Question

Let $f,g:R \to R$be two functions defined by f\left( x \right) = \left\\{ {\begin{array}{*{20}{c}} {x\sin \left( {\dfrac{1}{x}} \right)}&{{\text{if }}x \ne 0} \\\ {0,}&{{\text{ if }}x = 0} \end{array}} \right.
and $g\left( x \right) = xf\left( x \right)$
Statement I : f is a continuous function at x=0
Statement II : g is a differentiable function at x=0
A) Both statements I and II are false.
B) Both statements I and II are true.
C) Statement I is true, statement II is false.
D) Statement I is false, statement II is true.

Answer 1

Here first we will check the continuity of the given function at $x = 0$.
For the given function to be continuous$f\left( {{0^ - }} \right) = f\left( {{0^ + }} \right) = f\left( 0 \right)$. Then we will check for the differentiability of the given function by evaluating $f'\left( x \right)$ and then find its limit at $x = 0$. If the limit exists then it is differentiable and if the limit does not exist then it is not differentiable.

Complete step-by-step answer:
The given function is:-

{x\sin \left( {\dfrac{1}{x}} \right)}&{{\text{if }}x \ne 0} \\\ {0,}&{{\text{ if }}x = 0} \end{array}} \right.$$ Let us first evaluate the value of the function as$$x \to {0^ - }$$. Hence evaluating the limit of the function we get:- $\Rightarrow$$$f\left( {{0^ - }} \right) = \mathop {\lim }\limits_{x \to {0^ - }} x\sin \left( {\dfrac{1}{x}} \right)$$ Now we know that as $$x \to {0^ - }$$ $$ - 1 \leqslant \sin \left( {\dfrac{1}{x}} \right) \leqslant 1$$ i.e. the value of $$\sin \left( {\dfrac{1}{x}} \right)$$ from -1 to 1 Hence on evaluating the limit we get:-

f\left( {{0^ - }} \right) = \mathop {\lim }\limits_{x \to {0^ - }} x\sin \left( {\dfrac{1}{x}} \right) = 0 \\
\Rightarrow f\left( {{0^ - }} \right) = 0.................................\left( 1 \right) \\

Now we will evaluate the value of the function as $$x \to {0^ + }$$. Hence evaluating the limit of the function we get:- $\Rightarrow$$$f\left( {{0^ + }} \right) = \mathop {\lim }\limits_{x \to {0^ + }} x\sin \left( {\dfrac{1}{x}} \right)$$ Now we know that as $$x \to {0^ + }$$ $$ - 1 \leqslant \sin \left( {\dfrac{1}{x}} \right) \leqslant 1$$ i.e. the value of $$\sin \left( {\dfrac{1}{x}} \right)$$ from -1 to 1 Hence on evaluating the limit we get:-

f\left( {{0^ + }} \right) = \mathop {\lim }\limits_{x \to {0^ + }} x\sin \left( {\dfrac{1}{x}} \right) = 0 \\
\Rightarrow f\left( {{0^ + }} \right) = 0.................................\left( 2 \right) \\

Also, it is given that at $${\text{ }}x = 0$$ $$f\left( x \right) = 0$$………………………..(3) From 1, 2 and 3 we get:- $\Rightarrow$$$f\left( {{0^ - }} \right) = f\left( {{0^ + }} \right) = f\left( 0 \right)$$ Hence the given function is continuous. Now we will evaluate the derivative of the given function g(x). $\Rightarrow$$$g\left( x \right) = xf\left( x \right)$$ Putting the value of f(x) we get:- $\Rightarrow$$$g\left( x \right) = x\left[ {x\sin \left( {\dfrac{1}{x}} \right)} \right]$$ Simplifying it we get:- $\Rightarrow$$$g\left( x \right) = {x^2}\sin \left( {\dfrac{1}{x}} \right)$$ Now when we put x=0 we get:- $$g(0) = 0$$ Differentiating the function g(x) using first principle we get:- $$g\prime (x) = \mathop {lim}\limits_{h \to 0} \dfrac{{g(x + h) - g(x)}}{h}\;{\text{ }}\;{\text{ }}\;$$ Putting in x=0 we get:- $\Rightarrow$$$g\prime (0) = \mathop {lim}\limits_{h \to 0} \dfrac{{g(h) - g(0)}}{h}\;$$ Putting in the values we get:- $\Rightarrow$$$g\prime (0) = \mathop {lim}\limits_{h \to 0} \dfrac{{{h^2}\sin \left( {\dfrac{1}{h}} \right) - 0}}{h}\;{\text{ }}$$ Simplifying it further we get:- $\Rightarrow$$$g\prime (0) = \mathop {lim}\limits_{h \to 0} h\sin \left( {\dfrac{1}{h}} \right){\text{ }}$$ $\Rightarrow$$$ \Rightarrow g\prime (0) = \mathop {lim}\limits_{h \to 0} \dfrac{{\sin \left( {\dfrac{1}{h}} \right){\text{ }}}}{{\dfrac{1}{h}}}\;{\text{ }}\;$$ Now we know that, $\Rightarrow$$$\mathop {lim}\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$$ Applying this property we get:- $$g\prime (0) = 1\;$$ (finite) Therefore, g(x) is differentiable at x=0. **Hence option B is correct.** **Note:** Students should take note that: A function f is continuous when, for every value a in its domain the function is defined i.e. $$f\left( a \right)$$ is defined and $$\mathop {lim}\limits_{x \to a} f\left( x \right) = f(a)$$. Also, students should note that every differentiable function is continuous but the converse is not true i.e. every continuous function is not differentiable; it may or may not be differentiable.