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

Hint: 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:-
\[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
  {x\sin \left( {\dfrac{1}{x}} \right)}&{{\text{if }}x \ne 0} \\
  {0,}&{{\text{ if }}x = 0}
\end{array}} \right.\]
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.