5. A geometric sequence $\left\{ a _ { n } \right\}$ with all positive terms has the sum of its first 4 terms equal to 15, and $a _ { 5 } = 3 a _ { 3 } + 4 a _ { 1 }$ . Then $a _ { 3 } =$ A. 16 B. 8 C. 4 D. 2
6. The tangent line to the curve $y = a \mathrm { e } ^ { x } + x \ln x$ at the point $( 1 , a \mathrm { e } )$ has equation $y = 2 x + b$ . Then A. $a = \mathrm { e } , \quad b = - 1$ B. $a = \mathrm { e } , b = 1$ C. $a = \mathrm { e } ^ { - 1 } , b = 1$ D. $a = \mathrm { e } ^ { - 1 } , b = - 1$
7. The graph of the function $y = \frac { 2 x ^ { 3 } } { 2 ^ { x } + 2 ^ { - x } }$ on $[ - 6,6 ]$ is approximately A. [Figure] B. [Figure] C. [Figure] D. [Figure]
8. As shown in the figure, point $N$ is the center of square $A B C D$ , $\triangle E C D$ is an equilateral triangle, plane $E C D \perp$ plane $A B C D$ , and $M$ is the midpoint of segment $E D$ . Then [Figure] A. $B M = E N$ , and lines $B M$ and $E N$ are intersecting lines B. $B M \neq E N$ , and lines $B M$ and $E N$ are intersecting lines C. $B M = E N$ , and lines $B M$ and $E N$ are skew lines D. $B M \neq E N$ , and lines $B M$ and $E N$ are skew lines
10. For the hyperbola $C : \frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 2 } = 1$ with right focus $F$ , if point $P$ is on one of the asymptotes of $C$ , $O$ is the origin, and $| P O | = | P F |$ , then the area of $\triangle P F O$ is A. $\frac { 3 \sqrt { 2 } } { 4 }$ B. $\frac { 3 \sqrt { 2 } } { 2 }$ C. $2 \sqrt { 2 }$ D. $3 \sqrt { 2 }$
12. Let $f ( x ) = \sin \left( \omega x + \frac { \pi } { 5 } \right) ( \omega > 0 )$ . Given that $f ( x )$ has exactly 5 zeros on $[ 0,2 \pi ]$ , consider the following four conclusions: (1) $f ( x )$ has exactly 3 local maximum points on $( 0,2 \pi )$ (2) $f ( x )$ has exactly 2 local minimum points on $( 0,2 \pi )$ (3) $f ( x )$ is monotonically increasing on $\left( 0 , \frac { \pi } { 10 } \right)$ (4) The range of $\omega$ is $\left[ \frac { 12 } { 5 } , \frac { 29 } { 10 } \right)$ The numbers of all correct conclusions are A. (1)(4) B. (2)(3) C. (1)(2)(3) D. (1)(3)(4) II. Fill-in-the-Blank Questions: This section has 4 questions, each worth 5 points, for a total of 20 points.
13. Given that $\boldsymbol { a } , \boldsymbol { b }$ are unit vectors and $\boldsymbol { a } \cdot \boldsymbol { b } = 0$ , if $\boldsymbol { c } = 2 \boldsymbol { a } - \sqrt { 5 } \boldsymbol { b }$ , then $\cos \langle \boldsymbol { a } , \boldsymbol { c } \rangle =$ $\_\_\_\_$ .
14. Let $S _ { n }$ denote the sum of the first $n$ terms of an arithmetic sequence $\left\{ a _ { n } \right\}$ . If $a _ { 1 } \neq 0$ and $a _ { 2 } = 3 a _ { 1 }$ , then $\frac { S _ { 10 } } { S _ { 5 } } =$ $\_\_\_\_$ .
