2024安徽皖南8校12月高三第二次大联考：端点效应

已知函数\(f(x)=e^x+\sin x-kx\)

(1) 当\(k=0\)时，求\(f(x)\)在\(x=0\)处的切线方程；

(2) 当\(k=2\)时，求\(f(x)\)的单调性；

(3) 若\(x[f(x)-f(-x)]\geq 0\)恒成立，求\(k\)的范围.

解.(1) \(k=0,f(x)=e^x+\sin x,f^\prime(x)=e^x+\cos x,f^\prime(0)=2\)

而\(f(0)=1\)，则切线方程为\(y-1=2x\)

(2) \(k=2,f(x)=e^x+\sin x-2x,f^\prime(x)=e^x+\cos x-2\)

当\(x<0\)时，\(f^\prime(x)=e^x+\cos x-2<1+1-2=0\)

\(x\geq 0\)时，\(f^{\prime\prime}(x)=e^x-\sin x>1-1=0\)

则\(f^\prime(x)\)单调递增，从而\(f^\prime(x)>f^\prime(0)=1+1-2=0\)

则综上，\(f(x)\)在\(x<0\)单调递减，在\(x\geq 0\)单调递增

(3) 因\([f(x)-f(-x)]\)是奇函数，则需考虑\(x>0\)上即可

\(\varphi(x)=f(x)-f(-x)=e^x-e^{-x}+2\sin x-2kx,\varphi(0)=0\)

\(\varphi^\prime(x)=e^x+e^{-x}+2\cos x-2k,\varphi^\prime(0)=4-2k\)

\(\varphi^{\prime\prime}(x)=e^x-e^{-x}-2\sin x\)

\(\varphi^{\prime\prime\prime}(x)=e^x+e^{-x}-2\cos x\geq 2-2\cos x\geq 0\)

则\(\varphi^{\prime\prime}(x)\)单调递增，即\(\varphi^{\prime\prime}(x)\geq \varphi^{\prime\prime}(0)=0\)

即\(\varphi^{\prime}(x)\)单调递增

Case1 当 \(k\leq 2\)时，此时\(\varphi^{\prime}(x)\geq \varphi^{\prime}(0)=4-2k\geq 0\)

则\(\varphi(x)\geq \varphi(0)=0\)，合题

Case2 当\(k>2\)时，\(\varphi^{\prime}(0)<0\)，而\(\varphi^{\prime}(x)\)单调递增

从而必存在一个区间\((0,m)\)，使得在此区间上\(\varphi^{\prime}(x)\)小于\(0\)

即\(\varphi^{\prime}(x)\)单调递减，从而\(\varphi(x)\leq \varphi(0)=0\)不合题

综上，\(k\leq 2.\)