The coefficients a and b that make the function continuous and differentiable $f(x)=\left\{\begin{matrix}\frac{1}{|x|}&;&for\,|x|≥1\\ax^2+b&;&for\,|x|<1\end{matrix}\right.$
Answer & explanation
Correct answer: option 1
$f(x)=\left\{\begin{matrix}-\frac{1}{x}&;&x≤-1\\ax^2+b&;&-1<x<1\\\frac{1}{x}&;&x≥1\end{matrix}\right.$ …(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 ⇒ 1 = a + b = 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) ⇒ $b=\frac{3}{2}$