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Day 323: JEE Mains Maths

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

  1. Question 1

    Relations and Functions. Asked once: 3 Apr 2025 (same type asked 4 times) See all 4 of this type

    137
    If the domain of the function f(x)=log⁡7(1−log⁡4(x2−9x+18))f(x) =\log_{7}\left( 1 -\log_{4}\left( x^{2}- 9x+ 18 \right) \right) is (α,β)∪(γ,δ)(\alpha,\beta) \cup (\gamma,\delta), then α+β+γ+δ\alpha+\beta+\gamma+\delta is equal to.

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  2. Question 2

    Vector Algebra. Asked once: 2021 Paper 19 (same type asked 4 times) See all 4 of this type

    83
    If the projection of the vector i^+2j^+k^\widehat{i}+ 2\widehat{j}+\widehat{k} on the sum of the two vectors 2i^+4j^−5k^2\widehat{i}+ 4\widehat{j}- 5\widehat{k} and −λi^+2j^+3k^-\lambda\widehat{i}+ 2\widehat{j}+ 3\widehat{k} is 1 , then λ\lambda is equal to

  3. Question 3

    Conic Sections. Asked once: 2021 Paper 17 (same type asked 4 times) See all 4 of this type

    84
    Let EE be an ellipse whose axes are parallel to the co-ordinate's axes, having its center at (3,−4)(3, - 4), one focus at (4,−4)(4, - 4) and one vertex at (5,−4)(5, - 4). If mx−y=4, m>0mx - y = 4,\text{ }m > 0 is a tangent to the ellipse EE, then the value of 5 m25{\text{ }m}^{2} is equal to

  4. Question 4

    Conic Sections. Asked once: 2021 Paper 4 (same type asked 4 times) See all 4 of this type

    69
    A hyperbola passes through the foci of the ellipse x225+y216=1\frac{x^{2}}{25}+\frac{y^{2}}{16}= 1 and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is:

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  5. Question 5

    Conic Sections. Asked once: 8 Apr 2024 (same type asked 4 times) See all 4 of this type

    80
    Let H:−x2a2+y2 b2=1H:\frac{-x^{2}}{a^{2}}+\frac{y^{2}}{{\text{ }b}^{2}}= 1 be the hyperbola, whose eccentricity is 3\sqrt{3} and the length of the latus rectum is 434\sqrt{3}. Suppose the point (α,6),α>0(\alpha,6),\alpha> 0 lies on H. If β\beta is the product of the focal distances of the point (α,6)(\alpha,6), then α2+β\alpha^{2}+\beta is equal to:

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