Can there exist two functions \(f\) and \(g\) that take only integer values such that for any integer \(x\) the following relations hold:
a) \(f (f (x)) = x\), \(g (g (x)) = x\), \(f (g (x)) > x\), \(g (f (x)) > x\)?
b) \(f (f (x)) < x\), \(g (g (x)) < x\), \(f (g (x)) > x\), \(g (f (x)) > x\)?
The quadratic trinomials \(f (x)\) and \(g (x)\) are such that \(f' (x) g' (x) \geq | f (x) | + | g (x) |\) for all real \(x\). Prove that the product \(f (x) g (x)\) is equal to the square of some trinomial.
Calculate \(\int_0^{\pi/2} (\sin^2 (\sin x) + \cos^2 (\cos x))\,dx\).
Suppose that in each issue of our journal in the “Quantum” problem book there are five mathematics problems. We denote by \(f (x, y)\) the number of the first of the problems of the \(x\)-th issue for the \(y\)-th year. Write a general formula for \(f (x, y)\), where \(1 \geq x \geq 12\) and \(1970 \geq y \geq 1989\). Solve the equation \(f (x, y) = y\). For example, \(f (6, 1970) = 26\). Since \(1989\), the number of tasks has become less predictable. For example, in recent years, half the issues have 5 tasks, and in other issues there are 10. Even the number of magazine issues has changed, no longer being 12 but now 6.
Author: V.A. Popov
On the interval \([0; 1]\) a function \(f\) is given. This function is non-negative at all points, \(f (1) = 1\) and, finally, for any two non-negative numbers \(x_1\) and \(x_2\) whose sum does not exceed 1, the quantity \(f (x_1 + x_2)\) does not exceed the sum of \(f (x_1)\) and \(f (x_2)\).
a) Prove that for any number \(x\) on the interval \([0; 1]\), the inequality \(f (x_2) \leq 2x\) holds.
b) Prove that for any number \(x\) on the interval \([0; 1]\), the \(f (x_2) \leq 1.9x\) must be true?
The function \(f (x)\) for each real value of \(x\in (-\infty, + \infty)\) satisfies the equality \(f (x) + (x + 1/2) \times f (1 - x) = 1\).
a) Find \(f (0)\) and \(f (1)\). b) Find all such functions \(f (x)\).
\(a_1, a_2, a_3, \dots\) is an increasing sequence of natural numbers. It is known that \(a_{a_k} = 3k\) for any \(k\). Find a) \(a_{100}\); b) \(a_{2022}\).
Let \(n\) be some positive number. It is obvious that \[2n-1<2n.\] Take another positive number \(a\), and multiply both sides of the inequality by \((-a)\) \[-2na +a< -2na.\] Now, subtracting \((-2na)\) from both sides of the inequality we get \(a<0\).
Thus, ALL POSITIVE NUMBERS ARE NEGATIVE!
Suppose \(a \neq b\). We can write \[-a = b - (a+b)\] and \[-b = a - (a+b)\] Since \((-a)b = a(-b)\), then \[( b - (a+b))b = a(a - (a+b))\] Removing the brackets, we have \[b^2 - (a+b)b = a^2 - a(a+b)\] Adding \(\left(\frac{a+b}{2}\right)^2\) to each member of the equality we may complete the square of the differences of two numbers \[\left(b - \frac{a+b}{2}\right)^2 = \left(a - \frac{a+b}{2}\right)^2\] From the equality of the squares we conclude the equality of the bases \[b - \frac{a+b}{2} = a - \frac{a+b}{2}.\] Adding \(\frac{a+b}{2}\) to both sides of equality we get \(a=b\). Therefore, WE HAVE SHOWED THAT FROM \(a\neq b\) IT FOLLOWS \(a=b\).
Let \(x\) be equal to 1. Then we can write \(x^2=1\), or putting it differently \(x^2 -1 =0\). By using the difference of two squares formula we get \[(x+1)(x-1)=0\] Dividing both sides of the equality by \(x-1\) we obtain \[x+1=0,\] in other words \(x=-1\). But earlier we assumed \(x=1\). THUS \[-1=1\ !\]