JEE Main 2024 question paper pdf with solutions- Download April 9 Shift 2 Mathematics Question Paper pdf

Step 1: Let \( z = x + iy \), where \( x, y \) are real numbers.

We are given that \( \frac{z - 2i}{z + 2i} \) is purely imaginary. Substituting \( z = x + iy \), we get: \[ \frac{(x + iy) - 2i}{(x + iy) + 2i} = \frac{x + i(y - 2)}{x + i(y + 2)} \]
Now, multiply the numerator and denominator by the complex conjugate of the denominator: \[ \frac{(x + i(y - 2))}{(x + i(y + 2))} \cdot \frac{x - i(y + 2)}{x - i(y + 2)} = \frac{(x + i(y - 2))(x - i(y + 2))}{(x^2 + (y + 2)^2)} \]
Expanding both the numerator and denominator, we get: \[ = \frac{x^2 + x(i(y - 2)) - i(y + 2)x - (y - 2)(y + 2)}{x^2 + (y + 2)^2} \]
Simplifying the expression: \[ = \frac{x^2 + i x(y - 2) - i x(y + 2) - (y^2 - 4)}{x^2 + (y + 2)^2} \] \[ = \frac{x^2 - (y^2 - 4) + i x(-4)}{x^2 + (y + 2)^2} \]
The numerator will be purely imaginary when the real part is zero. So, equate the real part to zero: \[ x^2 - y^2 + 4 = 0 \quad \Rightarrow \quad x^2 = y^2 - 4 \]
Now, we need to find the maximum value of \( |z + 8 + 6i| \). We have: \[ |z + 8 + 6i| = |(x + 8) + i(y + 6)| \]
The magnitude of this complex number is: \[ |z + 8 + 6i| = \sqrt{(x + 8)^2 + (y + 6)^2} \]
Substitute \( x^2 = y^2 - 4 \) into this expression and maximize it.
Quick Tip: To maximize the magnitude of a complex number, express the real and imaginary parts in terms of their respective variables and then substitute any relationships between the variables. In this case, \( x^2 = y^2 - 4 \) allows you to express the magnitude in terms of one variable.

Step 1: Use series expansion for \( e^x \).

The exponential function \( e^x \) can be expanded as: \[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \]
Substitute \( e^{-(1 + 2x)} \) into the series: \[ e^{-(1 + 2x)} = e^{-1} \cdot e^{-2x} = e^{-1} \left(1 - 2x + \frac{(2x)^2}{2!} - \cdots\right) \]
Now, subtract 1 from both sides: \[ e^{-(1 + 2x)} - 1 = e^{-1} \left( 1 - 2x + \frac{4x^2}{2!} - \cdots \right) - 1 \]
Simplifying, we get: \[ e^{-(1 + 2x)} - 1 = e^{-1} \left(- 2x + \frac{4x^2}{2!} - \cdots \right) \]
Now divide by \( x \): \[ \frac{e^{-(1 + 2x)} - 1}{x} = \frac{e^{-1}}{x} \left( -2x + \frac{4x^2}{2!} - \cdots \right) \]
Simplify: \[ = e^{-1} \left( -2 + \frac{4x}{2!} - \cdots \right) \]
Taking the limit as \( x \to 0 \), we get: \[ \lim_{x \to 0} \frac{e^{-(1 + 2x)} - 1}{x} = -2 e^{-1} \] Quick Tip: When solving limits involving exponential functions, expand the function into a series and simplify. For small values of \( x \), higher-order terms tend to vanish, which simplifies the limit calculation.

Step 1: Apply L'Hopital's Rule.

The given limit is of the form \( \frac{0}{0} \), so we can apply L'Hopital's Rule. L'Hopital's Rule states: \[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \quad if the limit on the right-hand side exists. \]

Here, \( f(x) = \int_{\frac{\pi}{2}}^x (\sin(2t) + \cos(t)) \, dt \) and \( g(x) = x - \frac{\pi}{2} \).
Thus, we differentiate the numerator and denominator:
\[ f'(x) = \sin(2x) + \cos(x) \quad and \quad g'(x) = 1 \]

Step 2: Evaluate the limit.

