Mathematics · 2024
JEE Main · 29 January 2024, Shift 1 · Q6
In an A.P., the sixth term a_6=2. If the product a_1 a_4 a_5 is the greatest, then the common difference of the A.P. is equal to
In an A.P., the sixth term $\displaystyle a_6=2$. If the product $\displaystyle a_1 a_4 a_5$ is the greatest, then the common difference of the A.P. is equal to
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle \frac{8}{5}$
More from Applications of Derivatives
- Let the function f(x)=x/3+3/x+3, x ≠ 0 be strictly increasing in (-∞, α_1) ∪(α_2, ∞) and strictly decreasing in (α_3, α_4) ∪(α_4, α_5). Then Σ_i=1^5…2025
- If the function f(x)=2 x^3-9 a x^2+12 a^2 x+1, where a>0, attains its local maximum and local minimum values at p and q, respectively, such that…2025
- Let f: R → R be a twice differentiable function such that f^′ ′(x)>0 for all x ∈ R and f^′(a-1)=0, where a is a real number. Let g(x)=f( tan^2 x-2…2026
- Let f(x)=∫_0^x t(t^2-9 t+20) dt, 1 ≤ x ≤ 5. If the range of f is [α, β], then 4(α+β) equals:2025
- The sum of all local minimum values of the function f(x)= {1-2 x, x<-1; 1/3(7+2|x|), -1 ≤ x ≤ 2; 11/18(x-4)(x-5), x>2} is2025
- Let the sum of the maximum and the minimum values of the function f(x)=(2 x^2-3 x+8)/(2 x^2+3 x+8) be m/n, where gcd(m, n)=1. Then m+n is equal to:2024
- Let f(x) be a polynomial of degree 5, and have extrema at x=1 and x=-1. If lim_x → 0((f(x))/x^3)=-5, then f(2)-f(-2) is equal to:2026
- Consider the function f:[1/2, 1] → R defined by f(x)=4 √2 x^3-3 √2 x-1. Consider the statements (I) The curve y=f(x) intersects the x -axis exactly…2024
JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.