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Day 262: 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

    67
    Let the domain of the function f(x)=log⁡2log⁡4log⁡6(3+4x−x2)f(x) =\log_{2}\log_{4}\log_{6}\left( 3 + 4x -x^{2} \right) be (a,b)(a,b). If ∫0b−a[x2]dx=p−q−r,p,q,r∈N,gcd(p,q,r)=1\int_{0}^{b-a} \left\lbrack x^{2} \right\rbrack dx=p-\sqrt{q}-\sqrt{r},p,q,r\in\mathbb{N},gcd(p,q,r) = 1, where [⋅]\lbrack \cdot \rbrack is the greatest integer function, then p+q+rp + q + r is equal to.

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  2. Question 2

    Vector Algebra. Asked once: 4 Apr 2024 (same type asked 4 times) See all 4 of this type

    169
    Let a→=i^+j^+k^,b→=2i^+4j^−5k^\overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k},\overrightarrow{b}= 2\widehat{i}+ 4\widehat{j}- 5\widehat{k} and c→=xi^+2j^+3k^,x∈R\overrightarrow{c}= x\widehat{i}+ 2\widehat{j}+ 3\widehat{k},x \in R. If d→\overrightarrow{d} is the unit vector in the direction of b→+c→\overrightarrow{b}+\overrightarrow{c} such that a→⋅d→=1\overrightarrow{a}\cdot\overrightarrow{d}= 1, then (a→×b→)⋅c→(\overrightarrow{a}\times\overrightarrow{b}) \cdot\overrightarrow{c} is equal to

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  3. Question 3

    Conic Sections. Asked once: 2021 Paper 18 (same type asked 4 times) See all 4 of this type

    63
    On the ellipse x28+y24=1\frac{x^{2}}{8}+\frac{y^{2}}{4}= 1 let PP be a point in the second quadrant such that the tangent at PP to the ellipse is perpendicular to the line x+2y=0x + 2y = 0. Let SS and S′S^{'} be the foci of the ellipse and ee be its eccentricity. If A is the area of the triangle SPS' then, the value of (5−e2)\left( 5 -e^{2} \right). AA is:

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  4. Question 4

    Differentiation. Asked once: 24 Jan 2023 (same type asked 4 times) See all 4 of this type

    151
    If f(x)=x3−x2f′(1)+xf′′(2)−f′′′(3), x∈Rf(x)=x^{3}-x^{2}f'(1)+xf''(2)-f'''(3),\ x\in R, then

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  5. Question 5

    Complex Numbers. Asked once: 26 Jun 2022 (same type asked 4 times) See all 4 of this type

    180
    If z2+z+1=0,z∈Cz^{2}+ z + 1 = 0,z \in\mathbb{C}, then ∣∑n=115(Zn+(−1)n1Zn)2∣\left| \sum_{n = 1}^{15} \left( Z^{n}+ ( - 1)^{n}\frac{1}{Z^{n}} \right)^{2} \right| is equal to