Example 1. How many functions f : R → R are there satisfying the inequality |f(x + y) + sin x + sin y| < 2?
Example 2. Let ABCDEFGHIJ be a 10-digit number, where all the digits are distinct. Further, A > B > C, A + B + C = 9, D > E > F > G are consecutive odd numbers and H > I > J are consecutive even numbers. Find the value of A.
Of course, you can find the solution just below, but it is highly recommended that, you first try to solve it on your own.
Just remember the words of Paul Halmos, who says
the only way to learn mathematics is to do mathematics.
Solution 1. We will show that, there is no such function f for all x, y ∈ R exists satisfying the given conditions.
To show this, we need to consider a set of values of x, y for which the given condition fail to happen.
First, consider x = π/2 = y. From the given relation, we get
|f(π/2 + π/2) + sin (π/2) + sin (π/2)| < 2
or, |f(π) + 2| < 2
or, −2 < f(π) + 2 < 2
or, −4 < f(π) < 0
or, f(π) < 0 (1)
Next, consider x = 3π/2 and y = −π/2. From the given relation, we get
|f(3π/2 −π/2) + sin (3π/2) + sin (−π/2)| < 2
or, |f(π) −1 − 1| < 2
or, −2 < f(π) – 2 < 2
or, 0 < f(π) < 4
or, f(π) > 0 (2)
Clearly, (1) and (2) contradicts each other.
Solution 2. The only odd digits are 1, 3, 5, 7 and 9. Since A + B + C = 9, an odd number, either exactly one or all three of A, B, C are odd. Let A, B, C be all odd digits and given that D, E, F, G are consecutive odd numbers, this is not possible. So, only one of A, B, C is odd. Rest two are even.
The only way we can have four consecutive odd digits in a descending order is 9 > 7 > 5 > 3 and 7 > 5 > 3 > 1. If we consider the latter possibility, then 9 must be any one of A, B, C. Then the remaining two of these three will add to 0, since A + B + C = 9.
Hence both will be 0, which is not allowed as all the letters represent distinct digits. So we conclude that D = 9, E = 7, F = 5, G = 3 and 1 is any one of A, B, C. So the remaining two add to 8.
Next we focus on the last condition, that H, I, J are consecutive even numbers with H > I > J. This time there are three possibilities:
(i) 8 > 6 > 4 (ii) 6 > 4 > 2 and (iii) 4 > 2 > 0.
Recall that, any two of A, B and C are even. In each case, the missing two even digits are in A, B, C and add to 8. The even numbers left in (i) are 2, 0 and 2 + 0 = 2. Similarly, the even numbers left in (iii) are 8, 6 and 8 + 6 = 14. This rules out (i) and (iii). So we are left with (ii).
Now the missing two even digits 8 and 0 are in A, B, C. Hence A, B, C are 1, 0 and 8. As they are given to be in a descending order, we have A = 8.