2021 Zhejiang Gao Kao Calculus

Question

Let $a, b$ be real numbers, and $a > 1$ . Consider the function $f (x) = a^{x} - b x + e^{2}$ ( $x \in R$ ).

Find the monotonic intervals of the function $f (x)$ .
If for any $b > 2 e^{2}$ , the function $f (x)$ has two different zeros, find the range of $a$ .
When $a = e$ , prove that: for any $b > e^{4}$ , the function $f (x)$ has two different zeros $x_{1}, x_{2}$ , satisfying $x_{2} > \frac{b \ln b}{2 e^{2}} x_{1} + \frac{e^{2}}{b}$ .

Note:

$e = 2.71828 \dots$ is the base of the natural logarithm.
We define the natural logarithm as $\ln x$ , which is $\ln x = \log_{e} x$ .

中文原题

设 $a, b$ 为实数，且 $a > 1$ ，函数 $f (x) = a^{x} - b x + e^{2}$ ( $x \in R$ )

求函数 $f (x)$ 的单调区间；
若对任意 $b > 2 e^{2}$ ，函数 $f (x)$ 有两个不同的零点，求 $a$ 的取值范围；
当 $a = e$ 时，证明：对任意 $b > e^{4}$ ，函数 $f (x)$ 有两个不同的零点 $x_{1}, x_{2}$ ，满足 $x_{2} > \frac{b \ln b}{2 e^{2}} x_{1} + \frac{e^{2}}{b}$ 。

注： $e = 2.71828 \dots$ 是自然对数的底数。

Solution

Part 1

The derivative of $f (x)$ is

f^{'} (x) = a^{x} \ln a - b

$f^{″} (x) = a^{x} \ln^{2} a > 0$ , so $f^{'} (x)$ is an strictly increasing function.

When $b < 0$ , $f^{'} (x) > lim_{x \to - \infty} f^{'} (x) = 0$ , so $f (x)$ is an strictly increasing function.

When $b \geq 0$ , solving $f^{'} (x) = 0$ we can get $x = \log_{a} \frac{b}{\ln a}$ . Since $f^{'} (x)$ is increasing, we have:

$x$	$x \in (- \infty, \log_{a} \frac{b}{\ln a})$	$x = \log_{a} \frac{b}{\ln a}$	$x \in (\log_{a} \frac{b}{\ln a}, + \infty)$
$f^{'} (x)$	$f^{'} (x) < 0$	$f^{'} (x) = 0$	$f^{'} (x) > 0$
$f (x)$	$f (x)$ is decreasing	$f (x)$ has a local minimum	$f (x)$ is increasing

As a result, when $b < 0$ , $f (x)$ has a monotonically increasing interval $(- \infty, + \infty)$ ; when $b \geq 0$ , $f (x)$ has a monotonically decreasing interval $(- \infty, \log_{a} \frac{b}{\ln a})$ and a monotonically increasing interval $(\log_{a} \frac{b}{\ln a}, + \infty)$ .

Part 2

Given that $b > 2 e^{2} > 0$ , the function $f (x)$ has a local minimum at $x = \log_{a} \frac{b}{\ln a}$ .

Also, we have:

\begin{aligned} lim_{x \to - \infty} f (x) & = lim_{x \to - \infty} a^{x} - b x + e^{2} = \infty \\ lim_{x \to + \infty} f (x) & = lim_{x \to + \infty} a^{x} - b x + e^{2} = \infty \end{aligned}

Therefore, $f (x)$ has two distinct zeros if and only if $f (\log_{a} \frac{b}{\ln a}) < 0$ .

Let's simplify $f (\log_{a} \frac{b}{\ln a})$ :

\begin{aligned} f (\log_{a} \frac{b}{\ln a}) & = a^{\log_{a} \frac{b}{\ln a}} - b \log_{a} \frac{b}{\ln a} + e^{2} \\ = \frac{b}{\ln a} - b \cdot (\log_{a} b - \log_{a} \ln a) + e^{2} \\ = \frac{b}{\ln a} - \frac{b \ln b}{\ln a} + \frac{b \ln \ln a}{\ln a} + e^{2} \\ = \frac{b}{\ln a} (1 - \ln b + \ln \ln a) + e^{2} \end{aligned}

For $f (\log_{a} \frac{b}{\ln a}) < 0$ to hold, we must ensure:

lim_{b \to {(2 e^{2})}^{+}} f (\log_{a} \frac{b}{\ln a}) \leq 0

This implies:

\begin{aligned} 0 & \geq lim_{b \to {(2 e^{2})}^{+}} f (\log_{a} \frac{b}{\ln a}) \\ = lim_{b \to {(2 e^{2})}^{+}} \frac{b}{\ln a} (1 - \ln b + \ln \ln a) + e^{2} \\ = \frac{2 e^{2}}{\ln a} (1 - \ln (2 e^{2}) + \ln \ln a) + e^{2} \\ = \frac{2 e^{2}}{\ln a} (\ln \ln a - 1 - \ln 2) + e^{2} \end{aligned}

Hence, we should have:

2 e^{2} (\ln \ln a - 1 - \ln 2) + e^{2} \ln a \leq 0

This simplifies to:

2 \ln \ln a + \ln a \leq 2 \ln 2 + 2

Let's define $ϕ (x) = 2 \ln x + x$ . The inequality above is equivalent to:

ϕ (\ln a) \leq ϕ (2)

Given that $ϕ (x)$ is a strictly increasing function, we can have:

\ln a \leq 2 \Rightarrow a \leq e^{2}

However, this is not the final answer since we should consider the whole range of $b > 2 e^{2}$ instead of a specific value of $b \to {(2 e^{2})}^{+}$ . But we have a necessary condition for $a$ to satisfy.

From $f (\log_{a} \frac{b}{\ln a}) < 0$ , we can get:

\frac{b}{\ln a} (1 - \ln b + \ln \ln a) + e^{2} < 0

Let

g (a, b) = \frac{b}{\ln a} (1 - \ln b + \ln \ln a) + e^{2}

We have:

\frac{\partial g}{\partial b} = \frac{1}{\ln a} (1 - \ln b + \ln \ln a) - \frac{b}{\ln a} \cdot \frac{1}{b} = \frac{\ln \ln a - \ln b}{\ln a}

When $a \leq e^{2}, b > 2 e^{2}$ ,

\ln \ln a - \ln b \leq \ln \ln e^{2} - \ln 2 e^{2} = \ln 2 - \ln 2 - 2 = - 2 < 0

So $\frac{\partial g}{\partial b} < 0$ which means when $a \leq e^{2}$ ,

g (a, b) < g (a, 2 e^{2}) \leq 0

So when $a \leq e^{2}$ , $f (x) = 0$ must have two distinct solutions.

In conclusion, the range of $a$ is $(1, e^{2}]$ .

Part 3

When $a = e \in (1, e^{2}], b > e^{4} > 2 e^{2}$ , so $f (x) = 0$ has two distinct solutions.

Let $x_{1}, x_{2}$ be the two solutions, and $x_{1} < x_{2}$ :

{\begin{cases} e^{x_{1}} - b x_{1} + e^{2} = 0 & (1) \\ e^{x_{2}} - b x_{2} + e^{2} = 0 & (2) \end{cases}

$(1) - (2)$ , we have:

e^{x_{1}} - e^{x_{2}} - b (x_{1} - x_{2}) = 0 \Rightarrow \frac{e^{x_{1}} - e^{x_{2}}}{x_{1} - x_{2}} = b

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Question

Solution

Part 1

Part 2

Part 3

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