Mathematics · 2025
JEE Main · 24 January 2025, Shift 2 · Q20
Let f:(0, ∞) → R be a function which is differentiable at all points of its domain and satisfies the condition x^2 f^′(x)=2 x f(x)+3, with f(1)=4.…
Let $\displaystyle f:(0, \infty) \rightarrow \mathbf{R}$ be a function which is differentiable at all points of its domain and satisfies the condition $\displaystyle x^2 f^{\prime}(x)=2 x f(x)+3$, with $\displaystyle f(1)=4$. Then $\displaystyle 2 f(2)$ is equal to:
Official answer
From NTA’s final answer key for this paper.
(3)
$\displaystyle 39$
More from Differential Equations
- Let f(x)=x-1 and g(x)=e^x for x ∈ R. If (d y)/(d x)=(e^-2 √x g(f(f(x)))-y/(√x)), y(0)=0, then y(1) is2025
- Let f: R → R be a thrice differentiable odd function satisfying f^′(x) ≥ 0, f^′ ′(x)=f(x), f(0)=0, f^′(0)=3. Then 9 f( log_e 3) is equal to ____.2025
- Let for some function y=f(x), ∫_0^x t f(t) d t=x^2 f(x), x>0 and f(2)=3. Then f(6) is equal to2025
- Let f:[1, ∞) →[2, ∞) be a differentiable function. If 10 ∫_1^x f(t) dt=5 x f(x)-x^5-9 for all x ≥ 1, then the value of f(3) is:2025
- Let y=y(x) be the solution curve of the differential equation (1+ sin x) (d y)/(d x)+(y+1) cos x=0, y(0)=0. If the curve y=y(x) passes through the…2026
- Let y=y(x) be the solution of the differential equation x √(1-x^2) d y+(y √(1-x^2)-x cos^-1 x) d x=0, x ∈(0,1), lim_x → 1^- y(x)=1. Then y(1/2)…2026
- Let y=y(x) be the solution of the differential equation sec x (d y)/(d x)-2 y=2+3 sin x, x ∈(-(π)/2, (π)/2), y(0)=-7/4. Then y((π)/6) is equal to:2026
- Let y=y(x) be the solution curve of the differential equation (1+x^2) d y+(y- tan^-1 x) d x=0, y(0)=1. Then the value of y(1) is:2026
JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.