자연계열 1번

[문제 $\displaystyle 1 - 1$]

함수의 도함수를 구하면

$\displaystyle f ^{\prime} ( x ) = \frac{- \sin x ( 2 \sqrt{2} - 2 \sin x ) - \cos x ( - 2 \cos x )}{( 2 \sqrt{2} - 2 \sin x )^{2}} = \frac{2 - 2 \sqrt{2} \sin x}{( 2 \sqrt{2} - 2 \sin x )^{2}}\cdots \cdots ( 1 )$

$\displaystyle f ^{\prime} ( x ) = 0$을 만족시키는 $\displaystyle x$의 값은 $\displaystyle 2 - 2 \sqrt{2} \sin x = 0$에서 $\displaystyle x = \frac{\pi}{4}$또는 $\displaystyle x = \frac{3}{4} \pi\cdots\cdots ( 2 )$

함수 $\displaystyle f ( x )$의 증가와 감소를 표로 나타내면 다음과 같다.

이때 $\displaystyle \sin \frac{3}{4} \pi = \frac{1}{\sqrt{2}} , \cos \frac{3}{4} \pi = - \frac{1}{\sqrt{2}}$이므로

$\displaystyle \frac{- \frac{1}{\sqrt{2}}}{2 \sqrt{2} - 2 \frac{1}{\sqrt{2}}} = \frac{- \frac{1}{\sqrt{2}}}{\sqrt{2}} = - \frac{1}{2}\cdots \cdots ( 3 )$

따라서, 함수 $\displaystyle f ( x )$는 $\displaystyle x = \frac{3}{4} \pi$에서 최소이고, 최솟값은 $\displaystyle - \frac{1}{2}$이다.

[문제 $\displaystyle 1 - 2$]

$\displaystyle \sin \left( x - \frac{\pi}{2} \right) = - \cos x$이므로 함수 $\displaystyle f \left( x - \frac{\pi}{2} \right)$는 다음과 같다.

$\displaystyle f \left( x - \frac{\pi}{2} \right) = \begin{cases}0 & \left( x < \frac{\pi}{2} \right) \\ - \cos x & \left( \frac{\pi}{2} \le x \le \frac{3}{2} \pi \right) \\ 0 & \left( x > \frac{3}{2} \pi \right)\end{cases}\cdots \cdots ( 4 )$

이때, $\displaystyle \int_{0}^{2 \pi} \left\{ f ( x ) + f \left( x - \frac{\pi}{2} \right) \right\}^{2} d x= \int_{0}^{2 \pi} \left\{ f ( x ) \right\}^{2} d x + 2 \int_{0}^{2 \pi} f ( x ) f \left( x - \frac{\pi}{2} \right) d x + \int_{0}^{2 \pi} \left\{ f \left( x - \frac{\pi}{2} \right) \right\}^{2} d x$

이고, $\displaystyle \int_{0}^{2 \pi} \left\{ f ( x ) \right\}^{2} d x = \int_{0}^{\pi} \left\{ f ( x ) \right\}^{2} d x + \int_{\pi}^{2 \pi} \left\{ f ( x ) \right\}^{2} d x = \int_{0}^{\pi} \left\{ f ( x ) \right\}^{2} d x = \int_{0}^{\pi} \sin^{2} x d x = \int_{0}^{\pi} \frac{1 - \cos 2 x}{2} d x = \frac{1}{2} \Biggl[ x - \frac{1}{2} \sin 2 x \Biggr]_{0}^{\pi} = \frac{\pi}{2}\cdots \cdots ( 5 )$

이고,

$\displaystyle \int_{0}^{2 \pi} \left\{ f \left( x - \frac{\pi}{2} \right) \right\}^{2} d x = \int_{0}^{\frac{\pi}{2}} \left\{ f \left( x - \frac{\pi}{2} \right) \right\}^{2} d x + \int_{\frac{\pi}{2}}^{\frac{3}{2} \pi} \left\{ f \left( x - \frac{\pi}{2} \right) \right\}^{2} d x + \int_{\frac{3}{2} \pi}^{2 \pi} \left\{ f \left( x - \frac{\pi}{2} \right) \right\}^{2} d x$

$\displaystyle = \int_{\frac{\pi}{2}}^{\frac{3}{2} \pi} \left\{ f \left( x - \frac{\pi}{2} \right) \right\}^{2} d x = \int_{\frac{\pi}{2}}^{\frac{3}{2} \pi} ( - \cos x )^{2} d x$

$\displaystyle = \int_{\frac{\pi}{2}}^{\frac{3}{2} \pi} \frac{1 + \cos 2 x}{2} d x = \frac{1}{2} \Biggl[ x + \frac{1}{2} \sin 2 x \Biggr]_{\frac{\pi}{2}}^{\frac{3}{2} \pi} = \frac{\pi}{2}\cdots \cdots ( 6 )$

이다. 한편, $\displaystyle \int_{0}^{2 \pi} f ( x ) f \left( x - \frac{\pi}{2} \right) d x$

$\displaystyle = \int_{0}^{\frac{\pi}{2}} f ( x ) f \left( x - \frac{\pi}{2} \right) d x + \int_{\frac{\pi}{2}}^{\pi} f ( x ) f \left( x - \frac{\pi}{2} \right) d x + \int_{\pi}^{\frac{3}{2} \pi} f ( x ) f \left( x - \frac{\pi}{2} \right) d x + \int_{\frac{3}{2} \pi}^{2 \pi} f ( x ) f \left( x - \frac{\pi}{2} \right) d x$

$\displaystyle = \int_{0}^{\frac{\pi}{2}} \sin x \times 0 d x + \int_{\frac{\pi}{2}}^{\pi} \sin x \times ( - \cos x ) d x + \int_{\pi}^{\frac{3}{2} \pi} 0 \times ( - \cos x ) d x + \int_{\frac{3}{2} \pi}^{2 \pi} 0 \times 0 d x$

$\displaystyle = 0 + \int_{\frac{\pi}{2}}^{\pi} \sin x ( - \cos x ) d x + 0 + 0 = - \Biggl[ \frac{1}{2} \sin^{2} x \Biggr]_{\frac{\pi}{2}}^{\pi} = \frac{1}{2}\cdots \cdots ( 7 )$

이다. 따라서,

$\displaystyle \int_{0}^{2 \pi} \left\{ f ( x ) + f \left( x - \frac{\pi}{2} \right) \right\}^{2} d x$

$\displaystyle = \int_{0}^{2 \pi} \left\{ f ( x ) \right\}^{2} d x + 2 \int_{0}^{2 \pi} f ( x ) f \left( x - \frac{\pi}{2} \right) d x + \int_{0}^{2 \pi} \left\{ f \left( x - \frac{\pi}{2} \right) \right\}^{2} d x$

$\displaystyle = \frac{\pi}{2} + 2 \times \frac{1}{2} + \frac{\pi}{2} = \pi + 1\cdots \cdots ( 8 )$