If $f(x)=\left\{\begin{array}{ll}~~~x~~~~~~, & x \leq 1 \\ x^2+b x+c, & x>1\end{array}\right.$ and f'(x) exists finitely for all x ∈ R, then

b = -1, c = 1

Clearly, f(x) is a polynomial for all x < 1 and x > 1.

So, it is everywhere continuous and differentiable for all $x \in(-\infty, 1) \cup(1, \infty)$

For f(x) to be differentiable at x = 1, we must have

(LHD at x = 1) = (RHD at x = 1)

$\Rightarrow \left(\frac{d}{d x}(x)\right)_{\text {at } x=1}=\left(\frac{d}{d x}\left(x^2+b x+c\right)\right)_{\text {at } x=1}$

⇒ 1 = 2 + b ⇒ b = -1

Also, f(x) must be continuous at x = 1

∴  $\lim\limits_{x \rightarrow 1^{-}} f(x)=\lim\limits_{x \rightarrow 1^{+}} f(x)$

$\Rightarrow \lim\limits_{x \rightarrow 1} x=\lim\limits_{x \rightarrow 1} x^2+b x+c$

⇒ 1 = 1 + b + c ⇒ b + c = 0 ⇒ c = 1

Hence, f'(x) exists for all x ∈ R, if b = -1 and c = 1.