14. Let the function $f ( x ) = \left\{ \begin{array} { c c } 2 ^ { x } - a , & x < 1 , \\ 4 ( x - a ) ( x - 2 a ) , & x \geqslant 1 \text { .} \end{array} \right.$
(1) If $a = 1$, then the minimum value of $f ( x )$ is $\_\_\_\_$;
(2) If $f ( x )$ has exactly 2 zeros, then the range of the real number $a$ is $\_\_\_\_$.
III. Answer Questions (6 questions in total, 80 points. Solutions should include written explanations, calculation steps, or proof processes)

14. If the function $f ( x ) = \left\{ \begin{array} { l } - x + 6 , x \leq 2 , \\ 3 + \log _ { a } x , x > 2 , \end{array} ( a > 0 \right.$ and $a \neq 1 )$ has range $[ 4 , + \infty )$, then the range of the real number $a$ is $\_\_\_\_$.

20. Given functions $f ( x ) = \ln ( 1 + x )$ and $g ( x ) = k x$ ($k \in \mathbb{R}$),
(1) Prove that when $x > 0$, $f ( x ) < x$; Solution Method 1: (1) Stretch the vertical coordinates of all points on the graph of $g(x) = \cos x$ to 2 times the original (horizontal coordinates unchanged) to obtain the graph of $y = 2\cos x$, then shift the graph of $y = 2\cos x$ to the right by $\frac{p}{2}$ units to obtain the graph of $y = 2\cos\left(x - \frac{p}{2}\right)$, thus $f(x) = 2\sin x$.
Therefore, the equation of the axis of symmetry of the graph of function $f(x) = 2\sin x$ is $x = kp + \frac{p}{2}$ $(k \in \mathbb{Z})$.
(2) 1) $f(x) + g(x) = 2\sin x + \cos x = \sqrt{5}\left(\frac{2}{\sqrt{5}}\sin x + \frac{1}{\sqrt{5}}\cos x\right)$
$$= \sqrt{5}\sin(x + j) \quad \left(\text{where } \sin j = \frac{1}{\sqrt{5}}, \cos j = \frac{2}{\sqrt{5}}\right)$$
According to the problem, $\sin(x + j) = \frac{m}{\sqrt{5}}$ has two distinct solutions $a, b$ in the interval $[0, 2p)$ if and only if $\left|\frac{m}{\sqrt{5}}\right| < 1$, thus the range of $m$ is $(-\sqrt{5}, \sqrt{5})$.
2) Since $a, b$ are two distinct solutions of the equation $\sqrt{5}\sin(x + j) = m$ in the interval $[0, 2p)$,
we have $\sin(a + j) = \frac{m}{\sqrt{5}}, \sin(b + j) = \frac{m}{\sqrt{5}}$.
When $1 \leq m < \sqrt{5}$, $a + b = 2\left(\frac{p}{2} - j\right)$, i.e., $a - b = p - 2(b + j)$;
When $-\sqrt{5} < m < 1$, $a + b = 2\left(\frac{3p}{2} - j\right)$, i.e., $a - b = 3p - 2(b + j)$;
Therefore $\cos(a - b) = -\cos 2(b + j) = 2\sin^2(b + j) - 1 = 2\left(\frac{m}{\sqrt{5}}\right)^2 - 1 = \frac{2m^2}{5} - 1$.
Solution Method 2: (1) Same as Solution Method 1.
(2) 1) Same as Solution Method 1.
2) Since $a, b$ are two distinct solutions of the equation $\sqrt{5}\sin(x + j) = m$ in the interval $[0, 2p)$,
