Question

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

g(x)=$\{\begin{array}{cc}0& ifx<a,\\ {\int }_a^{x}dt& ifa\le x\le b,Then\\ {\int }_a^{b}f\left(t\right)dt& ifx>b\end{array}$

• A. g(x) is continuous but not differentiable at b
• B. g(x) is continuous but not different at a
• C. g(x) is continuous and differentiable at either a or b but not both
• D. g(x) is different on i.

Answer 1

Answer: (A), (B)

g(x) is continuous but not different at a

$g\left(a^{-}\right)0,g\left(a\right)={\int }_a^{a}f\left(t\right)dt=0,g\left(a^{+}\right)$

$={lim}_{h\to 0}{\int }_a^{a+h}f\left(t\right)dt=0$

$g\left(b^{-}\right){lim}_{h\to 0}{\int }_a^{b-h}f\left(t\right)dt={\int }_a^{b}f\left(t\right)dt$

$g\left(b^{-}\right)={lim}_{h\to 0}{\int }_a^{b-h}f\left(t\right)dt={\int }_a^{b}f\left(t\right)dt$

$g\left(b\right)={\int }_a^{b}f\left(t\right)dt=g\left(b^{+}\right)$

Hence g(x) is continuous at x = a as well as x = b

Now, $g^{{}^{\prime }}\left(a^{-}\right)={lim}_{h\to 0}\frac{g(a-h)-g\left(a\right)}{-h}=0$

$g^{\prime }\left(a^{+}\right)={lim}_{h\to 0}=\frac{g(a+h)-g\left(a\right)}{h}$

$={lim}_{h\to 0}\frac{O_a^{a+h}f\left(t\right)dt-0}{h}={lim}_{h\to 0}\frac{f(a+h)}{1}$

$=f\left(a\right)(\ne 0)$

Hence g (x) is not differentiable at x =a.

$g^{\prime }\left(b^{-}\right)={lim}_{h\to 0}\frac{g(b-h)-g\left(b\right)}{-h}$

$={lim}_{h\to 0}\frac{O_a^{b-h}f\left(t\right)dt=O_a^{b}f\left(t\right)dt}{-h}$

$={lim}_{h\to 0}\frac{f(b-h)}{1}=f\left(b\right){(}^{1}0)$

$g^{\prime }\left(b^{+}\right)={lim}_{h\to 0}\frac{g(b+h)-g\left(b\right)}{h}$

$={lim}_{h\to 0}\frac{O_a^{b}f\left(t\right)dt-O_a^{b}f\left(t\right)dt}{h}=0$

Hence g(x) is not differentiable at x = b.