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 \(\Bbb R\).

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TIFR 2019 GS Part A, Problem 3

Calculus

Hard

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)?

The derivative of a function may be used to determine whether the function is increasing or decreasing on any intervals in its domain. If f′(x) > 0 at each point in an interval I, then the function is said to be increasing on I. f′(x) < 0 at each point in an interval I, then the function is said to be decreasing on I. Because the derivative is zero or does not exist only at critical points of the function, it must be positive or negative at all other points where the function exists. In determining intervals where a function is increasing or decreasing, you first find domain values where all critical points will occur; then, test all intervals in the domain of the function to the left and to the right of these values to determine if the derivative is positive or negative. If f′(x) > 0, then f is increasing on the interval, and if f′(x) < 0, then f is decreasing on the interval. This and other information may be used to show a reasonably accurate sketch of the graph of the function. \(f'(1)=1+2\cos(\pi)+\pi \sin(\pi)=-1<0\)

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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