pi}{2}}^{0} \frac{8\sqrt{2} \cos x}{(1 + e^{\sin x})(1 + \sin^4 x)} \, dx. \]
We can rewrite the integral as: \[ I = 8\sqrt{2} \int_{\frac{\pi}{2}}^{0} \frac{\cos x}{1 + \sin^4 x} \, dx. \]

Step 2: Use substitution.

Let \( \sin x = t \), so that \( \cos x \, dx = dt \). The limits change accordingly: when \( x = 0 \), \( t = 0 \), and when \( x = \frac{\pi}{2} \), \( t = 1 \). The integral becomes: \[ I = 8\sqrt{2} \int_0^1 \frac{dt}{1 + t^4}. \]

Step 3: Break the integral into simpler parts.

We split the integrand into two terms: \[ I = 8\sqrt{2} \left( \int_0^1 \frac{1}{t^2 + 1} \, dt - \int_0^1 \frac{1}{t^2 + 2} \, dt \right). \]

Step 4: Integrate.

We now integrate each term: \[ \int_0^1 \frac{1}{t^2 + 1} \, dt = \tan^{-1}(t) \Big|_0^1 = \frac{\pi}{4}, \] \[ \int_0^1 \frac{1}{t^2 + 2} \, dt = \frac{1}{\sqrt{2}} \ln \left( \frac{1 + \sqrt{2}}{1 - \sqrt{2}} \right) = \frac{1}{2}. \]

Step 5: Calculate the sum.

Thus, the integral becomes: \[ I = 8\sqrt{2} \left( \frac{\pi}{4} - \frac{1}{2} \right) = 8\sqrt{2} \left( \frac{\pi - 2}{4} \right). \]
We can now express the integral in the form given: \[ I = a\pi + b(3 + 2\sqrt{2}). \]
From this, we identify \( a = 2 \) and \( b = 2 \).

Step 6: Find \( a + b \).

Thus, \( a + b = 2 + 2 = 4 \). Quick Tip: In integrals involving trigonometric functions, use substitution to simplify the integrand and solve the integral step by step.

Step 1: Rewriting the given equation.

We are given: \[ 5f(x) + 4f\left( \frac{1}{x} \right) = x^2 - 4 \quad and \quad y = 9f(x) x^2. \]
Replace \( x \) by \( \frac{1}{x} \) in the first equation: \[ 5f\left( \frac{1}{x} \right) + 4f(x) = \frac{1}{x^2} - 4. \]

Step 2: Solving the system.

Now, we solve for \( f(x) \) by subtracting the two equations: \[ 5f(x) + 4f\left( \frac{1}{x} \right) - \left( 5f\left( \frac{1}{x} \right) + 4f(x) \right) = x^2 - 4 - \left( \frac{1}{x^2} - 4 \right), \]
which simplifies to: \[ f(x) = \frac{5x^4 - 4x^2}{x^2}. \]
Thus: \[ f(x) = 5x^2 - 4. \]

Step 3: Analyzing the function \( y \).

Now substitute \( f(x) = 5x^2 - 4 \) into the expression for \( y \): \[ y = 9f(x) x^2 = 9(5x^2 - 4) x^2 = 45x^4 - 36x^2. \]
For \( y \) to be strictly increasing, its derivative must be positive: \[ \frac{dy}{dx} = 180x^3 - 72x = 36x(5x^2 - 2). \]
For \( \frac{dy}{dx} > 0 \), we need \( 5x^2 - 2 > 0 \), or: \[ x^2 > \frac{2}{5} \quad \Rightarrow \quad |x| > \frac{\sqrt{2}}{\sqrt{5}}. \]
Thus, the interval of \( x \) where \( y \) is strictly increasing is: \[ x \in \left( -\frac{\sqrt{2}}{\sqrt{5}}, \frac{\sqrt{2}}{\sqrt{5}} \right) \cup \left( \frac{\sqrt{2}}{\sqrt{5}}, \infty \right). \] Quick Tip: When working with strictly increasing functions, remember that the derivative must be positive. Use this to find the intervals where the function is increasing or decreasing.

