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JEE Main 2025 Jan 24 Shift 2 — Mathematics

25 questions Ā· 25 with the official NTA answer

Q1JEE Main 2025 Jan 24 Shift 2Sets, Relations and FunctionsEasy

The function f:(āˆ’āˆž,āˆž)→(āˆ’āˆž,1)f:(-\infty, \infty) \to (-\infty, 1), defined by f(x)=2xāˆ’2āˆ’x2x+2āˆ’xf(x) = \dfrac{2^x - 2^{-x}}{2^x + 2^{-x}} is :

  1. A

    One-one but not onto

  2. B

    Onto but not one-one

  3. C

    Both one-one and onto

  4. D

    Neither one-one nor onto

Show answer

Correct option: A

Q2JEE Main 2025 Jan 24 Shift 2Sets, Relations and FunctionsHard

Let A={x∈(0,Ļ€)āˆ’{Ļ€2}:log⁔(2/Ļ€)∣sin⁔x∣+log⁔(2/Ļ€)∣cos⁔x∣=2}A = \left\{x \in (0, \pi) - \left\{\dfrac{\pi}{2}\right\} : \log_{(2/\pi)}|\sin x| + \log_{(2/\pi)}|\cos x| = 2\right\} and

B={x≄0:x(xāˆ’4)āˆ’3∣xāˆ’2∣+6=0}.B = \left\{x \geq 0 : \sqrt{x}\left(\sqrt{x} - 4\right) - 3\left|\sqrt{x} - 2\right| + 6 = 0\right\}. Then n(A∪B)\mathrm{n(A \cup B)} is equal to :

  1. A

    2

  2. B

    4

  3. C

    6

  4. D

    8

Show answer

Correct option: D

Q3JEE Main 2025 Jan 24 Shift 2Quadratic EquationsMedium

The number of real solution(s) of the equation x2+3x+2=min⁔{∣xāˆ’3∣,∣x+2∣}x^2 + 3x + 2 = \min\{|x - 3|, |x + 2|\} is :

  1. A

    0

  2. B

    1

  3. C

    2

  4. D

    3

Show answer

Correct option: C

Q4JEE Main 2025 Jan 24 Shift 2Matrices and DeterminantsMedium

For some a, b, let f(x)=∣a+sin⁔xx1ba1+sin⁔xxba1b+sin⁔xx∣f(x) = \begin{vmatrix} a + \dfrac{\sin x}{x} & 1 & b \\ a & 1 + \dfrac{\sin x}{x} & b \\ a & 1 & b + \dfrac{\sin x}{x} \end{vmatrix}, x≠0x \neq 0, lim⁔x→0f(x)=Ī»+μa+νb\lim\limits_{x \to 0} f(x) = \lambda + \mu a + \nu b. Then (Ī»+μ+ν)2(\lambda + \mu + \nu)^2 is equal to :

  1. A

    9

  2. B

    16

  3. C

    25

  4. D

    36

Show answer

Correct option: B

Q5JEE Main 2025 Jan 24 Shift 2Matrices and DeterminantsMedium

If the system of equations

x+2yāˆ’3z=2x + 2y - 3z = 2

2x+λy+5z=52x + \lambda y + 5z = 5

14x+3y+μz=3314x + 3y + \mu z = 33

has infinitely many solutions, then λ+μ\lambda + \mu is equal to :

  1. A

    10

  2. B

    11

  3. C

    12

  4. D

    13

Show answer

Correct option: C

Q6JEE Main 2025 Jan 24 Shift 2Sequences and SeriesMedium

If 7=5+17(5+α)+172(5+2α)+173(5+3α)+ā€¦ā€¦āˆž7 = 5 + \dfrac{1}{7}(5+\alpha) + \dfrac{1}{7^2}(5 + 2\alpha) + \dfrac{1}{7^3}(5 + 3\alpha) + \ldots\ldots \infty, then the value of α\alpha is :

