 墨滴 Opsimath

2021/08/30  阅读：48  主题：默认主题

Mathematics Interview Questions (I)

1. Differentiate .

Solution 1. Consider the function . Taking the natural logarithm on both sides gives . Differentiate both sides, and we get . Hence, .

Solution 2. Consider the identity , which is trivial since and are inverse functions to each other. So . Using the chain rule,

.

2. Integrate and .

Solution. For , consider the identity . This gives where is an arbitrary constant.

For , consider and . This means

3. What is the square root of ?

Solution. The square root of must also be a complex number, so let . Squaring both sides gives . Equate both the real and the imaginary parts, and we get

where means or . If then gives , so because and are both positive. If then gives which is impossible since a square cannot be negative. In a word, .

4. If I had a cube and six colours, and painted each side a different colour, how many (different) ways could I paint the cube? What about if I had colours instead of 6?

Solution 1. Case : Colour one side with the ugliest colour, and put the cube on a table ugly side down. There are choices for the colour on top. For each of these choices, colour the side facing you with the nicest remaining colour. The last three sides can be coloured in ways, so the number of colourings is .

Case : First choose the colours, then use them. The number of colourings is .

Solution 2. Case : There are permutations in total but each 24 of them are isomorphic (6 sides, 4 rotations). Hence, there are different colourings.

Case : Same as before.

5. Prove that root 2 is irrational.

Solution. Proof by contradiction. Assume that is rational, so it can be written as where and . Squaring both sides gives , so . This means must be even, so is also even. (Otherwise, if is odd, then is odd; contradiction.) We can then write as for some . Therefore, , i.e., . This means must be even, so is also even, by the same logic above. and are both even, but ; contradiction. Opsimath

2021/08/30  阅读：48  主题：默认主题 