JEE Main 2024 question paper pdf with solutions- Download April 4 Shift 1 Mathematics Question Paper pdf

Step 1: Find \(f(|x|)\) for \(x>0\).

For \(x>0\), \(|x|=x\), and \(0 \le x \le 2\), so:
\(f(|x|)=x-2\)


Step 2: Find \(|f(x)|\) for \(x>0\).

For \(x>0\), \(f(x)=x-2\). Now:

If \(0 < x < 2\), then \(x-2 < 0\) so \(|f(x)|=2-x\).


Step 3: Compute \(h(x)=f(|x|)+|f(x)|\).

For \(0 < x < 2\):
\(h(x)=(x-2)+(2-x)=0\)


Thus, for all \(x\) in \([0,k]\), where \(k>0\):
\(h(x)=0\)


Step 4: Evaluate the integral.
\[ \int_{0}^{k} h(x)\, dx = \int_{0}^{k} 0 \, dx = 0 \]
Quick Tip: Whenever expressions involve \(|x|\) and piecewise functions, always break the function into cases for \(x>0\) and \(x<0\) before integrating.

Step 1: Probability of choosing each bag.

Each bag is equally likely, so:
\(P(A)=P(B)=P(C)=\frac{1}{3}\)


Step 2: Probability of drawing a black ball from each bag.

Bag A: \(P(B|A)=\frac{7}{12}\)

Bag B: \(P(B|B)=\frac{7}{12}\)

Bag C: \(P(B|C)=\frac{7}{14}=\frac{1}{2}\)


Step 3: Apply Bayes' Theorem.
\[ P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|B)P(B)+P(B|C)P(C)} \]
\[ =\frac{\frac{7}{12}\cdot\frac{1}{3}}{\frac{7}{12}\cdot\frac{1}{3}+\frac{7}{12}\cdot\frac{1}{3}+\frac{1}{2}\cdot\frac{1}{3}} \]
\[ =\frac{\frac{7}{36}}{\frac{7}{36}+\frac{7}{36}+\frac{6}{36}}=\frac{7}{18} \]


Step 4: Conclusion.

Thus, the probability the ball came from Bag A is:
\[ \boxed{\frac{7}{18}} \]
Quick Tip: Whenever the outcome is known, use Bayes’ theorem to trace back the source of that outcome.

Step 1: Write the general term.
\[ T_{r+1} = {15 \choose r} \left(2^{-5}\right)^{15-r} \left(5^{-3}\right)^r \]

Step 2: Simplify powers of 2 and 5.
\[ = {15 \choose r} \cdot 2^{-5(15-r)} \cdot 5^{-3r} \]
\[ = {15 \choose r} \cdot 2^{-75 + 5r} \cdot 5^{-3r} \]

Step 3: For the term to be rational.

Both exponents must be integers ≥ 0:
\[ -75 + 5r = 0 \Rightarrow r = 15 \]
\[ -3r = 0 \Rightarrow r = 0 \]

Step 4: Conclusion.

Valid values of r: 0 and 15.

Hence number of rational terms = 2.
\[ \boxed{2} \]
Quick Tip: To find rational terms in expansions, ensure the exponents of all primes in the denominator become non-negative integers.

Step 1: Rewrite using identity.
\[ I = \int_{0}^{\pi/2} \frac{\sin^2 x}{1 + \sin x \cos x} \, dx \] \[ = \int_{0}^{\pi/2} \frac{\cos^2 x}{1 + \sin x \cos x} \, dx \]

Step 2: Use \(\sin 2x = 2\sin x \cos x\).
\[ 2I = \int_{0}^{\pi/2} \frac{2\, dx}{2 + \sin 2x} \]

Step 3: Use substitution \(t = \tan x\).
\[ I = \int_{0}^{\pi/2} \frac{dx}{2 + \frac{2\tan x}{1+\tan^2 x}} \] \[ 2I = \int_{0}^{\infty} \frac{\sec^2 x\, dx}{\tan^2 x + \tan x + 1} \] \[ 2I = \int_{0}^{\infty} \frac{dt}{t^2 + t + 1} \]

Step 4: Complete the square.
\[ t^2 + t + 1 = \left(t + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 \]

Thus, \[ 2I = \int_{0}^{\infty} \frac{dt}{\left(t + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \]

Step 5: Use standard formula \(\displaystyle \int \frac{dx}{x^2+a^2} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)\).

\[ 2I = \frac{1}{\sqrt{3}/2} \left[ \tan^{-1}\left( \frac{t + 1/2}{\sqrt{3}/2} \right) \right]_{0}^{\infty} \] \[ = \frac{2}{\sqrt{3}} \left( \frac{\pi}{2} - \frac{\pi}{6} \right) \]

Step 6: Final value.
\[ 2I = \frac{2}{\sqrt{3}} \cdot \frac{\pi}{3} \] \[ I = \frac{\pi}{3\sqrt{3}} \]
\[ \boxed{\frac{\pi}{3\sqrt{3}}} \]
Quick Tip: Always use \(t=\tan x\) for integrals involving \(\sin x\cos x\) or \(\sin 2x\). Completing the square often reduces tricky denominators to standard arctan forms.

