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Geometrical Representation of Conjugate of a Complex Number

Let fx=|cos x...

Question

Let f (x) = |cos x|. Then,
(a) f (x) is everywhere differentiable
(b) f (x) is everywhere continuous but not differentiable at x = n π, n ∈ Z
(c) f (x) is everywhere continuous but not differentiable at $x = (2 n + 1) \frac{π}{2}, n \in Z$ .
(d) none of these

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Solution

(c) f (x) is everywhere continuous but not differentiable at $x = (2 n + 1) \frac{π}{2}, n \in Z$ .

$We have, f (x) = |\cos x| \Rightarrow f (x) = \{\begin{cases} \begin{array}{l} \begin{matrix} \cos x, 2 n π \leq x < (4 n + 1) \frac{π}{2} \\ 0, x = (4 n + 1) \frac{π}{2} \\ - \cos x, (4 n + 1) \frac{π}{2} < x < (4 n + 3) \frac{π}{2} \end{matrix} \\ 0, x = (4 n + 3) \frac{π}{2} \end{array} \\ \cos x, (4 n + 3) \frac{π}{2} < x \leq (2 n + 2) π \end{cases} When, x is in first quadrant, i . e . 2 n π \leq x < (4 n + 1) \frac{π}{2}, we have f (x) = \cos x which being a trigonometrical function is continuous and differentiable in (2 n π, (4 n + 1) \frac{π}{2}) When, x is in second quadrant or in third quadrant, i . e ., (4 n + 1) \frac{π}{2} < x < (4 n + 3) \frac{π}{2}, we have f (x) = - \cos x which being a trigonometrical function is continuous and differentiable in ((4 n + 1) \frac{π}{2}, (4 n + 3) \frac{π}{2}) When, x is in fourth quadrant, i . e ., (4 n + 3) \frac{π}{2} < x \leq (2 n + 2) π, we have f (x) = \cos x which being a trigonometrical function is continuous and differentiable in ((4 n + 3) \frac{π}{2}, (2 n + 2) π) Thus possible point of non - differentiability of f (x) are x = (4 n + 1) \frac{π}{2}, (4 n + 3) \frac{π}{2} Now, LHD [at x = (4 n + 1) \frac{π}{2}] = \lim_{x \to (4 n + 1) {\frac{π}{2}}^{-}} \frac{f (x) - f ((4 n + 1) \frac{π}{2})}{x - (4 n + 1) \frac{π}{2}} = \lim_{x \to (4 n + 1) {\frac{π}{2}}^{-}} \frac{\cos x - 0}{x - (4 n + 1) \frac{π}{2}} = \lim_{x \to (4 n + 1) {\frac{π}{2}}^{-}} \frac{- \sin x}{1 - 0} [By L' Hospital rule] = - 1 And RHD (at x = (4 n + 1) \frac{π}{2}) = \lim_{x \to (4 n + 1) {\frac{π}{2}}^{+}} \frac{f (x) - f ((4 n + 1) \frac{π}{2})}{x - (4 n + 1) \frac{π}{2}} = \lim_{x \to (4 n + 1) {\frac{π}{2}}^{+}} \frac{- \cos x - 0}{x - (4 n + 1) \frac{π}{2}} = \lim_{x \to (4 n + 1) {\frac{π}{2}}^{+}} \frac{\sin x}{1 - 0} [By L' Hospital rule] = 1 ∴ \lim_{x \to (4 n + 1) {\frac{π}{2}}^{-}} f (x) \neq \lim_{x \to (4 n + 1) {\frac{π}{2}}^{+}} f (x) So f (x) is not differentiable at x = (4 n + 1) \frac{π}{2} Now, LHD [at x = (4 n + 3) \frac{π}{2}] = \lim_{x \to (4 n + 1) {\frac{π}{2}}^{-}} \frac{f (x) - f ((4 n + 3) \frac{π}{2})}{x - (4 n + 3) \frac{π}{2}} = \lim_{x \to (4 n + 3) {\frac{π}{2}}^{-}} \frac{- \cos x - 0}{x - (4 n + 3) \frac{π}{2}} = \lim_{x \to (4 n + 3) {\frac{π}{2}}^{-}} \frac{\sin x}{1 - 0} [By L' Hospital rule] = 1 And RHD (at x = (4 n + 3) \frac{π}{2}) = \lim_{x \to (4 n + 3) {\frac{π}{2}}^{+}} \frac{f (x) - f ((4 n + 3) \frac{π}{2})}{x - (4 n + 3) \frac{π}{2}} = \lim_{x \to (4 n + 3) {\frac{π}{2}}^{+}} \frac{\cos x - 0}{x - (4 n + 3) \frac{π}{2}} = \lim_{x \to (4 n + 3) {\frac{π}{2}}^{+}} \frac{- \sin x}{1 - 0} [By L' Hospital rule] = - 1 ∴ \lim_{x \to (4 n + 3) {\frac{π}{2}}^{-}} f (x) \neq \lim_{x \to (4 n + 3) {\frac{π}{2}}^{+}} f (x) So f (x) is not differentiable at x = (4 n + 3) \frac{π}{2} Therefore, f (x) is neither differentiable at (4 n + 1) \frac{π}{2} nor at (4 n + 3) \frac{π}{2} i . e . f (x) is not differentiable at odd multiples of \frac{π}{2} i . e . f (x) is not differentiable at x = (2 n + 1) \frac{π}{2} Therefore, f (x) is everywhere continuous but not differentiable at (2 n + 1) \frac{π}{2} .$

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