we have $\sin(a + j) = \frac{m}{\sqrt{5}}, \sin(b + j) = \frac{m}{\sqrt{5}}$.
When $1 \leq m < \sqrt{5}$, $a + b = 2\left(\frac{p}{2} - j\right)$, i.e., $a + j = p - (b + j)$;
When $-\sqrt{5} < m < 1$, $a + b = 2\left(\frac{3p}{2} - j\right)$, i.e., $a + j = 3p - (b + j)$;
Therefore $\cos(a + j) = -\cos(b + j)$
Thus $\cos(a - b) = \cos[(a + j) - (b + j)] = \cos(a + j)\cos(b + j) + \sin(a + j)\sin(b + j)$
$$= -\cos^2(b + j) + \sin(a + j)\sin(b + j) = -\left[1 - \left(\frac{m}{\sqrt{5}}\right)^2\right] + \left(\frac{m}{\sqrt{5}}\right)^2 = \frac{2m^2}{5} - 1.$$
20. This problem mainly tests basic knowledge of derivatives and their applications, tests reasoning and proof ability, computational ability, and innovative thinking, tests function and equation ideas, transformation and conversion ideas, classification and integration ideas, finite and infinite ideas, and number-form combination ideas. Full marks: 14 points.
Solution Method 1: (1) Let $F(x) = f(x) - x = \ln(1 + x) - x, x \in [0, +\infty)$, then $F'(x) = \frac{1}{1+x} - 1 = -\frac{x}{1+x}$.
When $x \in [0, +\infty)$, $F'(x) < 0$, so $F(x)$ is monotonically decreasing on $[0, +\infty)$.
Therefore when $x > 0$, $F(x) < F(0) = 0$, i.e., when $x > 0$, $f(x) < x$.
(2) Let $G(x) = f(x) - g(x) = \ln(1 + x) - kx, x \in [0, +\infty)$, then $G'(x) = \frac{1}{1+x} - k = \frac{-kx + (1-k)}{1+x}$.
When $k \leq 0$, $G'(x) > 0$, so $G(x)$ is monotonically increasing on $[0, +\infty)$, $G(x) > G(0) = 0$.
Therefore the condition is satisfied for any positive real number $x_0$.
When $0 < k < 1$, let $G'(x) = 0$, we get $x = \frac{1-k}{k} = \frac{1}{k} - 1 > 0$.
Take $x_0 = \frac{1}{k} - 1$. For any $x \in (0, x_0)$, we have $G'(x) > 0$, so $G(x)$ is monotonically increasing on $[0, x_0)$, $G(x) > G(0) = 0$, i.e., $f(x) > g(x)$.
In summary, when $k < 1$, there always exists $x_0 > 0$ such that for any $x \in (0, x_0)$, we have $f(x) > g(x)$.
(3) When $k > 1$, from (1) we know that for $x \in [0, +\infty)$, $g(x) > x > f(x)$, thus $g(x) > f(x)$, $|f(x) - g(x)| = g(x) - f(x) = kx - \ln(1+x)$.
Let $M(x) = kx - \ln(1+x) - x^2, x \in [0, +\infty)$, then $M'(x) = k - \frac{1}{1+x} - 2x = \frac{-2x^2 + (k-2)x + k-1}{1+x}$.
Therefore when $x \in \left(0, \frac{k-2+\sqrt{(k-2)^2 + 8(k-1)}}{4}\right)$, $M'(x) > 0$, $M(x)$ is monotonically increasing on $\left[0, \frac{k-2+\sqrt{(k-2)^2 + 8(k-1)}}{4}\right)$, thus $M(x) > M(0) = 0$, i.e., $|f(x) - g(x)| > x^2$, so there is no $t$ satisfying the condition.
