19.在平面直角坐标系中,设△ABC的顶点分别为A(0,a),B(b,0),C(c,0),点P(0,p)在线段AO上(异于端点),若a,b,c,p均为非零实数,直线BP,CP分别交直线AC,AB于点E,F.某同学已正确算得直线OE的方程为($\frac{1}{b}$-$\frac{1}{c}$)x+($\frac{1}{p}$-$\frac{1}{a}$)y=0,则直线OF的方程为( )
| A. | ($\frac{1}{c}$-$\frac{1}{b}$)x+($\frac{1}{p}$-$\frac{1}{a}$)y=0 | B. | ($\frac{1}{b}$-$\frac{1}{c}$)x+($\frac{1}{p}$-$\frac{1}{a}$)y=0 | C. | (-$\frac{1}{b}$-$\frac{1}{c}$)x+($\frac{1}{p}$-$\frac{1}{a}$)y=0 | D. | ($\frac{1}{b}$+$\frac{1}{c}$)x+($\frac{1}{p}$-$\frac{1}{a}$)y=0 |
17.若函数f(x)=x+$\frac{1}{3}$e2x+aex在(-∞,+∞)单调递增,则a的取值范围是( )
| A. | $[-\frac{{2\sqrt{6}}}{3},+∞)$ | B. | $[\frac{{2\sqrt{6}}}{3},+∞)$ | C. | $[-\frac{{2\sqrt{6}}}{3},\frac{{2\sqrt{6}}}{3}]$ | D. | $(-\frac{{2\sqrt{6}}}{3},\frac{{2\sqrt{6}}}{3})$ |
11.已知命题p:?x0∈R,x02+(a-1)x0+1<0,命题q:?x∈R,x2+ax+1≥0,p∨(¬q)为假命题,则实数a的取值范围是( )
0 241255 241263 241269 241273 241279 241281 241285 241291 241293 241299 241305 241309 241311 241315 241321 241323 241329 241333 241335 241339 241341 241345 241347 241349 241350 241351 241353 241354 241355 241357 241359 241363 241365 241369 241371 241375 241381 241383 241389 241393 241395 241399 241405 241411 241413 241419 241423 241425 241431 241435 241441 241449 266669
| A. | [-2,-1] | B. | (-1,3) | C. | (-2,-1) | D. | [-1,2] |