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Day 529: JEE Mains Maths

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

  1. Question 1

    Sequences and Series. Asked once: 28 Jan 2025

    148
    The interior angles of a polygon with n sides, are in an A.P. with common difference 6∘6^{\circ}. If the largest interior angle of the polygon is 219∘219^{\circ}, then n is equal to ___\_\_\_

  2. Question 2

    Straight Lines. Asked once: 4 Apr 2024

    178
    Consider a triangle ABCABC having the vertices A(1,2),B(α,β)A(1,2),B(\alpha,\beta) and C(γ,δ)C(\gamma,\delta) and angles ∠ABC=π6\angle ABC =\frac{\pi}{6} and ∠BAC=2π3\angle BAC =\frac{2\pi}{3}. If the points BB and CC lie on the line y=x+4y = x + 4, then α2+γ2\alpha^{2}+\gamma^{2} is equal to .....

  3. Question 3

    Three Dimensional Geometry. Asked once: 29 Jul 2022

    83
    Let a line with direction ratios a,−4a,−7a, - 4a, - 7 be perpendicular to the lines with direction ratios 3 , −1,2b- 1,2b and b,a,−2b,a, - 2. If the point of intersection of the line x+1a2+b2=y−2a2−b2=z1\frac{x + 1}{a^{2}+b^{2}}=\frac{y - 2}{a^{2}-b^{2}}=\frac{z}{1} and the plane x−y+z=0x - y + z = 0 is (α,β,γ)(\alpha,\beta,\gamma), then α+β+γ\alpha + \beta + \gamma is equal to

  4. Question 4

    Trigonometric Identities. Asked once: 2021 Paper 8

    84
    Let tan⁡α,tan⁡β\tan\alpha,\tan\beta and tan⁡γ;α,β,γ≠(2n−1)π2\tan\gamma;\alpha,\beta,\gamma\neq\frac{(2n - 1)\pi}{2}, n∈Nn \in N be the slopes of three line segments OAOA, OBOB and OCOC, respectively, where OO is origin. If the circumcentre of △ABC\bigtriangleup ABC coincides with origin and its orthocentre lies on yy-axis, then the value of (cos⁡3α+cos⁡3β+cos⁡3γcos⁡αcos⁡βcos⁡γ)2\left( \frac{\cos3\alpha+ \cos3\beta+ \cos3\gamma}{\cos\alpha \cos\beta \cos\gamma} \right)^{2} is equal to:

  5. Question 5

    Vector Algebra. Asked once: 11 Apr 2023

    173
    Let a→=i^+2j^+3k^\overrightarrow{a}=\widehat{i}+ 2\widehat{j}+ 3\widehat{k} and b→=i^+j^−k^\overrightarrow{b}=\widehat{i}+\widehat{j}-\widehat{k}. If c→\overrightarrow{c} is a vector such that a→⋅c→=11,b→⋅(a→×c→)=27\overrightarrow{a}\cdot\overrightarrow{c}= 11,\overrightarrow{b}\cdot (\overrightarrow{a}\times\overrightarrow{c}) = 27 and b→⋅c→=−3∣b→∣\overrightarrow{b}\cdot\overrightarrow{c}= -\sqrt{3}|\overrightarrow{b}|, then ∣a→×c→∣2|\overrightarrow{a}\times\overrightarrow{c}|^{2} is equal to