JEE Main 202522 JanuaryMorning Shift
- Exam
- JEE Main
- Year
- 2025
- Date
- 22 January
- Shift
- Morning Shift
- Subject
- Mathematics
- Chapter
- Continuity and Differentiability
JEE Main 2025 · 22 January · Morning Shift · Mathematics · Continuity and Differentiability
Let the function, f(x)= cases -3 a x^2-2, & x 1 a^2+b x, & x ≥ 1...
Let the function,
\begin{cases}-3 a x^2-2, & x \lt 1 \\ a^2+b x, & x \geqslant 1\end{cases}$x \in \mathbf{R}\mathbf{a}\gt1, \mathbf{b} \in \mathbf{R}y=f(x)y=-20\alpha+\beta \sqrt{3}, \alpha, \beta \in Z\alpha+\beta$ is ________
Correct answer
0
Step-by-step solution
is continuous and differentiable
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