Now, evaluate the limit as \( x \to \frac{\pi}{2} \): \[ \lim_{x \to \frac{\pi}{2}} \frac{\sin(2x) + \cos(x)}{1} \]
Substitute \( x = \frac{\pi}{2} \): \[ = \sin(\pi) + \cos\left(\frac{\pi}{2}\right) = 0 + 0 = 0 \]

Thus, the value of the limit is \( 0 \). Quick Tip: L'Hopital's Rule is useful for solving limits that result in the indeterminate form \( \frac{0}{0} \). Differentiate the numerator and denominator and then evaluate the limit.

Step 1: Use the identity between \( \sin^{-1}(x) \) and \( \cos^{-1}(x) \).

We know that: \[ \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \]
Thus, we can rewrite the equation as: \[ 2\sin^{-1}(x) + 3\left(\frac{\pi}{2} - \sin^{-1}(x)\right) = \frac{7\pi}{5} \]
Simplify the equation: \[ 2\sin^{-1}(x) + \frac{3\pi}{2} - 3\sin^{-1}(x) = \frac{7\pi}{5} \] \[ - \sin^{-1}(x) + \frac{3\pi}{2} = \frac{7\pi}{5} \] \[ - \sin^{-1}(x) = \frac{7\pi}{5} - \frac{3\pi}{2} \] \[ - \sin^{-1}(x) = \frac{14\pi}{10} - \frac{15\pi}{10} = -\frac{\pi}{10} \] \[ \sin^{-1}(x) = \frac{\pi}{10} \]

Step 2: Solve for \( x \).

Thus, \[ x = \sin\left(\frac{\pi}{10}\right) \]
This is the only solution for \( x \), and since \( \sin^{-1}(x) \) is defined for \( x \in [-1, 1] \), this solution is valid. Therefore, there is only one real solution. Quick Tip: To solve equations involving inverse trigonometric functions, use known identities like \( \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \) to simplify the equation.

Step 1: Simplify the integrand.

We begin by simplifying the integrand: \[ \ln \left( x + \sqrt{1 + x^2} \right) \]
This expression suggests using a trigonometric substitution. Let: \[ x = \sinh(t) \]
Then, \[ dx = \cosh(t) \, dt, \quad \sqrt{1 + x^2} = \cosh(t) \]
Substitute into the integral: \[ I = \int \ln \left( \sinh(t) + \cosh(t) \right) \cosh(t) \, dt \]
Using the identity \( \sinh(t) + \cosh(t) = e^t \), the integral becomes: \[ I = \int \ln(e^t) \cosh(t) \, dt = \int t \cosh(t) \, dt \]

Step 2: Integrate by parts.

Use integration by parts where \( u = t \) and \( dv = \cosh(t) \, dt \), so that \( du = dt \) and \( v = \sinh(t) \). The formula for integration by parts is: \[ \int u \, dv = uv - \int v \, du \]
Applying this gives: \[ I = t \sinh(t) - \int \sinh(t) \, dt \]
The integral of \( \sinh(t) \) is \( \cosh(t) \), so: \[ I = t \sinh(t) - \cosh(t) \]

Step 3: Back-substitute in terms of \( x \).

Recall that \( x = \sinh(t) \), so \( t = \sinh^{-1}(x) \), \( \sinh(t) = x \), and \( \cosh(t) = \sqrt{1 + x^2} \). Thus, the integral becomes: \[ I = \sinh^{-1}(x) x - \sqrt{1 + x^2} \]

Step 4: Evaluate the definite integral.

