ce maths 95 Q.49 (20點)

👨🏻‍🏫 學術台數學

[匿名]

2008-04-30 5:51 pm

If 0°〈or = x〈 or = 360° ,the no. of points of intersection of graphs of
y = sin x and y = tan x is
A.1
B.2
C.3
D.4
E.5

更新1:

one more question when a polynomial p(x) is divided by x-1 , the remainders are 1 and 3 respectively.Find the remainder when p(x) is divided by x^2 -1. thanks

更新2:

sorry,the question should be: when a polynomial p(x) is divided by x-1 and x+1, the remainders are 1 and 3 respectively.Find the remainder when p(x) is divided by x^2 -1. A. 4 B. -x+2 C. x+3 D. 3x+1 thank you,sara_hoping

更新3:

sorry,may i ask one more question? Let a ,b and c be positive integers. If b=(ac)^1/2 ,which of the following must be true? 1. log a^2 , log b^2 , log c^2 is an arithmetic sequence 2. a^3 , b^3 , c^3 is an geometric sequence 3. 4^a , 4^b , 4^c is an geometric sequence

回答 (3)

魏王將張遼

2008-04-30 8:09 pm

✔ 最佳答案

To solve algebraically:
sin x = tan x
sin x = sin x/cos x
sin x (1 - 1/cos x) = 0
sin x = 0 or 1/cos x = 1
sin x = 0 or cos x = 1
x = 0, 180 or 360.
To view graphically:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Apr08/Crazytrigo7.jpg

So ans = C
答案 001 提供的 graph 不合比例, 以致 sin x 和 tan x 在另外兩個 positions 有了 points of intersection.

2008-04-30 17:01:45 補充：
From the given, p(1) = 1 and p(-1) = 3
Suppose the remainder required is Ax + B where A and B are constants. Then,
p(x) = f(x)(x^2 + 1) + (Ax + B)
p(1) = A + B, A + B = 1
p(-1) = -3A + B, -3A + B = 3
Solving we have A = -1/2 and B = 3/2
So the remainder is (1/2)(-x + 3)

2008-05-01 09:53:47 補充：
從頭做:
From the given, p(1) = 1 and p(-1) = 3
Suppose the remainder required is Ax + B where A and B are constants. Then,
p(x) = f(x)(x^2 - 1) + (Ax + B)
p(1) = A + B, A + B = 1
p(-1) = -A + B, -A + B = 3
Solving we have A = -1 and B = 2
So the remainder is -x + 2

參考: My Maths knowledge

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用戶7369

2008-04-30 8:11 pm

for points of intersection,
sinx=tanx
sinx=sinx/cosx
1=1/cosx
cosx=1
x=0 or 360
so the answer is B

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choy

2008-04-30 6:19 pm

Answer : E

2008-04-30 10:31:51 補充：
http://www.badongo.com/pic/3481232

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