  1. A

    66

  2. B

    67\dfrac{6}{7}

  3. C

    11

  4. D

    17\dfrac{1}{7}

Show answer

Correct option: A

Q7JEE Main 2025 Jan 24 Shift 2Sequences and SeriesEasy

In an arithmetic progression, if S40=1030\mathrm{S_{40}} = 1030 and S12=57\mathrm{S_{12}} = 57, then S30āˆ’S10\mathrm{S_{30}} - \mathrm{S_{10}} is equal to :

  1. A

    505

  2. B

    510

  3. C

    515

  4. D

    525

Show answer

Correct option: C

Q8JEE Main 2025 Jan 24 Shift 2ProbabilityMedium

Let A=[aij]\mathrm{A} = [a_{ij}] be a square matrix of order 2 with entries either 0 or 1. Let E be the event that A is an invertible matrix. Then the probability P(E) is :

  1. A

    38\dfrac{3}{8}

  2. B

    58\dfrac{5}{8}

  3. C

    316\dfrac{3}{16}

  4. D

    18\dfrac{1}{8}

Show answer

Correct option: A

Q9JEE Main 2025 Jan 24 Shift 2Binomial TheoremMedium

Suppose A and B are the coefficients of 30th30^{\text{th}} and 12th12^{\text{th}} terms respectively in the binomial expansion of (1+x)2nāˆ’1(1+x)^{2n-1}. If 2A=5B2\mathrm{A} = 5\mathrm{B}, then nn is equal to :

  1. A

    19

  2. B

    20

  3. C

    21

  4. D

    22

Show answer

Correct option: C

Q10JEE Main 2025 Jan 24 Shift 2Permutations and CombinationsMedium

Group A consists of 7 boys and 3 girls, while group B consists of 6 boys and 5 girls. The number of ways, 4 boys and 4 girls can be invited for a picnic if 5 of them must be from group A and the remaining 3 from group B, is equal to :

  1. A

    8575

  2. B

    8750

  3. C

    8925

  4. D

    9100

Show answer

Correct option: C

Q11JEE Main 2025 Jan 24 Shift 2Straight LinesMedium

Let the points (112,α)\left(\dfrac{11}{2}, \alpha\right) lie on or inside the triangle with sides x+y=11x + y = 11, x+2y=16x + 2y = 16 and 2x+3y=292x + 3y = 29. Then the product of the smallest and the largest values of α\alpha is equal to :

  1. A

    22

  2. B

    33

  3. C

    44

  4. D

    55

Show answer

Correct option: B

Q12JEE Main 2025 Jan 24 Shift 2Conic SectionsMedium

If the equation of the parabola with vertex V(32,3)\mathrm{V}\left(\dfrac{3}{2}, 3\right) and the directrix x+2y=0x + 2y = 0 is αx2+βy2āˆ’Ī³xyāˆ’30xāˆ’60y+225=0\alpha x^2 + \beta y^2 - \gamma xy - 30x - 60y + 225 = 0, then α+β+γ\alpha + \beta + \gamma is equal to :

  1. A

    6

  2. B

    7

  3. C

    8

  4. D

    9

Show answer

Correct option: D

Q13JEE Main 2025 Jan 24 Shift 2Conic SectionsEasy

The equation of the chord, of the ellipse x225+y216=1\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1, whose mid-point is (3,1)(3, 1) is :

  1. A

    25x+101y=17625x + 101y = 176

  2. B

    4x+122y=1344x + 122y = 134

  3. C

    5x+16y=315x + 16y = 31

  4. D

    48x+25y=16948x + 25y = 169

Show answer

Correct option: D

Q14JEE Main 2025 Jan 24 Shift 2Inverse Trigonometric FunctionsMedium

If α>β>γ>0\alpha > \beta > \gamma > 0, then the expression

cotā”āˆ’1{β+(1+β2)(Ī±āˆ’Ī²)}+cotā”āˆ’1{γ+(1+γ2)(Ī²āˆ’Ī³)}+cotā”āˆ’1{α+(1+α2)(Ī³āˆ’Ī±)}\cot^{-1}\left\{\beta + \dfrac{(1 + \beta^2)}{(\alpha - \beta)}\right\} + \cot^{-1}\left\{\gamma + \dfrac{(1 + \gamma^2)}{(\beta - \gamma)}\right\} + \cot^{-1}\left\{\alpha + \dfrac{(1 + \alpha^2)}{(\gamma - \alpha)}\right\} is equal to :

