정적분 함수 보기

[출제의도] 치환적분법을 이용하여 추론하기 $t - x = k$로 놓으면 $\displaystyle 1 = \frac{dk}{dt}$이고, $t = 0$일 때 $k = - x$, $t = x$일 때 $k = 0$이므로 $\displaystyle f ( x ) = \int _{0} ^{x} {} ( t + x ) \cos ( t - x ) dt = \int _{- x} ^{0} {} ( k + 2 x ) \cos k dk$ $\displaystyle = \left[ ( k + 2 x ) \sin k \right] _{- x} ^{0} - \int _{- x} ^{0} {} \sin k dk$ $= - x \sin x ( - x ) - \left[ - \cos k \right] _{- x} ^{0}$ $= x \sin x + 1 - \cos ( - x )$ $= x \sin x - \cos x + 1$ ㄱ. $f ( - x ) = - x \sin ( - x ) - \cos ( - x ) + 1 = x \sin x - \cos x + 1$ 이므로 $f ( x ) = f ( - x )$ (참) ㄴ. $f' ( x ) = \sin x + x \cos x + \sin x = 2 \sin x + x \cos x$ $f ( x )$가 미분가능하고 $x = \alpha$에서 극값을 가지므로 $f' ( \alpha ) = 2 \sin \alpha + \alpha \cos \alpha = 0$이다. $2 \tan \alpha + \alpha = 0$이므로 $\displaystyle \tan \alpha = - \frac{\alpha}{2}$ (참) ㄷ. $\displaystyle \tan \alpha = - \frac{\alpha}{2} < 0$이므로 $\displaystyle \frac{\pi}{2} < \alpha < \pi$ $\sec ^{2} \alpha = 1 + \tan ^{2} \alpha$이므로 $\displaystyle \sec ^{2} \alpha = 1 + \left( - \frac{\alpha}{2} \right) ^{2}$에서 $\displaystyle \frac{1}{\cos ^{2} \alpha} = \frac{\alpha ^{2} + 4}{4}$ $\displaystyle \cos ^{2} \alpha = \frac{4}{\alpha ^{2} + 4}$ $\displaystyle \frac{\pi}{2} < \alpha < \pi$일 때 $\cos \alpha < 0$이므로 $\displaystyle \cos \alpha = - \frac{2}{\sqrt{\alpha ^{2} + 4}}$ $\displaystyle \int _{- \alpha} ^{\alpha} {} f ( x ) dx = 2 \int _{0} ^{\alpha} {} f ( x ) dx$이므로 $\displaystyle \int _{0} ^{\alpha} {} f ( x ) dx = \int _{0} ^{\alpha} {} ( x \sin x - \cos x + 1 ) dx$ $= \left[ - x \cos x \right] _{0} ^{\alpha} + \alpha = - \alpha \cos \alpha + \alpha$ $\displaystyle = \frac{2 \alpha}{\sqrt{\alpha ^{2} + 4}} + \alpha$ $\displaystyle \int _{- \alpha} ^{\alpha} {} f ( x ) dx = \frac{4 \alpha}{\sqrt{\alpha ^{2} + 4}} + 2 \alpha$ (거짓) 따라서 옳은 것은 ㄱ, ㄴ