정답: $2$

$f(x)=\cos (4x)f(x)\; =\; \backslash cos(4x)$라 하면 $g(x)={\displaystyle \frac{1}{2}}f(x)g(x)\; =\; \backslash dfrac\{1\}\{2\}f(x)$이다.

$y=cf(x)y\; =\; cf(x)$이면 $y^{\mathrm{\prime }}=c{f}^{\mathrm{\prime }}(x)y\text{'}\; =\; cf\text{'}(x)$이므로
$g(x)={\displaystyle \frac{1}{2}}f(x)g(x)\; =\; \backslash dfrac\{1\}\{2\}f(x)$를 미분하면
$g^{\mathrm{\prime }}(x)={\displaystyle \frac{1}{2}}{f}^{\mathrm{\prime }}(x)g\text{'}(x)\; =\; \backslash dfrac\{1\}\{2\}f\text{'}(x)$⋯①이다.

$y=\cos (kx)y\; =\; \backslash cos(kx)$이면 $y^{\mathrm{\prime }}=-k\sin (kx)y\text{'}\; =\; -k\backslash sin(kx)$이므로
$f(x)=\cos (4x)f(x)\; =\; \backslash cos(4x)$를 미분하면 $f^{\mathrm{\prime }}(x)=-4\sin (4x)f\text{'}(x)\; =\; -4\backslash sin(4x)$이다.

이를 ①에 넣으면
$g^{\mathrm{\prime }}(x)={\displaystyle \frac{1}{2}}{f}^{\mathrm{\prime }}(x)g\text{'}(x)\; =\; \backslash dfrac\{1\}\{2\}f\text{'}(x)$에서
$g^{\mathrm{\prime }}(x)={\displaystyle \frac{1}{2}}\times (-4\sin (4x))g\text{'}(x)\; =\; \backslash dfrac\{1\}\{2\}\; \backslash times\; (-4\backslash sin(4x))$,
$g^{\mathrm{\prime }}(x)=-2\sin (4x)g\text{'}(x)\; =\; -2\backslash sin(4x)$이다.

$g^{\mathrm{\prime }}\left({\displaystyle \frac{\pi }{24}}\right)=-2\sin (4\times {\displaystyle \frac{\pi }{24}})g\text{'}\backslash left(\backslash dfrac\{\backslash pi\}\{24\}\backslash right)\; =\; -2\backslash sin\backslash left(4\; \backslash times\; \backslash dfrac\{\backslash pi\}\{24\}\backslash right)$,
$g^{\mathrm{\prime }}\left({\displaystyle \frac{\pi }{24}}\right)=-2\sin \left({\displaystyle \frac{\pi }{6}}\right)g\text{'}\backslash left(\backslash dfrac\{\backslash pi\}\{24\}\backslash right)\; =\; -2\backslash sin\backslash left(\backslash dfrac\{\backslash pi\}\{6\}\backslash right)$,
$g^{\mathrm{\prime }}\left({\displaystyle \frac{\pi }{24}}\right)=-2\times {\displaystyle \frac{1}{2}}g\text{'}\backslash left(\backslash dfrac\{\backslash pi\}\{24\}\backslash right)\; =\; -2\; \backslash times\; \backslash dfrac\{1\}\{2\}$,
$g^{\mathrm{\prime }}\left({\displaystyle \frac{\pi }{24}}\right)=-1g\text{'}\backslash left(\backslash dfrac\{\backslash pi\}\{24\}\backslash right)\; =\; -1$이다.

따라서 정답은 $22$번이다.

정답: $3$

$f(x)=\sin (2x)f(x)\; =\; \backslash sin(2x)$라 하면 $g(x)=4f(x)g(x)\; =\; 4f(x)$이다.

${\displaystyle \int kf(x)dx=k\int f(x)dx}\backslash displaystyle\backslash int\; kf(x)\backslash ,dx\; =\; k\backslash int\; f(x)\backslash ,dx$이므로
$g(x)=4f(x)g(x)\; =\; 4f(x)$를 적분하면
${\displaystyle \int g(x)dx=4\int f(x)dx}\backslash displaystyle\backslash int\; g(x)\backslash ,dx\; =\; 4\backslash int\; f(x)\backslash ,dx$⋯①이다.

