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, b a, c a, d b, c |
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. |