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JEE Main 2025 Jan 23 Shift 1 — Mathematics

23 questions Ā· 23 with the official NTA answer

Q1JEE Main 2025 Jan 23 Shift 1Sets, Relations and FunctionsMedium

Let f(x)=log⁔exf(x) = \log_e x and g(x)=x4āˆ’2x3+3x2āˆ’2x+22x2āˆ’2x+1g(x) = \frac{x^4 - 2x^3 + 3x^2 - 2x + 2}{2x^2 - 2x + 1}. Then the domain of fogfog is

  1. A

    R\mathbb{R}

  2. B

    [0,āˆž)[0, \infty)

  3. C

    (0,āˆž)(0, \infty)

  4. D

    [1,āˆž)[1, \infty)

Show answer

Correct option: A

Q2JEE Main 2025 Jan 23 Shift 1Sets, Relations and FunctionsMedium

Let R={(1,2),(2,3),(3,3)}R = \{(1, 2), (2, 3), (3, 3)\} be a relation defined on the set {1,2,3,4}\{1, 2, 3, 4\}. Then the minimum number of elements, needed to be added in RR so that RR becomes an equivalence relation, is:

  1. A

    77

  2. B

    88

  3. C

    99

  4. D

    1010

Show answer

Correct option: A

Q3JEE Main 2025 Jan 23 Shift 1Complex NumbersMedium

Let ∣zĖ‰āˆ’i2zˉ+i∣=13\left|\frac{\bar{z} - i}{2\bar{z} + i}\right| = \frac{1}{3}, z∈Cz \in \mathbb{C}, be the equation of a circle with center at CC. If the area of the triangle, whose vertices are at the points (0,0)(0, 0), CC and (α,0)(\alpha, 0) is 11 square units, then α2\alpha^2 equals:

  1. A

    5050

  2. B

    100100

  3. C

    8125\frac{81}{25}

  4. D

    12125\frac{121}{25}

Show answer

Correct option: B

Q4JEE Main 2025 Jan 23 Shift 1Matrices and DeterminantsMedium

If the system of equations

(Ī»āˆ’1)x+(Ī»āˆ’4)y+Ī»z=5(\lambda - 1)x + (\lambda - 4)y + \lambda z = 5

Ī»x+(Ī»āˆ’1)y+(Ī»āˆ’4)z=7\lambda x + (\lambda - 1)y + (\lambda - 4)z = 7

(Ī»+1)x+(Ī»+2)yāˆ’(Ī»+2)z=9(\lambda + 1)x + (\lambda + 2)y - (\lambda + 2)z = 9

has infinitely many solutions, then λ2+λ\lambda^2 + \lambda is equal to

  1. A

    66

  2. B

    1010

  3. C

    2020

  4. D

    1212

Show answer

Correct option: D

Q5JEE Main 2025 Jan 23 Shift 1Matrices and DeterminantsMedium

If AA, BB, and (adj(Aāˆ’1)+adj(Bāˆ’1))(\mathrm{adj}(A^{-1}) + \mathrm{adj}(B^{-1})) are non-singular matrices of same order, then the inverse of A(adj(Aāˆ’1)+adj(Bāˆ’1))āˆ’1BA(\mathrm{adj}(A^{-1}) + \mathrm{adj}(B^{-1}))^{-1}B, is equal to

  1. A

    ABāˆ’1+Aāˆ’1BAB^{-1} + A^{-1}B

  2. B

    adj(Bāˆ’1)+adj(Aāˆ’1)\mathrm{adj}(B^{-1}) + \mathrm{adj}(A^{-1})

  3. C

    1∣AB∣(adj(B)+adj(A))\frac{1}{|AB|}(\mathrm{adj}(B) + \mathrm{adj}(A))

  4. D

    ABāˆ’1∣A∣+BAāˆ’1∣B∣\frac{AB^{-1}}{|A|} + \frac{BA^{-1}}{|B|}

Show answer

Correct option: C

Q6JEE Main 2025 Jan 23 Shift 1Sequences and SeriesMedium

If the first term of an A.P. is 3 and the sum of its first four terms is equal to one-fifth of the sum of the next four terms, then the sum of the first 20 terms is equal to

  1. A

    āˆ’1080-1080

  2. B

    āˆ’1020-1020

  3. C

    āˆ’120-120

  4. D

    āˆ’1200-1200

Show answer

Correct option: A

Q7JEE Main 2025 Jan 23 Shift 1Permutations and CombinationsEasy

The number of words, which can be formed using all the letters of the word ā€œDAUGHTERā€, so that all the vowels never come together, is

  1. A

    3400034000

  2. B

    3500035000

  3. C

    3600036000

  4. D

    3700037000

Show answer

Correct option: C

Q8JEE Main 2025 Jan 23 Shift 1ProbabilityMedium

One die has two faces marked 1, two faces marked 2, one face marked 3 and one face marked 4. Another die has one face marked 1, two faces marked 2, two faces marked 3 and one face marked 4. The probability of getting the sum of numbers to be 4 or 5, when both the dice are thrown together, is

