Let f (x) = \ cases (1)/(x2) & : |x| ≥ 1 ^ 2 + β& : |x| < 1 cases...

The given function is clearly continuous at all points except possibly at $x = \pm 1 .$
For $f\left(x\right)$ to be continuous at $x = 1,$ we must have
$\lim_{x \to {1}^{-}}f\left(x\right) = \lim_{x \to {1}^{+}}f\left(x\right) = f\left(1\right)$
$⇒ \lim_{x \to {1}^{-}}\alpha{x}^{2} + \beta = \lim_{x \to {1}^{+}}\dfrac{1}{{x}^{2}}$
$⇒ \alpha + \beta = 1$ … \left(1\right)$<br />Now, for$f\left(x\right)$to be differentiable at$x = 1,$we must have<br />$\lim_{x \to {1}^{-}}\dfrac{f\left(x\right) - f\left(1\right)}{x - 1} = \lim_{x \to {1}^{+}}\dfrac{f\left(x\right) - f\left(1\right)}{x - 1}$<br />$⇒ \lim_{x \to {1}^{-}}\dfrac{\alpha{x}^{2} + \beta - 1}{x - 1} = \lim_{x \to {1}^{+}}\dfrac{\dfrac{1}{{x}^{2}} - 1}{x - 1}$\left(∵ \alpha + \beta = 1 ∴ \beta - 1 = - \alpha\right)$
$⇒ \lim_{x \to {1}^{-}}\dfrac{\alpha{x}^{2} - \alpha}{x - 1} = \lim_{x \to {1}^{+}}\dfrac{\dfrac{1}{{x}^{2}} - 1}{x - 1}$
$⇒ \lim_{x \to {1}^{-}}\alpha\left(x + 1\right) = \lim_{x \to {1}^{+}}\dfrac{- 2}{{x}^{3}} ⇒ 2\alpha = - 2$
$⇒ \alpha = - 1$
Putting $\alpha = - 1$ in \left(1\right),$we get$\beta = 2