Answer:
Option A,C
Explanation:
Plan A function f(x) is continuous at x=a, if
$\lim_{x \rightarrow a^{-}}$ f(x)= $\lim_{x \rightarrow a^{+}}$ f(x=f(a))
$\lim_{x \rightarrow a^{-}}\frac{f(x)-f(a)}{x-a}=\lim_{x \rightarrow a^{+}}\frac{f(x)-f(a)}{x-a}$
i.e, f '(a-)=f '(a+)
Given that $f:[a,b]\rightarrow[1,\infty]$
and $\begin{cases}0 & if & x <a &\\\int_{a}^{x}f(t)dt, &if& a\leq x\leq b & \\\int_{a}^{b}f(t)dt &if & x>b\end{cases}$
Now, g(a-)=0=g (a+)=g(a)
[as $g(a^{+})\lim_{x \rightarrow a^{+}}\int_{a}^{x} f(I) dt=0$
and $g(a)=\int_{a}^{a} f(t) dt=0$ ]
$g(b^{-})=g(b^{+})=g(b)=\int_{a}^{b} f(t) dt\Rightarrow $
is continuous , $\forall x\epsilon R$
Now, $g'(x)=\begin{cases}0 & & x <a &\\f(x), && a< x< b & \\0 & & x>b\end{cases}$
g'(a-)=0 but g'(a+)= $f(a)\geq1$
[ $\because $ Range of f(x) is $[1,\infty), \forall x\in[a,b]]$
$\Rightarrow$ g is non-differentiable at x=a
and g'(b+ )=0
but $g'(b^{-})=f(b)\geq 1$
$\Rightarrow$ g is not differentiable at x=b
