Question

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


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

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

Answer 1

Hint: Here we have f : [a, b]$ \to $[1,$\infty $] and f(x)$ \geqslant 1$,$x \in [a,b]$ start solving with left hand derivative and then do the right hand derivative. Always keep in mind the values given for f(x) function

Complete step-by-step answer:
Here we have f(x) $ \geqslant 1$,$x \in [a,b]$
And for g(x) we have
Left-hand derivative x = a is zero
Right-hand derivative at x = a is
We have to find the derivative of g(x) at x = a
$\mathop {\lim }\limits_{x \to {a^ + }} \dfrac{{\int\limits_a^x {f(t)dt - 0} }}{{x - a}} = \mathop {\lim }\limits_{x \to {a^ + }} f(x) \geqslant 1$
We can see that function g(x) is not differentiable at x = a
And as the same process we apply for doing the derivative of left hand at x = b we find that the derivative of right hand and left hand are not equal
LHD$ \ne $RHD
At both point x =a and x = b
Hence, we can say that g(x) is continuous but not differentiable at b

So, the correct answer is “Option C”.

Note: Here students get confused and do the mistake during the process of finding derivatives. Always first check the values given for f(x) and start solving g(x) with taking the help of f(x) values.