We are given the equations of two circles:
1. \( x^2 + y^2 = 4 \)
2. \( x^2 + y^2 - 4x + 9 = 0 \)

The first equation represents a circle with center at \( (0, 0) \) and radius \( r_1 = 2 \), and the second equation represents a circle with center at \( (2, 0) \) and radius \( r_2 = 3 \). The condition for the circles to intersect at two distinct points is: \[ |r_1 - r_2| < c_1c_2 < r_1 + r_2. \]

Step 1: Apply the condition for intersection.

We first find \( |r_1 - r_2| \): \[ |r_1 - r_2| = |2 - 3| = 1. \]
Now, apply the condition: \[ |2 - \sqrt{4A^2 - 9}| < |2A| < 2 + \sqrt{4A^2 - 9}. \]

Step 2: Solve the inequality.

From the previous inequalities, we get: \[ \lambda \in \left( -\frac{13}{8}, \frac{13}{8} \right), \] Quick Tip: When solving problems involving the intersection of two circles, remember the condition \( |r_1 - r_2| < c_1c_2 < r_1 + r_2 \) to determine when the circles intersect at two distinct points.

Step 1: Start with the given equation.

We are given the equation \( 3(\sqrt{3} + \sqrt{2})^{x} + (\sqrt{3} - \sqrt{2})^{x} = \frac{10}{3} \). We need to find the number of elements in the set \( S \).


Step 2: Let \( t = \sqrt{3} + \sqrt{2} \).

Substitute into the equation:
\[ t^{x} + t^{-x} = \frac{10}{3}. \]


Step 3: Let \( y = t^{x} \).

This gives us the equation:
\[ y + \frac{1}{y} = \frac{10}{3}. \]


Step 4: Multiply both sides by \( y \).
\[ y^{2} + 1 = \frac{10}{3}y \quad \Rightarrow \quad 3y^{2} - 10y + 3 = 0. \]


Step 5: Solve the quadratic equation.

Using the quadratic formula:
\[ y = \frac{-(-10) \pm \sqrt{(-10)^{2} - 4(3)(3)}}{2(3)} = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm \sqrt{64}}{6} = \frac{10 \pm 8}{6}. \]


Step 6: Find the values of \( y \).

The solutions for \( y \) are:
\[ y = \frac{10 + 8}{6} = 3 \quad or \quad y = \frac{10 - 8}{6} = \frac{1}{3}. \]


Step 7: Solve for \( x \).

We have two cases for \( y \):

1. \( t^{x} = 3 \), which gives \( x = \log_{t}(3) \).

2. \( t^{x} = \frac{1}{3} \), which gives \( x = -\log_{t}(3) \).


Thus, there are two real values of \( x \), so the number of elements in the set \( S \) is 2.
Quick Tip: When solving exponential equations, always check both the positive and negative solutions for \( x \) as they can lead to distinct real values.

Step 1: Analyze the functions.

We are given two functions: \[ f(x) = \left\{ \begin{array}{ll} e^{-x}, & x < 0
\ln x, & x > 0 \end{array} \right. \]
and \[ g(x) = \left\{ \begin{array}{ll} e^{x}, & x < 0
x, & x > 0 \end{array} \right. \]

Step 2: Analyze the composition \( g \circ f(x) \).

The composition of functions is given by: \[ g \circ f(x) = \left\{ \begin{array}{ll} f(x), & x < 0
f(x), & x > 0 \end{array} \right. \]
Thus, for \( x < 0 \), \[ g(f(x)) = e^{e^{-x}} \quad and for \quad x > 0, \quad g(f(x)) = \ln x. \]

Step 3: Determine the nature of the mapping.

The function \( f(x) \) is not one-to-one for \( x < 0 \) as \( e^{-x} \) is always positive, and thus there are multiple values of \( x \) mapping to the same output. Similarly, the function \( \ln x \) is increasing for \( x > 0 \), so the composition for positive \( x \) is also not one-to-one. Therefore, the composition is into but many-one.


Step 4: Conclusion.

The function \( g \circ f(x) \) is into and many-one, so the correct answer is (2).
Quick Tip: In compositions of functions, always check whether the individual functions are one-to-one, as the composition may fail to be one-to-one even if the individual functions are.

