Let **$$f : \mathbb{R} \to \mathbb{R}$$** be the function given by $$ f(x) = \begin{cases} \dfrac{1 - \cos 2x}{x^2}, & x 0 \end{cases} $$ where $$\alpha, \beta \in \mathbb{R}$$. If **$$f$$** is continuous at **$$x = 0$$**, then **$$\alpha^2 + \beta^2$$** is equal to:

Let f: ℝ → ℝ be the function given by f(x) = (1 - cos 2x)/x², x <… | JEE Main 2024 Mathematics PYQ

MCQMediumJEE Main 2024 · 4 April, Shift 1Continuity

Let $f : \mathbb{R} \to \mathbb{R}$ be the function given by

f(x) = \begin{cases} \dfrac{1 - \cos 2x}{x^2}, & x < 0 \\ \alpha, & x = 0 \\ \dfrac{\beta \sqrt{1 - \cos x}}{x}, & x > 0 \end{cases}

where $\alpha, \beta \in \mathbb{R}$ . If $f$ is continuous at $x = 0$ , then $\alpha^2 + \beta^2$ is equal to:

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