**(2025年天津卷)**
已知函数$f(x)=\mathrm{e}^x-\dfrac23\sin x$.

(1)求$f(x)$在点$(0,f(0))$处的切线方程;

(2)求证:当$x\geqslant-\dfrac13$时, $f(x)\geqslant 1+\dfrac x3$;

(3)求实数$\alpha$的最大可能值, 使得$f\left(\dfrac11\right)\cdot f\left(\dfrac12\right)\cdot f\left(\dfrac13\right)\cdots f\left(\dfrac1n\right)\geqslant (n+1)^\alpha$对任意的$n\in \mathbf{N}^*$都成立.

**解.** (1)由题意, $f(0)=1$, $f'(x)=\mathrm{e}^x-\dfrac23\cos x$,
则$f'(0)=\frac{1}{3}$,故$f(x)$在点$(0,f(0))$处的切线方程为$y=\frac{1}{3}x+1$.

(2)令$g\left( x \right) =\text{e}^x-\frac{2}{3}\sin x-\frac{x}{3}-1$,则$g'\left( x \right) =\text{e}^x-\frac{2}{3}\cos x-\frac{1}{3}$.

当$x\geqslant 0$时, $g'\left( x \right) =\text{e}^x-\frac{2}{3}\cos x-\frac{1}{3}\geqslant 1-\frac{2}{3}-\frac{1}{3}=0$.

当$-\dfrac13\leqslant x<0$时,
令$h\left( x \right) =\text{e}^x-\frac{2}{3}\cos x-\frac{1}{3}$,则$h'\left( x \right) =\text{e}^x+\frac{2}{3}\sin x$在$\left[-\dfrac13,0\right)$上单调递增.

利用$e^x\geqslant x+1$可知
$$
h'\left( x \right) \geqslant h'\left( -\frac{1}{3} \right) =\text{e}^{-\frac{1}{3}}+\frac{2}{3}\sin \left( -\frac{1}{3} \right) =\text{e}^{-\frac{1}{3}}-\frac{2}{3}\sin \frac{1}{3}>-\frac{1}{3}+1-\frac{2}{3}=0.
$$
故$h(x)$在$\left[-\dfrac13,0\right)$上单调递增,
则$g'(x)=h(x)\leqslant h(0)=0$.

故$g(x)$在$\left[-\dfrac13,0\right)$上单调递减;
在$[0,+\infty)$上单调递增.

故$g(x)\geqslant g(0)=0$.

(3)由二项式定理可知
$$
\left( 1+\frac{1}{3n} \right) ^3\geqslant \mathrm{C}_{3}^{0}+\mathrm{C}_{3}^{1}\frac{1}{3n}
=1+\frac{1}{n}=\frac{n+1}{n},
$$
由(2)可知
$$
f\left( \frac{1}{n} \right) \geqslant 1+\frac{1}{3n}\geqslant \left( \frac{n+1}{n} \right) ^{\frac{1}{3}},
$$
于是
$$
f\left( \frac{1}{1} \right) \cdot f\left( \frac{1}{2} \right) \cdot f\left( \frac{1}{3} \right) \cdots f\left( \frac{1}{n} \right) \geqslant \left( \frac{2}{1} \right) ^{\frac{1}{3}}\left( \frac{3}{2} \right) ^{\frac{1}{3}}\cdots \left( \frac{n+1}{n} \right) ^{\frac{1}{3}}=\left( n+1 \right) ^{\frac{1}{3}}.
$$
故$\alpha=\frac{1}{3}$时符合题意.

**证法一.** 当$\frac{1}{3}<\alpha<1$时,
记$\alpha_0=\frac{\alpha}{2}+\frac{1}{6}\in (0,1)$.

令$F\left( x \right) =\text{e}^x-\frac{2}{3}\sin x-\left( x+1 \right) ^{\alpha_0}$,则$F'\left( x \right) =\text{e}^x-\frac{2}{3}\cos x-\alpha_0 \left( x+1 \right) ^{\alpha_0 -1}$,

令$G\left( x \right) =\text{e}^x-\frac{2}{3}\cos x$,则$G(x)$在$(0,1)$上单调递增,则$G(x)>G(0)=\frac{1}{3}$.

取$m>0$,使得$\frac{1}{3}< G\left( m \right)<\alpha_0$.

当$0< x<\left( \frac{\alpha_0}{G\left( m \right)} \right) ^{\frac{1}{1-\alpha_0}}-1$时,
$$\begin{aligned}
F'\left( x \right) &=\text{e}^x-\frac{2}{3}\cos x-\alpha_0 \left( x+1 \right) ^{\alpha_0 -1}<\text{e}^m-\frac{2}{3}\cos m-\alpha_0 \left( x+1 \right) ^{\alpha_0 -1}\\
&=G\left( m \right) -\alpha_0 \left( x+1 \right) ^{\alpha_0 -1}<0.
\end{aligned}$$
此时$F(x)$单调递减, $F(x)<F(0)=0$,则.

