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Day 47: MHT-CET Maths

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

  1. Question 1

    Line and Plane. Asked 4 times: 3 May 2023 S1*, 3 May 2023 S2*, 12 May 2024 S2*, 23 Apr 2025 S1 (* = numbers or wording changed that year) See all 4 times it was asked

    149
    The equation of the plane containing the line x1=y2=z3\frac{x}{1}=\frac{y}{2}=\frac{z}{3} and perpendicular to the plane containing the lines x2=y3=z1\frac{x}{2}=\frac{y}{3}=\frac{z}{1} and x3=y2=z1\frac{x}{3}=\frac{y}{2}=\frac{z}{1} is

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  2. Question 2

    Limits. Asked 4 times: 10 May 2023 S1*, 10 May 2024 S1*, 12 May 2024 S1*, 22 Apr 2025 S1 (* = numbers or wording changed that year) See all 4 times it was asked

    120
    If the function f(x)={x+a2sin⁡x if 0≤x≤π42xcotx+b if π4<x≤π2acos2x−bsinx if π2<x≤πf(x) =\left\{ \begin{matrix} x+a\sqrt{2}\sin x & \text{~if~}0 \leq x\leq\frac{\pi}{4} \\ 2xcotx+b & \text{~if~}\frac{\pi}{4}<x\leq\frac{\pi}{2} \\ acos2x-bsinx & \text{~if~}\frac{\pi}{2}<x\leq\pi \end{matrix} \right. is continuous in [0,π]\lbrack 0,\pi\rbrack then a−b=a - b =

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  3. Question 3

    Indefinite Integration. Asked 4 times: 3 May 2023 S1, 9 May 2023 S1, 13 May 2024 S2*, 20 Apr 2025 S2 (* = numbers or wording changed that year) See all 4 times it was asked

    132
    ∫ex(x−1)(x+1)3 dx=\int e^{x}\frac{(x- 1)}{(x+ 1)^{3}}\text{ }dx=

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  4. Question 4

    Probability Distribution. Asked 4 times: 2 May 2023 S1, 11 May 2023 S2, 16 May 2023 S2, 20 Apr 2025 S2 See all 4 times it was asked

    145
    A random variable X takes the values 0, 1, 2, 3, ____\_\_\_\_ with probability P(X=x)=k(x+1)(15)xP(X =x) = k(x+ 1)\left( \frac{1}{5} \right)^{x}, where k is a constant.Then P(X=0)P(X = 0) is

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  5. Question 5

    Vectors. Asked 4 times: 3 May 2023 S2*, 11 May 2023 S2, 13 May 2024 S2, 20 Apr 2025 S2 (* = numbers or wording changed that year) See all 4 times it was asked

    147
    If aˉ=110(3i^+k^)\bar{a}=\frac{1}{\sqrt{10}}(3\widehat{i}+\widehat{k}) and bˉ=17(2i^+3j^−6k^)\bar{b}=\frac{1}{7}(2\widehat{i}+ 3\widehat{j}- 6\widehat{k}) then the value of (2aˉ−bˉ)⋅[(aˉ×bˉ)×(aˉ+2bˉ)]=(2\bar{a}-\bar{b}) \cdot \lbrack(\bar{a}\times\bar{b}) \times (\bar{a}+ 2\bar{b})\rbrack =

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