(1) 確率密度と期待値・分散 ∫ 0 π / 2 sin 2 y d y = 1 , Pr ( π 4 ≦ Y ≦ 3 π 8 ) = ∫ π / 4 3 π / 8 sin 2 y d y = [ − 1 2 cos 2 y ] π / 4 3 π / 8 = 2 4 . \displaystyle
\int_0^{\pi/2}\sin 2y\,dy=1,
\qquad
\Pr\!\left(\frac\pi4\le Y\le\frac{3\pi}8\right)
=\int_{\pi/4}^{3\pi/8}\sin2y\,dy
=\left[-\frac12\cos2y\right]_{\pi/4}^{3\pi/8}
=\frac{\sqrt2}{4}. ∫ 0 π /2 sin 2 y d y = 1 , Pr ( 4 π ≦ Y ≦ 8 3 π ) = ∫ π /4 3 π /8 sin 2 y d y = [ − 2 1 cos 2 y ] π /4 3 π /8 = 4 2 . E ( Y ) = ∫ 0 π / 2 y sin 2 y d y = [ − y 2 cos 2 y + 1 4 sin 2 y ] 0 π / 2 = π 4 . \displaystyle
E(Y)=\int_0^{\pi/2}y\sin2y\,dy
=\left[-\frac y2\cos2y+\frac14\sin2y\right]_0^{\pi/2}
=\frac\pi4. E ( Y ) = ∫ 0 π /2 y sin 2 y d y = [ − 2 y cos 2 y + 4 1 sin 2 y ] 0 π /2 = 4 π . E ( Y 2 ) = ∫ 0 π / 2 y 2 sin 2 y d y = [ − y 2 2 cos 2 y + y 2 sin 2 y + 1 4 cos 2 y ] 0 π / 2 = π 2 8 − 1 2 , \displaystyle
E(Y^2)=\int_0^{\pi/2}y^2\sin2y\,dy
=\left[-\frac{y^2}{2}\cos2y+\frac y2\sin2y+\frac14\cos2y\right]_0^{\pi/2}
=\frac{\pi^2}{8}-\frac12, E ( Y 2 ) = ∫ 0 π /2 y 2 sin 2 y d y = [ − 2 y 2 cos 2 y + 2 y sin 2 y + 4 1 cos 2 y ] 0 π /2 = 8 π 2 − 2 1 , ( y − m ) 2 = y 2 − 2 m y + m 2 (y-m)^2=y^2-2my+m^2 ( y − m ) 2 = y 2 − 2 m y + m 2 ∫ 0 π / 2 g ( y ) d y = 1 \displaystyle \int_0^{\pi/2}g(y)\,dy=1 ∫ 0 π /2 g ( y ) d y = 1 ∫ 0 π / 2 y g ( y ) d y = m \displaystyle \int_0^{\pi/2}yg(y)\,dy=m ∫ 0 π /2 y g ( y ) d y = m V ( Y ) = E ( Y 2 ) − m 2 V(Y)=E(Y^2)-m^2 V ( Y ) = E ( Y 2 ) − m 2 V ( Y ) = E ( Y 2 ) − { E ( Y ) } 2 = π 2 8 − 1 2 − π 2 16 = π 2 − 8 16 . \displaystyle
V(Y)=E(Y^2)-\{E(Y)\}^2
=\frac{\pi^2}{8}-\frac12-\frac{\pi^2}{16}
=\frac{\pi^2-8}{16}. V ( Y ) = E ( Y 2 ) − { E ( Y ) } 2 = 8 π 2 − 2 1 − 16 π 2 = 16 π 2 − 8 .
(2) 2区間が交わる条件 A = [ 3 a + 3 , 3 a + 6 ] , B = [ a 3 − 4 a , a 3 + 2 a 2 − 8 a + 3 ] . A=[3a+3,3a+6],\qquad
B=[a^3-4a,\ a^3+2a^2-8a+3]. A = [ 3 a + 3 , 3 a + 6 ] , B = [ a 3 − 4 a , a 3 + 2 a 2 − 8 a + 3 ] . 2 ( a − 1 ) 2 + 1 > 0 2(a-1)^2+1>0 2 ( a − 1 ) 2 + 1 > 0 3 a + 3 ≦ a 3 + 2 a 2 − 8 a + 3 , a 3 − 4 a ≦ 3 a + 6. 3a+3\le a^3+2a^2-8a+3,
\qquad a^3-4a\le3a+6. 3 a + 3 ≦ a 3 + 2 a 2 − 8 a + 3 , a 3 − 4 a ≦ 3 a + 6. a ( a 2 + 2 a − 11 ) ≧ 0 , ( a + 2 ) ( a + 1 ) ( a − 3 ) ≦ 0. a(a^2+2a-11)\ge0,
\qquad (a+2)(a+1)(a-3)\le0. a ( a 2 + 2 a − 11 ) ≧ 0 , ( a + 2 ) ( a + 1 ) ( a − 3 ) ≦ 0. − 1 − 2 3 ≦ a ≦ 0 または a ≧ − 1 + 2 3 -1-2\sqrt3\le a\le0\text{ または }a\ge-1+2\sqrt3 − 1 − 2 3 ≦ a ≦ 0 または a ≧ − 1 + 2 3 a ≦ − 2 または − 1 ≦ a ≦ 3 a\le-2\text{ または }-1\le a\le3 a ≦ − 2 または − 1 ≦ a ≦ 3 − 1 − 2 3 ≦ a ≦ − 2 または − 1 ≦ a ≦ 0 または − 1 + 2 3 ≦ a ≦ 3. -1-2\sqrt3\le a\le-2\quad\text{または}\quad-1\le a\le0\quad\text{または}\quad-1+2\sqrt3\le a\le3. − 1 − 2 3 ≦ a ≦ − 2 または − 1 ≦ a ≦ 0 または − 1 + 2 3 ≦ a ≦ 3.
