Exercise 1 (5 points)
The plane is equipped with an orthogonal coordinate system $(\mathrm{O}, \mathrm{I}, \mathrm{J})$.
We denote by $\Gamma$ the representative curve of the function $g$ defined on the interval $]0; 1]$ by $g(x) = \ln x$. Let $a$ be a real number in the interval $]0; 1]$. We denote by $M_a$ the point on the curve $\Gamma$ with abscissa $a$ and $d_a$ the tangent line to the curve $\Gamma$ at the point $M_a$. This line $d_a$ intersects the $x$-axis at point $N_a$ and the $y$-axis at point $P_a$. We are interested in the area of triangle $\mathrm{O}N_aP_a$ as the real number $a$ varies in the interval $]0; 1]$.
In this question, we study the particular case where $a = 0.2$.
a. Determine graphically an estimate of the area of triangle $\mathrm{O}N_{0.2}P_{0.2}$ in square units.
b. Determine an equation of the tangent line $d_{0.2}$.
c. Calculate the exact value of the area of triangle $\mathrm{O}N_{0.2}P_{0.2}$.

Find the maximum value $a$ of the function $f ( x ) = 2 \cos \left( x - \frac { \pi } { 3 } \right) + 2 \sqrt { 3 } \sin x$. Find the value of $a ^ { 2 }$. [3 points]

Given the function $\mathrm { f } ( \mathrm { x } ) = \frac { 1 } { 2 } \sin 2 \mathrm { x } - \sqrt { 3 } \cos ^ { 2 } x$ .
(I) Find the minimum positive period and minimum value of $\mathrm { f } ( \mathrm { x } )$;
(II) The graph of function $\mathrm { f } ( \mathrm { x } )$ is transformed by stretching each point's horizontal coordinate to twice the original length while keeping the vertical coordinate unchanged, resulting in the graph of function $\mathrm { g } ( \mathrm { x } )$. When $\mathrm { x } \in \left[ \frac { \pi } { 2 } , \pi \right]$, find the range of $\mathrm { g } ( \mathrm { x } )$.

15. Given $w > 0$, the two closest intersection points of the graphs of $y = 2 \sin w x$ and $y = 2 \cos w x$ have a distance of $2 \sqrt { 3 }$. Then $w =$ $\_\_\_\_$.
III. Solution Questions: This section has 6 questions, for a total of 75 points. Solutions should include written explanations, proofs, or calculation steps.

14. Given the function $f ( x ) = \sin \omega x + \cos \omega x ( \omega > 0 ) , x \in \mathbb{R}$. If the function $f ( x )$ is monotonically increasing on the interval $( - \omega , \omega )$, and the graph of $f ( x )$ is symmetric about the line $x = \omega$, then the value of $\omega$ is $\_\_\_\_$.

III. Solution Questions: This section has 6 questions, for a total of 80 points.

Given the function $f ( x ) = 2 \cos ^ { 2 } x - \sin ^ { 2 } x + 2$, then
A. The minimum positive period of $f ( x )$ is $\pi$, and the maximum value is 3
B. The minimum positive period of $f ( x )$ is $\pi$, and the maximum value is 4
C. The minimum positive period of $f ( x )$ is $2 \pi$, and the maximum value is 3
D. The minimum positive period of $f ( x )$ is $2 \pi$, and the maximum value is 4

Given the function $f ( x ) = 2 \sin x + \sin 2 x$, the minimum value of $f ( x )$ is $\_\_\_\_$

If $f ( x ) = \cos x - \sin x$ is decreasing on $[ 0 , a ]$, then the maximum value of $a$ is
A. $\frac { \pi } { 4 }$
B. $\frac { \pi } { 2 }$
C. $\frac { 3 \pi } { 4 }$
D. $\pi$

If $f ( x ) = \cos x - \sin x$ is an even function on $[ - a , a ]$, then the maximum value of $a$ is
A. $\frac { \pi } { 4 }$
B. $\frac { \pi } { 2 }$
C. $\frac { 3 \pi } { 4 }$
D. $\pi$

122. The figure shows the graph of the function with the formula $f(x) = \dfrac{a\sin 2x + b}{\sin x + \cos x}$, which has period one. What is $a$?
[Figure: Graph of a periodic function with amplitude 2]
(1) $-1$ (3) $\sqrt{2}$
(4) $2$ (1) $1$

112. The graph shown is the graph of $y = 1 + a\sin bx \cos bx$. What is $a + b$?
[Figure: Graph of $y = 1 + a\sin bx\cos bx$ showing a sinusoidal curve with amplitude markings at $\frac{3}{2}$ and $1$, and $x$-axis markings at $-\dfrac{\pi}{4}$ and $\dfrac{3\pi}{4}$]
(1) $1$ (2) $\dfrac{3}{2}$ (3) $2$ (4) $3$

12. If the figure below shows part of the graph of the function $f(x) = a + b\sin(cx - \dfrac{3\pi}{4})\cos(cx - \dfrac{3\pi}{4})$, what is the difference of the zeros of $f$ in the interval $[0, \pi]$?
[Figure: A sinusoidal curve with maximum value 3 and minimum value $-1$, with a visible point at $x = \pi$ on the x-axis]

• [(1)] $\dfrac{\pi}{6}$
• [(2)] $\dfrac{\pi}{4}$
• [(3)] $\dfrac{\pi}{2}$
• [(4)] $\dfrac{2\pi}{3}$

If the triangle $P Q R$ varies, then the minimum value of
$$\cos ( P + Q ) + \cos ( Q + R ) + \cos ( R + P )$$
is
[A] $- \frac { 5 } { 3 }$
[B] $- \frac { 3 } { 2 }$
[C] $\frac { 3 } { 2 }$
[D] $\frac { 5 } { 3 }$

If $A > 0 , B > 0$ and $A + B = \frac { \pi } { 6 }$, then the minimum positive value of $( \tan A + \tan B )$ is :
(1) $\sqrt { 3 } - \sqrt { 2 }$
(2) $4 - 2 \sqrt { 3 }$
(3) $\frac { 2 } { \sqrt { 3 } }$
(4) $2 - \sqrt { 3 }$

Let the range of the function $f ( x ) = 6 + 16 \cos x \cdot \cos \left( \frac { \pi } { 3 } - x \right) \cdot \cos \left( \frac { \pi } { 3 } + x \right) \cdot \sin 3 x \cdot \cos 6 x , x \in \mathbf { R }$ be $[ \alpha , \beta ]$. Then the distance of the point $( \alpha , \beta )$ from the line $3 x + 4 y + 12 = 0$ is :
(1) 11
(2) 8
(3) 10
(4) 9

Consider vectors $\vec{a}$ and $\vec{b}$ on the coordinate plane satisfying $|\vec{a}| + |\vec{b}| = 9$ and $|\vec{a} - \vec{b}| = 7$. Let $|\vec{a}| = x$, where $1 < x < 8$, and let the angle between $\vec{a}$ and $\vec{b}$ be $\theta$. Using the triangle formed by vectors $\vec{a}$, $\vec{b}$, and $\vec{a} - \vec{b}$, we can express $\cos\theta$ in terms of $x$ as $\frac{c}{9x - x^2} + d$, where $c$ and $d$ are constants with $c > 0$. Let this expression be $f(x)$, with domain $\{x \mid 1 < x < 8\}$. Find $f(x)$ and its derivative. (Non-multiple choice question, 4 points)