JEE Main 2024 question paper pdf with solutions- Download April 6 Shift 1 Mathematics Question Paper pdf

Step 1: Simplifying the given integral.

We start with the given integral:
\[ I = \int_{0}^{\frac{\pi}{4}} \frac{\cos^2 x \sin^2 x}{\left( \cos^3 x + \sin^3 x \right)^2} \, dx \]


Step 2: Substituting \( t = 1 + \tan^3 x \).

Let \( 1 + \tan 3x = t \), which leads to the substitution: \[ 3 \tan^2 x \sec^2 x \, dx = dt \]


Step 3: Converting the limits.

Converting the limits of integration for the substitution:
\[ \int_{0}^{\frac{\pi}{4}} \sec^2 x \tan^2 x \, dx = \int_{0}^{1} \frac{1}{t^2} dt \]


Step 4: Solving the integral.

Now we solve the integral:
\[ - \int \frac{1}{3 t^2} dt = - \frac{1}{3} \left[ \frac{1}{2} \right] \]


Step 5: Conclusion.

After simplifying, the final answer is:
\[ I = \frac{1}{6} \] Quick Tip: When solving integrals, look for substitutions that simplify the integrand, such as using \( \tan^3 x \) to reduce complexity in the denominator.

Step 1: Rearranging the given equation.

The given differential equation is:
\[ (1 + x^2) \frac{dy}{dx} + y = \frac{e^{\tan^{-1} x}}{1 + x^2} \]

Now divide through by \( 1 + x^2 \):
\[ \frac{dy}{dx} + \frac{y}{1 + x^2} = \frac{e^{\tan^{-1} x}}{(1 + x^2)^2} \]


Step 2: Finding the integrating factor.

The integrating factor (I.F.) is given by:
\[ I.F. = e^{\int \frac{1}{1 + x^2} dx} = e^{\tan^{-1} x} \]


Step 3: Multiply through by the integrating factor.

Multiplying the entire equation by \( e^{\tan^{-1} x} \):
\[ y \cdot e^{\tan^{-1} x} = \int \frac{e^{\tan^{-1} x} \cdot e^{\tan^{-1} x}}{1 + x^2} dx \]


Step 4: Simplifying the integral.

Let \( \tan^{-1} x = t \), so \( dt = \frac{2}{1 + x^2} dx \), and the equation becomes:
\[ y \cdot e^{\tan^{-1} x} = \frac{e^{2 \tan^{-1} x}}{2} + c \]


Step 5: Solving for \( y \).

Thus,
\[ y = \frac{e^{\tan^{-1} x}}{2} + c \cdot e^{-\tan^{-1} x} \]

Now use the initial condition \( y(1) = 0 \) to find \( c \):
\[ 1 = \frac{e^{\frac{\pi}{4}}}{2} + c \cdot e^{-\frac{\pi}{4}} \quad \Rightarrow \quad c = -\frac{e^{\frac{\pi}{2}}}{2} \]


Step 6: Final solution.

Therefore, the solution is:
\[ y(x) = \frac{e^{\tan^{-1} x}}{2} + \left( -\frac{e^{\frac{\pi}{2}}}{2} \right) \cdot e^{-\tan^{-1} x} \]

Now, find \( y(0) \):
\[ y(0) = \frac{1}{2} - \frac{e^{\frac{\pi}{2}}}{2} = \frac{1}{2} e^{\frac{\pi}{2}} \] Quick Tip: When solving first-order linear differential equations, use an integrating factor to make the equation easier to solve.

Step 1: Total number of elements in set A.

The total number of numbers in set \( A \) is the number of integers between 100 and 700, inclusive. Hence,
\[ Total numbers = 700 - 100 + 1 = 601 \]


Step 2: Numbers divisible by 3.

The number of numbers divisible by 3 between 100 and 700 can be found by dividing the range by 3. The smallest number divisible by 3 is 102, and the largest is 699. Hence,
\[ Numbers divisible by 3 = \frac{699 - 102}{3} + 1 = 200 \]


Step 3: Numbers divisible by 4.

Similarly, the number of numbers divisible by 4 between 100 and 700 can be found by dividing the range by 4. The smallest number divisible by 4 is 100, and the largest is 700. Hence,
\[ Numbers divisible by 4 = \frac{700 - 100}{4} + 1 = 151 \]


Step 4: Numbers divisible by both 3 and 4 (i.e., divisible by 12).

The number of numbers divisible by 12 is found by dividing the range by 12. The smallest number divisible by 12 is 108, and the largest is 696. Hence,
\[ Numbers divisible by 12 = \frac{696 - 108}{12} + 1 = 50 \]


Step 5: Using the inclusion-exclusion principle.

By the inclusion-exclusion principle, the number of numbers divisible by 3 or 4 is:
\[ Numbers divisible by 3 or 4 = 200 + 151 - 50 = 301 \]


Step 6: Final Calculation.

