Question
Let a differentiable function f satisfy the equation _ 0 ^ 36 f ( t x 36 ) d t=4 f(x). If y=f(x) is a standard parabola passing through the points (2,1) and (-4, ), then ^ is equal to \_\_\_\_.
JEE Main
Mathematics
Definite Integration
2026
JEE Main 2026 (24 January Shift 1)
JEE Main Mathematics Question (2026) — Solution
Options
- A. A
- B. B
- C. C
- D. D
Answer
A. A
Step-by-step solution
From _0^ 36 f ( tx 36 ) dt = 4 f(x), substitute u = tx 36 to get 36 x _0^x f(u) du = 4 f(x). Differentiating: f(x) = 9 [f(x) + xf'(x)], which simplifies to (9- )f(x) = x f'(x). This gives f'(x) f(x) = 9- x , integrating to f(x) = Kx^ 9- . For f to be a standard parabola: 9- = 2 = 3. Thus f(x) = Kx^2. Using point (2,1): K = 1 4 , so f(x) = x^2 4 . At (-4, ): = 16 4 = 4. Therefore ^ = 4^3 = 64
Related: Mathematics — Definite Integration · All PYQ Banks