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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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Opsimath