If α and β are such that the function f ( x ) defined by f x = α x 2 - β , &#160…

It is given that the function is differentiable at all points,
By the first principle,

$\mathrm{LHD}\left(\text{at} x=1\right)=\mathrm{RHD}\left(\text{at} x=1\right)$ 

$\underset{x\to 1^{-}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1}=\underset{x\to 1^{+}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1}$

$\underset{x\to 1^{-}}{\lim }\frac{\alpha x^2-\beta -\alpha +\beta }{x-1}=\underset{x\to 1^{+}}{\lim }\frac{-\frac{1}{x}+1}{x-1}$

$\underset{x\to 1^{-}}{\lim }\frac{\alpha \left(x^2-1\right)}{x-1}=\underset{x\to 1^{+}}{\lim }\frac{\left(\frac{x-1}{x}\right)}{x-1}$

$\underset{x\to 1^{-}}{\lim }\frac{\alpha (x+1)(x-1)}{x-1}=\underset{x\to 1^{+}}{\lim }\frac{1}{x}$

$\underset{x\to 1^{-}}{\lim }\alpha \left(x+1\right)=\underset{x\to 1^{+}}{\lim }\frac{1}{x}$

$\alpha (1+1)=1$

$\alpha =1/2$

Every differentiable function is also continuous,

$\mathrm{RHL}=\mathrm{LHL}$

$\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }f\left(x\right)$

$\frac{1}{2}(1{)}^2-\beta =-\frac{1}{1}$

$\beta =\frac{1}{2}+1$

$\beta =3/2$