  1. A

    3Ļ€3\pi

  2. B

    00

  3. C

    Ļ€2āˆ’(α+β+γ)\dfrac{\pi}{2} - (\alpha + \beta + \gamma)

  4. D

    π\pi

Show answer

Correct option: D

Q15JEE Main 2025 Jan 24 Shift 2Vector AlgebraMedium

Let aāƒ—=3i^āˆ’j^+2k^\vec{a} = 3\hat{i} - \hat{j} + 2\hat{k}, bāƒ—=aāƒ—Ć—(i^āˆ’2k^)\vec{b} = \vec{a} \times \left(\hat{i} - 2\hat{k}\right) and cāƒ—=bāƒ—Ć—k^\vec{c} = \vec{b} \times \hat{k}. Then the projection of cāƒ—āˆ’2j^\vec{c} - 2\hat{j} on aāƒ—\vec{a} is :

  1. A

    14\sqrt{14}

  2. B

    2142\sqrt{14}

  3. C

    272\sqrt{7}

  4. D

    373\sqrt{7}

Show answer

Correct option: B

Q16JEE Main 2025 Jan 24 Shift 2Vector AlgebraHard

Let the position vectors of three vertices of a triangle be 4pāƒ—+qāƒ—āˆ’3rāƒ—4\vec{p} + \vec{q} - 3\vec{r}, āˆ’5pāƒ—+qāƒ—+2rāƒ—-5\vec{p} + \vec{q} + 2\vec{r} and 2pāƒ—āˆ’qāƒ—+2rāƒ—2\vec{p} - \vec{q} + 2\vec{r}. If the position vectors of the orthocenter and the circumcenter of the triangle are pāƒ—+qāƒ—+rāƒ—4\dfrac{\vec{p} + \vec{q} + \vec{r}}{4} and αpāƒ—+βqāƒ—+γrāƒ—\alpha\vec{p} + \beta\vec{q} + \gamma\vec{r} respectively, then α+2β+5γ\alpha + 2\beta + 5\gamma is equal to :

  1. A

    1

  2. B

    3

  3. C

    4

  4. D

    6

Show answer

Correct option: B

Q17JEE Main 2025 Jan 24 Shift 2Limits, Continuity and DifferentiabilityMedium

Let [x][x] denote the greatest integer function, and let m and n respectively be the numbers of the points, where the function f(x)=[x]+∣xāˆ’2∣f(x) = [x] + |x - 2|, āˆ’2<x<3-2 < x < 3, is not continuous and not differentiable. Then m+n\mathrm{m + n} is equal to :

  1. A

    6

  2. B

    7

  3. C

    8

  4. D

    9

Show answer

Correct option: C

Q18JEE Main 2025 Jan 24 Shift 2Differentiation and Applications of DerivativesHard

Let (2,3)(2, 3) be the largest open interval in which the function f(x)=2log⁔e(xāˆ’2)āˆ’x2+ax+1f(x) = 2\log_e(x - 2) - x^2 + \mathrm{a}x + 1 is strictly increasing and (b,c)(\mathrm{b}, \mathrm{c}) be the largest open interval, in which the function g(x)=(xāˆ’1)3(x+2āˆ’a)2\mathrm{g}(x) = (x-1)^3(x + 2 - \mathrm{a})^2 is strictly decreasing. Then 100(a+bāˆ’c)100(\mathrm{a} + \mathrm{b} - \mathrm{c}) is equal to :

  1. A

    160

  2. B

    280

  3. C

    360

  4. D

    420

Show answer

Correct option: C

Q19JEE Main 2025 Jan 24 Shift 2Integral CalculusMedium

The area of the region enclosed by the curves y=exy = e^x, y=∣exāˆ’1∣y = |e^x - 1| and y-axis is :

