<img src="http://edu-image.nosdn.127.net/2507E32A7888F1F05F34CD6088FE894F.png?imageView&thumbnail=890x0&quality=100" />? AC+AB×cosθ1=BC×cosθ3; AB×sinθ1=BC×sinθ3<br >|AC+AB×cosθ1=BC×cosθ3; AB×cosθ1=BCcos×θ3<br >|AB×sinθ1=BC×cosθ3; AC+AB×cosθ1=BC×sinθ3|;AB×cosθ1=BC×cosθ3; AC+AB×sinθ1=BC×sinθ3<br >

$\int_{0}^{\frac{\text{ }\!\!\pi\!\!\text{ }}{4}}{&#91;\cos (2t)\mathbf{i}+\sin (2t)\mathbf{j}+t\sin t\mathbf{k}&#93;}\operatorname{dt}=$( ) 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: $(1,1,\frac{4-\text{ }\!\!\pi\!\!\text{ }}{4\sqrt{2}})$

Text 1

已知齐次方程$(x-1){{y}^{''}}-x{{y}^{'}}+y=0$的通解为$Y={{C}_{1}}x+{{C}_{2}}{{e}^{x}}$，则方程$(x-1){{y}^{''}}-x{{y}^{'}}+y={{(x-1)}^{2}}$的通解是( ) A: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-({{x}^{2}}+1)$ B: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-({{x}^{3}}+1)$ C: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-{{x}^{2}}$ D: ${{\text{C}}_{1}}x+{{\text{C}}_{2}}{{e}^{x}}-{{x}^{2}}+1$

Skim the text and answer the following questions. 1) What type of writing is the text?

（1）AꞌBꞌCꞌ+A(B+C)+BC=(ABꞌCꞌ+AꞌBꞌC+AꞌBCꞌ) ꞌ

图示变截面短杆，AB段的压应力σAB与BC段压应力σBC的关系是：（） A: σAB比σBC大1／4 B: σAB比σBC小1／4 C: σAB是σBC的2倍 D: σAB是σBC的1／2

Part 1 ： The text begins with an _________.

Fast reading text 1

从原点向曲线$$y=1-\ln x$$作切线，则由切线、曲线和$$x$$轴围成图形的面积为()． A: $$\frac{1}{2}{{\text{e}}^{2}}+\text{e}$$ B: $$\frac{1}{2}{{\text{e}}^{2}}-\text{e}$$ C: $${{\text{e}}^{2}}+\text{e}$$ D: $${{\text{e}}^{2}}-\text{e}$$