优优班--学霸训练营 > 知识点挑题
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            • 1.

              已知函数,\(f(x)=\begin{cases} \dfrac{1}{{e}^{x}},x < 0 \\ {e}^{x},x > 0\end{cases} \),\(g(x)=m{x}^{2} \),若关于\(x\)的方程\(f(x)+g(x)=0\)有四个不同的实数解,则实数\(m\)的取值范围是        

            • 2.

              函数\(f\left(x\right)={x}^{2}-2ax+\ln x\left(a∈R\right) \).

              \((I)\)函数\(y=f(x)\)在点\((1,f(1))\)处的切线与直线\(x-2y+1=0\)垂直,求\(a\)的值;

              \((II)\)讨论函数\(f(x)\)的单调性;

              \((III)\)不等式\(2x\ln x\geqslant -{x}^{2}+ax-3 \)在区间\(\left( 0,e \right]\)上恒成立,求实数\(a\)的取值范围.

            • 3.

              已知函数\(f(x)=\dfrac{ax+b}{x}e^{x}\),\(a\),\(b\in R\),且\(a > 0\).

              \((1)\)若\(a=2\),\(b=1\),求函数\(f(x)\)的极值;

              \((2)\)设\(g(x)=a (x-1)e^{x}-f(x).\)当\(a=1\)时,对任意\(x\in \) \((0,+∞)\),都有\(g(x)\geqslant 1\)成立,求\(b\)的最大值;

            • 4.
              设函数\(C\):\(f(x)=2ax- \dfrac {b}{x}+\ln x\),若\(f(x)\)在\(x=1\),\(x= \dfrac {1}{2}\)处取得极值,
              \((i)\)求\(a\),\(b\)的值;
              \((ii)\)在\([ \dfrac {1}{4},2]\)存在\(x_{0}\),使得不等式\(f(x_{0})-c\leqslant 0\),求\(c\)的最小值.
            • 5.

              设函数\(f\left( x \right)={\ln }\left( x+1 \right)+a\left( {{x}^{2}}-x \right)\),其中\(a\in R\).

              \((1)\)讨论函数\(f\left( x \right)\)极值点的个数,并说明理由;

              \((2)\)若\(\forall x > 0,f\left( x \right)\geqslant 0\)成立,求\(a\)的取值范围.

            • 6. 已知函数\(f(x){=}\dfrac{1}{2}ax^{2}{-}(2a{+}1)x{+}2\ln x(\dfrac{1}{2}{ < }a{ < }1)\).
              \(\mathbf{(}\)Ⅰ\(\mathbf{)}\)求函数\(f(x)\)的单调区间;并判断函数\(f(x)\)在区间\({[}1{,}2{]}\)上是否有零点,若有,求出零点,若没有,请说明理由;
              \(\mathbf{(}\)Ⅱ\(\mathbf{)}\)若任意的\(x_{1}{,}x_{2}{∈}(1{,}2)\)且\(x_{1}{\neq }x_{2}\),证明:\({|}f(x_{2}){-}f(x_{1}){| < }\dfrac{1}{2}{.}(\)注:\(\ln 2{≈}0{.}693)\)
            • 7.

              若函数\(f(x)= \dfrac{x^{3}}{3}- \dfrac{a}{2}x^{2}+x+1\)在区间\(( \dfrac{1}{2},3)\)上有极值点,则实数\(a\)的取值范围是________.

            • 8. 已知函数\(f\left(x\right)=\ln x- \dfrac{1}{2}a{x}^{2}+x,a∈R \).
              \((1)\)当\(a=0\)时,求函数\(f(x)\)在\(\left(1,f\left(1\right)\right) \)处的切线方程;
              \((2)\)令\(g\left(x\right)=f\left(x\right)-\left(ax-1\right) \),求函数\(g(x)\)的极值;
              \((3)\)若\(a=-2\),正实数\(x_{1}\),\(x_{2}\)满足\(f\left({x}_{1}\right)+f\left({x}_{2}\right)+{x}_{1}{x}_{2}=0 \),证明:\({x}_{1}+{x}_{2}\geqslant \dfrac{ \sqrt{5}-1}{2} \).
            • 9.

              已知函数\(f(x)=\ln (1+x^{2})+ax(a\leqslant 0)\).

              \((1)\)若\(f(x)\)在\(x=0\)处取得极值,求\(a\)的值;

              \((2)\)讨论\(f(x)\)的单调性;

              \((3)\)证明:\(\left( 1+\dfrac{1}{4} \right)\left( 1+\dfrac{1}{16} \right)\cdots \left( 1+\dfrac{1}{{{4}^{n}}} \right) < {{e}^{1-\frac{1}{{{2}^{n}}}}}(n\in {{N}_{+}},e)\)为自然对数的底数\()\)

            • 10.

              若直线\(l\)\(y\)\(=\)\(kx\)\(-1\)与曲线\(C\)\(y\)\(=\)\(x\)\(-1+ \dfrac{1}{e^{x}}\)没有公共点,则实数\(k\)的取值范围为_____.

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