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JEE Main 2025 Apr 8 Shift 2 — Mathematics

25 questions Ā· 25 with the official NTA answer

Q1JEE Main 2025 Apr 8 Shift 2Sets, Relations and FunctionsMedium

Let A={0,1,2,3,4,5}A = \{0, 1, 2, 3, 4, 5\}. Let RR be a relation on AA defined by (x,y)∈R(x, y) \in R if and only if max⁔{x,y}∈{3,4}\max\{x, y\} \in \{3, 4\}. Then among the statements

(S1)(\mathrm{S}_1) : The number of elements in RR is 18, and

(S2)(\mathrm{S}_2) : The relation RR is symmetric but neither reflexive nor transitive

  1. A

    both are true

  2. B

    both are false

  3. C

    only (S1)(\mathrm{S}_1) is true

  4. D

    only (S2)(\mathrm{S}_2) is true

Show answer

Correct option: D

Q2JEE Main 2025 Apr 8 Shift 2Quadratic EquationsMedium

The sum of the squares of the roots of ∣xāˆ’2∣2+∣xāˆ’2āˆ£āˆ’2=0|x-2|^2 + |x-2| - 2 = 0 and the squares of the roots of x2āˆ’2∣xāˆ’3āˆ£āˆ’5=0x^2 - 2|x-3| - 5 = 0, is

  1. A

    24

  2. B

    26

  3. C

    30

  4. D

    36

Show answer

Correct option: D

Q3JEE Main 2025 Apr 8 Shift 2Complex NumbersMedium

Let A={θ∈[0,2Ļ€]:1+10 Re(2cos⁔θ+isin⁔θcosā”Īøāˆ’3isin⁔θ)=0}A = \left\{\theta \in [0, 2\pi] : 1 + 10\,\mathrm{Re}\left(\frac{2\cos\theta + i\sin\theta}{\cos\theta - 3i\sin\theta}\right) = 0\right\}. Then āˆ‘ĪøāˆˆAĪø2\sum_{\theta \in A} \theta^2 is equal to

  1. A

    8Ļ€28\pi^2

  2. B

    274Ļ€2\frac{27}{4}\pi^2

  3. C

    6Ļ€26\pi^2

  4. D

    214Ļ€2\frac{21}{4}\pi^2

Show answer

Correct option: D

Q4JEE Main 2025 Apr 8 Shift 2Matrices and DeterminantsMedium

Let A=[22+p2+p+q46+2p8+3p+2q612+3p20+6p+3q]\mathrm{A} = \begin{bmatrix} 2 & 2+p & 2+p+q \\ 4 & 6+2p & 8+3p+2q \\ 6 & 12+3p & 20+6p+3q \end{bmatrix}. If det⁔(adj (adj (3A)))=2mā‹…3n\det(adj\,(adj\,(3A))) = 2^m \cdot 3^n, m,n∈Nm, n \in \mathbb{N}, then m+nm + n is equal to

  1. A

    20

  2. B

    22

  3. C

    24

  4. D

    26

Show answer

Correct option: C

Q5JEE Main 2025 Apr 8 Shift 2Complex NumbersMedium

Let α\alpha be a solution of x2+x+1=0x^2 + x + 1 = 0, and for some aa and bb in R\mathbb{R}, [4ab][11613āˆ’1āˆ’12āˆ’2āˆ’14āˆ’8]=[000]\begin{bmatrix} 4 & a & b \end{bmatrix} \begin{bmatrix} 1 & 16 & 13 \\ -1 & -1 & 2 \\ -2 & -14 & -8 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix}. If 4α4+mαa+nαb=3\frac{4}{\alpha^4} + \frac{m}{\alpha^a} + \frac{n}{\alpha^b} = 3, then m+nm + n is equal to ________

  1. A

    8

  2. B

    11

  3. C

    7

  4. D

    3

Show answer

Correct option: B

Q6JEE Main 2025 Apr 8 Shift 2Sequences and SeriesEasy

If 114+124+134+ā€¦āˆž=Ļ€490\frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \ldots \infty = \frac{\pi^4}{90},

114+134+154+ā€¦āˆž=α\frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \ldots \infty = \alpha,