When $k < 1$, from (2) we know there exists $x_0 > 0$ such that for any $x \in (0, x_0)$, we have $f(x) > g(x)$.
At this time $|f(x) - g(x)| = f(x) - g(x) = \ln(1+x) - kx$.
Let $N(x) = \ln(1+x) - kx - x^2, x \in [0, +\infty)$, then $N'(x) = \frac{1}{1+x} - k - 2x = \frac{-2x^2 - (k+2)x - k+1}{1+x}$.
Therefore when $x \in \left(0, \frac{-(k+2)+\sqrt{(k+2)^2 + 8(1-k)}}{4}\right)$, $N'(x) > 0$, $N(x)$ is monotonically increasing on $\left[0, \frac{-(k+2)+\sqrt{(k+2)^2 + 8(1-k)}}{4}\right)$, thus $N(x) > N(0) = 0$, i.e., $f(x) - g(x) > x^2$. Let $x_1$ be the smaller of $x_0$ and $\frac{-(k+2)+\sqrt{(k+2)^2 + 8(1-k)}}{4}$.
Then when $x \in (0, x_1)$, we have $|f(x) - g(x)| > x^2$, thus there is no $t$ satisfying the condition.
When $k = 1$, from (1) we know that for $x \in (0, +\infty)$, $|f(x) - g(x)| = g(x) - f(x) = x - \ln(1+x)$.
Let $H(x) = x - \ln(1+x) - x^2, x \in [0, +\infty)$, then $H'(x) = 1 - \frac{1}{1+x} - 2x = \frac{-2x^2 - x}{1+x}$.
When $x > 0$, $H'(x) < 0$, so $H(x)$ is monotonically decreasing on $[0, +\infty)$, thus $H(x) < H(0) = 0$.
Therefore when $x > 0$, we have $|f(x) - g(x)| < x^2$, at this time any real number $t$ satisfies the condition.
In summary, $k = 1$.
Solution Method 2: (1) (2) Same as Solution Method 1.
(3) When $k > 1$, from (1) we know that for $x \in [0, +\infty)$, $g(x) > x > f(x)$,
thus $|f(x) - g(x)| = g(x) - f(x) = kx - \ln(1+x) > kx - x = (k-1)x$.
Let $(k-1)x > x^2$, we get $0 < x < k-1$.
Therefore when $k > 1$, for $x \in (0, k-1)$ we have $|f(x) - g(x)| > x^2$, so there is no $t$ satisfying the condition.
When $k < 1$, take $k_1 = \frac{k+1}{2}$, thus $k < k_1 < 1$.
From (2) we know there exists $x_0 > 0$ such that for any $x \in (0, x_0)$, we have $f(x) > k_1 x > kx = g(x)$.
At this time $|f(x) - g(x)| = f(x) - g(x) > (k_1 - k)x = \frac{1-k}{2}x$.
Let $\frac{1-k}{2}x > x^2$, we get $0 < x < \frac{1-k}{2}$. At this time $f(x) - g(x) > x^2$.
Let $x_1$ be the smaller of $x_0$ and $\frac{1-k}{2}$. Then when $x \in (0, x_1)$, we have $|f(x) - g(x)| > x^2$,
thus there is no $t$ satisfying the condition.
When $k = 1$, from (1) we know that for $x \in (0, +\infty)$, $|f(x) - g(x)| = g(x) - f(x) = x - \ln(1+x)$.
Let $M(x) = x - \ln(1+x) - x^2, x \in [0, +\infty)$, then $M'(x) = 1 - \frac{1}{1+x} - 2x = \frac{-2x^2 - x}{1+x}$.
When $x > 0$, $M'(x) < 0$, so $M(x)$ is monotonically decreasing on $[0, +\infty)$, thus $M(x) < M(0) = 0$.
Therefore when $x > 0$, we have $|f(x) - g(x)| < x^2$, at this time any real number $t$ satisfies the condition.
In summary, $k = 1$.

If $a > b > 1 , ~ 0 < c < 1$, then
(A) $a ^ { c } < b ^ { c }$
(B) $a b ^ { c } < b a ^ { c }$
(C) $a \log _ { b } c < b \log _ { a } c$
(D) $\log _ { a } c < \log _ { b } c$

Given sets $A = \{ x \mid x < 1 \} , B = \left\{ x \mid 3 ^ { x } < 1 \right\}$, then
A. $A \cap B = \{ x \mid x < 0 \}$
B. $A \cup B = \mathbf { R }$
C. $A \cup B = \{ x \mid x > 1 \}$
D. $A \cap B = \varnothing$

Let $x, y, z$ be positive numbers such that $2 ^ { x } = 3 ^ { y } = 5 ^ { z }$, then
A. $2x < 3y < 5z$
B. $5z < 2x < 3y$
C. $3y < 5z < 2x$
D. $3y < 2x < 5z$

5. The sum of the zeros of the function $f ( x ) = \left\{ \begin{array} { l } 6 ^ { x } - 2 , x > 0 , \\ x + \log _ { 6 } 12 , x \leq 0 \end{array} \right.$ is
A. $-1$ B. $1$ C. $-2$ D. $2$

If $a > b$, then
A． $\ln ( a - b ) > 0$
B． $3 ^ { a } < 3 ^ { b }$
C．$a ^ { 3 } - b ^ { 3 } > 0$
D． $| a | > | b |$

6. If $a > b$, then
A. $\ln ( a - b ) > 0$
B. $3 ^ { a } < 3 ^ { b }$
C. $a ^ { 3 } - b ^ { 3 } > 0$
D. $| a | > | b |$

Let the function $f ( x ) = \mathrm { e } ^ { x } + a \mathrm { e } ^ { - x }$ ($a$ is a constant). If $f ( x )$ is an odd function, then $a =$ $\_\_\_\_$; if $f ( x )$ is an increasing function on $\mathbb { R }$, then the range of $a$ is $\_\_\_\_$.

16. A line $l$ with slope $k ( k < 0 )$ passes through point $F ( 0,1 )$ and intersects the curve $y = \frac { 1 } { 4 } x ^ { 2 } ( x \geq 0 )$ and the line $y = - 1$ at points $A$ and $B$ respectively. If $| F B | = 6 | F A |$, then $k = $ \_\_\_\_.
III. Solution Questions: This section contains 6 questions, totaling 70 points. Show your working, proofs, or calculation steps. Questions 17-21 are required questions that all candidates must answer. Questions 22 and 23 are optional questions; candidates should answer according to the requirements. (I) Required Questions: Total 60 points.

7. D

Solution: By the concavity of the function, the point $(a, b)$ cannot be above the curve $y = e ^ { x }$. Since $y = 0$ is an asymptote, the point lies between the curve and the asymptote, so $0 < b < e ^ { a }$. The answer is $D$.

Given $9 ^ { m } = 10 , a = 10 ^ { m } - 11 , b = 8 ^ { m } - 9$ , then
A. $a > 0 > b$
B. $a > b > 0$
C. $b > a > 0$
D. $b > 0 > a$

Let $a \in ( 0,1 )$. If the function $f ( x ) = a ^ { x } + ( 1 + a ) ^ { x }$ is monotonically increasing on $( 0 , + \infty )$, then the range of $a$ is \_\_\_\_

Given that $\left( x _ { 1 } , y _ { 1 } \right) , \left( x _ { 2 } , y _ { 2 } \right)$ are points on $y = 2 ^ { x }$, which of the following is correct?
A. $\log _ { 2 } \frac { y _ { 1 } + y _ { 2 } } { 2 } > \frac { x _ { 1 } + x _ { 2 } } { 2 }$
B. $\log _ { 2 } \frac { y _ { 1 } + y _ { 2 } } { 2 } < \frac { x _ { 1 } + x _ { 2 } } { 2 }$
C. $\log _ { 2 } \frac { y _ { 1 } + y _ { 2 } } { 2 } > x _ { 1 } + x _ { 2 }$
D. $\log _ { 2 } \frac { y _ { 1 } + y _ { 2 } } { 2 } < x _ { 1 }$

Let $a$ and $b$ be two reals satisfying $a < b$. Show that $\forall \lambda \in [0,1], \mathrm{e}^{\lambda a+(1-\lambda) b} \leqslant \lambda \mathrm{e}^{a}+(1-\lambda) \mathrm{e}^{b}$.

We define the function $\theta : \mathbb { R } \rightarrow \mathbb { C }$ by $$\begin{cases} \theta ( x ) = 0 & \text { if } x \leqslant 0 \\ \theta ( x ) = \exp \left( - \frac { \ln ^ { 2 } x } { 4 \pi ^ { 2 } } + \mathrm { i } \frac { \ln x } { 2 \pi } \right) & \text { if } x > 0 \end{cases}$$
Show that $\lim _ { \substack { x \rightarrow 0 \\ x > 0 } } | \theta ( x ) | = 0$.

