In this problem set we present four questions:

**(Middle School Students)**

The irrational number is formed

by using alternating blocks of and zeros, where the th block

of zeros following the decimal contains zeros. What is the digit

in the 2550th place following the decimal?

**(High School Students)** Consider the function given by

Find the maximum and minimum value of the function.

**(Undergraduate Students)** Let be an infinitely differentiable function, such that

- whenever
- and where and is the -th derivative of

Let is given by

Decide whether is infinitely differentiable on .

**(Masters Students and above)** If is a measurable subset of such that

then has measure

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Undergraduate : No, it may not be.Consider the function $$Since as for each , it can be shown inductively that its k-th derivative as and for each .Hence is a function and satisfies the hypothesis of the question.Now we show that the second derivative of the function is discontinuous at .For , . At the second term as . But the first term as . Hence as but So is not a function.

On the other hand, the function $$ is such that for each and clearly its square root is also a function

the digit in the 2550th place is 4. The first 71 blocks consist of 2546 digits. the 4th digit of the 72 th block is 4, which is the 2550 digit. we can guess that 2550 th digit occured in the 72 th block by using 5+6+…….+n=2550, which gives n=71.05.

the digit in the 2550th place is 4. The first 71 blocks consist of 2546 digits. the 4th digit of the 72 th block is 4, which is the 2550 digit. we can guess that 2550 th digit occured in the 72 th block by using 5+6+…….+n=2550, which gives n=71.05.

the digit in the 2550th place is 4. The first 71 blocks consist of 2546 digits. the 4th digit of the 72 th block is 4, which is the 2550 digit. we can guess that 2550 th digit occured in the 72 th block by using 5+6+…….+n=2550, which gives n=71.05.

For the undergraduate problem, I guess something surprising is going to happen…

For the graduate problem, if , take a rectangle almost occupied by . Choose a point sufficiently close to its center and consider reflection about .

For the high school problem, the maximum of should be attained at the solutions of in the interval . Numerically it is

with .

The value at 4.6 (slightly smaller than 3*pi/2 exceeds the value at 3*pi/2.

Kavi Rama Murthy

I am trying to give solution to the problem for high school students

f(x)=1-(sinx/x) for x not equal to 0

=0 otherwise

as x-sinx>0, 1>sinx/x

therefore 1-sinx/x will be positive for x belongs to (0,pi/2] & as x takes values greater the numerator of sinx/x will be greater hence the 1-sinx/x will be positive

therefore minimum value of f(x)=0

in [pi,3pi/2]

sinx is negative

hence at 3pi/2 the valiue of sinx/x will be -2/(3pi)

hence value of f(x) becomes 1-(-2/(3pi)) i.e. 1+2/3(pi)

as x takes value greater the value of sinx/x will be very very small

hence at x=3pi/2 f(x) is maximum

hence maximum of f(x)=1+2/3(pi)