15. Given functions $f ( x ) = 2 ^ { x }$ and $g ( x ) = x ^ { 2 } + a x$ (where $a \in \mathbb{R}$). For unequal real numbers $x _ { 1 }$ and $x _ { 2 }$, let $m = \frac { f \left( x _ { 1 } \right) - f \left( x _ { 2 } \right) } { x _ { 1 } - x _ { 2 } }$ and $n = \frac { g \left( x _ { 1 } \right) - g \left( x _ { 2 } \right) } { x _ { 1 } - x _ { 2 } }$.
The following propositions are given:
(1) For any unequal real numbers $x _ { 1 }$ and $x _ { 2 }$, we have $m > 0$;
(2) For any $a$ and any unequal real numbers $x _ { 1 }$ and $x _ { 2 }$, we have $n > 0$;
(3) For any $a$, there exist unequal real numbers $x _ { 1 }$ and $x _ { 2 }$ such that $m = n$;
(4) For any $a$, there exist unequal real numbers $x _ { 1 }$ and $x _ { 2 }$ such that $m = - n$. The true propositions are $\_\_\_\_$ (write out the numbers of all true propositions).

III. Solution Questions

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) $\log _ { a } c < \log _ { b } c$

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

11. If the range of the function $f ( x ) = a \cdot \left( \frac { 1 } { 3 } \right) ^ { x } \left( \frac { 1 } { 2 } \leq x \leq 1 \right)$ is a subset of the range of the function $g ( x ) = \frac { x ^ { 2 } - 1 } { x ^ { 2 } + x + 1 } ( x \in \mathbb{R} )$, then the range of positive number $a$ is
A. $(0,2]$
B. $(0,1]$
C. $( 0,2 \sqrt { 3 } ]$
D. $( 0 , \sqrt { 3 } ]$

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 $\_\_\_\_$.

14. Given that $f ( x )$ is an odd function, and when $x < 0$, $f ( x ) = - \mathrm { e } ^ { a x }$. If $f ( \ln 2 ) = 8$, then $a =$ $\_\_\_\_$ .

The Logistic model is one of the commonly used mathematical models and can be applied in epidemiology. Based on published data, scholars established a Logistic model for the cumulative confirmed cases of COVID-19 in a certain region $I ( t )$ (where $t$ is measured in days): $I ( t ) = \frac { K } { 1 + \mathrm { e } ^ { -0.23 ( t - 53 ) } }$, where $K$ is the maximum number of confirmed cases. When $I \left( t ^ { * } \right) = 0.95 K$, it indicates that the epidemic has been initially controlled. Then $t ^ { * }$ is approximately (given $\ln 19 \approx 3$)
A. 60
B. 63
C. 66
D. 69

The Logistic model is one of the commonly used mathematical models and can be applied in epidemiology. Based on published data, scholars established a Logistic model for the cumulative confirmed cases $I ( t )$ of COVID-19 in a certain region ($t$ in days): $I ( t ) = \frac { K } { 1 + \mathrm { e } ^ { - 0.23 ( t - 53 ) } }$ , where $K$ is the maximum number of confirmed cases. When $I \left( t ^ { * } \right) = 0.95 K$ , it indicates that the epidemic has been initially controlled. Then $t ^ { * }$ is approximately ( $\ln 19 \approx 3$ )
A. 60
B. 63
C. 66
D. 69

If $2 ^ { x } - 2 ^ { y } < 3 ^ { - x } - 3 ^ { - y }$ , then
A． $\ln ( y - x + 1 ) > 0$
B． $\ln ( y - x + 1 ) < 0$
C． $\ln | x - y | > 0$
D． $\ln | x - y | < 0$

4. Which of the following functions is an increasing function?
A. $f ( x ) = - x$
B. $f ( x ) = \left( \frac { 2 } { 3 } \right) ^ { x }$
C. $f ( x ) = x ^ { 2 }$
D. $f ( x ) = \sqrt [ 3 ] { x }$

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 $D$.

7. Let $a = 0.1 \mathrm { e } ^ { 0.1 }$, $b = \frac { 1 } { 9 }$, $c = - \ln 0.9$. Then
A. $a < b < c$
B. $c < b < a$
C. $c < a < b$
D. $a < c < b$

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. $b > 0 > a$

Given that the function $f(x)=a\mathrm{e}^x-\ln x$ is monotonically increasing on the interval $(1,2)$, the minimum value of $a$ is
A. $\mathrm{e}^2$
B. $\mathrm{e}$
C. $\mathrm{e}^{-1}$
D. $\mathrm{e}^{-2}$

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 $\forall \lambda \in [0,1], \mathrm{e}^{\lambda a+(1-\lambda) b} \leqslant \lambda \mathrm{e}^{a}+(1-\lambda) \mathrm{e}^{b}$.

