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            • 1.
              若点\(M\)是\(\triangle ABC\)所在平面内一点,且满足:\( \overrightarrow{AM}= \dfrac {3}{4} \overrightarrow{AB}+ \dfrac {1}{4} \overrightarrow{AC}\).
              \((1)\)求\(\triangle ABM\)与\(\triangle ABC\)的面积之比.
              \((2)\)若\(N\)为\(AB\)中点,\(AM\)与\(CN\)交于点\(O\),设\( \overrightarrow{BO}=x \overrightarrow{BM}+y \overrightarrow{BN}\),求\(x\),\(y\)的值.
            • 2.

              如图,在\(Δ\) \(OBC\)中, \(A\)\(BC\)的中点, \(D\)\(OB\)的靠近 \(B\)点的一个三等分点, \(CD\)\(OA\)交于点 \(E\)\(.\)若 \(\overrightarrow{OE}=\lambda \overrightarrow{OA}\),求实数 \(\lambda \)的值.


            • 3.

              如图,在\(\Delta ABC\)中,已知点\(D,E\)分别在边\(AB,BC\)上,且\(AB=3AD,BC=2BE\)


              \((1)\)用向量\(\overrightarrow{AB},\overrightarrow{AC}\)表示\(\overrightarrow{DE}\);

              \((2)\)设\(AB=6,AC=4,A={{60}^{\circ }}\),求线段\(DE\)的长.

            • 4.
              如图,在平面直角坐标系\(xOy\)上,点\(A(1,0)\),点\(B\)在单位圆上,\(∠AOB=θ(0 < θ < π)\).
              \((1)\)若点\(B(- \dfrac {3}{5}, \dfrac {4}{5})\),求\(\tan (θ+ \dfrac {π}{4})\)的值;
              \((2)\)若\( \overrightarrow{OA}+ \overrightarrow{OB}= \overrightarrow{OC}\),\( \overrightarrow{OB}\cdot \overrightarrow{OC}= \dfrac {18}{13}\),求\(\cos ( \dfrac {π}{3}-θ)\).
            • 5.

              已知椭圆\({C}_{1}\;:\; \dfrac{{x}^{2}}{{a}^{2}}+ \dfrac{{y}^{2}}{{b}^{2}}=1\left(a > b > 0\right) \) 经过点\(M\left(1, \dfrac{3}{2}\right) \),且其右焦点与抛物线\({C}_{2}\;:\;{y}^{2}=4x \)的焦点\(F\)重合,过点\(F\)且与坐标轴不垂直的直线与椭圆交于\(P\),\(Q\)两点.

              \((1)\)求椭圆\({C}_{1} \)的方程;

              \((2)\)设\(O\)为坐标原点,线段\(OF\)上是否存在点\(N\left(n,0\right) \),使得\( \overrightarrow{QP}· \overrightarrow{NP}= \overrightarrow{PQ}· \overrightarrow{NQ} \)?若存在,求出\(n\)的取值范围;若不存在,说明理由;

              \((3)\)过点\({P}_{0}\left(4,0\right) \)且不垂直于\(x\)轴的直线与椭圆交于\(A\),\(B\)两点,点\(B\)关于\(x\)轴的对称点为\(E\),试证明:直线\(AE\)过定点.

            • 6.

              \((1)\)如图,已知\(\vartriangle ABC\)中,\(D\)为边\(BC\)上靠近\(B\)点的三等分点,连接\(AD\),\(E\)为线段\(AD\)的中点,若\(\overrightarrow{CE}=m\overrightarrow{AB}+n\overrightarrow{AC}\),则\(m+n=\)      

              \((2)\)方程\(\left| \dfrac{2x+3}{x+1} \right|={{(x+2)}^{2}}\) 解的个数为      

              \((3)\)已知\(\tan (\theta +\dfrac{\pi }{2})=2,\) 则\(\sin \theta \cos \theta =\)       

              \((4)\)已知\(\omega > 0,A > 0,a > 0,0 < \varphi < \pi ,y=\sin x\) 的图象按照以下次序变换:\(①\)纵坐标不变,横坐标变为原来的\(\dfrac{1}{\omega }\) ;\(②\)向左移动\(\varphi \) 个单位;\(③\)向上移动\(a\) 个单位;\(④\)纵坐标变为\(A\) 倍\(.\)得到\(y=3\sin (2x-\dfrac{\pi }{6})+1\) 的图象,则\(A+a+\omega +\varphi =\)       

            • 7. 如图,在\(\triangle ABC\)中,已知\(∠BAC= \dfrac{π}{3} \),\(AB=2\),\(AC=3\),\(D\)在线段\(BC\)上.

              \((1)\)若\( \overrightarrow{BD}= \overrightarrow{DC} \),\( \overrightarrow{AE}=3 \overrightarrow{ED} \),且\( \overrightarrow{BE}=x \overrightarrow{AB}+y \overrightarrow{AC} \),求 \(x\)\(+\) \(y\)
              \((2)\)若\( \overrightarrow{AD}· \overrightarrow{BC}=0 \),求\(\left| \overrightarrow{AD}\right| \).
            • 8.

              如图,直三棱柱\(ABC\)\(-\)\(A\)\({\,\!}_{1}\)\(B\)\({\,\!}_{1}\)\(C\)\({\,\!}_{1}\)中,\(AB\)\(=\)\(AC\)\(=\)\(AA\)\({\,\!}_{1}=4\),\(BC\)\(=2 \sqrt{2}\).\(BD\)\(⊥\)\(AC\),垂足为\(D\)\(E\)为棱\(BB\)\({\,\!}_{1}\)上一点,\(BD\)\(/\!/\)平面\(AC\)\({\,\!}_{1}\)\(E\)



              \((I)\)求线段\(B\)\({\,\!}_{1}\)\(E\)的长;

              \((II)\)求二面角\(C\)\({\,\!}_{1}-\)\(AC\)\(-\)\(E\)的余弦值.

            • 9.

              \((\)本小题满分\(12\)分\()\)

              平面内给定三个向量\( \overrightarrow{a}=(3,2)\),\( \overrightarrow{b}=(-1,2)\),\( \overrightarrow{c}=(4,1)\)
              \((1)\)求\(3 \overrightarrow{a}+ \overrightarrow{b}-2 \overrightarrow{c}\);
              \((2)\)求满足\( \overrightarrow{a}=m \overrightarrow{b}+n \overrightarrow{c}\)的实数\(m\)、\(n\).

            • 10.

              在平面直角坐标系中,\(O\)为坐标原点,\(A\)、\(B\)、\(C\)三点满足\( \overset{→}{OC}= \dfrac{1}{3} \overset{→}{OA}+ \dfrac{2}{3} \overset{→}{OB} \).

               \((1)\)求证:\(A\)、\(B\)、\(C\)三点共线;
               \((2)\)已知\(A(1, \)\(\cos x\)\()\)、\(B(2 \)\(\cos \)\({\,\!}^{2} \dfrac{x}{2} \), \(\cos x\)\()\), \(x\)\(∈[0, \dfrac{x}{2} ]\),若 \(f\)\(( \)\(x\)\()= \overset{→}{OA} ⋅ \overset{→}{OC} -(2\) \(m\)\(+ \dfrac{2}{3} )| \overset{→}{AB} |\)的最小值为\(-1\),求实数 \(m\)值\(.\)
               \((3)\)若点\(A(2,0)\),在 \(y\)轴正半轴上是否存在点\(B\)满足\({ \overset{→}{OC}}^{2} = \overset{→}{AC} ⋅ \overset{→}{CB} \),若存在求出点\(B\);若不存在,请说明理由.
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