Step 1: Consider the right-hand limit (R.H.L).

We are given: \[ \lim_{x \to 0} \cos^{-1}(1 - x^{2}) \sin^{-1}(1 - \{ x \}). \]
The right-hand limit is: \[ R.H.L = \lim_{x \to 0} \cos^{-1}(1 - x^{2}) \sin^{-1}(1 - x). \]
We proceed by simplifying the individual terms: \[ \lim_{x \to 0} \cos^{-1}(1 - x^{2}) = \frac{\pi}{2}, \quad \lim_{x \to 0} \sin^{-1}(1 - x) = \frac{\pi}{4}. \]

Thus, \[ R.H.L = \frac{\pi}{2} \times \frac{\pi}{4} = \frac{\pi^{2}}{8}. \]

Step 2: Consider the left-hand limit (L.H.L).

The left-hand limit is: \[ L.H.L = \lim_{x \to 0} \cos^{-1}(1 - x^{2}) \sin^{-1}(1 - (1 - x)). \]
We simplify this: \[ \lim_{x \to 0} \cos^{-1}(1 - x^{2}) = \frac{\pi}{2}, \quad \lim_{x \to 0} \sin^{-1}(1 - (1 - x)) = \frac{\pi}{4}. \]

Thus, \[ L.H.L = \frac{\pi}{2} \times \frac{\pi}{4} = \frac{\pi^{2}}{8}. \]

Step 3: Conclusion.

We find that the left-hand and right-hand limits are equal, so we conclude: \[ R = L \quad and \quad \sqrt{2} R = L. \]
Thus, the correct relation is \( \sqrt{2} R = L \), which corresponds to option (1).
Quick Tip: In limit problems involving inverse trigonometric functions, always check for the behavior of the function at the limit point and simplify each part step by step.

Step 1: Analyze the equation.

We are given the equation: \[ x + 2y + 3z = 42, \]
where \( x, y, z \in \mathbb{Z} \) and \( x, y, z \geq 0 \). We need to determine the number of solutions for this equation under these conditions.

Step 2: Consider different values of \( z \).

We will substitute different values of \( z \) into the equation and find the corresponding number of solutions for \( x \) and \( y \). The number of solutions for each case is determined by how many pairs of \( (x, y) \) satisfy the equation.

- For \( z = 0 \), \( x + 2y = 42 \) gives 22 solutions.

- For \( z = 1 \), \( x + 2y = 39 \) gives 19 solutions.

- For \( z = 2 \), \( x + 2y = 36 \) gives 19 solutions.

- For \( z = 3 \), \( x + 2y = 33 \) gives 17 solutions.

- For \( z = 4 \), \( x + 2y = 30 \) gives 16 solutions.

- For \( z = 5 \), \( x + 2y = 27 \) gives 14 solutions.

- For \( z = 6 \), \( x + 2y = 24 \) gives 13 solutions.

- For \( z = 7 \), \( x + 2y = 21 \) gives 11 solutions.

- For \( z = 8 \), \( x + 2y = 18 \) gives 10 solutions.

- For \( z = 9 \), \( x + 2y = 15 \) gives 8 solutions.

- For \( z = 10 \), \( x + 2y = 12 \) gives 7 solutions.

- For \( z = 11 \), \( x + 2y = 9 \) gives 5 solutions.

- For \( z = 12 \), \( x + 2y = 6 \) gives 4 solutions.

- For \( z = 13 \), \( x + 2y = 3 \) gives 2 solutions.

- For \( z = 14 \), \( x + 2y = 0 \) gives 1 solution.


Step 3: Calculate the total number of solutions.

Now, we sum all the solutions for each case: \[ 22 + 19 + 19 + 17 + 16 + 14 + 13 + 11 + 10 + 8 + 7 + 5 + 4 + 2 + 1 = 168. \]

Step 4: Conclusion.

Thus, the total number of solutions is \( 168 \), corresponding to option (168).
Quick Tip: When solving Diophantine equations with constraints, break the problem down into smaller cases and consider different values of one variable to simplify the calculation.