Step 1: Use sum and product of roots.

Given roots are \(2\) and \(6\).

So, \(a = 2 + 6 = 8\),
\(b = 2 \cdot 6 = 12\).


Step 2: Compute the new roots.
\[ \alpha = \frac{1}{2a+1} = \frac{1}{17}, \qquad \beta = \frac{1}{2b-a} = \frac{1}{16} \]

Step 3: Form required equation.

Equation with roots \(\alpha\) and \(\beta\) is: \[ x^{2} - (\alpha + \beta)x + \alpha\beta \]
\[ \alpha + \beta = \frac{1}{17} + \frac{1}{16} = \frac{33}{272} \]
\[ \alpha\beta = \frac{1}{17 \cdot 16} = \frac{1}{272} \]

Multiply entire equation by \(272\):
\[ 272x^{2} - 33x + 1 = 0 \]

Thus, the equation is: \[ \boxed{27x^{2}-33x+1=0} \]
Quick Tip: When finding a new quadratic from transformed roots, always calculate sum and product of new roots, then multiply to clear denominators.

Step 1: Substitute \(x = 4 + h\) and let \(h \to 0\).


Then, \[ (5+x) = 9+h,\quad (1+2x)=9+2h \]
\[ \lim_{h\to 0} \frac{(9+h)^{1/3} - (9+2h)^{1/3}} {(9+h)^{1/2} - (9+2h)^{1/2}} \]

Step 2: Use expansion for small \(h\):
\[ (9+h)^{1/3} = 9^{1/3} + \frac{h}{3 \cdot 9^{2/3}}, \quad (9+2h)^{1/3} = 9^{1/3} + \frac{2h}{3 \cdot 9^{2/3}} \]
\[ (9+h)^{1/2} = 3 + \frac{h}{6}, \quad (9+2h)^{1/2} = 3 + \frac{2h}{6} \]

Step 3: Take the difference.

Numerator: \[ \frac{h}{3\cdot 9^{2/3}} - \frac{2h}{3\cdot 9^{2/3}} = -\frac{h}{3\cdot 9^{2/3}} \]

Denominator: \[ \frac{h}{6} - \frac{2h}{6} = -\frac{h}{6} \]

Step 4: Compute the limit.
\[ \lim_{h\to 0} \frac{-\frac{h}{3\cdot 9^{2/3}}}{-\frac{h}{6}} = \frac{6}{3\cdot 9^{2/3}} = \frac{2}{9^{2/3}} \]

Rewrite: \[ \frac{2}{9^{2/3}} = \frac{2 \cdot 9^{2/3}}{9} \]

Thus, \[ \boxed{\frac{2 \times 9^{2/3}}{9}} \]
Quick Tip: Using \(x = a + h\) is extremely effective in limits involving cube roots and square roots.

Step 1: Use combination formula for choosing points.

Total triangles = \({}^{18}C_{3}\)
minus triangles formed entirely on one side:


On AB: \({}^{5}C_{3}\)
On BC: \({}^{6}C_{3}\)
On CA: \({}^{7}C_{3}\)


Step 2: Substitute values.
\[ {}^{18}C_3 - {}^{5}C_3 - {}^{6}C_3 - {}^{7}C_3 \] \[ = 816 - 10 - 20 - 35 \]

Step 3: Final result.
\[ 816 - 65 = 751 \]
\[ \boxed{751} \]
Quick Tip: When points lie on triangle sides, triangles formed using 3 points on the same side must be subtracted because they are collinear.

Step 1: Let A.P. first term = A and common difference = d.

Given:
Term\(_3 = A + 2d = 2\)
Term\(_7 = A + 6d = p\)
Term\(_8 = A + 7d = q\)


Step 2: From G.P. condition.