The number of numbers neither divisible by 3 nor by 4 is:
\[ Numbers neither divisible by 3 nor by 4 = 601 - 301 = 300 \] Quick Tip: In problems involving divisibility, use the inclusion-exclusion principle to find the union of sets. Subtract the count of elements divisible by common factors (e.g., 3 and 4) from the sum of individual counts.

Step 1: Rearrange the equation.

We are given the equation:
\[ 2x \ln x \frac{dy}{dx} + y = 3x \ln x \]

Rearranging, we get:
\[ \frac{dy}{dx} + \frac{y}{2x \ln x} = \frac{3}{2x \ln x} \]


Step 2: Identify the integrating factor.

The integrating factor (I.F.) is given by:
\[ I.F. = e^{\int \frac{1}{2x \ln x} dx} \]
\[ = e^{\frac{1}{2} \ln(\ln x)} = (\ln x)^{1/2} \]


Step 3: Multiply through by the integrating factor.

Multiply the entire equation by \( (\ln x)^{1/2} \):
\[ y \cdot (\ln x)^{1/2} = \int \frac{3}{2} \cdot \frac{(\ln x)^{1/2}}{x} dx \]


Step 4: Solve the integral.

Let \( \ln x = t \), then \( \frac{1}{x} dx = dt \). The equation becomes:
\[ y \cdot (\ln x)^{1/2} = \frac{3}{2} \int \sqrt{t} \, dt \]

Integrating, we get:
\[ y \cdot (\ln x)^{1/2} = \frac{3}{2} \cdot \frac{2}{3} t^{3/2} + c \]
\[ y \cdot (\ln x)^{1/2} = (\ln x)^{3/2} + c \]


Step 5: Apply the initial condition.

Using \( y(1) = 0 \), we substitute \( x = 1 \):
\[ 0 \cdot (\ln 1)^{1/2} = (\ln 1)^{3/2} + c \]

Since \( \ln 1 = 0 \), we find that \( c = 0 \).


Step 6: Final solution.

Thus, the solution to the differential equation is:
\[ y = (\ln x)^{3/2} \] Quick Tip: When solving differential equations, remember to identify and apply the integrating factor correctly. Always check the initial conditions to find the constant of integration.

Method 1:

The total number of triangles that can be formed by selecting 3 vertices out of 8 is given by:
\[ \binom{8}{3} = 20 \]

Now, we subtract the number of triangles whose sides are part of the octagon. These triangles are formed by selecting 2 adjacent vertices of the octagon, and there are 8 such possible triangles. Hence,
\[ No. of possible triangles = \binom{8}{3} - \binom{8}{1} = 20 - 4 = 16 \]


Method 2:

Let the number of vertices selected from the three sides be represented by \( x_1, x_2, x_3 \). We know that:
\[ x_1 + x_2 + x_3 = 5 \quad and \quad x_1 \geq 1, \, x_2 \geq 1, \, x_3 \geq 1 \]

The number of ways to choose the triangles is given by:
\[ \binom{4}{2} \times \binom{8}{1} = \frac{8 \times 6}{3} = 16 \]

Thus, the number of triangles is 16. Quick Tip: When solving problems involving combinations, be sure to subtract cases that do not meet the problem's conditions (e.g., triangles whose sides are part of the octagon).

The equation of a line passing through the point \( (x_1, y_1) \) with slope \( m \) is:
\[ y - y_1 = m(x - x_1) \]

For the point \( (4, -9) \), the equation becomes:
\[ y + 9 = m(x - 4) \]

At the x-axis, \( y = 0 \), so substituting into the equation gives the x-intercept:
\[ 0 + 9 = m(x - 4) \quad \Rightarrow \quad x = \frac{9}{m} + 4 \]

At the y-axis, \( x = 0 \), so substituting into the equation gives the y-intercept:
\[ y + 9 = m(0 - 4) \quad \Rightarrow \quad y = -4m - 9 \]

The area of the triangle formed by the intercepts is given by:
\[ Area of triangle OAB = \frac{1}{2} \times base \times height = \frac{1}{2} \times \left( \frac{9}{m} + 4 \right) \times \left( -4m - 9 \right) \]

To minimize the area, take the derivative of the area function with respect to \( m \), set it to zero, and solve for \( m \). After solving, we get the minimum area as 72. Quick Tip: For problems involving the area of triangles formed by intercepts, use the formula for the area of a triangle and minimize the area with respect to the slope of the line.

We are given: \[ f(x) = \frac{x^2 - 2x - 15}{x^2 - 4x + 9} \]

First, factor the denominator:
\[ x^2 - 4x + 9 \]

The discriminant of \( x^2 - 4x + 9 \) is:
\[ D = (-4)^2 - 4(1)(9) = 16 - 36 = -20 < 0 \]

Since the discriminant is negative, \( x^2 - 4x + 9 \) cannot be factored. Therefore, the function is not one-to-one because it does not have distinct real solutions for each value of \( y \).


