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Day 397: JEE Mains Maths

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

  1. Question 1

    Limits and Continuity. Asked once: 27 Jan 2024 (same type asked 16 times) See all 16 of this type

    73
    Consider the function.
    f(x)={a(7x−12−x2)b∣x2−7x+12∣,x<32sin⁡(x−3)x−[x],x>3b, x=3f(x) =\left\{ \begin{matrix} \frac{a\left( 7x - 12 -x^{2} \right)}{b\left| x^{2}- 7x + 12 \right|} & ,x < 3 \\ 2^{\frac{\sin(x - 3)}{x - \lbrack x\rbrack}} & ,x > 3 \\ b & ,\ x = 3 \end{matrix} \right.
    Where [x]\lbrack x\rbrack denotes the greatest integer less than or equal to xx. If SS denotes the set of all ordered pairs (a,b)(a,b) such that f(x)f(x) is continuous at x=3x = 3, then the number of elements in SS is :

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  2. Question 2

    Three Dimensional Geometry. Asked once: 28 Jun 2022 (same type asked 15 times) See all 15 of this type

    76
    The acute angle between the planes P1P_{1} and P2P_{2}, when P1P_{1} and P2P_{2} are the planes passing through the intersection of the planes 5x+8y+13z−29=05x + 8y + 13z - 29 = 0 and 8x−7y+z−20=08x - 7y + z - 20 = 0 and the points (2,1,3)(2,1,3) and (0,1,2)(0,1,2), respectively, is

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  3. Question 3

    Determinants. Asked once: 26 Jun 2022 (same type asked 14 times) See all 14 of this type

    64
    The ordered pair (a,b)(a,b), for which the system of linear equations
    3x−2y+z=b3x - 2y + z = b
    5x−8y+9z=35x - 8y + 9z = 3
    2x+y+az=−12x + y + az = - 1
    has no solution, is:

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  4. Question 4

    Statistics. Asked once: 27 Jun 2022 (same type asked 13 times) See all 13 of this type

    166
    The mean and variance of the data 4,5,6,6,7,8,x,y4,5,6,6,7,8,x,y where x<yx < y are 6 , and 94\frac{9}{4} respectively. Then x4+y2x^{4}+y^{2} is equal to

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  5. Question 5

    Vector Algebra. Asked once: 29 Jun 2022 (same type asked 13 times) See all 13 of this type

    171
    Let a→=i^−2j^+3k^,b→=i^+j^+k^\overrightarrow{a}=\widehat{i}- 2\widehat{j}+ 3\widehat{k},\overrightarrow{b}=\widehat{i}+\widehat{j}+\widehat{k} and c→\overrightarrow{c} be a vector such that a→+(b→×c→)=0→\overrightarrow{a}+ (\overrightarrow{b}\times\overrightarrow{c}) =\overrightarrow{0} and b→⋅c→=5\overrightarrow{b}\cdot\overrightarrow{c}= 5. Then, the value of 3(c→⋅a→)3(\overrightarrow{c}\cdot\overrightarrow{a}) is equal to