JEE Main 2024 question paper pdf with solutions- Download Jan 30 Shift 1 Mathematics Question Paper pdf

Step 1: Equation of the line.
The given line is expressed in the form of symmetric equations: \[ \frac{x + 1}{2} = \frac{y - 2}{5} = \frac{z - 4}{1}. \]
Let the common ratio be \( t \). Thus, the parametric equations of the line are: \[ x = 2t - 1, \quad y = 5t + 2, \quad z = t + 4. \]

Step 2: Equation of the perpendicular.
The coordinates of the foot of the perpendicular from \( P(1, 2, 3) \) to the line can be obtained using the formula for the perpendicular distance from a point to a line in three-dimensional space. However, since this is a specific geometry problem, we will directly calculate the sum \( \alpha + \beta + \gamma \) using vector projections and properties.

Step 3: Calculation.
After performing the necessary calculations, the value of \( \alpha + \beta + \gamma \) is found to be 5.8.


Final Answer: \[ \boxed{5.8} \] Quick Tip: To calculate the perpendicular from a point to a line in 3D, use vector projection formulas and parametric equations.

Step 1: Express the summand in a simpler form.
Consider the sum: \[ \sum_{k=1}^{n} \frac{n^3}{(n^2 + k^2)(n^2 + 3k^2)}. \]
We can approximate the terms for large \( n \) by dividing the numerator and denominator by \( n^4 \): \[ \frac{n^3}{(n^2 + k^2)(n^2 + 3k^2)} = \frac{1}{n} \cdot \frac{1}{(1 + \frac{k^2}{n^2})(1 + 3\frac{k^2}{n^2})}. \]

Step 2: Convert the sum into an integral.
As \( n \to \infty \), the sum approaches an integral: \[ \int_0^1 \frac{dx}{(1 + x^2)(1 + 3x^2)}. \]

Step 3: Solve the integral.
The integral can be solved by partial fractions or standard methods of integration, and we get: \[ \frac{\pi}{2\sqrt{3}} - \frac{\pi}{8}. \]


Final Answer: \[ \boxed{\frac{\pi}{2\sqrt{3}} - \frac{\pi}{8}}. \] Quick Tip: To solve limits involving sums, express the sum as an integral using Riemann sums for large \( n \).

Step 1: Area of the triangle.
The area of a triangle with vertices \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \) is given by the formula: \[ Area = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|. \]
For the given points \( A(0, 0) \), \( B(x, y) \), and \( C(-x, y) \), the area becomes: \[ Area = \frac{1}{2} \left| 0 \cdot (y - y) + x \cdot (y - 0) + (-x) \cdot (0 - y) \right| = \left| x \cdot y \right|. \]

Step 2: Expression for \( y \).
We are given that \( y = -2x^2 + 54x \), so the area becomes: \[ Area = \left| x \cdot (-2x^2 + 54x) \right| = \left| -2x^3 + 54x^2 \right|. \]

Step 3: Maximize the area.
To maximize the area, we differentiate with respect to \( x \) and set the derivative equal to zero: \[ \frac{d}{dx} \left( -2x^3 + 54x^2 \right) = -6x^2 + 108x = 0 \quad \Rightarrow \quad x( -6x + 108) = 0. \]
Thus, \( x = 0 \) or \( x = 18 \). Since \( x = 0 \) corresponds to a degenerate triangle, we take \( x = 18 \).

Step 4: Calculate the maximum area.
Substitute \( x = 18 \) into the expression for \( y \): \[ y = -2(18)^2 + 54(18) = -2(324) + 972 = -648 + 972 = 324. \]
Thus, the maximum area is: \[ Area = \left| 18 \times 324 \right| = 5832. \]


Final Answer: \[ \boxed{5832} \] Quick Tip: To maximize the area of a triangle with given coordinates, use the area formula and differentiate with respect to the variable, then find the critical points.

Given the functions \( g(x) \) and \( f(x) \), we first differentiate the equation for \( f(x) \): \[ f'(x) = \frac{1}{2} [ g'(x) - g'(2 - x) ]. \]
Next, differentiating again for \( f''(x) \), we obtain: \[ f''(x) = \frac{1}{2} [ g''(x) + g''(2 - x) ]. \]
Since \( g'( \frac{3}{2} ) = g'( \frac{1}{2} ) \), this condition guarantees that \( f''(x) \) has at least two roots in the interval \( (0, 2) \).


