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# Understand the problem

Define a function: $$f (x) =\begin{cases}\frac{x+x^2\cos(\pi/x)}{x}, & \text{if \(x\neq 0$$} \\ 0, & \text{if $$x=0$$}\end{cases}\)

#### Consider the following statements:

1. $$f'(0)$$ exists and is equal to $$1$$.
2. $$f$$ is not increasing in any neighborhood of $$0$$
3. $$f'(0)$$ does not exist.
4. $$f$$ is increasing on $$\Bbb R$$.
How many of the above statements is/are true?
• 0
• 1
• 2
• 3
##### Source of the problem
TIFR 2019 GS Part A, Problem 3
Calculus
Hard

### Introduction to Real Analysis by Donald R. Sherbert and Robert G. Bartle

Do you really need a hint? Try it first!

$$f'(0)=\lim_{h \to 0}\frac{f(h)-f(0)}{h}=\lim_{h \to 0} \frac{h+h^2\cos(\pi/h)}{h}=\lim_{h \to 0}1+h\cos(\pi/h)=1$$ $$f'(0)=1>0$$
Can you comment something about ii)?
$$f'(1)=1+2\cos(\pi)+\pi \sin(\pi)=-1<0$$ Can you coment something about iv)?

$$\frac{\pi}{x}\in \{-\frac{\pi}{2}+2\pi k| k\in \Bbb Z\}$$ So, $$x\in \{\frac{2}{4k-1}: k \in \Bbb Z\}$$.
$$f'(x_n)=-\pi<0$$ for $$x_n \to 0$$.   Hence, every interval around zero contains such a point in $$x$$, where $$f'(x)<0$$. Can you comment something about ii)?

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