  1. A

    23\frac{2}{3}

  2. B

    12\frac{1}{2}

  3. C

    49\frac{4}{9}

  4. D

    35\frac{3}{5}

Show answer

Correct option: B

Q9JEE Main 2025 Jan 23 Shift 1Straight LinesMedium

Let the area of a ā–³PQR\triangle PQR with vertices P(5,4)P(5, 4), Q(āˆ’2,4)Q(-2, 4) and R(a,b)R(a, b) be 35 square units. If its orthocenter and centroid are O(2,145)O\left(2, \frac{14}{5}\right) and C(c,d)C(c, d) respectively, then c+2dc + 2d is equal to

  1. A

    22

  2. B

    73\frac{7}{3}

  3. C

    83\frac{8}{3}

  4. D

    33

Show answer

Correct option: D

Q10JEE Main 2025 Jan 23 Shift 1Inverse Trigonometric FunctionsMedium

If Ļ€2≤x≤3Ļ€4\frac{\pi}{2} \le x \le \frac{3\pi}{4}, then cosā”āˆ’1(1213cos⁔x+513sin⁔x)\cos^{-1}\left(\frac{12}{13}\cos x + \frac{5}{13}\sin x\right) is equal to

  1. A

    xāˆ’tanā”āˆ’143x - \tan^{-1}\frac{4}{3}

  2. B

    x+tanā”āˆ’145x + \tan^{-1}\frac{4}{5}

  3. C

    xāˆ’tanā”āˆ’1512x - \tan^{-1}\frac{5}{12}

  4. D

    x+tanā”āˆ’1512x + \tan^{-1}\frac{5}{12}

Show answer

Correct option: C

Q11JEE Main 2025 Jan 23 Shift 1Trigonometric Ratios and EquationsMedium

The value of (sin⁔70°)(cot⁔10°cot⁔70Ā°āˆ’1)(\sin 70°)(\cot 10° \cot 70° - 1) is

  1. A

    00

  2. B

    11

  3. C

    3/23/2

  4. D

    2/32/3

Show answer

Correct option: B

Q12JEE Main 2025 Jan 23 Shift 1Conic SectionsMedium

If the line 3xāˆ’2y+12=03x - 2y + 12 = 0 intersects the parabola 4y=3x24y = 3x^2 at the points AA and BB, then at the vertex of the parabola, the line segment ABAB subtends an angle equal to

  1. A

    tanā”āˆ’1(45)\tan^{-1}\left(\frac{4}{5}\right)

  2. B

    Ļ€2āˆ’tanā”āˆ’1(32)\frac{\pi}{2} - \tan^{-1}\left(\frac{3}{2}\right)

  3. C

    tanā”āˆ’1(97)\tan^{-1}\left(\frac{9}{7}\right)

  4. D

    tanā”āˆ’1(119)\tan^{-1}\left(\frac{11}{9}\right)

Show answer

Correct option: C

Q13JEE Main 2025 Jan 23 Shift 1Vector AlgebraMedium

Let the arc ACAC of a circle subtend a right angle at the centre OO. If the point BB on the arc ACAC, divides the arc ACAC such that lengthĀ ofĀ arcĀ ABlengthĀ ofĀ arcĀ BC=15\frac{\text{length of arc } AB}{\text{length of arc } BC} = \frac{1}{5}, and OCāƒ—=αOAāƒ—+βOBāƒ—\vec{OC} = \alpha \vec{OA} + \beta \vec{OB}, then α+2(3āˆ’1)β\alpha + \sqrt{2}(\sqrt{3} - 1)\beta is equal to

  1. A

    232\sqrt{3}

  2. B

    2+32 + \sqrt{3}

  3. C

    2āˆ’32 - \sqrt{3}

  4. D

    535\sqrt{3}

Show answer

Correct option: C

Q14JEE Main 2025 Jan 23 Shift 1Vector AlgebraHard

Let the position vectors of the vertices AA, BB and CC of a tetrahedron ABCDABCD be i^+2j^+k^\hat{i} + 2\hat{j} + \hat{k}, i^+3j^āˆ’2k^\hat{i} + 3\hat{j} - 2\hat{k} and 2i^+j^āˆ’k^2\hat{i} + \hat{j} - \hat{k} respectively. The altitude from the vertex DD to the opposite face ABCABC meets the median line segment through AA of the triangle ABCABC at the point EE. If the length of ADAD is 1103\frac{\sqrt{110}}{3} and the volume of the tetrahedron is 80562\frac{\sqrt{805}}{6\sqrt{2}}, then the position vector of EE is [Note: stem recovered from a low-resolution embedded thumbnail; the English and Hindi stem images were destroyed by PDF corruption — numeric values should be verified against another copy of this paper]

  1. A

    112(7i^+4j^+3k^)\frac{1}{12}(7\hat{i} + 4\hat{j} + 3\hat{k})

  2. B

    16(7i^+12j^+k^)\frac{1}{6}(7\hat{i} + 12\hat{j} + \hat{k})

  3. C

    16(12i^+12j^+k^)\frac{1}{6}(12\hat{i} + 12\hat{j} + \hat{k})

