## Problem 4

Determine the primitives of the function $f:[0, \pi] \rightarrow \mathbf{R}$, which is primitive-able, and satisfies the relation $f(x) \sin x - f(\pi - x) = \cos^2 x, \forall x \in [0, \pi]$.
## Solution:

From the relation in the statement, for $x:=\pi-x$, we deduce: $f(\pi-x) \sin x-f(x)=\cos ^{2} x$.

From the relation in the statement, by multiplying with $\sin x$ we deduce:

$$
f(x) \sin ^{2} x-f(\pi-x) \sin x=\sin x \cos ^{2} x
$$

Adding (1) and (2), we deduce $[1+\sin x+f(x)] \cdot \cos ^{2} x=0$.

From (3), we obtain:

Case 1. $\cos x=0 \Rightarrow x=\frac{\pi}{2}$. From the relation in the hypothesis we obtain $f(x)=f\left(\frac{\pi}{2}\right)=a$.

Case 2. For $x \in[0, \pi] \backslash\left\{\frac{\pi}{2}\right\}, 1+\sin x+f(x)=0$.

Thus: $\quad f(x)=\left\{\begin{array}{l}-1-\sin x, x \in[0, \pi] \backslash\left\{\frac{\pi}{2}\right\} \\ a, \quad x=\frac{\pi}{2}\end{array}, a \in \mathbf{R}\right.$.

If $a \in \mathbf{R} \backslash\{-2\}, f$ has a discontinuity point of the first kind at $x=\frac{\pi}{2} \Rightarrow f$ does not have the Darboux property and therefore $f$ does not admit primitives.

According to the statement, $f$ admits primitives and thus $a=-2$ and consequently, $f(x)=-1-\sin x$ for any $x \in[0, \pi]$ and therefore $F(x)=-x+\cos x+C$.

From the exam: $1 \mathbf{p}$

## S.S.M.R - MURES BRANCH

## Mathematics Olympiad

Local phase 9.02.2013

12th Grade