${\displaystyle \int \sin (kx)dx=-\frac{1}{k}\cos (kx)+C}\backslash displaystyle\backslash int\backslash sin(kx)\backslash ,dx\; =\; -\backslash dfrac\{1\}\{k\}\backslash cos(kx)\; +\; C$이므로
$f(x)=\sin (2x)f(x)\; =\; \backslash sin(2x)$를 적분하면 ${\displaystyle \int f(x)dx=-\frac{1}{2}\cos (2x)+C_1}\backslash displaystyle\backslash int\; f(x)\backslash ,dx\; =\; -\backslash dfrac\{1\}\{2\}\backslash cos(2x)\; +\; C\_\{1\}$이다.

이를 ①에 넣으면
${\displaystyle \int g(x)dx=4\int f(x)dx}\backslash displaystyle\backslash int\; g(x)\backslash ,dx\; =\; 4\backslash int\; f(x)\backslash ,dx$에서
${\displaystyle \int g(x)dx=4(-\frac{1}{2}\cos (2x)+C_1)}\backslash displaystyle\backslash int\; g(x)\backslash ,dx\; =\; 4\backslash left(-\backslash dfrac\{1\}\{2\}\backslash cos(2x)\; +\; C\_\{1\}\backslash right)$,
${\displaystyle \int g(x)dx=-2\cos (2x)+4C_1}\backslash displaystyle\backslash int\; g(x)\backslash ,dx\; =\; -2\backslash cos(2x)\; +\; 4C\_\{1\}$이다.

여기서 $C=4C_{1}C\; =\; 4C\_\{1\}$이라 하면,
${\displaystyle \int g(x)dx=-2\cos (2x)+4C_1}\backslash displaystyle\backslash int\; g(x)\backslash ,dx\; =\; -2\backslash cos(2x)\; +\; 4C\_\{1\}$은
${\displaystyle \int g(x)dx=-2\cos (2x)+C}\backslash displaystyle\backslash int\; g(x)\backslash ,dx\; =\; -2\backslash cos(2x)\; +\; C$이다.

$G(x)=-2\cos (2x)+CG(x)\; =\; -2\backslash cos(2x)\; +\; C$에서 $G\left({\displaystyle \frac{\pi }{6}}\right)=3G\backslash left(\backslash dfrac\{\backslash pi\}\{6\}\backslash right)\; =\; 3$이므로
$-2\cos (2\times {\displaystyle \frac{\pi }{6}})+C=3-2\backslash cos\backslash left(2\; \backslash times\; \backslash dfrac\{\backslash pi\}\{6\}\backslash right)\; +\; C\; =\; 3$,
$-2\cos \left({\displaystyle \frac{\pi }{3}}\right)+C=3-2\backslash cos\backslash left(\backslash dfrac\{\backslash pi\}\{3\}\backslash right)\; +\; C\; =\; 3$,
$-2\times {\displaystyle \frac{1}{2}}+C=3-2\; \backslash times\; \backslash dfrac\{1\}\{2\}\; +\; C\; =\; 3$,
$-1+C=3-1\; +\; C\; =\; 3$,
$C=3+1C\; =\; 3\; +\; 1$,
$C=4C\; =\; 4$이다.

$G(x)=-2\cos (2x)+4G(x)\; =\; -2\backslash cos(2x)\; +\; 4$이므로
$G(0)=-2\cos (2\times 0)+4G(0)\; =\; -2\backslash cos(2\; \backslash times\; 0)\; +\; 4$,
$G(0)=-2\cos 0+4G(0)\; =\; -2\backslash cos\{0\}\; +\; 4$,
$G(0)=-2\times 1+4G(0)\; =\; -2\; \backslash times\; 1\; +\; 4$,
$G(0)=-2+4G(0)\; =\; -2\; +\; 4$,
$G(0)=2G(0)\; =\; 2$이다.

따라서 정답은 $33$번이다.