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Day 96: JEE Mains Maths

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

  1. Question 1

    Binomial Theorem. Asked 2 times: 2021 Paper 8*, 2021 Paper 20 (* = numbers or wording changed that year) See all 2 times it was asked

    73
    ∑k=020( 20Ck)2\sum_{k = 0}^{20} \left( \ ^{20}C_{k} \right)^{2} is equal to:

    Tap an option to check your answer.

  2. Question 2

    Inverse Trigonometric Functions. Asked 2 times: 2021 Paper 7*, 2021 Paper 19 (* = numbers or wording changed that year) See all 2 times it was asked

    70
    If ∑r=150tan⁡−112r2=p\sum_{r = 1}^{50} \tan^{- 1}\frac{1}{2r^{2}}= p, then the value of tan⁡p\tan p is :

    Tap an option to check your answer.

  3. Question 3

    Vector Algebra. Asked 2 times: 2021 Paper 7*, 2021 Paper 17 (* = numbers or wording changed that year) See all 2 times it was asked

    81
    Let a→=i^−αj^+βk^\overrightarrow{a}=\widehat{i}-\alpha\widehat{j}+\beta\widehat{k}, b→=3i^+βj^−αk^\overrightarrow{b}=3\widehat{i}+\beta\widehat{j}-\alpha\widehat{k} and c→=−αi^−2j^+k^\overrightarrow{c}=-\alpha\widehat{i}-2\widehat{j}+\widehat{k}, where α\alpha and β\beta are integers. If a→⋅b→=−1\overrightarrow{a}\cdot\overrightarrow{b}=-1 and b→⋅c→=10\overrightarrow{b}\cdot\overrightarrow{c}=10, then (a→×b→)⋅c→(\overrightarrow{a}\times\overrightarrow{b})\cdot\overrightarrow{c} is equal to:

  4. Question 4

    Determinants. Asked 2 times: 2021 Paper 6*, 2021 Paper 16 (* = numbers or wording changed that year) See all 2 times it was asked

    54
    Let f(x)=∣sin⁡2x−2+cos⁡2xcos⁡2x2+sin⁡2xcos⁡2xcos⁡2xsin⁡2xcos⁡2x1+cos⁡2x∣;x∈(0,π]f(x) =\left| \begin{matrix} \sin^{2}x & - 2 +\cos^{2}x & \cos2x \\ 2 +\sin^{2}x & \cos^{2}x & \cos2x \\ \sin^{2}x & \cos^{2}x & 1 + \cos2x \end{matrix} \right|;x\in (0,\pi\rbrack. Then the maximum value of f(x)f(x) is?

  5. Question 5

    Binomial Theorem. Asked 2 times: 2021 Paper 5*, 2021 Paper 15 (* = numbers or wording changed that year) See all 2 times it was asked

    57
    Find the number of irrational terms in (213+314)12\left( 2^{\frac{1}{3}}+3^{\frac{1}{4}} \right)^{12}.