  4. D

    12(i^+4j^+7k^)\frac{1}{2}(\hat{i} + 4\hat{j} + 7\hat{k})

Show answer

Correct option: B

Q15JEE Main 2025 Jan 23 Shift 1Three Dimensional GeometryMedium

Let PP be the foot of the perpendicular from the point Q(10,āˆ’3,āˆ’1)Q(10, -3, -1) on the line xāˆ’37=yāˆ’2āˆ’1=z+1āˆ’2\frac{x-3}{7} = \frac{y-2}{-1} = \frac{z+1}{-2}. Then the area of the right angled triangle PQRPQR, where RR is the point (3,āˆ’2,1)(3, -2, 1), is

  1. A

    30\sqrt{30}

  2. B

    8158\sqrt{15}

  3. C

    9159\sqrt{15}

  4. D

    3303\sqrt{30}

Show answer

Correct option: D

Q16JEE Main 2025 Jan 23 Shift 1Limits, Continuity and DifferentiabilityMedium

If the function

f(x)={2x{sin⁔(k1+1)x+sin⁔(k2āˆ’1)x},x<04,x=02xlog⁔e(2+k1x2+k2x),x>0f(x) = \begin{cases} \frac{2}{x}\{\sin(k_1 + 1)x + \sin(k_2 - 1)x\}, & x < 0 \\ 4, & x = 0 \\ \frac{2}{x}\log_e\left(\frac{2 + k_1 x}{2 + k_2 x}\right), & x > 0 \end{cases}

is continuous at x=0x = 0, then k12+k22k_1^2 + k_2^2 is equal to

  1. A

    55

  2. B

    88

  3. C

    1010

  4. D

    2020

Show answer

Correct option: C

Q17JEE Main 2025 Jan 23 Shift 1StatisticsMedium

The marks obtained by all the students of class 12 are presented in a frequency distribution with classes of equal width. Let the median class interval be 12-18 with median class frequency 12, and the median of these grouped data be 14. If the number of students whose marks are less than 12 is 18, then the total number of students is

  1. A

    4040

  2. B

    4444

  3. C

    4848

  4. D

    5252

Show answer

Correct option: B

Q18JEE Main 2025 Jan 23 Shift 1Integral CalculusMedium

Let I(x)=∫dx(xāˆ’11)1113(x+15)1513I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}}. If I(37)āˆ’I(24)=14(1b113āˆ’1c113)I(37) - I(24) = \frac{1}{4}\left(\frac{1}{b^{\frac{1}{13}}} - \frac{1}{c^{\frac{1}{13}}}\right), b,c∈Nb, c \in \mathbb{N}, then 3(b+c)3(b + c) is equal to

  1. A

    2222

  2. B

    2626

  3. C

    3939

  4. D

    4040

Show answer

Correct option: C

Q19JEE Main 2025 Jan 23 Shift 1Integral CalculusMedium

The value of

∫e2e41x(e((log⁔ex)2+1)āˆ’1e((log⁔ex)2+1)āˆ’1+e((6āˆ’log⁔ex)2+1)āˆ’1)dx\int_{e^2}^{e^4} \frac{1}{x}\left(\frac{e^{((\log_e x)^2 + 1)^{-1}}}{e^{((\log_e x)^2 + 1)^{-1}} + e^{((6 - \log_e x)^2 + 1)^{-1}}}\right) dx is

  1. A

    11

  2. B

    22

  3. C

    log⁔e2\log_e 2

  4. D

    e2e^2

Show answer

Correct option: A

Q20JEE Main 2025 Jan 23 Shift 1Differential EquationsMedium

Let a curve y=f(x)y = f(x) pass through the points (0,5)(0, 5) and (log⁔e2,k)(\log_e 2, k). If the curve satisfies the differential equation 2(3+y)e2xdxāˆ’(7+e2x)dy=02(3 + y)e^{2x}dx - (7 + e^{2x})dy = 0, then kk is equal to

  1. A

    44

  2. B

    88

  3. C

    1616

  4. D

    3232

Show answer

Correct option: B

Q21JEE Main 2025 Jan 23 Shift 1Quadratic EquationsMediumNumerical

If the equation a(bāˆ’c)x2+b(cāˆ’a)x+c(aāˆ’b)=0a(b - c)x^2 + b(c - a)x + c(a - b) = 0 has equal roots, where a+c=15a + c = 15 and b=365b = \frac{36}{5}, then a2+c2a^2 + c^2 is equal to____

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Answer: 117

Q22JEE Main 2025 Jan 23 Shift 1Integral CalculusMediumNumerical

If the area of the larger region bounded between the curves x2+y2=25x^2 + y^2 = 25 and y=∣xāˆ’1∣y = |x - 1| is 14(bĻ€+c)\frac{1}{4}(b\pi + c), b,c∈Nb, c \in \mathbb{N}, then b+cb + c is equal to ________.

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Answer: 77

Q23JEE Main 2025 Jan 23 Shift 1Binomial TheoremMediumNumerical

The sum of all rational terms in the expansion of (1+213+312)6(1 + 2^{\frac{1}{3}} + 3^{\frac{1}{2}})^6 is equal to ______

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Answer: 612