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Day 465: JEE Mains Maths

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

  1. Question 1

    Three Dimensional Geometry. Asked once: 28 Jan 2025

    56
    Let A(x,y,z)A(x,y,z) be a point in xy-plane, which is equidistant from three points (0,3,2),(2,0,3)(0,3,2),(2,0,3) and (0,0,1)(0,0,1). Let B=(1,4,−1)B = (1,4, - 1) and C=(2,0,−2)C = (2,0, - 2). Then among the statements (S1): △ABC\bigtriangleup ABC is an isosceles right angled triangle and (S2) : the area of △ABC\bigtriangleup ABC is 922\frac{9\sqrt{2}}{2}.

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  2. Question 2

    Trigonometric Equations. Asked once: 29 Jan 2023

    71
    Let f(θ)=3(sin⁡4(3π2−θ)+sin⁡4(3π+θ))−2(1−sin⁡22θ)f(\theta) = 3\left( \sin^{4}\left( \frac{3\pi}{2}- \theta \right)+\sin^{4}(3\pi + \theta) \right)- 2\left( 1 -\sin^{2}2\theta \right) and S={θ∈[0,π]:f′(θ)=S =\left\{ \theta \in \lbrack 0,\pi\rbrack:f^{'}(\theta) = \right.  −32}\left. \ -\frac{\sqrt{3}}{2} \right\}. If 4β=∑θ∈Sθ4\beta =\sum_{\theta \in S} \theta, then f(β)f(\beta) is equal to

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  3. Question 3

    Trigonometric Identities. Asked once: 22 Jan 2026 S2

    73
    Let cos⁡(α+β)=−110\cos(\alpha + \beta) = - \frac{1}{10} and sin⁡(α−β)=38\sin(\alpha - \beta) = \frac{3}{8}, where 0<α<π30 < \alpha < \frac{\pi}{3} and 0<β<π40 < \beta < \frac{\pi}{4}. If tan⁡2α=3(1−r5)11(s+5),r,s∈N\tan2\alpha = \frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})},r,s \in \mathbb{N}, then r+sr + s is equal to ____\_\_\_\_.

  4. Question 4

    Vector Algebra. Asked once: 23 Jan 2026 S2

    60
    Let a→,b→,c→\overrightarrow{a},\overrightarrow{b},\overrightarrow{c} be three vectors such that a→×b→=2(a→×c→)\overrightarrow{a}\times\overrightarrow{b}= 2(\overrightarrow{a}\times\overrightarrow{c}).If ∣a→∣=1,∣ b→∣=4,∣c→∣=2|\overrightarrow{a}| = 1,|\overrightarrow{\text{ }b}| = 4,|\overrightarrow{c}| = 2, and the angle between b→\overrightarrow{b} and c→\overrightarrow{c} is 60∘60^{\circ}, then ∣a→.c→∣|\overrightarrow{a}.\overrightarrow{c}| is :

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  5. Question 5

    Application of Derivatives. Asked once: 28 Jan 2025

    64
    The sum of all local minimum values of the function f(x)={1−2x,x<−113(7+2∣x∣),−1≤x≤21118(x−4)(x−5),x>2f(x)=\begin{cases} 1-2x, & x<-1 \\ \frac{1}{3}(7+2|x|), & -1\leq x\leq 2 \\ \frac{11}{18}(x-4)(x-5), & x>2 \end{cases} is

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