試題分析:(1)由圖可知
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146369765.png)
,因此為了求
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146385430.png)
,可通過求
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146400646.png)
和
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146416630.png)
,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146432966.png)
,下面關(guān)鍵要求
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146400646.png)
,為止作
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146463595.png)
,垂足為
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146478318.png)
,這時會發(fā)現(xiàn)隨
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146478275.png)
的取值不同,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146478318.png)
點(diǎn)可能在線段
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146541403.png)
上,也可能在線段
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146541403.png)
外,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146572533.png)
可能為銳角也可能為鈍角,這里出現(xiàn)了分類討論,作
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146588577.png)
交
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146603391.png)
延長線于
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146619302.png)
,由已知可求出
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146650579.png)
,這就是分類的分界點(diǎn);(2)由(1)求得
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240501466661303.png)
,要求它的最大值,可以采取兩種方法,一種是由于分子是一次,分母是二次的,可把分子
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146681377.png)
作為整體,分子分母同時除以
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146681377.png)
(當(dāng)然分母也已經(jīng)化為
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146681377.png)
的多項(xiàng)式了),再用基本不等式求解,也可用導(dǎo)數(shù)知識求得最大值.
(1)過A分別作直線CD,BC的垂線,垂足分別為E,F(xiàn).
由題知,AB=4.5,BC=4
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146291344.png)
,∠ABF=90
o-60
o=30
o,
所以CE=AF=4.5×sin30
o=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146759382.png)
,BF=4.5×cos30
o=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146790534.png)
,
AE=CF=BC+BF=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146806433.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146291344.png)
.
因?yàn)镃D=x(x>0),所以tan∠BDC=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146853526.png)
=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146868521.png)
.
當(dāng)x>
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146759382.png)
時,ED=x-
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146759382.png)
,tan∠ADC=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146915552.png)
=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146946810.png)
=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146962708.png)
(如圖1);
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240501469784411.png)
當(dāng)0<x<
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146759382.png)
時,ED=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146759382.png)
-x,tan∠ADC=-
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146915552.png)
=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146962708.png)
(如圖2). 4分
所以tanq=tan∠ADB=tan(∠ADC-∠BDC)=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240501470711184.png)
=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240501470871280.png)
=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240501471021127.png)
,其中x>0且x≠
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146759382.png)
.
當(dāng)x=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050146759382.png)
時tanq=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050147149556.png)
=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050147165616.png)
,符合上式.
所以tanq=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240501471021127.png)
( x>0) 8分
(2)(方法一)tanq==
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240501471021127.png)
=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240501472121141.png)
,x>0. 11分
因?yàn)?(x+4)+
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050147227535.png)
-41≥2
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050147243864.png)
-41=39,
當(dāng)且僅當(dāng)4(x+4)=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050147227535.png)
,即x=6時取等號.
所以當(dāng)x=6時,4(x+4)+
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050147227535.png)
-41取最小值39.
所以當(dāng)x=6時,tanq取最大值
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050147290488.png)
. 13分
由于y=tanx在區(qū)間(0,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050147321420.png)
)上是增函數(shù),所以當(dāng)x=6時,q取最大值.
答:在海灣一側(cè)的海岸線CT上距C點(diǎn)6km處的D點(diǎn)處觀看飛機(jī)跑道的視角最大 14分
(方法二)tanq=f(x)=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240501471021127.png)
=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240501471021127.png)
.
f ¢(x)=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240501473682010.png)
=-
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240501473991475.png)
,x>0.
由f ¢(x)=0得x=6. 11分
當(dāng)x∈(0,6)時,f ¢(x)>0,函數(shù)f(x)單調(diào)遞增;當(dāng)x∈(6,+∞)時,f ¢(x)<0,此時函數(shù)f(x)單調(diào)遞減.
所以函數(shù)f(x)在x=6時取得極大值,也是最大值f(6)=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050147290488.png)
. 13分
由于y=tanx在區(qū)間(0,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824050147321420.png)
)上是增函數(shù),所以當(dāng)x=6時,q取最大值.
答:在海灣一側(cè)的海岸線CT上距C點(diǎn)6km處的D點(diǎn)處觀看飛機(jī)跑道的視角最大. 14分