124+144+164+ā€¦āˆž=β\frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + \ldots \infty = \beta,

then αβ\frac{\alpha}{\beta} is equal to

  1. A

    14

  2. B

    15

  3. C

    18

  4. D

    23

Show answer

Correct option: B

Q7JEE Main 2025 Apr 8 Shift 2Permutations and CombinationsEasy

There are 12 points in a plane, no three of which are in the same straight line, except 5 points which are collinear. Then the total number of triangles that can be formed with the vertices at any three of these 12 points is

  1. A

    210

  2. B

    220

  3. C

    200

  4. D

    230

Show answer

Correct option: A

Q8JEE Main 2025 Apr 8 Shift 2Binomial TheoremEasy

The number of integral terms in the expansion of (512+718)1016\left(5^{\frac{1}{2}} + 7^{\frac{1}{8}}\right)^{1016} is

  1. A

    129

  2. B

    127

  3. C

    128

  4. D

    130

Show answer

Correct option: C

Q9JEE Main 2025 Apr 8 Shift 2ProbabilityMedium

If AA and BB are two events such that P(A)=0.7P(A) = 0.7, P(B)=0.4P(B) = 0.4 and P(A∩B‾)=0.5P\left(A \cap \overline{B}\right) = 0.5, where B‾\overline{B} denotes the complement of BB, then P(B∣(A∪B‾))P\left(B \mid \left(A \cup \overline{B}\right)\right) is equal to

  1. A

    16\frac{1}{6}

  2. B

    14\frac{1}{4}

  3. C

    12\frac{1}{2}

  4. D

    13\frac{1}{3}

Show answer

Correct option: B

Q10JEE Main 2025 Apr 8 Shift 2Straight LinesMedium

A line passing through the point P(a,0)P(a, 0) makes an acute angle α\alpha with the positive xx-axis. Let this line be rotated about the point PP through an angle α2\frac{\alpha}{2} in the clock-wise direction. If in the new position, the slope of the line is 2āˆ’32 - \sqrt{3} and its distance from the origin is 12\frac{1}{\sqrt{2}}, then the value of 3a2tan⁔2Ī±āˆ’233a^2 \tan^2\alpha - 2\sqrt{3} is

  1. A

    8

  2. B

    5

  3. C

    6

  4. D

    4

Show answer

Correct option: D

Q11JEE Main 2025 Apr 8 Shift 2Straight LinesMedium

Let aa be the length of a side of a square OABC with O being the origin. Its side OA makes an acute angle α\alpha with the positive xx-axis and the equations of its diagonals are (3+1)x+(3āˆ’1)y=0\left(\sqrt{3}+1\right)x + \left(\sqrt{3}-1\right)y = 0 and (3āˆ’1)xāˆ’(3+1)y+83=0\left(\sqrt{3}-1\right)x - \left(\sqrt{3}+1\right)y + 8\sqrt{3} = 0. Then a2a^2 is equal to

  1. A

    16

  2. B

    24

  3. C

    48

  4. D

    32

Show answer

Correct option: C

Q12JEE Main 2025 Apr 8 Shift 2Conic SectionsMedium

Let the ellipse 3x2+py2=43x^2 + py^2 = 4 pass through the centre CC of the circle x2+y2āˆ’2xāˆ’4yāˆ’11=0x^2 + y^2 - 2x - 4y - 11 = 0 of radius rr. Let f1,f2f_1, f_2 be the focal distances of the point CC on the ellipse. Then 6f1f2āˆ’r6f_1 f_2 - r is equal to

  1. A

    74

  2. B

    70

  3. C

    68

  4. D

    78

Show answer

Correct option: B

Q13JEE Main 2025 Apr 8 Shift 2Inverse Trigonometric FunctionsMedium

The value of cotā”āˆ’1(1+tan⁔2(2)āˆ’1tan⁔(2))āˆ’cotā”āˆ’1(1+tan⁔2(12)+1tan⁔(12))\cot^{-1}\left(\frac{\sqrt{1+\tan^2(2)}-1}{\tan(2)}\right) - \cot^{-1}\left(\frac{\sqrt{1+\tan^2\left(\frac{1}{2}\right)}+1}{\tan\left(\frac{1}{2}\right)}\right) is equal to

