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Question
The set of points where the function f (x) given by f (x) = |x − 3| cos x is differentiable, is
Options
R
R − {3}
(0, ∞)
none of these
Solution
(b)
\[\left(\text { LHD at x } = 3 \right) = \lim_{x \to 3^-} \frac{f\left( x \right) - f\left( 3 \right)}{x - 3}\]
\[\left( \text { LHD at x = 3 } \right) = \lim_{h \to 0} \frac{f\left( 3 - h \right) - f\left( 3 \right)}{3 - h - 3}\]
\[\left( \text { LHD at x = 3 } \right) = \lim_{h \to 0} \frac{f\left( 3 - h \right) - f\left( 3 \right)}{- h}\]
\[\left( \text { LHD at x = 3} \right) = \lim_{h \to 0} \frac{\left| 3 - h - 3 \right|\cos\left( 3 - h \right) - f\left( 3 \right)}{- h}\]
\[\left(\text{ LHD at x } = 3 \right) = \lim_{h \to 0} \frac{h\cos\left( 3 - h \right) - 0}{- h} = - \cos3\]
\[\left( \text { RHD at x } = 3 \right) = \lim_{x \to 3^+} \frac{f\left( x \right) - f\left( 3 \right)}{x - 3}\]
\[\left( \text { RHD at x = 3 } \right) = \lim_{h \to 0} \frac{f\left( 3 + h \right) - f\left( 3 \right)}{3 + h - 3}\]
\[\left( \text { RHD at x } = 3 \right) = \lim_{h \to 0} \frac{f\left( 3 + h \right) - f\left( 3 \right)}{h}\]
\[\left( \text { RHD at x = 3 } \right) = \lim_{h \to 0} \frac{\left| 3 + h - 3 \right|\cos\left( 3 + h \right) - f\left( 3 \right)}{h}\]
\[\left( \text { RHD at x } = 3 \right) = \lim_{h \to 0} \frac{h\cos\left( 3 + h \right) - 0}{h} = \cos3\]
So, f(x) is not differentiable at x = 3.
Also, f(x) is differentiable at all other points because both modulus and cosine functions are differentiable and the product of two differentiable function is differentiable.
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