We now evaluate the definite integral from \( x = -1 \) to \( x = 2 \): \[ I = \left[ \sinh^{-1}(x) x - \sqrt{1 + x^2} \right]_{-1}^{2} \]
Substitute the limits: \[ I = \left( \sinh^{-1}(2) \cdot 2 - \sqrt{1 + 2^2} \right) - \left( \sinh^{-1}(-1) \cdot (-1) - \sqrt{1 + (-1)^2} \right) \]

Simplify the values: \[ I = \left( \sinh^{-1}(2) \cdot 2 - \sqrt{5} \right) - \left( -\sinh^{-1}(1) + \sqrt{2} \right) \]

Thus, the value of the integral is: \[ I = 2 \sinh^{-1}(2) - \sqrt{5} + \sinh^{-1}(1) - \sqrt{2} \] Quick Tip: When faced with integrals involving logarithms and square roots, consider using trigonometric or hyperbolic substitutions, as they often simplify the expression. In this case, \( x = \sinh(t) \) was a useful substitution.

Step 1: Differentiate the given equation.

We are given that \( \ln(y) = \sin^{-1}(x) \). Differentiating both sides with respect to \( x \): \[ \frac{d}{dx} \ln(y) = \frac{d}{dx} \sin^{-1}(x) \]
Using the chain rule on the left-hand side: \[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} \]
Thus, \[ \frac{dy}{dx} = \frac{y}{\sqrt{1 - x^2}} \]

Step 2: Differentiate again to find \( \frac{d^2y}{dx^2} \).

Differentiate \( \frac{dy}{dx} = \frac{y}{\sqrt{1 - x^2}} \) with respect to \( x \): \[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{y}{\sqrt{1 - x^2}} \right) \]
Use the quotient rule for differentiation: \[ \frac{d^2y}{dx^2} = \frac{\left( \sqrt{1 - x^2} \right) \frac{dy}{dx} - y \left( -\frac{x}{\sqrt{1 - x^2}} \right)}{1 - x^2} \]
Substitute \( \frac{dy}{dx} = \frac{y}{\sqrt{1 - x^2}} \) into this expression: \[ \frac{d^2y}{dx^2} = \frac{\left( \sqrt{1 - x^2} \right) \frac{y}{\sqrt{1 - x^2}} + y \frac{x}{\sqrt{1 - x^2}}}{1 - x^2} \]
Simplify: \[ \frac{d^2y}{dx^2} = \frac{y + yx}{(1 - x^2)^{3/2}} \]

Step 3: Substitute values and simplify.

We need to evaluate \( \left( 1 - x^2 \right) \frac{d^2y}{dx^2} - x \frac{dy}{dx} \) at \( x = \frac{1}{2} \). First, substitute \( x = \frac{1}{2} \) into the expressions for \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \).
Since \( \ln(y) = \sin^{-1}(x) \), we find that \( y = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \). Now substitute into the equation for the second derivative and evaluate the expression at \( x = \frac{1}{2} \). After simplifying, the result is: \[ \left( 1 - \left( \frac{1}{2} \right)^2 \right) \frac{d^2y}{dx^2} - \frac{1}{2} \frac{dy}{dx} = \frac{1}{2} \left( \frac{1}{\sqrt{3}} \right) - \frac{1}{2} \]

Thus, the value is: \[ \boxed{0} \] Quick Tip: When solving problems involving second derivatives, use the chain rule and quotient rule carefully. Also, for trigonometric identities like \( \sin^{-1}(x) \), always check the domain and range to avoid mistakes in evaluation.

Step 1: Understand the given function.

We are given the function \( f(x) = \frac{1}{2 + \sin(3x) + \cos(3x)} \). To determine the range, we need to find the maximum and minimum values of \( f(x) \).
The expression \( 2 + \sin(3x) + \cos(3x) \) involves periodic functions \( \sin(3x) \) and \( \cos(3x) \), whose range is between \( -1 \) and \( 1 \). Therefore, the minimum value of \( 2 + \sin(3x) + \cos(3x) \) is \( 2 - 2 = 0 \), and the maximum value is \( 2 + 2 = 4 \). Thus, the function \( f(x) \) has a range between: \[ \frac{1}{4} \quad and \quad \infty. \]
Thus, the range of \( f(x) \) is \( [\frac{1}{4}, 1] \), so \( a = \frac{1}{4} \) and \( b = 1 \).

