Question:medium
Let \( f:(-1,1)\to(-1,1) \) be continuous, \( f(x)=f(x^2) \), and \( f(0)=\frac{1}{2} \). Find \( 4f\left(\frac{1}{4}\right) \)
Let \( f:(-1,1)\to(-1,1) \) be continuous, \( f(x)=f(x^2) \), and \( f(0)=\frac{1}{2} \). Find \( 4f\left(\frac{1}{4}\right) \)
Show Hint
Repeated squaring inside (-1,1) converges to 0.
Updated On: May 1, 2026
- \( 1 \)
- \( 2 \)
- \( 3 \)
- \( 4 \)
- \( 5 \)
Show Solution
The Correct Option is B
Solution and Explanation
Was this answer helpful?
0
Top Questions on Continuity and differentiability
- Given that: \[ x = a \sin(2t) (1 + \cos(2t)), \quad y = a \cos(2t) (1 - \cos(2t)) \] Find \(\frac{dy}{dx}\).
- MHT CET - 2025
- Mathematics
- Continuity and differentiability
- Let \( f(x) = |1 - 2x| \), then:
- WBJEE - 2025
- Mathematics
- Continuity and differentiability
- A function \( f \) is defined by \( f(x) = 2 + (x - 1)^{2/3} \) on \( [0, 2] \). Which of the following statements is incorrect?
- WBJEE - 2025
- Mathematics
- Continuity and differentiability
- Let \( f : \mathbb{R} \to \mathbb{R} \) be a function given by
\[f(x) = \begin{cases} \frac{1 - \cos 2x}{x^2}, & x < 0 \\\alpha, & x = 0, \text{ where } \alpha, \beta \in \mathbb{R}. \\\beta \sqrt{1 - \cos x} / x, & x > 0 \end{cases} \]
If \( f \) is continuous at \( x = 0 \), then \( \alpha^2 + \beta^2 \) is equal to:- JEE Main - 2024
- Mathematics
- Continuity and differentiability
- Want to practice more? Try solving extra ecology questions todayView All Questions
