# 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:

- \(f'(0)\) exists and is equal to \(1\).
- \(f\) is not increasing in any neighborhood of \(0\)
- \(f'(0)\) does not exist.
- \(f\) is increasing on \(\Bbb R\).

- 0
- 1
- 2
- 3

##### Source of the problem

TIFR 2019 GS Part A, Problem 3

##### Topic

Calculus

##### Difficulty Level

Hard

##### Suggested Book

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

# Start with hints

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)?

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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