사차함수와 접선 넓이

[출제의도] 정적분을 활용하여 문제 해결하기 두 점 ${\mathrm{A}}$, ${\mathrm{B}}$의 좌표를 각각 $\displaystyle \left( a , \frac{a}{2} \right)$, $\displaystyle \left( b , \frac{b}{2} \right) \left( 0 < a < b \right)$라 하자. 곡선 $y = f ( x )$와 직선 $\displaystyle y = \frac{1}{2} x$가 원점 ${\mathrm{O}}$에서 접하고 두 점 ${\mathrm{A}}$, ${\mathrm{B}}$에서 만나므로 $f ( x ) -$$\displaystyle \frac{1}{2} x$$= x ^{2} \left( x - a \right) \left( x - b \right)$ $= x ^{4} - ( a + b ) x ^{3} + a b x ^{2}$ $S _{1} - S _{2}$$\displaystyle = \int _{0} ^{a} \left| f ( x ) - \frac{1}{2} x \right| d x - \int _{a} ^{b} \left| f ( x ) - \frac{1}{2} x \right| d x$ $\displaystyle = \int _{0} ^{a} \left\{ f ( x ) - \frac{1}{2} x \right\} d x + \int _{a} ^{b} \left\{ f ( x ) - \frac{1}{2} x \right\} d x$ $\displaystyle = \int _{0} ^{b} \left\{ f ( x ) - \frac{1}{2} x \right\} d x$ $\displaystyle = \int _{0} ^{b} \left\{ x ^{4} - ( a + b ) x ^{3} + a b x ^{2} \right\} d x$ $\displaystyle = \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} \frac{1}{5} x ^{5} - \frac{a + b}{4} x ^{4} + \frac{a b}{3} x ^{3} \end{array} \right] _{0} ^{b}$ $\displaystyle = - \frac{b ^{5}}{20} + \frac{ab ^{4}}{12} = 0$ 에서 $5 a - 3 b = 0$이고, $\displaystyle \overline{{\mathrm{AB}}} = \sqrt{( b - a ) ^{2} + \left( \frac{b}{2} - \frac{a}{2} \right) ^{2}} = \frac{\sqrt{5}}{2} ( b - a ) = \sqrt{5}$ 에서 $b - a = 2$이므로 두 식을 연립하여 계산하면 $a = 3$, $b = 5$ 그러므로 $\displaystyle f ( x ) = x ^{4} - 8 x ^{3} + 15 x ^{2} + \frac{1}{2} x$ 따라서 $\displaystyle f ( 1 ) = \frac{17}{2}$