We denote by $\mathcal{E}$ the set of functions $f : \mathbb{C} \rightarrow \mathbb{C}$ expandable as a power series with radius of convergence infinity. Justify that if $(f, g) \in \mathcal{E}^{2}$ and $(\lambda, \mu) \in \mathbb{C}^{2}$, then $\lambda f + \mu g \in \mathcal{E}$ and $fg \in \mathcal{E}$.

Let $f : \mathbb { R } \rightarrow [ 0 , \infty )$ be a continuous function such that $$f ( x + y ) = f ( x ) f ( y )$$ for all $x , y \in \mathbb { R }$. Suppose that $f$ is differentiable at $x = 1$ and $$\left. \frac { d f ( x ) } { d x } \right| _ { x = 1 } = 2$$ Then, the value of $f ( 1 ) \log _ { e } f ( 1 )$ is
(A) $e$.
(B) 2.
(C) $\log _ { e } 2$.
(D) 1 .

The set of all real numbers $x$ for which $3^{2^{1-x^2}}$ is an integer has
(A) 3 elements
(B) 15 elements
(C) 24 elements
(D) infinitely many elements

If $f : \mathbb { R } \rightarrow \mathbb { R }$ is a differentiable function such that $f ^ { \prime } ( x ) > 2 f ( x )$ for all $x \in \mathbb { R }$, and $f ( 0 ) = 1$, then
[A] $f ( x )$ is increasing in $( 0 , \infty )$
[B] $f ( x )$ is decreasing in $( 0 , \infty )$
[C] $f ( x ) > e ^ { 2 x }$ in $( 0 , \infty )$
[D] $f ^ { \prime } ( x ) < e ^ { 2 x }$ in $( 0 , \infty )$

Let $f : \mathbb { R } \rightarrow \mathbb { R }$ and $g : \mathbb { R } \rightarrow \mathbb { R }$ be two non-constant differentiable functions. If $$f ^ { \prime } ( x ) = \left( e ^ { ( f ( x ) - g ( x ) ) } \right) g ^ { \prime } ( x ) \text { for all } x \in \mathbb { R }$$ and $f ( 1 ) = g ( 2 ) = 1$, then which of the following statement(s) is (are) TRUE?
(A) $f ( 2 ) < 1 - \log _ { \mathrm { e } } 2$
(B) $f ( 2 ) > 1 - \log _ { \mathrm { e } } 2$
(C) $g ( 1 ) > 1 - \log _ { \mathrm { e } } 2$
(D) $g ( 1 ) < 1 - \log _ { e } 2$

Let $f : \mathbb { R } \rightarrow \mathbb { R }$ be a differentiable function with $f ( 0 ) = 1$ and satisfying the equation
$$f ( x + y ) = f ( x ) f ^ { \prime } ( y ) + f ^ { \prime } ( x ) f ( y ) \text { for all } x , y \in \mathbb { R }$$
Then, the value of $\log _ { e } ( f ( 4 ) )$ is $\_\_\_\_$ .

Let $b$ be a nonzero real number. Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is a differentiable function such that $f(0) = 1$. If the derivative $f'$ of $f$ satisfies the equation $$f'(x) = \frac{f(x)}{b^{2} + x^{2}}$$ for all $x \in \mathbb{R}$, then which of the following statements is/are TRUE?
(A) If $b > 0$, then $f$ is an increasing function
(B) If $b < 0$, then $f$ is a decreasing function
(C) $f(x)f(-x) = 1$ for all $x \in \mathbb{R}$
(D) $f(x) - f(-x) = 0$ for all $x \in \mathbb{R}$

The sum of the series $\frac { 1 } { 2 ! } - \frac { 1 } { 3 ! } + \frac { 1 } { 4 ! } - \ldots$ upto infinity is
(1) $e ^ { - 2 }$
(2) $e ^ { - 1 }$
(3) $e ^ { - 1 / 2 }$
(4) $e ^ { 1 / 2 }$