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Day 78: JEE Mains Maths

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

  1. Question 1

    Conic Sections. Asked 2 times: 7 Apr 2025*, 24 Jan 2026 S2 (* = numbers or wording changed that year) See all 2 times it was asked

    52
    Let the length of the latus rectum of an ellipse x2a2+y2 b2=1,(a>b)\frac{x^{2}}{a^{2}}+\frac{y^{2}}{{\text{ }b}^{2}}= 1,(a > b), be 30 . If its eccentricity is the maximum value of the function f(t)=−34+2t−t2f(t) = -\frac{3}{4}+ 2t -t^{2}, then (a2+b2)\left( a^{2}+b^{2} \right) is equal to -

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  2. Question 2

    Sequences and Series. Asked 2 times: 15 Apr 2023*, 24 Jan 2026 S2 (* = numbers or wording changed that year) See all 2 times it was asked

    57
    (13+47)+(132+13×47+4272)+\left( \frac{1}{3}+\frac{4}{7} \right)+\left( \frac{1}{3^{2}}+\frac{1}{3}\times\frac{4}{7}+\frac{4^{2}}{7^{2}} \right)+ (133+132×47+13×4272+4373)+……\left( \frac{1}{3^{3}}+\frac{1}{3^{2}}\times\frac{4}{7}+\frac{1}{3}\times\frac{4^{2}}{7^{2}}+\frac{4^{3}}{7^{3}} \right)+ \ldots\ldots upto infinite terms is equal to -

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  3. Question 3

    Binomial Theorem. Asked 2 times: 1 Feb 2023*, 24 Jan 2026 S1 (* = numbers or wording changed that year) See all 2 times it was asked

    64
    Let S=125!+13!23!+15!21!+…S =\frac{1}{25!}+\frac{1}{3!23!}+\frac{1}{5!21!}+ \ldots up to 13 terms. if 13 S=2kn!,k∈N13\text{ }S =\frac{2^{k}}{n!},k \in N, then n+kn + k is equal to

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  4. Question 4

    Relations and Functions. Asked 2 times: 15 Apr 2023*, 24 Jan 2026 S1 (* = numbers or wording changed that year) See all 2 times it was asked

    62
    Let R be a relation defined on the set {1,2,3,4}×{1,2,3,4}\mathbf{\{ 1,2,3,4\} \times \{ 1,2,3,4\}} by R={((a,b),(c,d)):2a+3b=3c+4d}\mathbf{R = \{((a,b),(c,d)):2a + 3b = 3c + 4d\}}. Then the number of elements in R is

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  5. Question 5

    Trigonometric Identities. Asked 2 times: 23 Jan 2026 S2*, 24 Jan 2026 S1 (* = numbers or wording changed that year) See all 2 times it was asked

    60
    If cot⁡x=512\cot x =\frac{5}{12} for some x∈(π,3π2)x \in\left( \pi,\frac{3\pi}{2} \right), then sin⁡7x(cos⁡13x2+sin⁡13x2)+cos⁡7x(cos⁡13x2−sin⁡13x2)\sin7x\left( \cos\frac{13x}{2}+ \sin\frac{13x}{2} \right)+ \cos7x\left( \cos\frac{13x}{2}- \sin\frac{13x}{2} \right) is equal to

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