Let f : R → be a function given by $f(x)=\{\begin{array}{ccc}\frac{1-\cos 2x}{x^2}& ,& x<0\\ \alpha & ,& x=0\\ \frac{\beta \sqrt{1-\cos x}}{x}& ,& x>0\end{array}$, where α, β ∈ R. If f is continuous at x = 0, then α² + β² is equal to: 

Concept:

A function y = f(x) is said to be continuous at a point x = a if $\underset{x\to a^{-}}{lim}f(x)=\underset{x\to a^{+}}{lim}f(x)$ = f(a)

$\underset{x\to a}{lim}\frac{\sin x}{x}$ = 1

Explanation:

LHL = f(0^-) = $\underset{x\to 0}{lim}\frac{1-\cos 2x}{x^2}$ = ${\displaystyle \underset{x\to 0}{lim}\frac{2{\sin }^{2}x}{x^2}}$ = 2${\displaystyle \underset{x\to 0}{lim}(\frac{\sin x}{x}{)}^2}$ = 2

RHL = f(0⁺) = $\underset{x\to 0}{lim}\frac{\beta \surd 1-\cos x}{x}$${\displaystyle \underset{x\to 0^{+}}{lim}\beta \surd 2\frac{\sin \frac{x}{2}}{2\frac{x}{2}}=\frac{\beta }{\surd 2}}$

Since f(x) is continuous at x = 0

So, LHL = RHL = f(0)

i.e., 2 = $\frac{\beta }{\surd 2}$ = α

So, α = 2 and β = 2√2

∴ ${\alpha }^2+{\beta }^2=4+8=12$

Option (2) is true.