Week 2 Challenging Problems

Problems for the week

Deadline 26.04.2020

Problem 1.

Determine all positive integers K such that,

    \[\frac{\tau{(n^2)}}{\tau{(n)}}=k\]

for some n, where \tau(n) is the number of divisors of n.

Problem 2.

Find the first digit before and after the decimal point in \bigg(\sqrt{2}+\sqrt{3}\bigg)^{1980}. (without using a calculator)

Problem 3.

2n points are chosen on a circle. In how many ways can you join pairs of points by nonintersecting chords?

Problem 4.

Let f:[a,b] \to \mathbb{R} be a function, continuous on [a,b] and differentiable on (a,b). Prove that if there exists c\in (a,b) such that

    \[\frac{f(b)-f(c)}{f(c)-f(a)}<0\]

then there exists \xi \in (a,b) such that f'(\xi)=0.

To check week 1 challenging problems and discussion: Click here

We will post the solutuions next week along with a new set of problem.

1 Comment

  • Answer for problem 4:
    case 1:: f(b)<f(c) and f(a)f(b)-f(c)f'(j)=[f(b)-f(c)]+[f(c)-f(a)]/ (b-a)
    Clearly if l f(b)-f(c) l 0 else f'(j)<0

    Case 2:: f(c)f(c)

    By similar way we can say that there exist k belongs to (a, b) such that f'(k)<0 when l f(b)-f(c) l 0.

    Combining cases we say that there exist two points j and k such that f'(j)f'(k) there exist i in (a, b) such that f'(i)=0.

    Note :::if l f(b)-f(c) l = l f(c)-f(a) l then f'(j)=f'(k)=0.

    This complete the proof.

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