  1. A

    Ļ€āˆ’54\pi - \frac{5}{4}

  2. B

    Ļ€āˆ’32\pi - \frac{3}{2}

  3. C

    π+52\pi + \frac{5}{2}

  4. D

    π+32\pi + \frac{3}{2}

Show answer

Correct option: A

Q14JEE Main 2025 Apr 8 Shift 2Vector AlgebraMedium

Let aāƒ—=i^+2j^+k^\vec{a} = \hat{i} + 2\hat{j} + \hat{k} and bāƒ—=2i^+j^āˆ’k^\vec{b} = 2\hat{i} + \hat{j} - \hat{k}. Let c^\hat{c} be a unit vector in the plane of the vectors aāƒ—\vec{a} and bāƒ—\vec{b} and be perpendicular to aāƒ—\vec{a}. Then such a vector c^\hat{c} is :

  1. A

    13(i^āˆ’j^+k^)\frac{1}{\sqrt{3}}\left(\hat{i} - \hat{j} + \hat{k}\right)

  2. B

    13(āˆ’i^+j^āˆ’k^)\frac{1}{\sqrt{3}}\left(-\hat{i} + \hat{j} - \hat{k}\right)

  3. C

    12(āˆ’i^+k^)\frac{1}{\sqrt{2}}\left(-\hat{i} + \hat{k}\right)

  4. D

    15(j^āˆ’2k^)\frac{1}{\sqrt{5}}\left(\hat{j} - 2\hat{k}\right)

Show answer

Correct option: C

Q15JEE Main 2025 Apr 8 Shift 2Three Dimensional GeometryHard

Let the values of Ī»\lambda for which the shortest distance between the lines xāˆ’12=yāˆ’23=zāˆ’34\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} and xāˆ’Ī»3=yāˆ’44=zāˆ’55\frac{x-\lambda}{3} = \frac{y-4}{4} = \frac{z-5}{5} is 16\frac{1}{\sqrt{6}} be Ī»1\lambda_1 and Ī»2\lambda_2. Then the radius of the circle passing through the points (0,0)(0, 0), (Ī»1,Ī»2)(\lambda_1, \lambda_2) and (Ī»2,Ī»1)(\lambda_2, \lambda_1) is

  1. A

    4

  2. B

    23\frac{\sqrt{2}}{3}

  3. C

    3

  4. D

    523\frac{5\sqrt{2}}{3}

Show answer

Correct option: D

Q16JEE Main 2025 Apr 8 Shift 2Limits, Continuity and DifferentiabilityMedium

Given below are two statements:

Statement I : lim⁔x→0(tanā”āˆ’1x+log⁔e1+x1āˆ’xāˆ’2xx5)=25\lim\limits_{x \to 0}\left(\frac{\tan^{-1}x + \log_e\sqrt{\frac{1+x}{1-x}} - 2x}{x^5}\right) = \frac{2}{5}

Statement II : lim⁔x→1(x21āˆ’x)=1e2\lim\limits_{x \to 1}\left(x^{\frac{2}{1-x}}\right) = \frac{1}{e^2}

In the light of the above statements, choose the correct answer from the options given below

  1. A

    Both Statement I and Statement II are true

  2. B

    Both Statement I and Statement II are false

  3. C

    Statement I is true but Statement II is false

  4. D

    Statement I is false but Statement II is true

Show answer

Correct option: A

Q17JEE Main 2025 Apr 8 Shift 2Differentiation and Applications of DerivativesMedium

Let the function f(x)=x3+3x+3f(x) = \frac{x}{3} + \frac{3}{x} + 3, x≠0x \neq 0 be strictly increasing in (āˆ’āˆž,α1)∪(α2,āˆž)(-\infty, \alpha_1) \cup (\alpha_2, \infty) and strictly decreasing in (α3,α4)∪(α4,α5)(\alpha_3, \alpha_4) \cup (\alpha_4, \alpha_5). Then āˆ‘i=15αi2\sum\limits_{i=1}^{5} \alpha_i^2 is equal to

