🛠️ JEE➗ Maths

Let \(f:R\to R\) be a function given by \(f\left(x\right)=\left{\begin{matrix}\frac{1-\cos 2x}{{x}^{2}},x 0 \\ \end{matr…

Q1

Let \(f:R\to R\) be a function given by
\(f\left(x\right)=\left{\begin{matrix}\frac{1-\cos 2x}{{x}^{2}},x<0\\ \alpha x=0,\\ \frac{\beta \sqrt{1-\cos x}}{x},x>0 \\ \end{matrix}\right.\)
where \(\alpha ,\beta \in R\). If \(f\) is continuous at \(x=0\), then \({\alpha }^{2}+{\beta }^{2}\) is equal to :

[JEE Main 2024, 04 Apr (Shift 1)]

a

6

b

3

c

12

d

48

🔒
Answer & explanation — PYQ Pass

Options are free to see. Unlock the correct answer and full explanation with Pass.

Unlock · ₹149

Practice more JEE Maths PYQs

See every question on Continuity and Differentiability, or browse the full JEE question bank.

See all questions on Continuity and Differentiability →