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.
Answer & explanation
Correct answer: option 2
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.