Let $f:[a, b] \rightarrow[1, \infty)$ be continuous function and let $g: R → R$ be defined as

$g(x)=\left\{\begin{array}{l}0, \text { if } x<a \\ \int\limits_a^x f(t) d t, \text { if } a \leq x \leq b \\ \int\limits_a^b f(t) d t, \text { if } x>b\end{array}\right.$

Then,

a. $g(x)$ is continuous but not differentiable at $x=a$

b. $g(x)$ is differentiable on $R$

c. $g(x)$ is continuous but not differentiable at $x=b$

d. $g(x)$ is continuous and differentiable at either $x=a$ or $x=b$ but not both.

a, c

Continuity of g(x) :

(LHL at x = a) = $\lim\limits_{x \rightarrow a^{-}} g(x)=\lim\limits_{x \rightarrow a^{-}} 0=0$

(RHL at x = a) = $\lim\limits_{x \rightarrow a^{+}} g(x) \lim\limits_{x \rightarrow a^{+}} \int\limits_a^x f(t) d t=0$

and, $g(a)=\int\limits_a^a f(t) d t=0$

∴  $\lim\limits_{x \rightarrow a^{-}} g(x)=\lim\limits_{x \rightarrow a^{+}} g(x)=g(a)$

So, $g(x)$ is continuous at $x=a$.

(LHL at x = b) = $\lim\limits_{x \rightarrow b^{-}} g(x)=\lim\limits_{x \rightarrow b^{-}} \int\limits_a^x f(t) d t=\int\limits_a^b f(t) d t$

(RHL at x = b) = $\lim\limits_{x \rightarrow b^{+}} g(x)=\lim\limits_{x \rightarrow b^{+}} \int\limits_a^b f(t) d t=\int\limits_a^b f(t) d t$

and,  $g(b)=\int\limits_a^b f(t) d t$

∴  $\lim\limits_{x \rightarrow b^{-}} g(x)=\lim\limits_{x \rightarrow b^{+}} g(x)=g(b)$

So, g(x) is continuous at x = b.

Thus, g(x) is continuous at x = a and x = b.

Differentiability of g(x) :

(LHD of g(x) at x = a) = $\lim\limits_{x \rightarrow a^{-}} \frac{g(x)-g(a)}{x-a}$

$=\lim\limits_{x \rightarrow a^{-}} \frac{0-\int\limits_a^a f(t) d t}{x-a}=0$

(RHD of g(x) at x = a) = $\lim\limits_{x \rightarrow a^{+}} \frac{g(x)-g(a)}{x-a}$

$=\lim\limits_{x \rightarrow a^{+}} \frac{\int\limits_a^x f(t) d t-\int\limits_a^a f(t) d t}{x-a}$

$=\lim\limits_{x \rightarrow a^{+}} \frac{\int\limits_a^x f(t) d t-0}{x-a}$

$=\lim\limits_{x \rightarrow a^{+}} \frac{\int\limits_a^x f(t) d t}{x-a}$

$=\lim\limits_{x \rightarrow a^{+}} \frac{f(x)}{1}$            [By L'Hospital's rule]

$=\lim\limits_{x \rightarrow a^{+}} f(x) \geq 1$            [∵ Codomain f(x) = [1, ∞)]

∴  (LHD of g(x) at x = a) ≠ (RHD of g(x) at x = a)

So, g(x) is not differentiable at x = a.

(LHD of g(x) at x = b)

$=\lim\limits_{x \rightarrow b^{-}} \frac{g(x)-g(b)}{x-b}$

$=\lim\limits_{x \rightarrow b^{-}} \frac{\int\limits_a^x f(t) d t-\int\limits_a^b f(t) d t}{x-b}$

$=\lim\limits_{x \rightarrow b^{-}} \frac{f(x)-0}{1-0}=\lim\limits_{x \rightarrow b^{-}} f(x) \geq 1$          [∵ Range f(x) ⊆ [1, ∞)]

(RHD of g(x) at x = b) = $\lim\limits_{x \rightarrow b^{+}} \frac{g(x)-g(b)}{x-b}$

$=\lim\limits_{x \rightarrow b^{+}} \frac{\int\limits_a^b f(t) d t-\int\limits_a^b f(t) d t}{x-b}$

$=\lim\limits_{x \rightarrow b^{+}} \frac{0}{x-b}=0$

∴  (LHD of g(x) at x = b) ≠ (RHD of g(x) at x = b)

So, g(x) is not differentiable at x = b.