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The set of points where the functions f given by $f(x)=|2x-1|\sin x$ is differentiable is

\begin{array}{1 1}(A)\;R & (B)\;R-\left \{\frac{1}{2}\right \}\\(C)\;(0,\infty) & (D)\;none\;of\;these.\end{array}

1 Answer

  • Every differentiable function is continuous,but the converse is not true.
  • A function is said to be differentiable at every point in its domain.
Step 1:
$f(x)=|2x-1|\sin x$
Consider $2x-1$,when $x=\large\frac{1}{2}$,then $f(x)=0$
Hence the function is not differentiable .
Therefore the function is differentiable at the set of points (i.e)$R$-{$\large\frac{1}{2}$}
Hence $B$ is the correct option.
answered Jul 4, 2013 by sreemathi.v