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Day 114: JEE Mains Maths

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

  1. Question 1

    Definite Integration. This type asked 2 times: 27 Jan 2024, 31 Jan 2024. This one: 31 Jan 2024 See all 2 of this type

    90
    Let f:R→Rf:R \rightarrow R be a function defined by f(x)=4x4x+2f(x) =\frac{4^{x}}{4^{x}+ 2} and M=∫f(a)f(1−a)xsin⁡4(x(1−x))dxM =\int_{f(a)}^{f(1 - a)} x\sin^{4}(x(1 - x))dx, N=∫f(a)f(1−a)sin⁡4(x(1−x))dx;a≠12N =\int_{f(a)}^{f(1 - a)} \sin^{4}(x(1 - x))dx;a \neq\frac{1}{2}. If αM=βN,α,β∈N\alpha M = \beta N,\alpha,\beta \in N, then the least value of α2+β2\alpha^{2}+\beta^{2} is equal to

  2. Question 2

    Straight Lines. This type asked 2 times: 29 Jan 2024, 31 Jan 2024. This one: 31 Jan 2024 See all 2 of this type

    152
    Let A(a,b),B(3,4)A(a,b),B(3,4) and (−6,−8)( - 6, - 8) respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point P(2a+3,7 b+5)P(2a + 3,7\text{ }b + 5) from the line 2x+3y−4=02x + 3y - 4 = 0 measured parallel to the line x−2y−1=0x - 2y - 1 = 0 is

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  3. Question 3

    Relations and Functions. This type asked 2 times: 30 Jan 2023, 30 Jan 2024. This one: 30 Jan 2024 See all 2 of this type

    84
    Let A={1,2,3,….7}\mathbf{A = \{ 1,2,3,\ldots.7\}} and let P(a)P(a) denote the power set of A. If the number of functions f:A→P(A)f:A \rightarrow P(A) such that a∈f(a),∀a∈Aa \in f(a),\forall a \in A is mn,mm^{n},m and n∈Nn\in N and mm is least, then m+nm+n is equal to

  4. Question 4

    Trigonometric Equations. This type asked 2 times: 27 Jan 2024, 30 Jan 2024. This one: 30 Jan 2024 See all 2 of this type

    79
    If 2sin⁡3x+sin⁡2xcos⁡x+4sin⁡x−4=02\sin^{3}x+ \sin2x\cos x+ 4\sin x- 4 = 0 has exactly 3 solutions in the interval [0,nπ2],n∈N\left\lbrack 0,\frac{n\pi}{2} \right\rbrack,n \in N, then the roots of the equation x2+nx+(n−3)=0x^{2}+nx+ (n- 3) = 0 belong to :

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  5. Question 5

    Complex Numbers. This type asked 2 times: 8 Apr 2023, 29 Jan 2024. This one: 29 Jan 2024 See all 2 of this type

    65
    If z=12−2iz =\frac{1}{2}- 2i, is such that ∣z+1∣=αz+β(1+i),i=−1|z + 1| = \alpha z + \beta(1 + i),i =\sqrt{- 1} and  α,β∈R ,\ \alpha,\beta \in R\ , then α+β\alpha + \beta is equal to

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