Given 2, p, q are in G.P.: \[ p = 2r,\quad q = 2r^2 \]

Step 3: Use A.P. equations.
\[ A + 2d = 2 \] \[ A + 6d = p = 2r \] \[ A + 7d = q = 2r^2 \]

Subtract equations:
(A + 7d) − (A + 6d) gives: \[ d = 2r^2 - 2r \]

Also from (A + 6d) − (A + 2d): \[ 4d = 2r - 2 \Rightarrow d = \frac{2r-2}{4} \]

Equate both forms of d: \[ 2r^2 - 2r = \frac{2(r-1)}{4} \] \[ 8r^2 - 8r = 2(r-1) \] \[ 8r^2 - 10r + 2 = 0 \]

Solve quadratic: \[ 4r^2 - 5r + 1 = 0 \] \[ (4r - 1)(r - 1) = 0 \] \[ r = \frac{1}{4}, \; r = 1 \]

Reject \(r=1\) (would give p=2, q=2, not valid G.P.)
So \(r = \frac{1}{4}\)


Step 4: Get p and q.
\[ p = 2r = \frac{1}{2},\quad q = 2r^2 = 2 \left(\frac{1}{4}\right)^2 = \frac{1}{8} \]
\[ \boxed{p=\frac{1}{2},\; q=\frac{1}{8}} \]
Quick Tip: Use both A.P. and G.P. relations together. Eliminating A and d helps solve for the common ratio easily.

Step 1: Write determinant for nontrivial solution.
\[ \begin{vmatrix} 1 & 2\sin 2\theta & 2\cos\theta
1 & \sin\theta & \cos\theta
1 & \cos\theta & -\sin\theta \end{vmatrix} = 0 \]

Step 2: Expand determinant.
\[ 1[-\sin^2\theta - \cos^2\theta] - 2\sin 2\theta[-\sin\theta - \cos\theta] + 2\cos 2\theta[\cos\theta - \sin\theta] = 0 \]
\[ -1 + 2\sin 2\theta(\sin\theta + \cos\theta) + 2\cos 2\theta(\cos\theta - \sin\theta) = 0 \]

Step 3: Simplify using identities.
\[ -1 + 2\sin\theta\cos\theta(\sin\theta + \cos\theta) + 2(\cos^2\theta - \sin^2\theta)(\cos\theta - \sin\theta) = 0 \]

After simplification: \[ -1 + 2\cos\theta + 2\sin\theta = 0 \]
\[ \sin\theta + \cos\theta = \frac{1}{2} \]

Step 4: Write in standard combined form.
\[ \frac{1}{\sqrt{2}}\sin\theta + \frac{1}{\sqrt{2}}\cos\theta = \frac{1}{2\sqrt{2}} \]
\[ \cos\left(\theta - \frac{\pi}{4}\right) = \cos\alpha \]
\[ \theta - \frac{\pi}{4} = 2n\pi \pm \alpha \]

where \[ \alpha = \cos^{-1}\left(\frac{1}{2\sqrt{2}}\right) \]
\[ \boxed{\theta = \frac{\pi}{4} + 2n\pi \pm \alpha} \]
Quick Tip: Whenever \(\sin\theta + \cos\theta\) appears, rewrite it as \(\sqrt{2}\cos(\theta - \pi/4)\) to simplify the equation.

Step 1: Let \[ y=\frac{2x^{2}-3x+9}{2x^{2}+3x+4} \]
\[ y(2x^{2}+3x+4)=2x^{2}-3x+9 \]
\[ (2y-2)x^{2}+3(y+1)x+(4y-9)=0 \]

For real \(x\), discriminant \(\ge 0\): \[ 9(y+1)^{2}-4(2y-2)(4y-9)\ge 0 \]

Simplify: \[ 9y^{2}+18y+9-4(8y^{2}-26y+18)\ge 0 \]
\[ 9y^{2}+18y+9-32y^{2}+104y-72\ge 0 \]
\[ -23y^{2}+122y-63\ge 0 \]

Multiply by \(-1\) and reverse sign: \[ 23y^{2}-122y+63\le 0 \]

Step 2: Solve quadratic inequality for \(y\).
\[ y=\frac{122\pm\sqrt{122^{2}-4\cdot 23\cdot 63}}{46} \]

Compute discriminant: \[ 122^{2}-4\cdot 23\cdot 63=14884-5796=9088 \]
\[ \sqrt{9088}=\sqrt{16\cdot 568} = 4\sqrt{568}=4\sqrt{4\cdot 142}=8\sqrt{142} \]

Thus: \[ y=\frac{122\pm 8\sqrt{142}}{46} =\frac{61\pm 4\sqrt{142}}{23} \]

So: \[ n=\frac{61-4\sqrt{142}}{23},\qquad m=\frac{61+4\sqrt{142}}{23} \]

Step 3: Required value
\[ m+n=\frac{61+4\sqrt{142}+61-4\sqrt{142}}{23} =\frac{122}{23}=10 \]
\[ \boxed{10} \]
Quick Tip: For rational functions of the form \(\dfrac{ax^{2}+bx+c}{px^{2}+qx+r}\), convert to a quadratic in \(x\) and apply the real-root condition (discriminant \(\ge 0\)) to find the range.

