Opsimath

2021/09/16  阅读：27  主题：默认主题

# Mathematics Interview Questions (XVI)

### Incidentally ...

Yesterday someone told me that my questions are a little too easy. Actually, they are from a (very old) The Student Room thread. I modified the wording slightly to fix some grammatical mistakes and mathematical ambiguities. However, in order to make them more interesting (!), I will be attaching an "extension" section after each solution. It will give the more advanced readers an opportunity to think beyond the questions themelves. Enjoy it!

### 81. Draw the graphs of , , and .

Solution. For , direct integration gives where is an arbitrary constant.

For , consider the identity . This gives .

For , consider and . This means

Extension. What about ? There are actually several methods of doing it. Try at least two!

### 82. Prove the infinity of primes. Then prove the infinity of primes of the form .

Solution. If there are only finite primes, we may list them in ascending order as due to the well-ordering principle. Now consider . This number is greater than any and must have at least one prime factor, say . If for some , must also divide which is clearly absurd. is therefore another distinct prime, leading to a contradiction.

Similarly, if there are only finite primes of the form , we may list them in ascending order as due to the well-ordering principle. Now consider . This number is of the form and is greater than any so must have at least one prime factor of the form . Using the fact that any number of the form has only prime factors of the form (see below), division into by each prime factor of the form leaves a remainder 1, so cannot be composite, leading to a contradiction.

Proof of the aforementioned fact. Using Lagrange's theorem of group theory, , so has order 4 and so 4 divides .

Extension. There is a book named Proofs from THE BOOKS of which the first chapter has 6 different proofs of the infinity of primes, ranging from analytical to topological methods. You may get a PDF version of it here. Have fun!

### 83. Differentiate .

Solution. Using chain rule, .

Extension. Draw its graph. What about ? What about their graphs for ? What about ? ?

### 84. Show that has no real roots if .

Solution 1. By direct expansion, . Its determinant should be . Then , giving which is true iff .

Solution 2. The original equation implies ; the LHS is while the RHS is which are equal iff LHS = RHS = 0. Hence .

Extension. See the next question.

### 85. Hence show that, if , (i) has 1 real root; (ii) has no real roots; and (iii) has 2 real roots.

Solution. (i) This gives so , giving . (ii) The original equation implies ; the LHS is while the RHS is which are equal iff LHS = RHS = 0. Hence , which is impossible. (iii) and are obviously roots. Differentiating the equation with respect to gives , which, from (i), has only one root. This means that this function can only "turn" once, giving only two roots. These two roots are distinct since .

Extension. Consider the number of roots of and . What do you notice? Can you prove a general pattern?

Opsimath

2021/09/16  阅读：27  主题：默认主题