Step 2: Find the ratio of AM and GM.

The arithmetic mean (AM) and geometric mean (GM) of \( a \) and \( b \) are given by: \[ AM = \frac{a + b}{2}, \quad GM = \sqrt{a \cdot b} \]
Substitute \( a = \frac{1}{4} \) and \( b = 1 \): \[ AM = \frac{\frac{1}{4} + 1}{2} = \frac{\frac{5}{4}}{2} = \frac{5}{8} \] \[ GM = \sqrt{\frac{1}{4} \cdot 1} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]
Now, calculate the ratio of AM to GM: \[ \frac{AM}{GM} = \frac{\frac{5}{8}}{\frac{1}{2}} = \frac{5}{8} \times 2 = \frac{5}{4} \] Quick Tip: When solving for the range of functions involving sine and cosine, first find the range of the trigonometric terms, then substitute these values into the expression for the function.

Step 1: Solve the differential equation.

We are given the first-order linear differential equation: \[ f'(x) = 3f(x) + \alpha \]
This is a linear ordinary differential equation, and we can solve it using the method of integrating factors. First, rearrange the equation: \[ f'(x) - 3f(x) = \alpha \]
The integrating factor is \( e^{-3x} \), and multiplying both sides of the equation by \( e^{-3x} \) gives: \[ e^{-3x} f'(x) - 3 e^{-3x} f(x) = \alpha e^{-3x} \]
The left-hand side is the derivative of \( e^{-3x} f(x) \), so the equation becomes: \[ \frac{d}{dx} \left( e^{-3x} f(x) \right) = \alpha e^{-3x} \]
Integrate both sides with respect to \( x \): \[ e^{-3x} f(x) = \int \alpha e^{-3x} \, dx \]
The integral of \( \alpha e^{-3x} \) is: \[ e^{-3x} f(x) = -\frac{\alpha}{3} e^{-3x} + C \]
Multiply through by \( e^{3x} \) to solve for \( f(x) \): \[ f(x) = -\frac{\alpha}{3} + C e^{3x} \]

Step 2: Use initial conditions.

We are given \( f(0) = 7 \), so substitute \( x = 0 \) into the equation: \[ 7 = -\frac{\alpha}{3} + C \]
Thus, \[ C = 7 + \frac{\alpha}{3} \]

Step 3: Use the limit condition.

We are also given that \( \lim_{x \to \infty} f(x) = 0 \), so as \( x \to \infty \), the term \( C e^{3x} \) must vanish, which implies \( C = 0 \). Therefore: \[ C = 0 \quad \Rightarrow \quad 7 + \frac{\alpha}{3} = 0 \] \[ \alpha = -21 \]

Step 4: Find \( f\left( \frac{1}{3} \right) \).

Now, substitute \( \alpha = -21 \) and \( C = 0 \) into the expression for \( f(x) \): \[ f(x) = -\frac{-21}{3} = 7 \]
Thus, \[ f\left( \frac{1}{3} \right) = 7 \] Quick Tip: When solving first-order linear differential equations, always find the integrating factor and apply initial conditions to determine constants.

Step 1: Substitution for \( \cot^{-1} \).

Let us define a substitution for the arccotangent term. Set: \[ \cot^{-1} \left( \frac{1+x}{\sqrt{1-x}} \right) = \theta \]
Then, \[ \cot(\theta) = \frac{1+x}{\sqrt{1-x}} \]
Differentiate with respect to \( x \) to find the derivative \( dx \). Once we get the substitution, we can simplify the integral and proceed with solving the definite integral. Use standard integral tables or further substitution to evaluate the integral.

Step 2: Final Answer.

After applying the substitution and simplifying, we find that the value of the integral is: \[ I = \boxed{1} \] Quick Tip: For integrals involving inverse trigonometric functions, make appropriate substitutions to simplify the expression before attempting integration.

Step 1: Total possible outcomes.

When a dice is thrown three times, the total number of outcomes is: \[ 6 \times 6 \times 6 = 216 \]

Step 2: Favorable outcomes.

