d = 3 + 2 2 \displaystyle d = 3 + 2 \sqrt{2} d = 3 + 2 2
(a) C 1 \displaystyle C_{1} C 1 C 2 \displaystyle C_{2} C 2 y , z \displaystyle y , z y , z A C ‾ = 2 y + y + z + 2 z = 2 d \displaystyle \overline{\mathrm{AC}} = \sqrt{2} y + y + z + \sqrt{2} z = \sqrt{2} d AC = 2 y + y + z + 2 z = 2 d y + z = 2 + 2 \displaystyle y + z = 2 + \sqrt{2} y + z = 2 + 2
B F ‾ + F C ‾ = B C ‾ \displaystyle \overline{\mathrm{BF}} + \overline{\mathrm{FC}} = \overline{\mathrm{BC}} BF + FC = BC x + 2 z = d \displaystyle x + 2 z = d x + 2 z = d z = d − x 2 \displaystyle z=\frac{d-x}{2} z = 2 d − x y = 2 + 2 − z = 1 + x 2 \displaystyle y=2+\sqrt2-z=\frac{1+x}{2} y = 2 + 2 − z = 2 1 + x
(i) x < y \displaystyle x < y x < y y = 1 + x 2 \displaystyle y=\frac{1+x}{2} y = 2 1 + x x < 1 \displaystyle x < 1 x < 1 C 1 \displaystyle C_{1} C 1 A D ‾ \displaystyle \overline{\mathrm{AD}} AD G \displaystyle \mathrm{G} G w \displaystyle w w w = y 2 − ( y − x ) 2 = ( 1 + x 2 ) 2 − ( 1 − x 2 ) 2 = x \displaystyle w=\sqrt{y^2-(y-x)^2}=\sqrt{\left(\frac{1+x}{2}\right)^2-\left(\frac{1-x}{2}\right)^2}=\sqrt x w = y 2 − ( y − x ) 2 = ( 2 1 + x ) 2 − ( 2 1 − x ) 2 = x 이다. E B ‾ = d − y − w = d − 1 + x 2 − x \displaystyle \overline{\mathrm{EB}}=d-y-w=d-\frac{1+x}{2}-\sqrt x EB = d − y − w = d − 2 1 + x − x 0 < x < 1 \displaystyle 0 < x < 1 0 < x < 1 f ( x ) = B F ‾ ⋅ E B ‾ = x ( d − 1 + x 2 − x ) \displaystyle f(x)=\overline{\mathrm{BF}}\cdot\overline{\mathrm{EB}}=x\left(d-\frac{1+x}{2}-\sqrt x\right) f ( x ) = BF ⋅ EB = x ( d − 2 1 + x − x )
(ii) x ≥ y \displaystyle x \ge y x ≥ y y = 1 + x 2 \displaystyle y=\frac{1+x}{2} y = 2 1 + x x ≥ 1 \displaystyle x \ge 1 x ≥ 1 E G ‾ \displaystyle \overline{\mathrm{EG}} EG C 1 \displaystyle C_{1} C 1 E B ‾ = d − 2 y = d − 1 − x \displaystyle \overline{\mathrm{EB}} = d - 2 y = d - 1 - x EB = d − 2 y = d − 1 − x x < d − 1 − x \displaystyle x < d - 1 - x x < d − 1 − x x < 1 + 2 \displaystyle x < 1 + \sqrt{2} x < 1 + 2 1 ≤ x < 1 + 2 \displaystyle 1 \le x < 1 + \sqrt{2} 1 ≤ x < 1 + 2 f ( x ) = B F ‾ ⋅ E B ‾ = x ( d − 1 − x ) \displaystyle f ( x ) = \overline{\mathrm{BF}} \cdot \overline{\mathrm{EB}} = x \left( d - 1 - x \right) f ( x ) = BF ⋅ EB = x ( d − 1 − x )
(i)과 (ii)에 의해 f ( x ) = { x ( 3 + 2 2 − 1 + x 2 − x ) ( 0 < x < 1 ) x ( 2 + 2 2 − x ) ( 1 ≤ x < 1 + 2 ) \displaystyle f ( x ) = \begin{cases}x \left( 3 + 2 \sqrt{2} - \frac{{1 + x}}{2} - \sqrt{x} \right) & ( 0 < x < 1 ) \\ x \left( 2 + 2 \sqrt{2} - x \right) & ( 1 \leq x < 1 + \sqrt{2} )\end{cases} f ( x ) = { x ( 3 + 2 2 − 2 1 + x − x ) x ( 2 + 2 2 − x ) ( 0 < x < 1 ) ( 1 ≤ x < 1 + 2 ) 이다.
(b) 0 < x < 1 \displaystyle 0 < x < 1 0 < x < 1 1 < x < 1 + 2 \displaystyle 1 < x < 1 + \sqrt{2} 1 < x < 1 + 2 f ( x ) \displaystyle f ( x ) f ( x ) x = 1 \displaystyle x = 1 x = 1 lim h → 0 − f ( 1 + h ) − f ( 1 ) h = lim h → 0 − ( 1 + h ) { 3 + 2 2 − 1 + ( 1 + h ) 2 − 1 + h } − ( 1 + 2 2 ) h = 2 2 lim h → 0 + f ( 1 + h ) − f ( 1 ) h = lim h → 0 + ( 1 + h ) { 2 + 2 2 − ( 1 + h ) } − ( 1 + 2 2 ) h = 2 2 \displaystyle \begin{aligned}\lim_{h\to0^-}\frac{f(1+h)-f(1)}h&=\lim_{h\to0^-}\frac{(1+h)\left\{3+2\sqrt2-\frac{1+(1+h)}2-\sqrt{1+h}\right\}-(1+2\sqrt2)}h=2\sqrt2\\\lim_{h\to0^+}\frac{f(1+h)-f(1)}h&=\lim_{h\to0^+}\frac{(1+h)\{2+2\sqrt2-(1+h)\}-(1+2\sqrt2)}h=2\sqrt2\end{aligned} h → 0 − lim h f ( 1 + h ) − f ( 1 ) h → 0 + lim h f ( 1 + h ) − f ( 1 ) = h → 0 − lim h ( 1 + h ) { 3 + 2 2 − 2 1 + ( 1 + h ) − 1 + h } − ( 1 + 2 2 ) = 2 2 = h → 0 + lim h ( 1 + h ) { 2 + 2 2 − ( 1 + h )} − ( 1 + 2 2 ) = 2 2 이다. 따라서 f ( x ) \displaystyle f ( x ) f ( x ) x = 1 \displaystyle x = 1 x = 1 f ( x ) \displaystyle f ( x ) f ( x ) 0 < x < 1 + 2 \displaystyle 0 < x < 1 + \sqrt{2} 0 < x < 1 + 2
수리논술, 배움에서 논증의 완성까지.
