The coefficients a and b that make the function continuous and differentiable $f(x)=\left\{\begin{matrix}\frac{1}{|x|}&;&for\,|x|≥1\\ax^2+b&;&for\,|x|<1\end{matrix}\right.$

$a=-\frac{1}{2},\,b=\frac{3}{2}$

$f(x)=\left\{\begin{matrix}-\frac{1}{x}&;&x≤-1\\2ax^2+b&;&-1<x<1\\\frac{1}{x}&;&x>1\end{matrix}\right.$   …(i)

$f(x)=\left\{\begin{matrix}\frac{1}{x^2}&;&x≤-1\\2ax&;&-1<x<1\\-\frac{1}{x^2}&;&x>1\end{matrix}\right.$   …(ii)

For the function to be continuous

LHL = RHL = value of function

at x = -1 ⇒ 1 = a + b = 1   ….(iii)

at x = 1 ⇒ a + b = 1 = 1   …(iv)

For the function to be differentiable

LHD = RHD

at x = -1 ⇒ 1 = -2a ….(v)

at x =1 ⇒ 2a = −1 …(vi)

From (v) or (vi) ⇒ $a=-\frac{1}{2}$

Put $a=-\frac{1}{2}$ in (iii) or (iv) ⇒ $b=\frac{3}{2}$