(3) 平行四辺形 頂点の順がO P Q R OPQR O P QR O Q → = O P → + O R → \overrightarrow{OQ}=\overrightarrow{OP}+\overrightarrow{OR} O Q = O P + O R Q = ( 0 , 1 , − 1 ) + ( 1 , b , 1 / b ) = ( 1 , b + 1 , 1 / b − 1 ) , Q=(0,1,-1)+(1,b,1/b)=(1,b+1,1/b-1), Q = ( 0 , 1 , − 1 ) + ( 1 , b , 1/ b ) = ( 1 , b + 1 , 1/ b − 1 ) , c = b + 1 , d = 1 / b − 1 c=b+1,\ d=1/b-1 c = b + 1 , d = 1/ b − 1 u = O P → \boldsymbol{u}=\overrightarrow{OP} u = O P v = O R → \boldsymbol{v}=\overrightarrow{OR} v = O R θ \theta θ u \boldsymbol{u} u v \boldsymbol{v} v 0 < θ < π 0<\theta<\pi 0 < θ < π ∣ u ∣ |\boldsymbol{u}| ∣ u ∣ ∣ v ∣ sin θ |\boldsymbol{v}|\sin\theta ∣ v ∣ sin θ S = ∣ u ∣ ∣ v ∣ sin θ , u ⋅ v = ∣ u ∣ ∣ v ∣ cos θ . S=|\boldsymbol{u}||\boldsymbol{v}|\sin\theta,
\qquad \boldsymbol{u}\cdot\boldsymbol{v}=|\boldsymbol{u}||\boldsymbol{v}|\cos\theta. S = ∣ u ∣∣ v ∣ sin θ , u ⋅ v = ∣ u ∣∣ v ∣ cos θ . sin 2 θ = 1 − cos 2 θ \sin^2\theta=1-\cos^2\theta sin 2 θ = 1 − cos 2 θ S 2 = ∣ u ∣ 2 ∣ v ∣ 2 − ( u ⋅ v ) 2 S^2=|\boldsymbol{u}|^2|\boldsymbol{v}|^2-(\boldsymbol{u}\cdot\boldsymbol{v})^2 S 2 = ∣ u ∣ 2 ∣ v ∣ 2 − ( u ⋅ v ) 2 ∣ u ∣ 2 = 2 , ∣ v ∣ 2 = 1 + b 2 + 1 b 2 , u ⋅ v = b − 1 b . \displaystyle
|\boldsymbol{u}|^2=2,\qquad
|\boldsymbol{v}|^2=1+b^2+\frac1{b^2},\qquad
\boldsymbol{u}\cdot\boldsymbol{v}=b-\frac1b. ∣ u ∣ 2 = 2 , ∣ v ∣ 2 = 1 + b 2 + b 2 1 , u ⋅ v = b − b 1 . S ( b ) 2 = 2 ( 1 + b 2 + 1 b 2 ) − ( b − 1 b ) 2 = b 2 + 1 b 2 + 4. \displaystyle
S(b)^2=2\left(1+b^2+\frac1{b^2}\right)
-\left(b-\frac1b\right)^2
=b^2+\frac1{b^2}+4. S ( b ) 2 = 2 ( 1 + b 2 + b 2 1 ) − ( b − b 1 ) 2 = b 2 + b 2 1 + 4. S ( b ) = b 2 + 1 b 2 + 4 . \displaystyle
S(b)=\sqrt{b^2+\frac1{b^2}+4}. S ( b ) = b 2 + b 2 1 + 4 . b 2 + 1 / b 2 ≧ 2 b^2+1/b^2\ge2 b 2 + 1/ b 2 ≧ 2 ( ∣ b ∣ − 1 / ∣ b ∣ ) 2 ≧ 0 (|b|-1/|b|)^2\ge0 ( ∣ b ∣ − 1/∣ b ∣ ) 2 ≧ 0 S ( b ) ≧ 6 S(b)\ge\sqrt6 S ( b ) ≧ 6 b = ± 1 b=\pm1 b = ± 1
解答 あ = 2 4 , い = π 4 , う = π 2 − 8 16 , ( え , お ) = ( − 1 − 2 3 , − 2 ) , ( か , き ) = ( − 1 , 0 ) , ( く , け ) = ( − 1 + 2 3 , 3 ) , こ = b + 1 , さ = 1 b − 1 , し = b 2 + 1 b 2 + 4 , す = 6 . \displaystyle
\begin{gathered}
\text{あ}=\frac{\sqrt2}{4},\quad \text{い}=\frac\pi4,\quad \text{う}=\frac{\pi^2-8}{16},\\
(\text{え},\text{お})=(-1-2\sqrt3,-2),\quad
(\text{か},\text{き})=(-1,0),\quad
(\text{く},\text{け})=(-1+2\sqrt3,3),\\
\text{こ}=b+1,\quad \text{さ}=\frac1b-1,\quad
\text{し}=b^2+\frac1{b^2}+4,\quad \text{す}=\sqrt6.
\end{gathered} あ = 4 2 , い = 4 π , う = 16 π 2 − 8 , ( え , お ) = ( − 1 − 2 3 , − 2 ) , ( か , き ) = ( − 1 , 0 ) , ( く , け ) = ( − 1 + 2 3 , 3 ) , こ = b + 1 , さ = b 1 − 1 , し = b 2 + b 2 1 + 4 , す = 6 .
