JEE Main 2024 · MathsNumerical answerNumerical valueMediumMulti-step

JEE Main 5 April 2024, Shift 1, Maths Q24

Question 24 of 90 in this shift, Maths question 24 of 30, Section B.

If the constant term in the expansion of

(1+2x-3x^3)\left(\frac{3}{2}x^2-\frac{1}{3x}\right)^9

p

, then

108p

is equal to ________.

Official answer

NTA final key.

Chapter: Binomial Theorem
Topic: Binomial Coefficients
Idea tested: Coefficient Extraction from Products

Same idea in other shifts

Asked 6× in all

1 Feb 2024, Shift 1 · Q23If the Coefficient of $x^{30}$ in the expansion of $\left(1+\frac{1}{x}\right)^6(1+x^2)^7(1-x^3)^8$ ; $x\neq0$ is $\alpha$ , then…HardNumerical value
31 Jan 2024, Shift 1 · Q24In the expansion of $(1+x)(1-x^2)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5,\ x\neq0$ , the sum of the coefficients of $x^3$ …MediumNumerical value
26 Jul 2022, Shift 1 · Q25If the coefficients of $x$ and $x^2$ in the expansion of $(1 + x)^{\mathrm{p}}\ (1 - x)^{\mathrm{q}}$ , $\mathrm{p},\ \mathrm{q} \leq 15$ ,…MediumNumerical value
30 Jul 2022, Shift 2 · Q82If $p_1$ and $p_2$ , respectively, are the second largest and the fourth largest prime numbers in the factorization of the coefficient of…HardNumerical value
28 Jun 2022, Shift 2 · Q4The term independent of $x$ in the expansion of $(1-x^2+3x^3)\left(\frac{5}{2}x^3-\frac{1}{5x^2}\right)^{11}$ , $x \neq 0$ is :MediumSingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.