Question:medium
Let $f(x)$ and $g(x)$ be twice differentiable functions satisfying $f''(x) = g''(x)$ for all $x \in \mathbb{R}$, $f'(1) = 2g'(1) = 4$ and $g(2) = 3f(2) = 9$. Then $f(25) - g(25)$ is equal to :
Let $f(x)$ and $g(x)$ be twice differentiable functions satisfying $f''(x) = g''(x)$ for all $x \in \mathbb{R}$, $f'(1) = 2g'(1) = 4$ and $g(2) = 3f(2) = 9$. Then $f(25) - g(25)$ is equal to :
Updated On: Apr 12, 2026
- 20
- 40
- $-20$
- $-40$
Show Solution
The Correct Option is B
Solution and Explanation
Was this answer helpful?
0
Top Questions on Differentiation
- The number of points, at which the function \(f(x) = \max\{6x, 2 + 3x^2\} + |x - 1| \cos|x^2 - \frac{1}{4}|\), \(x \in (-\pi, \pi)\), is not differentiable, is ____.
- JEE Main - 2026
- Mathematics
- Differentiation
- Let \(f\) be a real polynomial of degree \(n\) such that \(f(x) = f'(x)f''(x)\), for all \(x \in \mathbb{R}\). If \(f(0) = 0\), then \(36(f''(2) + f''(2) + \int_0^2 f(x)\,dx)\) is equal to:
- JEE Main - 2026
- Mathematics
- Differentiation
- Let \( f(x) \) be a polynomial of degree 5, and have extrema at \( x = 1 \) and \( x = -1 \). If \( \lim_{x \to 0} \frac{f(x)}{x^3} = -5 \), then \( f(2) - f(-2) \) is equal to:
- JEE Main - 2026
- Mathematics
- Differentiation
- The number of critical points of the function} \[ f(x)= \begin{cases} \dfrac{|\sin x|}{x}, & x\neq0\\ 1, & x=0 \end{cases} \] in the interval \((-2\pi,2\pi)\) is equal to:
- JEE Main - 2026
- Mathematics
- Differentiation
- Want to practice more? Try solving extra ecology questions todayView All Questions