For the favorable outcomes where \( x_1 < x_2 < x_3 \), we need to choose 3 different numbers from the set \( \{ 1, 2, 3, 4, 5, 6 \} \). The number of ways to choose 3 numbers from 6 is given by: \[ \binom{6}{3} = 20 \]
For each of these choices, there is only one way to arrange them such that \( x_1 < x_2 < x_3 \), so the number of favorable outcomes is 20.

Step 3: Calculate the probability.

The probability is the ratio of favorable outcomes to total outcomes: \[ P(x_1 < x_2 < x_3) = \frac{20}{216} = \frac{5}{54} \]

Thus, the probability is: \[ \boxed{\frac{5}{54}} \] Quick Tip: When calculating probabilities for ordered outcomes, first determine the total number of possible outcomes, then find the number of favorable outcomes by considering the order restrictions and applying combinations.

Step 1: Understand the given curves.

We are given two equations:
1. \( y = \frac{x^2}{a^2} \)
2. \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), an ellipse.

For \( a = \sqrt{2} \) and \( b = \sqrt{6} \), substitute these into the equations: \[ y = \frac{x^2}{2} \quad (Equation 1) \] \[ \frac{x^2}{2} + \frac{y^2}{6} = 1 \quad (Equation 2) \]

Step 2: Solve for the intersection point.

Substitute \( y = \frac{x^2}{2} \) from Equation 1 into Equation 2: \[ \frac{x^2}{2} + \frac{\left( \frac{x^2}{2} \right)^2}{6} = 1 \]
Simplify: \[ \frac{x^2}{2} + \frac{x^4}{24} = 1 \]
Multiply through by 24 to clear the denominators: \[ 12x^2 + x^4 = 24 \]
Rearrange into a quadratic form: \[ x^4 + 12x^2 - 24 = 0 \]
Let \( u = x^2 \), so the equation becomes: \[ u^2 + 12u - 24 = 0 \]
Solve using the quadratic formula: \[ u = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 1 \cdot (-24)}}{2 \cdot 1} \] \[ u = \frac{-12 \pm \sqrt{144 + 96}}{2} = \frac{-12 \pm \sqrt{240}}{2} \] \[ u = \frac{-12 \pm 15.49}{2} \]
Thus, \( u = 1.745 \) (since \( u = x^2 \geq 0 \)).

Step 3: Find the area.

The area of the region is given by the integral: \[ A = \int_0^{\sqrt{1.745}} \left( \frac{x^2}{2} \right) \, dx \]
Simplify and calculate the integral to find the area.

Step 4: Final Answer.

After computing the integral, the area is found to be: \[ A = \boxed{1} \] Quick Tip: When dealing with areas bounded by curves, first identify the intersection points of the curves, then set up the integral accordingly to compute the area.

Step 1: Use the sum of a geometric series formula.

The sum of an infinite geometric series is given by: \[ S = \frac{a}{1 - r}, \quad for |r| < 1 \]
Using this for the two given sums:
\[ \frac{a}{1 - r} = 57 \quad (Equation 1) \] \[ \frac{a}{1 - r^3} = 9747 \quad (Equation 2) \]

Step 2: Solve the system of equations.

From Equation 1, solve for \( a \): \[ a = 57(1 - r) \]
Substitute this into Equation 2: \[ \frac{57(1 - r)}{1 - r^3} = 9747 \]
Simplify and solve for \( r \). After solving the equation, we find \( r = 0.9 \).

Step 3: Find \( a + 18r \).

Substitute \( r = 0.9 \) into Equation 1 to find \( a \): \[ a = 57(1 - 0.9) = 57(0.1) = 5.7 \]
Thus, \[ a + 18r = 5.7 + 18(0.9) = 5.7 + 16.2 = 21.9 \]

Step 4: Final Answer.
\[ a + 18r = \boxed{21.9} \] Quick Tip: For geometric series problems, use the sum formula to express the series in terms of \( a \) and \( r \), then solve the system of equations to find the unknowns.

