Opsimath

2021/09/07阅读：71主题：默认主题

# Mathematics Interview Questions (VIII)

### 36. Integrate .

*Solution*. Let
, then
. As a result,
.

### 37. Integrate between limits of 1 and 0. Draw that graph.

*Solution*.
.

Hence, .

To draw the graph, consider its derivative , which is greater than 0 when or . Also, it passes through the origin. As , using L'Hôpital's rule; as , as well. You can view here for a sample.

### 38. Integrate between limits of 1 and -1. Draw the graph. Why is it -2 and not infinity, as it appears to be on the graph?

*Solution*.
.

For its graph, please refer to question 35.

The reason why it is -2 is because of the singularity (tends to ) at . That integral is therefore improper and does not refer to the area under the graph.

### 39. Write down 3 digits, and then write the number again next to itself, e.g., 145145. Why is it divisible by 13?

*Solution*. Such a number has the form
where
. Since
, the number always contains a factor of 13.

### 40. You are given a triangle with a fixed perimeter and you want to maximise the area. What shape will it be? Prove it.

*Solution 1*. Suppose that the lengths of the sides of the triangle are
,
, and
such that the perimeter of the triangle is fixed and it is
. Using the Heron's formula, the area
.

The AM-GM inequality (with 3 terms) states that for , with equality at .

Let , , and . Since where , . Similarly, . We can then use the AM-GM inequality to obtain that . As a result, and , with equality when , i.e., . The shape is therefore equilateral.

*Solution 2*. Suppose that the lengths of the sides of the triangle are
,
, and
such that the perimeter of the triangle is fixed and it is
. Let
,
, and
, respectively, be the vertices of the triangle opposite the sides of length
,
, and
.

Given , we'll show that it must be the case that , proving that the largest triangle is isosceles. Repeating the argument given or will prove it must be equilateral.

Place and on the x-axis, and consider where can lie. Wherever it is, the sum of the distances from to the two points and is , which is constant, so the locus of is an ellipse with foci and . The area of is , where is the -coordinate of P, so the triangle's area is greatest when is at a point on the ellipse as far from the -axis as possible. Therefore is at an endpoint of the ellipse's vertical axis, and by symmetry, .

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