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Day 19: JEE Mains Maths

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

  1. Question 1

    Permutations and Combinations. This type asked 6 times: 2021 Paper 25, 25 Jun 2022, 13 Apr 2023, 9 Apr 2024, 28 Jan 2025, 29 Jan 2025. This one: 29 Jan 2025 See all 6 of this type

    56
    Let PP be the set of seven -digit numbers with sum of their digits equal to 11 . If the numbers in P are formed by using the digits 1,2 and 3 only, then the number of elements in the set PP is :

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  2. Question 2

    Inverse Trigonometric Functions. This type asked 6 times: 2021 Paper 19, 2021 Paper 23, 29 Jun 2022, 28 Jul 2022, 10 Apr 2023, 28 Jan 2025. This one: 28 Jan 2025 See all 6 of this type

    140
    Let [x]\lbrack x\rbrack denote the greatest integer less than or equal to x . Then domain of f(x)=sec⁡−1(2[x]+1)f(x) =\sec^{- 1}(2\lbrack x\rbrack + 1) is :

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  3. Question 3

    Relations and Functions. This type asked 6 times: 31 Jan 2023, 29 Jan 2024, 1 Feb 2024, 4 Apr 2024, 8 Apr 2024, 23 Jan 2025. This one: 23 Jan 2025 See all 6 of this type

    139
    Let X=R×RX = R \times R. Define a relation R on X as:
    (a1,b1)R(a2,b2)⇔b1=b2.\left( a_{1},b_{1} \right)R\left( a_{2},b_{2} \right)\Leftrightarrow b_{1}=b_{2}.
    Statement-I: R is an equivalence relation. Statement-II: For some (a,b)∈X(a,b) \in X, the set S={(x,y)∈X:(x,y)R(a,b)}\mathbf{S = \{(x,y) \in X:(x,y)R(a,b)\}} represents a line parallel to y=xy = x. In the light of the above statements, choose the correct answer from the options given below:

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  4. Question 4

    Differential Equations. This type asked 6 times: 2021 Paper 21, 14 Jun 2022, 27 Jun 2022, 8 Apr 2023, 15 Apr 2023, 22 Jan 2025. This one: 22 Jan 2025 See all 6 of this type

    135
    If x=f(y)x = f(y) is the solution of the differential equation
    (1+y2)+(x−2etan⁡−1y)dydx=0,y∈(−π2,π2)\left( 1 +y^{2} \right)+\left( x - 2e^{\tan^{- 1}y} \right)\frac{dy}{dx}= 0,y \in\left( -\frac{\pi}{2},\frac{\pi}{2} \right)
    with f(0)=1f(0) = 1, then f(13)f\left( \frac{1}{\sqrt{3}} \right) is equal to :

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  5. Question 5

    Definite Integration. This type asked 6 times: 2021 Paper 5, 14 Jun 2022, 26 Jul 2022, 30 Jan 2023, 30 Jan 2024, 9 Apr 2024. This one: 9 Apr 2024 See all 6 of this type

    84
    Let
    lim⁡n→∞(nn4+1−2n(n2+1)n4+1+nn4+16−8n(n2+4)n4+16\lim_{n \rightarrow \infty} \left( \frac{n}{\sqrt{n^{4}+ 1}}-\frac{2n}{\left( n^{2}+ 1 \right)\sqrt{n^{4}+ 1}}+\frac{n}{\sqrt{n^{4}+ 16}}-\frac{8n}{\left( n^{2}+ 4 \right)\sqrt{n^{4}+ 16}} \right.
    +……+nn4+n4−2n⋅n2(n2+n2)n4+n4+ \ldots\ldots +\frac{n}{\sqrt{n^{4}+n^{4}}}-\frac{2n \cdot n^{2}}{\left( n^{2}+n^{2} \right)\sqrt{n^{4}+n^{4}}} be πk\frac{\pi}{k}, using only the principal values of the inverse trigonometric functions. Then k2k^{2} is equal to............