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, $k = 1$.