Mathematics Interview Questions (V)

21. Integrate and differentiate .


22. Draw .

Solution. is undefined because it is oscillating with constant amplitude 1.

Since , , so the graph is bounded by .

, so iff , i.e., . The leftmost (rightmost) stationary point is .

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23. Differentiate .

Solution 1. , so . From question 1, . Hence, .

Solution 2. Consider the function . Taking the natural logarithm on both sides gives . Differentiate both sides, and we get . Hence, .

24. What do you know about triangles?

(Partial) Solution. It is the simplest polygon, with 3 edges, 3 vertices, and 3 angles; the sum of the three angles is 180°; the edges and angles obey the sine and cosine rules. It can be used to define trigonometric functions for acute angles. The area can be calculated using Heron's formula.

An equilateral triangle has three sides of the same length. An isosceles triangle has two sides of equal length. A scalene triangle has all its sides of different lengths. A right triangle has one of its interior angles measuring 90° (a right angle). It obeys the Pythagorean theorem.

An altitude of a triangle is a straight line through a vertex and perpendicular to (i.e. forming a right angle with) the opposite side. The three altitudes intersect in a single point, called the orthocenter of the triangle, usually denoted by H.

An angle bisector of a triangle is a straight line through a vertex which cuts the corresponding angle in half. The three angle bisectors intersect in a single point, the incenter, usually denoted by I, the center of the triangle's incircle.

A median of a triangle is a straight line through a vertex and the midpoint of the opposite side, and divides the triangle into two equal areas. The intersection of the medians is the centroid. The three medians intersect in a single point, the centroid, usually denoted by G.

The intersection of the perpendicular bisectors of each edge is called the circumcenter, usually denoted by O, the center of the triangle's circumcircle.

25. Find a series of consecutive integers such that the sum of the series is a power of 2.

Solution. Let the integers be . Then their sum is and therefore should be a power of 2.

This is only possible if and is a power of 2, because if , then it is either even or odd. If even, is odd. If odd, is odd. So, in either case, the sum has an odd factor.

If the integers are allowed to be odd, then noticing that gives another representation.