Let $ f: \mathbb{R} \to \mathbb{R} $ and $ g: \mathbb{R} \to \mathbb{R} $ be defined by $$ f(x) = \begin{cases} x (\sin x) \cos \frac{1}{x}, & x \ne 0 \\ 0, & x = 0 \end{cases} \text{ and } g(x) = \begin{cases} x \cos \frac{1}{x}, & x \ne 0 \\ 0, & x = 0 \end{cases} $$ where $ \mathbb{R} $ denotes the set of real numbers. Then, at $ x = 0, $
Differentiability Analysis at $ x = 0 $
We need to determine if the functions $ f(x) $ and $ g(x) $ are differentiable at $ x = 0 $ using the definition of the derivative.
The definition of the derivative of a function $ F(x) $ at $ x = a $ is:
$ F'(a) = \lim_{h \to 0} \frac{F(a+h) - F(a)}{h} $In our case, $ a = 0 $, and $ F(0) = 0 $ for both functions.
Derivative of $ f(x) $ at $ x = 0 $
- Apply the definition for $ f(x) $: $ f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} $
- Substitute the function definition for $ x \ne 0 $: $ f'(0) = \lim_{h \to 0} \frac{h (\sin h) \cos \frac{1}{h} - 0}{h} $
- Simplify the expression: $ f'(0) = \lim_{h \to 0} (\sin h) \cos \frac{1}{h} $
- Analyze the limit using the Squeeze Theorem. We know that $ -1 \le \cos \frac{1}{h} \le 1 $. Therefore: $ -|\sin h| \le (\sin h) \cos \frac{1}{h} \le |\sin h| $
- As $ h \to 0 $, $ \sin h \to 0 $. Thus, $ \lim_{h \to 0} (-|\sin h|) = 0 $ and $ \lim_{h \to 0} |\sin h| = 0 $.
- By the Squeeze Theorem, the limit exists: $ \lim_{h \to 0} (\sin h) \cos \frac{1}{h} = 0 $
- Conclusion: $ f'(0) = 0 $. So, $ f $ is differentiable at $ x = 0 $.
Derivative of $ g(x) $ at $ x = 0 $
- Apply the definition for $ g(x) $: $ g'(0) = \lim_{h \to 0} \frac{g(0+h) - g(0)}{h} $
- Substitute the function definition for $ x \ne 0 $: $ g'(0) = \lim_{h \to 0} \frac{h \cos \frac{1}{h} - 0}{h} $
- Simplify the expression: $ g'(0) = \lim_{h \to 0} \cos \frac{1}{h} $
- Analyze the limit. As $ h \to 0 $, the argument $ \frac{1}{h} $ approaches $ \pm \infty $. The cosine function oscillates between -1 and 1 infinitely often as its argument approaches infinity.
- Conclusion: The limit $ \lim_{h \to 0} \cos \frac{1}{h} $ does not exist. Therefore, $ g $ is NOT differentiable at $ x = 0 $.
Final Conclusion
Based on the analysis:
- $ f $ is differentiable at $ x = 0 $.
- $ g $ is NOT differentiable at $ x = 0 $.
This corresponds to the first option.
Important Questions from Differentiability
What is the value of f'(x) at x = 4 from the following table of values?
x 1 2 3 4 f(x) 20 22 27 35 The set of all points, where the function \({\rm{f}}\left( {\rm{x}} \right) = \sqrt {1 - {{\rm{e}}^{ - {{\rm{x}}^2}}}} \) is differentiable, is
Let f be a differentiable function defined for all x ∈ R such that f(x3) = x5 for all x ∈ R, x ≠ 0. Then the value of \(\dfrac{df}{dx} (8)\) is:
If \(f(x)=\displaystyle\sum_{n-0}^{2k}\left(a_n|x|^n+b_n\ \sin^2x\right)\), where \(a_i^{'}\)s and \(b_i^{'}\)s (0 ≤ i ≤ k) are real constants, then f(x) is:
The set of all point where the function f(x) = 2x|x| is differentiable, is:
