Question 73

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

As x→0⁻: (1-cos2x)/x² = 2sin²x/x² → 2. So α = 2.

As x→0⁺: β√(1-cosx)/x = β·√(2sin²(x/2))/x = β·√2|sin(x/2)|/x ≈ β·√2·(x/2)/x = β/(√2)

Wait: √(1-cosx) = √(2sin²(x/2)) = √2|sin(x/2)| ≈ √2·x/2 for small x>0

So β√(1-cosx)/x ≈ β·√2·x/(2x) = β/√2

For continuity: β/√2 = 2, so β = 2√2

α² + β² = 4 + 8 = 12

The correct answer is Option 2: 12.

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