令$N_0=\left[ \frac{1}{\left( \frac{\alpha_0}{G\left( m \right)} \right) ^{\frac{1}{1-\alpha_0}}-1} \right] +1$,
则当$n>N_0$时, $0<\frac{1}{n}<\left( \frac{\alpha_0}{G\left( m \right)} \right) ^{\frac{1}{1-\alpha_0}}-1$,此时
$$
f\left( \frac{1}{n} \right) <\left( \frac{n+1}{n} \right) ^{\alpha_0}=\left( \frac{n+1}{n} \right) ^{\frac{\alpha}{2}+\frac{1}{6}}.
$$
则当$n>\left[ \left( \frac{1}{N_0+1} \right) ^{\frac{\alpha}{2}+\frac{1}{6}}\prod_{k=1}^{N_0}{f\left( \frac{1}{k} \right)} \right] ^{\frac{1}{\frac{\alpha}{2}-\frac{1}{6}}}-1$时,
$$\begin{aligned}
f\left( \frac{1}{1} \right) \cdot f\left( \frac{1}{2} \right) \cdot f\left( \frac{1}{3} \right) \cdots f\left( \frac{1}{n} \right) &=\prod_{k=1}^{N_0}{f\left( \frac{1}{k} \right)}\cdot \prod_{k=N_0+1}^n{f\left( \frac{1}{k} \right)}<\prod_{k=1}^{N_0}{f\left( \frac{1}{k} \right)}\cdot \prod_{k=N_0+1}^n{\left( \frac{k+1}{k} \right) ^{\frac{\alpha}{2}+\frac{1}{6}}}\\
&=\left( \frac{n+1}{N_0+1} \right) ^{\frac{\alpha}{2}+\frac{1}{6}}\prod_{k=1}^{N_0}{f\left( \frac{1}{k} \right)}<\left( n+1 \right) ^{\alpha},
\end{aligned}$$
矛盾!

当$\alpha\geqslant 1$时,由上面讨论可知当$n$足够大时,
$f\left( \frac{1}{1} \right) \cdot f\left( \frac{1}{2} \right) \cdot f\left( \frac{1}{3} \right) \cdots f\left( \frac{1}{n} \right)<\left( n+1 \right) ^{0.5}<\left( n+1 \right) ^{\alpha}$.
故$\alpha>\frac{1}{3}$不符合题意.

综上所述, 实数$\alpha$的最大可能值为$\frac{1}{3}$.

**分析:** 由泰勒公式可得
$$
f\left( x \right) =\text{e}^x-\frac{2}{3}\sin x=1+\frac{x}{3}+\frac{x^2}{2}+\frac{5x^3}{18}
+\frac{x^4}{24}+o\left( x^4 \right).
$$

**证法二.** 下面证明:当$0< x\leqslant\frac{1}{2}$时, $\text{e}^x-\frac{2}{3}\sin x<1+\frac{x}{3}+x^2$.

令$g\left( x \right) =\text{e}^x-\frac{2}{3}\sin x-1-\frac{x}{3}-x^2$,则$g'\left( x \right) =\text{e}^x-\frac{2}{3}\cos x-\frac{1}{3}-2x$,

令$h\left( x \right) =\text{e}^x-\frac{2}{3}\cos x-\frac{1}{3}-2x$,则$h'\left( x \right) =\text{e}^x+\frac{2}{3}\sin x-2$在$\left(0,\frac{1}{2}\right]$上单调递增,
又$h'\left( x \right)\leqslant h'\left( \frac{1}{2} \right) =\sqrt{\text{e}}+\frac{2}{3}\sin \frac{1}{2}-2<\sqrt{\text{e}}+\frac{2}{3}\times \frac{1}{2}-2=\sqrt{\text{e}}-\frac{5}{3}<0$.

故函数$h(x)$在$\left(0,\frac{1}{2}\right]$单调递减,
则$g'(x)=h(x)< h(0)=0$.
于是$g(x)$在$\left(0,\frac{1}{2}\right]$单调递减,
则$g(x)< g(0)=0$.

因此,当$0< x\leqslant\frac{1}{2}$时, $\text{e}^x-\frac{2}{3}\sin x<1+\frac{x}{3}+x^2$.

又$g\left( 1 \right) =\text{e}-\frac{2}{3}\sin 1-\frac{7}{3}<\text{e}-\frac{2}{3}\sin \frac{\pi}{4}-\frac{7}{3}<0$,
故
$$
f\left( \frac{1}{n} \right) <1+\frac{1}{3n}+\frac{1}{n^2}
<\mathrm{e}^{\frac{1}{3n}+\frac{1}{n^2}},
$$
由于
$$
\sum_{k=2}^n{\left( \frac{1}{3k}+\frac{1}{k^2} \right)}<\frac{1}{3}\sum_{k=2}^n{\ln \frac{k}{k-1}}+\sum_{k=2}^n{\frac{1}{k\left( k-1 \right)}}=\frac{\ln n}{3}+1,
$$
于是当$n>\mathrm{e}^{\frac{2}{\alpha -\frac{1}{3}}}-1$时,
$$\begin{aligned}
&f\left( \frac{1}{1} \right) \cdot f\left( \frac{1}{2} \right) \cdot f\left( \frac{1}{3} \right) \cdots f\left( \frac{1}{n} \right) <\text{e}\cdot \text{e}^{\sum_{k=2}^n{\left( \frac{1}{3k}+\frac{1}{k^2} \right)}}\\
&<\text{e}^{\frac{\ln n}{3}+2}=\text{e}^2n^{\frac{1}{3}}<\text{e}^2\left( n+1 \right) ^{\frac{1}{3}}<\left( n+1 \right) ^{\alpha}.
\end{aligned}$$
故$\alpha>\frac{1}{3}$不符合题意.

综上所述, 实数$\alpha$的最大可能值为$\frac{1}{3}$.