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

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