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Day 418: JEE Mains Maths

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

  1. Question 1

    Determinants. Asked once: 24 Jan 2023 (same type asked 19 times) See all 19 of this type

    158
    Let AA be a 3×33 \times 3 matrix such that ∣adj(adj(adjA))∣=124|adj(adj(adjA))| =12^{4} Then ∣A−1adjA∣\left| A^{- 1}adjA \right| is equal to

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  2. Question 2

    Vector Algebra. Asked once: 8 Apr 2023 (same type asked 18 times) See all 18 of this type

    83
    Let a→=6i^+9j^+12k^,b→=αi^+11j^−2k^\overrightarrow{a}= 6\widehat{i}+ 9\widehat{j}+ 12\widehat{k},\overrightarrow{b}= \alpha\widehat{i}+ 11\widehat{j}- 2\widehat{k} and c→\overrightarrow{c} be vectors such that a→×c→=a→×b→\overrightarrow{a}\times\overrightarrow{c}=\overrightarrow{a}\times\overrightarrow{b}. If a→⋅c→=−12\overrightarrow{a}\cdot\overrightarrow{c}= - 12, c→⋅(i^−2j^+k^)=5\overrightarrow{c}\cdot (\widehat{i}- 2\widehat{j}+\widehat{k}) = 5, then c→⋅(i^+j^+k^)\overrightarrow{c}\cdot (\widehat{i}+\widehat{j}+\widehat{k}) is equal to

  3. Question 3

    Limits and Continuity. Asked once: 2021 Paper 18 (same type asked 16 times) See all 16 of this type

    90
    Let a,b∈R,b≠0a,b \in R,b \neq 0, Define a function
    f(x)={asinπ2(x−1), for x≤0tan⁡2x−sin⁡2xbx3, for x>0.f(x) =\left\{ \begin{matrix} asin\frac{\pi}{2}(x- 1), & \text{~for~}x\leq 0 \\ \frac{\tan2x- \sin2x}{bx^{3}}, & \text{~for~}x> 0. \end{matrix} \right.
    If ff is continuous at x=0x = 0, then 10−ab10 - ab is equal to

  4. Question 4

    Three Dimensional Geometry. Asked once: 2021 Paper 18 (same type asked 15 times) See all 15 of this type

    75
    A plane PP contains the line x+2y+3z+1=0=x−y−z−6x + 2y + 3z + 1 = 0 = x - y - z - 6, and is perpendicular to the plane −2x+y+z+8=0- 2x+y+z+ 8 = 0. Then which of the following points lies on PP ?

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  5. Question 5

    Determinants. Asked once: 2021 Paper 21 (same type asked 14 times) See all 14 of this type

    70
    Let [λ]\lbrack\lambda\rbrack be the greatest integer less than or equal to λ\lambda. The set of all values of λ\lambda for which the system of linear equations x+y+z=4,3x+2y+5z=3x + y + z = 4,3x + 2y + 5z = 3, 9x+4y+(28+[λ])z=[λ]9x + 4y + (28 + \lbrack\lambda\rbrack)z = \lbrack\lambda\rbrack has a solution is:

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