1. **Problem:** Find the point on the curve $y = \sqrt{x} + 3$ closest to the origin $(0,0)$. 2. The distance squared from the origin is $D^2 = x^2 + y^2 = x^2 + (\sqrt{x} + 3)^2 = x^2 + (x + 6\sqrt{x} + 9)$. 3. Simplify: $$D^2 = x^2 + x + 6\sqrt{x} + 9$$ with $x \ge 0$ (since $y = \sqrt{x} + 3$). 4. To minimize $D^2$, differentiate with respect to $x$ using $\sqrt{x} = x^{1/2}$: $$\frac{dD^2}{dx} = 2x + 1 + 6 \cdot \frac{1}{2} x^{-1/2} = 2x + 1 + 3 x^{-1/2}.$$ 5. Set derivative to zero for critical points: $$2x + 1 + \frac{3}{\sqrt{x}} = 0.$$ Multiply both sides by $\sqrt{x}$ to clear denominator: $$2x \sqrt{x} + \sqrt{x} + 3 = 0.$$ Let $t = \sqrt{x}$, then $x = t^2$: $$2 t^3 + t + 3 = 0.$$ 6. Solve for $t$: Possible rational roots do not yield a root; check negative values because $t = \sqrt{x} \ge 0$ means $t \ge 0$. Since $t \ge 0$, and $2t^3 + t + 3 > 0$ for $t\ge0$, no zero here. Hence, derivative never zero; check endpoint $x=0$. 7. At $x=0$, $D^2 = 0 + 0 + 0 + 9 = 9$; $D=3$. 8. As $x \to \infty$, $D^2 \to \infty$. Check if any minimum closer by testing values: At $x= \frac{1}{2}$ (from choice), check $D^2$: $$D^2 = \left(\frac{1}{2}\right)^2 + \frac{1}{2} + 6 \sqrt{\frac{1}{2}} + 9 = \frac{1}{4} + \frac{1}{2} + 6 \cdot \frac{1}{\sqrt{2}} + 9$$ $$= 0.75 + 6 \cdot 0.707 + 9 = 0.75 + 4.242 + 9 = 13.992.$$ 9. This is larger than 9, so the closest point is at $x=0$, $y=\sqrt{0}+3=3$; point $(0,3)$. 10. Check provided answers: (D) $(0, \sqrt{3})$ is not $(0,3)$, so none match exactly. Possibly a misprint. 11. If the question means $y=\sqrt{x}+3$, the closest point to origin is $(0,3)$ which matches none exactly; closest option is (E) $(\frac{1}{2}, \frac{3}{2})$, but as shown distances don't fit. So the correct point is $(0,3)$. --- 12. **Problem:** Find the second derivative of the composition $f \circ g$. 13. By chain rule, first derivative: $$ (f \circ g)' = (f' \circ g) g' $$ 14. Differentiating again: $$ (f \circ g)'' = \frac{d}{dx}[(f' \circ g) g'] = (f'' \circ g)(g')^2 + (f' \circ g) g'' $$ 15. Thus the correct expression is option (D). --- 16. **Problem:** Solve the system modulo 13: $$3x + 2y \equiv 5 \pmod{13}$$ $$x + 7y \equiv 1 \pmod{13}$$ 17. Solve for $x$ in second: $$x \equiv 1 - 7y \pmod{13}$$ 18. Substitute in first: $$3(1 - 7y) + 2y \equiv 5 \pmod{13}$$ $$3 - 21y + 2y \equiv 5$$ $$3 - 19y \equiv 5$$ $$-19y \equiv 2$$ 19. Modulo 13, $-19 \equiv -19 + 26 = 7$, so: $$7y \equiv 2 \pmod{13}$$ 20. Find inverse of 7 mod 13: $$7 \times 2 = 14 \equiv 1 \pmod{13}$$ 21. So: $$y \equiv 2 \times 2 = 4 \pmod{13}$$ 22. Substitute back: $$x \equiv 1 - 7 \times 4 = 1 - 28 = 1 - 28 + 26 = -1 \equiv 12 \pmod{13}$$ 23. Evaluate: $$5x + 3y = 5 \times 12 + 3 \times 4 = 60 + 12 = 72$$ 24. Modulo 13: $$72 \equiv 72 - 65 = 7$$ 25. So the answer is (C) 7.