Q3
Completing the square and sketching
Optimization on a constrained domain via completing the square
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3. Execute the flowchart shown in Figure 1. If the input is $n = 3$, then the output is
A. $\frac { 6 } { 7 }$
B. $\frac { 3 } { 7 }$
C. $\frac { 8 } { 9 }$
D. $\frac { 4 } { 9 }$ [Figure]
(4) If variables $x , y$ satisfy the constraint conditions $\left\{ \begin{array} { c } x + y \geq - 1 \\ 2 x - y \leq 1 \\ y \leq 1 \end{array} \right.$, then the minimum value of $z = 3 x - y$ is
(A) $- 7$
(B) $- 1$
(C) $1$
(D) $2$
(5) Let the function $f ( x ) = \ln ( 1 + x ) - \ln ( 1 - x )$. Then $f ( x )$ is
(A) an odd function and increasing on $( 0,1 )$
(B) an odd function and decreasing on $(0,1)$
(C) an even function and increasing on $( 0,1 )$
(D) an even function and decreasing on $(0,1)$ (6) Given that the expansion of $\left( \sqrt { \mathrm { x } } - \frac { \mathrm { a } } { \sqrt { \mathrm { x } } } \right) ^ { 5 }$ contains a term with $\mathrm { x } ^ { \frac { 3 } { 2 } }$ whose coefficient is 30, then $\mathrm { a } =$
(A) $\sqrt { 3 }$
(B) $- \sqrt { 3 }$
(C) $6$
(D) $- 6$ (7) In the square shown in Figure 2, 10000 points are randomly thrown. The estimated number of points falling in the shaded region (where curve C is the density curve of the normal distribution $N ( 0,1 )$) is
[Figure]
Figure 2
(A) $2386$
(B) $2718$
(C) $3413$
(D) $4772$
Attachment: If $X \sim N \left( \mu , \sigma ^ { 2 } \right)$, then
$$\begin{aligned}
& P ( \mu - \sigma < x \leq \mu + \sigma ) = 0.6826 \\
& P ( \mu - 2 \sigma < x \leq \mu + 2 \sigma ) = 0.9544
\end{aligned}$$
(8) Points $\mathrm { A } , \mathrm { B } , \mathrm { C }$ move on the circle $\mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } = 1$, and $\mathrm { AB } \perp \mathrm { BC }$. If point P has coordinates $( 2,0 )$, then the maximum value of $| \overrightarrow { P A } + \overrightarrow { P B } + \overrightarrow { P C } |$ is
A. $6$
B. $7$
C. $8$
D. $9$ (9) The graph of the function $f ( x ) = \sin 2 x$ is shifted to the right by $\varphi \left( 0 < \varphi < \frac { \pi } { 2 } \right)$ units to obtain the graph of function $g ( x )$. If for $x _ { 1 }$ and $x _ { 2 }$ satisfying $\left| f \left( x _ { 1 } \right) - g \left( x _ { 2 } \right) \right| = 2$, we have $\left| x _ { 1 } - x _ { 2 } \right| _ { \min } = \frac { \pi } { 3 }$, then $\varphi =$
A. $\frac { 5 \pi } { 12 }$
B. $\frac { \pi } { 3 }$
C. $\frac { \pi } { 4 }$
D. $\frac { \pi } { 6 }$