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Day 30: JEE Mains Maths

5 questions. Try each one first, then open its answer.

  1. Question 1

    Probability. This type asked 5 times: 2021 Paper 19, 11 Apr 2023, 27 Jan 2024, 29 Jan 2024, 24 Jan 2025. This one: 24 Jan 2025 See all 5 of this type

    61
    A and B alternately throw a pair of dice. A wins if he throws a sum of 5 before B throws a sum of 8 , and BB wins if he throws a sum of 8 before AA throws a sum of 5 . The probability, that A wins if A makes the first throw, is

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  2. Question 2

    Relations and Functions. This type asked 5 times: 24 Jan 2023, 30 Jan 2023, 29 Jan 2024, 31 Jan 2024, 23 Jan 2025. This one: 23 Jan 2025 See all 5 of this type

    61
    Let R={(1,2),(2,3),(3,3)}\mathbf{R = \{(1,2),(2,3),(3,3)\}} be a relation defined on the set {1,2,3,4}\mathbf{\{ 1,2,3,4\}}. Then the minimum number of elements, needed to be added in R so the R becomes an equivalence relation, is :

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  3. Question 3

    Application of Derivatives. This type asked 5 times: 2021 Paper 5, 2021 Paper 22, 14 Jun 2022, 26 Jul 2022, 9 Apr 2024. This one: 9 Apr 2024 See all 5 of this type

    172
    Let the set of all values of pp, for which f(x)=(p2−6p+8)(sin⁡22x−cos⁡22x)+2(2−p)x+7f(x) =\left( p^{2}- 6p + 8 \right)\left( \sin^{2}2x -\cos^{2}2x \right)+ 2(2 - p)x + 7 does not have any critical point, be the interval (a,b)(a,b). Then 16ab16ab is equal to

  4. Question 4

    Complex Numbers. This type asked 5 times: 2021 Paper 12, 2021 Paper 17, 27 Jul 2022, 8 Apr 2023, 8 Apr 2024. This one: 8 Apr 2024 See all 5 of this type

    156
    The sum of all possible values of θ∈[−π,2π]\theta \in \lbrack - \pi,2\pi\rbrack, for which 1+icosθ1−2icosθ\frac{1 + icos\theta}{1 - 2icos\theta} is purely imaginary, is equal to

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  5. Question 5

    Binomial Theorem. This type asked 5 times: 2021 Paper 8, 29 Jan 2023, 8 Apr 2023, 11 Apr 2023, 6 Apr 2024. This one: 6 Apr 2024 See all 5 of this type

    157
    Let 0≤r≤n0 \leq r\leq n. If  n+1Cr+1: nCr: n−1Cr−1=55:35:21\ ^{n+ 1}C_{r+ 1}:\ ^{n}C_{r}:\ ^{n- 1}C_{r- 1}= 55:35:21, then 2n+5r2n+ 5r is equal to:

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