  1. A

    28

  2. B

    36

  3. C

    40

  4. D

    48

Show answer

Correct option: B

Q18JEE Main 2025 Apr 8 Shift 2Integral CalculusMedium

Let f(x)f(x) be a positive function and I1=āˆ«āˆ’1212x f(2x(1āˆ’2x))dxI_1 = \int\limits_{-\frac{1}{2}}^{1} 2x\, f\left(2x(1-2x)\right) dx and I2=āˆ«āˆ’12f(x(1āˆ’x))dxI_2 = \int\limits_{-1}^{2} f\left(x(1-x)\right) dx. Then the value of I2I1\frac{I_2}{I_1} is equal to __________

  1. A

    12

  2. B

    9

  3. C

    6

  4. D

    4

Show answer

Correct option: D

Q19JEE Main 2025 Apr 8 Shift 2Integral CalculusMedium

The integral āˆ«āˆ’132(āˆ£Ļ€2xsin⁔(Ļ€x)∣)dx\int\limits_{-1}^{\frac{3}{2}} \left(\left|\pi^2 x \sin(\pi x)\right|\right) dx is equal to :

  1. A

    1+3Ļ€1 + 3\pi

  2. B

    2+3Ļ€2 + 3\pi

  3. C

    3+2Ļ€3 + 2\pi

  4. D

    4+Ļ€4 + \pi

Show answer

Correct option: A

Q20JEE Main 2025 Apr 8 Shift 2Differential EquationsMedium

Let f(x)=xāˆ’1f(x) = x - 1 and g(x)=exg(x) = e^x for x∈Rx \in \mathbb{R}. If dydx=(eāˆ’2x g(f(f(x)))āˆ’yx)\frac{dy}{dx} = \left(e^{-2\sqrt{x}}\, g\left(f\left(f(x)\right)\right) - \frac{y}{\sqrt{x}}\right), y(0)=0y(0) = 0, then y(1)y(1) is

  1. A

    eāˆ’1e4\frac{e-1}{e^4}

  2. B

    2eāˆ’1e3\frac{2e-1}{e^3}

  3. C

    1āˆ’e2e4\frac{1-e^2}{e^4}

  4. D

    1āˆ’e3e4\frac{1-e^3}{e^4}

Show answer

Correct option: A

Q21JEE Main 2025 Apr 8 Shift 2Sets, Relations and FunctionsMediumNumerical

Let the domain of the function f(x)=cosā”āˆ’1(4x+53xāˆ’7)f(x) = \cos^{-1}\left(\frac{4x+5}{3x-7}\right) be [α,β][\alpha, \beta] and the domain of g(x)=log⁔2(2āˆ’6log⁔27(2x+5))g(x) = \log_2\left(2 - 6\log_{27}(2x+5)\right) be (γ,Ī“)(\gamma, \delta).

Then ∣7(α+β)+4(γ+Γ)∣|7(\alpha + \beta) + 4(\gamma + \delta)| is equal to __________

Show answer

Answer: 96

Q22JEE Main 2025 Apr 8 Shift 2Binomial TheoremMediumNumerical

The product of the last two digits of (1919)1919(1919)^{1919} is __________

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Answer: 63

Q23JEE Main 2025 Apr 8 Shift 2CirclesHardNumerical

Let rr be the radius of the circle, which touches xx-axis at point (a,0)(a, 0), a<0a < 0 and the parabola y2=9xy^2 = 9x at the point (4,6)(4, 6). Then rr is equal to __________

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Answer: 30

Q24JEE Main 2025 Apr 8 Shift 2Three Dimensional GeometryHardNumerical

Let the area of the triangle formed by the lines x+2=yāˆ’1=zx + 2 = y - 1 = z, xāˆ’35=yāˆ’1=zāˆ’11\frac{x-3}{5} = \frac{y}{-1} = \frac{z-1}{1} and xāˆ’3=yāˆ’33=zāˆ’21\frac{x}{-3} = \frac{y-3}{3} = \frac{z-2}{1} be AA. Then A2A^2 is equal to __________

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Answer: 56

Q25JEE Main 2025 Apr 8 Shift 2Integral CalculusMediumNumerical

Let the area of the bounded region {(x,y):0≤9x≤y2,Ā y≄3xāˆ’6}\{(x, y) : 0 \leq 9x \leq y^2,\ y \geq 3x - 6\} be AA. Then 6A6A is equal to __________

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Answer: 15