Let $f(x) = xe^x$, and let $V$ and $W$ denote the inverses of $f|_{]-\infty,-1]}$ and $f|_{[-1,+\infty[}$ respectively. For a real parameter $m$, we consider the inequality with unknown $x \in \mathbb { R }$
$$x \mathrm { e } ^ { x } \leqslant m \tag{I.2}$$
Using the functions $V$ and $W$, determine, according to the values of $m$, the solutions of (I.2). Illustrate graphically the different cases.

Prove that, for every real number $t$, $\operatorname { ch } ( t ) \leqslant \exp \left( \frac { t ^ { 2 } } { 2 } \right)$.

1. Let $a$ and $b$ be two real numbers with $a > 0$. Choose without justification the correct expression for $a ^ { b }$: $(A) : \mathrm { e } ^ { b \ln ( a ) }$ $(B) : \mathrm { e } ^ { a \ln ( b ) }$ $(C) : \mathrm { e } ^ { \ln ( a ) \ln ( b ) }$.
2. Let $x$ and $y$ be two real numbers such that $x < y$ and $t$ a real number in $]0,1[$. Compare $t ^ { x }$ and $t ^ { y }$.
3. Give, without proof, the power series expansion of the real exponential function and give its domain of validity.
4. We consider the function $\Gamma$ defined by $\Gamma ( x ) = \int _ { 0 } ^ { + \infty } t ^ { x - 1 } e ^ { - t } \mathrm {~d} t$. We admit that this function is defined on $]0 , + \infty[$ and that, for all strictly positive real $x$: $$\Gamma ( x + 1 ) = x \Gamma ( x )$$ Calculate $\Gamma ( 1 )$ and deduce, by using a proof by induction, the value of $\Gamma ( n + 1 )$ for $n \in \mathbb { N }$.
5. For $x \in \mathbb { R }$, we denote, when it makes sense: $$F ( x ) = \int _ { 0 } ^ { 1 } t ^ { t ^ { x } } \mathrm { d } t$$ where, as is customary, $t ^ { t ^ { x } } = t^{(t^{x})}$.
   1. [5.1.] Determine the domain of definition of $F$.
   2. [5.2.] Determine the monotonicity of $F$.
   3. [5.3.] Prove that for all non-negative real $x$, we have: $F ( x ) \geqslant \frac { 1 } { 2 }$.
   4. [5.4.] Prove that $F$ is continuous on its domain of definition.
   5. [5.5.] Determine $\lim _ { x \rightarrow + \infty } F ( x )$ and $\lim _ { x \rightarrow - \infty } F ( x )$. The theorems used will be cited with precision and we will ensure that their hypotheses are well verified.
   6. [5.6.] Then carefully draw up the table of variations of $F$ and give a general sketch of its representative curve in an orthonormal coordinate system. We will admit that $F ^ { \prime } ( 0 ) = \frac { 1 } { 4 }$ and we will draw the tangent line at the point with abscissa $x = 0$.
6. Let $x$ be a strictly positive real number. For every natural number $n$, we denote by $g _ { n }$ the function defined on $]0,1]$ by $g _ { n } ( t ) = \frac { t ^ { n x } \ln ^ { n } ( t ) } { n ! }$.
   1. [6.1.] Prove that the series of functions $\sum _ { n \in \mathbb { N } } g _ { n }$ converges pointwise on $]0,1]$ and give its sum.
   2. [6.2.] Prove that, for every natural number $n , \int _ { 0 } ^ { 1 } \left| g _ { n } ( t ) \right| \mathrm { d } t = \frac { 1 } { n ! } \frac { \Gamma ( n + 1 ) } { ( n x + 1 ) ^ { n + 1 } }$.
   3. [6.3.] Finally establish that we have: $$F ( x ) = \sum _ { n = 0 } ^ { + \infty } \frac { ( - 1 ) ^ { n } } { ( 1 + n x ) ^ { n + 1 } }$$

103- If $f(x)=1-\left(\dfrac{1}{2}\right)^x$, and the domain of $y=\sqrt{x\,f(x)}$ is given, what is the range?
(1) $[-1,1]$ (2) $(-\infty,0)$ (3) $(-\infty,+\infty)$ (4) $(0,+\infty)$

111. Find the range of the function $f(x) = 2^{\frac{1}{3}\sqrt{9\cos^2(x)-1}} - 2^{\frac{1}{3}\sqrt{1-9\cos^2(x)}}$ in the form $[a, b]$. The value of $b - a$ is:
(1) $\dfrac{9}{5}$ (2) $\dfrac{15}{4}$ (3) $\dfrac{9}{2}$ (4) $\dfrac{21}{4}$

115. The graph of the function $f(x) = 9^{\log_3 x}$ is:
[Figure: Four graph options labeled (1), (2), (3), (4) showing different curve shapes in coordinate planes]
\rule{\textwidth}{0.4pt}
Calculation Space
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111. Suppose $5^x = 10$ and $5^x = 20$. If $2^{f(x)} = 2$ holds, what is the function $f$?
(1) $\dfrac{2x+1}{x+1}$ (2) $\dfrac{x-1}{2x-1}$ (3) $\dfrac{2x-1}{x-1}$ (4) $\dfrac{x+1}{2x+1}$