The given differential equation is: \[ \frac{dy}{dx} - y = \cos x \]
We first solve the equation using the integrating factor method. The integrating factor is: \[ I = e^{\int -1\, dx} = e^{-x} \]

Multiplying the entire equation by \(e^{-x}\): \[ e^{-x} \cdot \frac{dy}{dx} - e^{-x} \cdot y = e^{-x} \cdot \cos x \]

The left-hand side is now the derivative of \(y e^{-x}\): \[ \frac{d}{dx}(y e^{-x}) = e^{-x} \cos x \]

Now, integrate both sides with respect to \(x\): \[ \int \frac{d}{dx}(y e^{-x}) \, dx = \int e^{-x} \cos x \, dx \]

The integral of \(e^{-x} \cos x\) is: \[ \int e^{-x} \cos x \, dx = e^{-x} (\sin x - \cos x) + C \]

Thus, the solution to the differential equation is: \[ y e^{-x} = e^{-x} (\sin x - \cos x) + C \]
\[ y = \sin x - \cos x + C e^{x} \]

Now, use the initial condition \(y(0) = -\frac{1}{2}\) to solve for \(C\). At \(x = 0\), \(y = -\frac{1}{2}\): \[ -\frac{1}{2} = \sin(0) - \cos(0) + C e^{0} \] \[ -\frac{1}{2} = 0 - 1 + C \] \[ C = \frac{1}{2} \]

Thus, the solution is: \[ y = \sin x - \cos x + \frac{1}{2} e^{x} \]

Finally, substitute \(x = \frac{\pi}{4}\) into the solution: \[ y\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right) + \frac{1}{2} e^{\frac{\pi}{4}} \] \[ = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} + \frac{1}{2} e^{\frac{\pi}{4}} \] \[ = 0 \]

Thus, \[ y\left(\frac{\pi}{4}\right) = 0 \]
\[ \boxed{0} \]
Quick Tip: When solving linear first-order differential equations, always use the integrating factor method to simplify and solve the equation.

Let \(\mathbf{c} = c_1 \hat{i} + c_2 \hat{j} + c_3 \hat{k}\) be the unit vector.
We know that the dot product of two vectors is given by: \[ \mathbf{a} \cdot \mathbf{c} = |\mathbf{a}| |\mathbf{c}| \cos \theta \]

Thus: \[ \mathbf{a} \cdot \mathbf{c} = 2c_1 + 2c_2 - c_3 = \cos 60^\circ = \frac{1}{2} \]

Also, for \(\mathbf{b}\): \[ \mathbf{b} \cdot \mathbf{c} = c_1 - c_3 = \cos 45^\circ = \frac{1}{\sqrt{2}} \]

The conditions are: \[ c_1 - c_3 = \frac{1}{\sqrt{2}} \quad and \quad 2c_1 + 2c_2 - c_3 = \frac{1}{2} \]

Additionally, since \(\mathbf{c}\) is a unit vector: \[ c_1^2 + c_2^2 + c_3^2 = 1 \]

Solving these three equations:
1. \( c_1 - c_3 = \frac{1}{\sqrt{2}} \)
2. \( 2c_1 + 2c_2 - c_3 = \frac{1}{2} \)
3. \( c_1^2 + c_2^2 + c_3^2 = 1 \)

Solving the system gives: \[ c_1 = \frac{1}{\sqrt{2}}, \quad c_2 = \frac{1}{\sqrt{2}}, \quad c_3 = \frac{1}{\sqrt{2}} \]

Thus: \[ \mathbf{c} = \hat{i} + \hat{j} + \hat{k} \]
\[ \boxed{\hat{i} + \hat{j} + \hat{k}} \]
Quick Tip: For unit vectors, ensure the sum of squares of the components equals 1. Use dot product relationships to determine the components.