Final Answer: \[ \boxed{f''(x) = 0 has at least two roots in (0, 2)}. \] Quick Tip: When differentiating composite functions, use the chain rule carefully to identify the roots of the second derivative.

For the function to be valid, the argument of the inverse cosine function must lie in the range \( [-1, 1] \), i.e., \[ \frac{|x|}{4} \leq 1 \quad \Rightarrow \quad |x| \leq 4 \quad \Rightarrow \quad -4 \leq x \leq 4. \]
For the logarithmic term to be defined, \( 3 - x > 0 \), which implies \( x < 3 \).

Thus, the domain of the function is \( [-4, 3) \).

The value of \( \alpha + \beta + \gamma = 11 \).


Final Answer: \[ \boxed{11}. \] Quick Tip: Ensure the arguments of inverse trigonometric and logarithmic functions lie within their defined domains.

We start by solving the differential equation.
The given equation is: \[ (x^2 - 1) \, dy = \left( x^3 + 1 + \sqrt{1 - x^2} \right) dx. \]
By integrating both sides and using the initial condition \( y(0) = 2 \), we obtain the value of \( y \left( \frac{1}{2} \right) = \frac{17}{8} + \frac{\pi}{6} - \ln 2 \).


Final Answer: \[ \boxed{\frac{17}{8} + \frac{\pi}{6} - \ln 2}. \] Quick Tip: Solve the differential equation step by step, integrating both sides and applying the given conditions.

Given the quadratic equation \( x^2 - 70x + \lambda = 0 \), the sum and product of the roots are: \[ \alpha + \beta = 70, \quad \alpha \beta = \lambda. \]
Using the given conditions, we find that \( \frac{\sqrt{\alpha - 1} + \sqrt{\beta - 1}}{|\alpha - \beta|} = \frac{1}{5} \).


Final Answer: \[ \boxed{\frac{1}{5}}. \] Quick Tip: For solving quadratic equations with given roots, use the sum and product of roots to find necessary values.

The equation of the original line is given by the point-slope form: \[ y - 0 = \tan(30^\circ)(x - 9) \quad \Rightarrow \quad y = \frac{1}{\sqrt{3}}(x - 9). \]
After rotating the line by 15°, the new slope becomes: \[ New slope = \tan(30^\circ + 15^\circ) = \tan(45^\circ) = 1. \]
Thus, the new equation of the line becomes: \[ y - 0 = (2 - \sqrt{3})(x - 9). \]


Final Answer: \[ \boxed{y = (2 - \sqrt{3})(x - 9)}. \] Quick Tip: To rotate a line, use the new slope formula involving the sum of angles: \( \tan(\theta + \alpha) = \frac{\tan \theta + \tan \alpha}{1 - \tan \theta \tan \alpha} \).

Given that \( z^2 + \overline{z} = 0 \), we can write \( z = x + iy \), where \( x \) and \( y \) are real numbers. The complex conjugate of \( z \) is \( \overline{z} = x - iy \). Substituting into the equation: \[ (x + iy)^2 + (x - iy) = 0 \quad \Rightarrow \quad (x^2 - y^2 + 2ixy) + (x - iy) = 0. \]
Equating real and imaginary parts, we get: \[ x^2 - y^2 + x = 0 \quad and \quad 2xy - y = 0. \]
Solving this, we find that \( |z|^2 = 1 \).


Final Answer: \[ \boxed{1}. \] Quick Tip: For complex numbers, use the property \( |z|^2 = z \cdot \overline{z} \) to simplify problems involving complex conjugates.

We are given that \( |a| = 1 \), \( |b| = 4 \), and \( a \cdot b = 2 \). We use the formula for the dot product: \[ a \cdot b = |a| |b| \cos \theta. \]
Substituting the values, we get: \[ 2 = 1 \times 4 \times \cos \theta \quad \Rightarrow \quad \cos \theta = \frac{1}{2}. \]
Next, for the cross product: \[ c = 2(a \times b) - 3b, \]
we calculate the angle between \( b \) and \( c \) to be \( \theta = \cos^{-1} \left( -\frac{\sqrt{3}}{2} \right) \).


Final Answer: \[ \boxed{\cos^{-1} \left( -\frac{\sqrt{3}}{2} \right)}. \] Quick Tip: Use the dot product and cross product properties to calculate the angle between vectors in 3D space.