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Day 229: JEE Mains Maths

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

  1. Question 1

    Three Dimensional Geometry. Asked once: 10 Apr 2023 (same type asked 11 times) See all 11 of this type

    76
    Let two vertices of triangle ABCABC be (2,4,6)(2,4,6) and (0,−2,−5)(0, - 2, - 5), and its centroid be (2,1,−1)(2,1, - 1). If the image of third vertex in the plane x+2y+4z=11x + 2y + 4z = 11 is (α,β,γ)(\alpha,\beta,\gamma), then αβ+βγ+γα\alpha\beta + \beta\gamma + \gamma\alpha is equal to

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  2. Question 2

    Three Dimensional Geometry. Asked once: 2 Apr 2026 S1 (same type asked 10 times) See all 10 of this type

    65
    Let a line L passing through the point (1,1,1)(1,1,1) be perpendicular to both the vectors 2i^+2j^+k^2\widehat{i}+ 2\widehat{j}+\widehat{k} and i^+2j^+k^\widehat{i}+ 2\widehat{j}+\widehat{k}. If P(a,b,c)P(a,b,c) is the foot of perpendicular from the origin on the line L , then the value of 34(a+b+c)34(a + b + c) is :

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  3. Question 3

    Inverse Trigonometric Functions. Asked once: 24 Jan 2025 (same type asked 10 times) See all 10 of this type

    128
    If α>β>γ>0\alpha > \beta > \gamma > 0, then the expression
    cot⁡−1{β+(1+β2)(α−β)}+cot⁡−1{γ+(1+γ2)(β−γ)}\cot^{- 1}\left\{ \beta +\frac{\left( 1 +\beta^{2} \right)}{(\alpha - \beta)} \right\}+\cot^{- 1}\left\{ \gamma +\frac{\left( 1 +\gamma^{2} \right)}{(\beta - \gamma)} \right\}
    +cot⁡−1{α+(1+α2)(γ−α)}+\cot^{- 1}\left\{ \alpha +\frac{\left( 1 +\alpha^{2} \right)}{(\gamma - \alpha)} \right\} is equal to:

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  4. Question 4

    Quadratic Equations. Asked once: 7 Apr 2025 (same type asked 10 times) See all 10 of this type

    142
    The number of real roots of the equation x∣x−2∣+3∣x−3∣+1=0x|x - 2| + 3|x - 3| + 1 = 0 is :

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  5. Question 5

    Vector Algebra. Asked once: 30 Jan 2023 (same type asked 10 times) See all 10 of this type

    154
    Let λ∈R,a→=λı^+2ȷ^−3k^,b→=ı^−λȷ^+2k^\lambda \in R,\overrightarrow{a}= \lambda\widehat{\imath}+ 2\widehat{\jmath}- 3\widehat{k},\overrightarrow{b}=\widehat{\imath}- \lambda\widehat{\jmath}+ 2\widehat{k} If ((a→+b→)×(a→×b→))×(a→−b→)=8ı^−40ȷ^−24k^((\overrightarrow{a}+\overrightarrow{b}) \times (\overrightarrow{a}\times\overrightarrow{b})) \times (\overrightarrow{a}-\overrightarrow{b}) = 8\widehat{\imath}- 40\widehat{\jmath}- 24\widehat{k}, then ∣λ(a→+b→)×(a→−b→)∣2|\lambda(\overrightarrow{a}+\overrightarrow{b}) \times (\overrightarrow{a}-\overrightarrow{b})|^{2} is equal to

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