1. **Problem 14:** Given a differentiable function $f$ on $(0,2)$ with derivative $f'(x) = x^2 - x + 1$. We analyze the truth of the statements: 2. **Analyze the derivative:** The function $f'(x) = x^2 - x + 1$ can be rewritten as $f'(x) = (x - \frac{1}{2})^2 + \frac{3}{4}$. Since squares are nonnegative and $\frac{3}{4} > 0$, we find $f'(x) > 0$ for all $x \in (0,2)$. 3. **Interpret $f'(x) > 0$:** This means $f$ is strictly increasing on $(0,2)$. 4. **Check options using this information:** - (A) $f(1) = f(2-1) = f(1)$ is trivially true but does not reflect an inherent property for every $x$, so statement's phrasing is ambiguous. - (B) $f(1) = -f(2-1)$ cannot hold since $f(1) = f(1)$, not its negative. - (C) If $0 < x < y < 2$, then $f(x) < f(y)$ because $f$ is strictly increasing. This statement is true. - (D) Concavity is determined by $f''(x)$. Compute $f''(x)$: $$f''(x) = \frac{d}{dx} f'(x) = 2x - 1.$$ On $(0,2)$, $f''(x)$ is negative from $0$ to $\frac{1}{2}$, positive from $\frac{1}{2}$ to $2$, so concavity changes; hence (D) is false. - (E) Since $f''(x)$ changes sign, graph is not concave downward throughout $(0,2)$, so (E) is false. --- 5. **Problem 15:** Given the system: $$ x^2 + y^2 = a $$ $$ xy = b $$ We seek necessary and sufficient condition on positive $a,b$ for existence of real $(x,y)$ solving both. 6. **Rewrite problem:** From $xy = b$, express $y = \frac{b}{x}$ (assuming $x\neq 0$). Substitute into the circle equation: $$ x^2 + \left(\frac{b}{x}\right)^2 = a \implies x^2 + \frac{b^2}{x^2} = a. $$ Multiply both sides by $x^2 >0$: $$ x^4 - a x^2 + b^2 = 0. $$ Let $t = x^2 \geq 0$, then: $$ t^2 - a t + b^2 = 0. $$ 7. **Condition for real $x$:** Real $x$ exists if $t$ is a nonnegative root of above quadratic. Discriminant of quadratic in $t$: $$ \Delta = a^2 - 4 b^2. $$ For real $t$, $\Delta \geq 0$, thus: $$ a^2 \geq 4 b^2 \implies a \geq 2b $$ (since $a,b>0$). 8. **Checking roots $t \geq 0$:** Both roots positive or at least one positive if this holds. Thus, the necessary and sufficient condition for a real solution $(x,y)$ is: $$ a \geq 2b. $$ **Answer for 15 is (A).**