Question 7:

Write the points where f (x) = |log_e x| is not differentiable.

Answer:

Given: $f (x) = |\log_{e} x| = \{\begin{cases} - \log_{e} x, 0 < x < 1 \\ \log_{e} x, x \geq 1 \end{cases}$
Clearly $f (x)$ is differentiable for all $x > 1$ and for all $x < 1$ . So, we have to check the differentiability at $x = 1$ .
We have,
(LHD at x = 1)

$\lim_{x \to 1^{-}} \frac{f (x) - f (1)}{x - 1}$

$= \lim_{x \to 1^{-}} \frac{- \log x - \log 1}{x - 1} = - \lim_{x \to 1^{-}} \frac{\log x}{x - 1} = - \lim_{h \to 0} \frac{\log (1 - h)}{1 - h - 1} = - \lim_{h \to 0} \frac{\log (1 - h)}{- h} = - 1$

(RHD at x=1)
= $\lim_{x \to 1^{+}} \frac{f (x) - f (1)}{x - 1}$

$= \lim_{x \to 1^{+}} \frac{\log x - \log 1}{x - 1} = \lim_{x \to 1^{+}} \frac{\log x}{x - 1} \underset{}{= \lim_{h \to 0}} \frac{\log (1 + h)}{1 + h - 1} = \lim_{h \to 0} \frac{\log (1 + h)}{h} = 1$

Thus, (LHD at x =1) ≠ (RHD at x =1)
So, $f (x)$ is not differentiable at $x = 1 .$

Page No 10.17:

Question 8:

Write the points of non-differentiability of $f (x) = |\log |x|| .$

(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} .$

Page No 10.19:

Question 18:

The function f (x) = 1 + |cos x| is
(a) continuous no where
(b) continuous everywhere
(c) not differentiable at x = 0
(d) not differentiable at x = n π, n ∈ Z

Answer:

(b) continuous everywhere

Graph of the function f (x) = 1 + |cos x| is as shown below:

From the graph, we can see that f (x) is everywhere continuous but not differentiable at $x = (2 n + 1) \frac{π}{2}, n \in Z$

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