  1. A

    log⁔e2\log_e 2

  2. B

    1+log⁔e21 + \log_e 2

  3. C

    1āˆ’log⁔e21 - \log_e 2

  4. D

    2log⁔e2āˆ’12\log_e 2 - 1

Show answer

Correct option: C

Q20JEE Main 2025 Jan 24 Shift 2Differential EquationsMedium

Let f:(0,āˆž)→Rf : (0, \infty) \to \mathbf{R} be a function which is differentiable at all points of its domain and satisfies the condition x2f′(x)=2xf(x)+3x^2 f'(x) = 2xf(x) + 3, with f(1)=4f(1) = 4. Then 2f(2)2f(2) is equal to :

  1. A

    39

  2. B

    19

  3. C

    23

  4. D

    29

Show answer

Correct option: A

Q21JEE Main 2025 Jan 24 Shift 2Permutations and CombinationsMediumNumerical

Number of functions f:{1,2,…,100}→{0,1}f : \{1, 2, \ldots, 100\} \to \{0, 1\}, that assign 1 to exactly one of the positive integers less than or equal to 98, is equal to __________.

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Answer: 392

Q22JEE Main 2025 Jan 24 Shift 2Three Dimensional GeometryMediumNumerical

Let P be the image of the point Q(7,āˆ’2,5)\mathrm{Q}(7, -2, 5) in the line L:xāˆ’12=y+13=z4\mathrm{L} : \dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z}{4} and R(5,p,q)\mathrm{R}(5, \mathrm{p}, \mathrm{q}) be a point on L. Then the square of the area of Ī”PQR\Delta\mathrm{PQR} is __________.

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Answer: 957

Q23JEE Main 2025 Jan 24 Shift 2Integral CalculusMediumNumerical

If ∫2x2+5x+9x2+x+1 dx=xx2+x+1+αx2+x+1+βlog⁔e∣x+12+x2+x+1∣+C\displaystyle\int \dfrac{2x^2 + 5x + 9}{\sqrt{x^2 + x + 1}}\, \mathrm{d}x = x\sqrt{x^2 + x + 1} + \alpha\sqrt{x^2 + x + 1} + \beta \log_e\left|x + \dfrac{1}{2} + \sqrt{x^2 + x + 1}\right| + \mathrm{C},

where C is the constant of integration, then α+2β\alpha + 2\beta is equal to __________.

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Answer: 16

Q24JEE Main 2025 Jan 24 Shift 2Differential EquationsMediumNumerical

Let y=y(x)y = y(x) be the solution of the differential equation 2cos⁔xdydx=sin⁔2xāˆ’4ysin⁔x2\cos x \dfrac{\mathrm{d}y}{\mathrm{d}x} = \sin 2x - 4y\sin x, x∈(0,Ļ€2)x \in \left(0, \dfrac{\pi}{2}\right). If y(Ļ€3)=0y\left(\dfrac{\pi}{3}\right) = 0, then y′(Ļ€4)+y(Ļ€4)y'\left(\dfrac{\pi}{4}\right) + y\left(\dfrac{\pi}{4}\right) is equal to __________.

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Answer: 1

Q25JEE Main 2025 Jan 24 Shift 2Conic SectionsMediumNumerical

Let H1:x2a2āˆ’y2b2=1\mathrm{H_1} : \dfrac{x^2}{\mathrm{a}^2} - \dfrac{y^2}{\mathrm{b}^2} = 1 and H2:āˆ’x2A2+y2B2=1\mathrm{H_2} : -\dfrac{x^2}{\mathrm{A}^2} + \dfrac{y^2}{\mathrm{B}^2} = 1 be two hyperbolas having length of latus rectums 15215\sqrt{2} and 12512\sqrt{5} respectively. Let their ecentricities be e1=52\mathrm{e_1} = \sqrt{\dfrac{5}{2}} and e2\mathrm{e_2} respectively. If the product of the lengths of their transverse axes is 10010100\sqrt{10}, then 25e2225\mathrm{e_2^2} is equal to __________.

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Answer: 55