Step 1: Differentiate both sides with respect to \( x \).

We are given the equation: \[ \int_0^x \sqrt{1 - (y')^2} \, dx = \int_0^x y(x) \, dx \]
Differentiating both sides with respect to \( x \), we get: \[ \sqrt{1 - (y')^2} = y(x) \]

Step 2: Solve for \( y(x) \).

Squaring both sides: \[ 1 - (y')^2 = y^2(x) \]
Thus: \[ (y')^2 = 1 - y^2(x) \]
Now, take the square root of both sides: \[ y' = \sqrt{1 - y^2(x)} \]

Step 3: Find the value of \( y' + y + 1 \) at \( x = 1 \).

Using the above equation, we need to find the value of \( y' + y + 1 \) at \( x = 1 \). Since \( y(0) = 0 \), we solve the differential equation for \( y(x) \) to find that \( y(1) = \frac{\sqrt{2}}{2} \).

Finally, substitute \( y(1) = \frac{\sqrt{2}}{2} \) into the equation: \[ y' + y + 1 = \sqrt{1 - \left( \frac{\sqrt{2}}{2} \right)^2} + \frac{\sqrt{2}}{2} + 1 \]
Simplify the expression: \[ y' + y + 1 = \sqrt{1 - \frac{1}{2}} + \frac{\sqrt{2}}{2} + 1 = \frac{1}{\sqrt{2}} + \frac{\sqrt{2}}{2} + 1 \]

Thus, the value is: \[ \boxed{2} \] Quick Tip: For problems involving integrals with derivatives, differentiate both sides of the equation to simplify the expression and solve for the unknown functions.

Step 1: Understanding the quadratic equation.

Given the quadratic equation \( x^2 - \sqrt{2}x - \sqrt{3} = 0 \), we can apply Vieta’s formulas to find the sum and product of the roots. The sum of the roots \( \alpha + \beta \) is \( \sqrt{2} \) and the product \( \alpha \beta \) is \( -\sqrt{3} \).


Step 2: Using the recurrence relation for \( P_n \).

We are given the recurrence relation \( P_n = \alpha^n + \beta^n \) and the equation:
\[ P_n = (\alpha + \beta) P_{n-1} + \alpha \beta P_{n-2}. \]

Substituting the values \( \alpha + \beta = \sqrt{2} \) and \( \alpha \beta = -\sqrt{3} \), we get:
\[ P_n = \sqrt{2} P_{n-1} + \sqrt{3} P_{n-2}. \]

This gives us a linear recurrence relation for \( P_n \).

Step 3: Calculation of the first few terms of \( P_n \).

Using the recurrence relation, we calculate the following terms:

- \( P_0 = 2 \) (since \( \alpha^0 + \beta^0 = 1 + 1 = 2 \))
- \( P_1 = \alpha + \beta = \sqrt{2} \)
- \( P_2 = \sqrt{2}P_1 + \sqrt{3}P_0 = \sqrt{2} \cdot \sqrt{2} + \sqrt{3} \cdot 2 = 2 + 2\sqrt{3} \)
- \( P_3 = \sqrt{2}P_2 + \sqrt{3}P_1 = \sqrt{2}(2 + 2\sqrt{3}) + \sqrt{3} \cdot \sqrt{2} = 2\sqrt{2} + 2\sqrt{6} + \sqrt{6} = 2\sqrt{2} + 3\sqrt{6} \)

Now, using this recurrence, we can compute higher values for \( P_4, P_5, \dots, P_{12} \).

Step 4: Substituting into the given equation.

We substitute the known values of \( P_9, P_{10}, P_{11}, P_{12} \) into the given equation:
\[ 11P_{12} + (10 - 11\sqrt{2})P_{11} - (11\sqrt{3} + 10\sqrt{2})P_{10} - \lambda = 0. \]

After solving this equation, we find that:
\[ \lambda = \sqrt{3}P_9. \] Quick Tip: When solving recurrence relations involving powers of roots, always use Vieta’s formulas to relate the sum and product of the roots to the given equation.