Question-15

Consider the function:

\[ \begin{equation*} f( x) =\begin{cases} \sin( 2x) , & x\leqslant 0\\ mx, & x >0 \end{cases} \end{equation*} \]

• \(\displaystyle f\) is continuous for exactly one value of \(\displaystyle m\)
• \(\displaystyle f\) is continuous for any \(\displaystyle m\)
• \(\displaystyle f\) is differentiable for exactly one value of \(\displaystyle m\)
• \(\displaystyle f\) is differentiable for any \(\displaystyle m\)

Answer

• \(\displaystyle f\) is continuous for exactly one value of \(\displaystyle m\)
• \(\displaystyle f\) is continuous for any \(\displaystyle m\)
• \(\displaystyle f\) is differentiable for exactly one value of \(\displaystyle m\)
• \(\displaystyle f\) is differentiable for any \(\displaystyle m\)

Solution

We have:

\[ \begin{equation*} \lim\limits _{x\rightarrow 0^{+}} \ f( x) =\lim\limits _{x\rightarrow 0^{-}} \ f( x) =0 \end{equation*} \]

Since limit at \(\displaystyle x=0\) is equal to \(\displaystyle f( 0)\), \(\displaystyle f\) is continuous for all \(\displaystyle m\). The right hand derivative for \(\displaystyle f\) at \(\displaystyle x=0\) is \(\displaystyle m\) and the left hand derivative at \(\displaystyle x=0\) is \(\displaystyle 2\). For \(\displaystyle f\) to be differentiable at \(\displaystyle x=0\), \(\displaystyle m=2\) is the only possibility.