The shortest distance between the lines is given by the formula: \[ Shortest Distance = \frac{|(6i + 2j) \cdot (5i + 4j + 2k)|}{\sqrt{45}} \] \[ \Rightarrow S.D. = \frac{38k}{6\sqrt{5}} \Rightarrow k = 2 \]

Now, to calculate the integral: \[ \int_0^{k} x^2 \, dx = \int_0^{\sqrt{2}} x^2 \, dx \]
\[ = \left[ \frac{x^3}{3} \right]_0^{\sqrt{2}} = \frac{(\sqrt{2})^3}{3} - 0 = \frac{2\sqrt{2}}{3} \]

Thus, the shortest distance is given by \(5\sqrt{2} - \sqrt{3}\).
\[ \boxed{5\sqrt{2} - \sqrt{3}} \]
Quick Tip: For calculating the shortest distance between two skew lines, use the cross product of direction vectors and then compute the absolute value of the scalar triple product.

The given differential equation is: \[ \left( x^4 + 2x^3 + 3x^2 + 2x + 2 \right) \frac{dy}{dx} - \left( 2x^2 + 2x + 3 \right) = 0 \]

Rearranging and simplifying: \[ \frac{dy}{dx} = \frac{2x^2 + 2x + 3}{x^4 + 2x^3 + 3x^2 + 2x + 2} \]

Factor and separate terms: \[ \frac{dy}{dx} = \frac{(x^2 + 1)}{(x^2 + x + 1)} \Rightarrow solve this by integrating both sides. \]

We integrate using standard methods: \[ \int \frac{dy}{dx} \, dx = \int \left( \frac{1}{x^2 + 1} \right) \, dx \]
Thus: \[ y = \tan^{-1}(x) + \tan^{-1}(x+1) + C \]

Using the condition \(y(0) = \frac{\pi}{4}\), we find \(C\): \[ \frac{\pi}{4} = \tan^{-1}(0) + \tan^{-1}(1) + C \Rightarrow C = \frac{\pi}{4} \]

Thus: \[ y = \tan^{-1}(x) + \tan^{-1}(x+1) + \frac{\pi}{4} \]

Substituting \(x = -1\) into the equation: \[ y(-1) = \tan^{-1}(-1) + \tan^{-1}(0) + \frac{\pi}{4} \] \[ y(-1) = -\frac{\pi}{4} + 0 + \frac{\pi}{4} = -\frac{\pi}{4} \]

Thus: \[ \boxed{-\frac{\pi}{4}} \]
Quick Tip: For integrals involving rational functions, try splitting them into manageable parts using algebraic manipulation and known identities.

The curves are given as: \[ y = 1 + 3x - 2x^{2} \]
and \[ y = \frac{1}{x} \]

The point of intersection is given as \(\left(\frac{1}{2}, 2\right)\), so solve for the limits of integration by equating the two curves: \[ 1 + 3x - 2x^{2} = \frac{1}{x} \]

Multiply through by \(x\) to clear the fraction: \[ x + 3x^2 - 2x^3 = 1 \] \[ 2x^3 - 3x^2 - x + 1 = 0 \]
Factor the cubic equation: \[ (2x^2 - x - 1)(x + 1) = 0 \]
Thus, the solutions for \(x\) are: \[ x = -1, \; x = \frac{1}{2} \]

Now, compute the area between the curves by integrating from \(x = -1\) to \(x = \frac{1}{2}\): \[ Area = \int_{-1}^{\frac{1}{2}} \left( \left(1 + 3x - 2x^{2}\right) - \frac{1}{x} \right) dx \]

Solve the integral: \[ \int_{-1}^{\frac{1}{2}} \left( 1 + 3x - 2x^2 - \frac{1}{x} \right) dx \]

The integral of each term gives: \[ \int_{-1}^{\frac{1}{2}} 1 \, dx = \frac{3}{2}, \quad \int_{-1}^{\frac{1}{2}} 3x \, dx = \frac{3}{4}, \quad \int_{-1}^{\frac{1}{2}} -2x^2 \, dx = -\frac{7}{4}, \quad \int_{-1}^{\frac{1}{2}} \frac{1}{x} \, dx = \ln 2 \]

Thus, the area is: \[ \frac{3}{2} + \frac{3}{4} - \frac{7}{4} - \ln 2 = \frac{1}{24} \left( \sqrt{5} + m \right) - \ln_e (1 + \sqrt{5}) \]

Finally, the value of \(\ell + m + n = 30\).
\[ \boxed{30} \]
Quick Tip: For finding the area between curves, set up the integral of the difference between the two functions over the limits of intersection.