函数$y=\sqrt{x}$在区间$[0,4]$内满足微分中值定理的中值点是哪个点？
A: $(2,1)$
B: $(1,1)$
C: $(2,\sqrt{2})$
D: $(\frac{1}{2},\frac{1}{\sqrt{2}})$
A: $(2,1)$
B: $(1,1)$
C: $(2,\sqrt{2})$
D: $(\frac{1}{2},\frac{1}{\sqrt{2}})$
举一反三
- 函数$f(x,y)=\sqrt{1+{{y}^{2}}}\cos x$在点$(0,1)$处的1次Taylor多项式为 A: $\sqrt{2}-\frac{1}{\sqrt{2}}(y-1)$ B: $\frac{\sqrt{2}}{2}+\frac{1}{\sqrt{2}(}y-1)$ C: $2\sqrt{2}+\frac{1}{\sqrt{2}}(y-1)$ D: $\sqrt{2}+\frac{1}{\sqrt{2}}(y-1)$
- 设随机变量服从区间（0,2）上的均匀分布，则$Y=X^{2}$在（0,4）上的密度函数为（） A: $\frac{1}{3\sqrt{y}}$ B: $\frac{1}{\sqrt{y}}$ C: $\frac{1}{2\sqrt{y}}$ D: $\frac{1}{4\sqrt{y}}$
- $\int_{0}^{\frac{\text{ }\!\!\pi\!\!\text{ }}{4}}{[\cos (2t)\mathbf{i}+\sin (2t)\mathbf{j}+t\sin t\mathbf{k}]}\operatorname{dt}=$( ) A: $(\frac{1}{2},\frac{1}{2},\frac{4-\text{ }\!\!\pi\!\!\text{ }}{4\sqrt{2}})$ B: $(1,\frac{1}{2},\frac{4-\text{ }\!\!\pi\!\!\text{ }}{4\sqrt{2}})$ C: $(\frac{1}{2},1,\frac{4-\text{ }\!\!\pi\!\!\text{ }}{4\sqrt{2}})$ D: $(1,1,\frac{4-\text{ }\!\!\pi\!\!\text{ }}{4\sqrt{2}})$
- \(已知L是抛物线y=x^2上点O(0,0)与点A(1,1)之间的一段弧,则\int_{L}\sqrt{y}ds=(\,)\) A: \[\frac{1}{12}(5\sqrt{5}-1)\] B: \[\frac{1}{12}(3\sqrt{3}-1)\] C: \[\frac{1}{13}(5\sqrt{5}-1)\] D: \[\frac{1}{13}(3\sqrt{3}-1)\]
- \(函数f(x,y)=\ln(x+y+1)在点(0,0)处的带佩亚诺余项的三阶泰勒公式为(\,)\) A: \(x+y-\frac{1}{2}(x+y)^2+\frac{1}{6}(x+y)^3+o((\sqrt{x^2+y^2})^3)\) B: \(x+y-\frac{1}{2}(x+y)^2+\frac{1}{3}(x+y)^3+o((\sqrt{x^2+y^2})^3)\) C: \(x+y-\frac{1}{3}(x+y)^2+\frac{1}{6}(x+y)^3+o((\sqrt{x^2+y^2})^3)\) D: \(x+y-\frac{1}{2}(x+y)^2-\frac{1}{3}(x+y)^3+o((\sqrt{x^2+y^2})^3)\)