If the function f(x) given by $f(x)= \begin{cases}2^{1 /(x-1)}, & x<1 \\ a x^2+b x, & x \geq 1\end{cases}$ is everywhere differentiable, then

a = 0, b = 0

Clearly, f(x) is everywhere continuous and differentiable except possibly at x = 1.

At x = 1, we have

$\lim\limits_{x \rightarrow 1^{-}} f(x)=\lim\limits_{x \rightarrow 1^{-}} 2^{1 /(x-1)}=\lim\limits_{h \rightarrow 0} 2^{-1 / h}=0$

$\lim\limits_{x \rightarrow 1^{+}} f(x)=\lim\limits_{x \rightarrow 1^{+}}\left(a x^2+b x\right)=a+b$

and,

$f(1)=a+b$

For f(x) to be differentiable at x = 1, it must be continuous there at.

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

⇒  0 = a + b              .........(i)

Also,

(LHD at x = 1) = $\lim\limits_{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}$

⇒ (LHD at x = 1) = $\lim\limits_{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h}$

⇒ (LHD at x = 1) = $\lim\limits_{h \rightarrow 0} \frac{2^{-1 / h}-(a+b)}{-h}$

⇒ (LHD at x = 1) = $\lim\limits_{h \rightarrow 0} \frac{2^{-1 / h}}{-h}$                  [∵ a + b = 0]

⇒ (LHD at x = 1) = $-\lim\limits_{h \rightarrow 0} \frac{1 / h}{2^{1 / h}}$             ${\left[\frac{\infty}{\infty} \text { form }\right]}$

⇒ (LHD at x = 1) = $-\lim\limits_{h \rightarrow 0} \frac{-1 / h^2}{2^{1 / h} \log _e 2\left(-\frac{1}{h^2}\right)}$

⇒ (LHD at x = 1) = $-\lim\limits_{h \rightarrow 0} \frac{2^{-1 / h}}{\log _e 2}=\frac{0}{\log _e 2}=0$

and,

⇒ (RHD at x = 1) = $\lim\limits_{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}$

⇒ (RHD at x = 1) = $\lim\limits_{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}$

⇒ (RHD at x = 1) = $\lim\limits_{h \rightarrow 0} \frac{a(1+h)^2+b(1+h)-(a+b)}{h}$

⇒ (RHD at x = 1) = $\lim\limits_{h \rightarrow 0} \frac{a h^2+2 a h+b h}{h}$

⇒ (RHD at x = 1) = $\lim\limits_{h \rightarrow 0} a h+2 a+b=2 a+b$

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

⇒ 0 = 2a + b                   .............(ii)

Solving (i) and (ii), we get a = b = 0.