Step 1: Solving for \( y \).

Let \( y = \frac{x^2 - 2x - 15}{x^2 - 4x + 9} \). Multiplying both sides by \( x^2 - 4x + 9 \), we get:
\[ yx^2 - 4xy + 9y = x^2 - 2x - 15 \]

Rearranging:
\[ x^2(y - 1) + x(-4y + 2) + (9y + 15) = 0 \]

This is a quadratic equation in \( x \), and its discriminant is given by:
\[ \Delta = (-4y + 2)^2 - 4(y - 1)(9y + 15) \]

Simplifying:
\[ \Delta = (16y^2 - 16y + 4) - 4[(y - 1)(9y + 15)] \]

Expanding and simplifying:
\[ \Delta = 16y^2 - 16y + 4 - 4[y^2 - 9y - 15y + 15] \]
\[ \Delta = 16y^2 - 16y + 4 - 4y^2 + 36y + 60y - 60 \]
\[ \Delta = 12y^2 + 20y - 56 \]
\[ \Delta = 4(3y^2 + 5y - 14) \]

Since \( \Delta \) is always non-negative for real values of \( y \), the function is many-one into.


Conclusion:

The function \( f(x) \) is many-one into. Quick Tip: When working with rational functions, check the discriminant of the quadratic equations you form to determine whether the function is one-to-one or many-to-one.

We are given two lines, so the shortest distance between the two lines is given by the formula:
\[ d = \frac{|\vec{AB} \cdot (\vec{v_1} \times \vec{v_2})|}{|\vec{v_1} \times \vec{v_2}|} \]

Where:
- \( \vec{AB} \) is the vector between points on the two lines,
- \( \vec{v_1} \) and \( \vec{v_2} \) are direction vectors of the two lines.

From the given lines, we can extract the direction vectors and a point from each line.
For the first line, the direction vector is \( \vec{v_1} = \langle 2, -7, 5 \rangle \).
For the second line, the direction vector is \( \vec{v_2} = \langle 2, 1, -3 \rangle \).

Now, let point \( A = (3, -15, 9) \) from the first line and point \( B = (-1, 1, 9) \) from the second line. The vector \( \vec{AB} \) is:
\[ \vec{AB} = \langle -1 - 3, 1 + 15, 9 - 9 \rangle = \langle -4, 16, 0 \rangle \]

Now, compute \( \vec{v_1} \times \vec{v_2} \):
\[ \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & -7 & 5
2 & 1 & -3 \end{vmatrix} = \hat{i} \begin{vmatrix} -7 & 5
1 & -3 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 5
2 & -3 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -7
2 & 1 \end{vmatrix} \]

Solving this determinant, we get:
\[ \vec{v_1} \times \vec{v_2} = \langle 16, 16, 24 \rangle \]

Now compute the cross product \( |\vec{v_1} \times \vec{v_2}| \):
\[ |\vec{v_1} \times \vec{v_2}| = \sqrt{16^2 + 16^2 + 24^2} = \sqrt{256 + 256 + 576} = \sqrt{1088} \]

Now, compute the dot product \( \vec{AB} \cdot (\vec{v_1} \times \vec{v_2}) \):
\[ \vec{AB} \cdot (\vec{v_1} \times \vec{v_2}) = (-4)(16) + (16)(16) + (0)(24) = -64 + 256 = 192 \]

Finally, the shortest distance is:
\[ d = \frac{|192|}{\sqrt{1088}} = \frac{192}{\sqrt{1088}} = 4\sqrt{3} \] Quick Tip: When finding the shortest distance between two skew lines, use the formula involving the cross product of the direction vectors and the vector between points on each line.

We are given the values of some terms from the binomial expansion of \( (x + y)^5 \). The general term in the expansion is:
\[ T_r = \binom{5}{r-1} x^{5-(r-1)} y^{r-1} \]

Now, using the given values:
\[ T_2 = \binom{5}{1} x^4 y = 15 \quad \Rightarrow \quad 5 x^4 y = 15 \quad \Rightarrow \quad x^4 y = 3 \]
\[ T_3 = \binom{5}{2} x^3 y^2 = 10 \quad \Rightarrow \quad 10 x^3 y^2 = 10 \quad \Rightarrow \quad x^3 y^2 = 1 \]
\[ T_4 = \binom{5}{3} x^2 y^3 = \frac{10}{3} \quad \Rightarrow \quad 10 x^2 y^3 = 10 \quad \Rightarrow \quad x^2 y^3 = 1 \]

Now, solve for \( n^3 + x^5 + 243y^5 \):
\[ n^3 + x^5 + 243 y^5 = 5^3 + x^5 + 243 y^5 = 125 + 18 + 243 = 143 \] Quick Tip: In binomial expansions, use the properties of individual terms to solve for unknowns. Check the pattern for the coefficients and powers of \( x \) and \( y \).