15. Let $F _ { 1 } , F _ { 2 }$ be the two foci of the ellipse $C : \frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 20 } = 1$ , and let $M$ be a point on $C$ in the first quadrant. If $\triangle M F _ { 1 } F _ { 2 }$ is an isosceles triangle, then the coordinates of $M$ are $\_\_\_\_$ .
18. (12 points) In $\triangle A B C$ , the sides opposite to angles $A , B , C$ are $a , b , c$ respectively. Given that When $t = 0$, $S = 3$; when $t = \pm 1$, $S = 4\sqrt{2}$. Therefore, the area of quadrilateral $ADBE$ is $3$ or $4\sqrt{2}$.
22. Solution: (1) From the given conditions, the polar coordinate equations of the circles containing arcs $AB$, $BC$, $CD$ are $\rho = 2\cos\theta$, $\rho = 2\sin\theta$, $\rho = -2\cos\theta$ respectively. Thus the polar coordinate equation of $M_1$ is $\rho = 2\cos\theta \left(0 \leq \theta \leq \frac{\pi}{4}\right)$, the polar coordinate equation of $M_2$ is $\rho = 2\sin\theta \left(\frac{\pi}{4} \leq \theta \leq \frac{3\pi}{4}\right)$, the polar coordinate equation of $M_3$ is $\rho = -2\cos\theta \left(\frac{3\pi}{4} \leq \theta \leq \pi\right)$. (2) Let $P(\rho, \theta)$. From the given conditions and (1), we have If $0 \leq \theta \leq \frac{\pi}{4}$, then $2\cos\theta = \sqrt{3}$, solving gives $\theta = \frac{\pi}{6}$; If $\frac{\pi}{4} \leq \theta \leq \frac{3\pi}{4}$, then $2\sin\theta = \sqrt{3}$, solving gives $\theta = \frac{\pi}{3}$ or $\theta = \frac{2\pi}{3}$; If $\frac{3\pi}{4} \leq \theta \leq \pi$, then $-2\cos\theta = \sqrt{3}$, solving gives $\theta = \frac{5\pi}{6}$. In summary, the polar coordinates of $P$ are $\left(\sqrt{3}, \frac{\pi}{6}\right)$ or $\left(\sqrt{3}, \frac{\pi}{3}\right)$ or $\left(\sqrt{3}, \frac{2\pi}{3}\right)$ or $\left(\sqrt{3}, \frac{5\pi}{6}\right)$.
23. Solution: (1) Since $[(x-1) + (y+1) + (z+1)]^2$ $= (x-1)^2 + (y+1)^2 + (z+1)^2 + 2[(x-1)(y+1) + (y+1)(z+1) + (z+1)(x-1)]$ $\leq 3[(x-1)^2 + (y+1)^2 + (z+1)^2]$, from the given condition we have $(x-1)^2 + (y+1)^2 + (z+1)^2 \geq \frac{4}{3}$, with equality if and only if $x = \frac{5}{3}$, $y = -\frac{1}{3}$, $z = -\frac{1}{3}$. Therefore, the minimum value of $(x-1)^2 + (y+1)^2 + (z+1)^2$ is $\frac{4}{3}$. (2) Since $[(x-2) + (y-1) + (z-a)]^2$ $= (x-2)^2 + (y-1)^2 + (z-a)^2 + 2[(x-2)(y-1) + (y-1)(z-a) + (z-a)(x-2)]$ $\leq 3[(x-2)^2 + (y-1)^2 + (z-a)^2]$, from the given condition $(x-2)^2 + (y-1)^2 + (z-a)^2 \geq \frac{(2+a)^2}{3}$, with equality if and only if $x = \frac{4-a}{3}$, $y = \frac{1-a}{3}$, $z = \frac{2a-2}{3}$. Therefore, the minimum value of $(x-2)^2 + (y-1)^2 + (z-a)^2$ is $\frac{(2+a)^2}{3}$. From the given condition, $\frac{(2+a)^2}{3} \geq \frac{1}{3}$, solving gives $a \leq -3$ or $a \geq -1$.