Qing asked in 科學及數學數學 · 7 years ago

# 5條MATHS MOCK MC急

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• yc
Lv 5
7 years ago

suggested solutions:

• 7 years ago

8.y=x^2+bx+c=0

Ax=[-b+√(b^2-4ac)]/2

Dx=[-b-√(b^2-4ac)]/2

Mx=(Ax+Dx)/2=-2b/4a=-b/2

y(0)=0+0+c=c => Cy=By=c

Bx=2Mx-Cx=-b/a-0=-b

area(MBC)=|Bx*Cy/2|

=|(-b)*c/2|

=|bc/2|.........ans

18.螺旋=最短路線: w=角速度, Q=角位移=wt, v=rw

(1) 基本公式:

x=r*cosQ=r*cos(wt) => dx=-rw*sin(wt)dt

y=r*sinQ=r*sin(wt) => dy=rw*cos(wt)dt

z=vt=rwt => dz=rwdt

(2) 求弧線長s=?

(ds)^2=(dx)^2+(dy)^2+(dz)^2

=[(rw)^2*(cosQ^2+sinQ^2)+(rw)^2]dt^2

=[(rw)^2+(rw)^2]dt^2

=2*(rw)^2*dt^2

ds=rw√2dt

s=∫rw√2dt

=∫r√2d(wt)

=∫r√2dQ........Q=0~3π

=r√2*Q

=3r√2π

=6√(π^2+1)

(3) 求半徑r=?

兩邊平方: 18r^2*π^2=36(π^2+1)

r^2=2(π^2+1)/π^2

r=√[2(π^2+1)]/π

=√2*a/π

where a=√(π^2+1)

(4) 求柱高h=?:

繞1.5圈 => s=6a

繞1圈 => s=4a

商高定理:

h^2=s^2-(2πr)^2

=16a^2-4*2a^2

=8a^2

h=2√2*a

area(ABCD)=2πrh

=2π*(√2*a/π)*2√2*a

=8a^2

=8(π^2+1)

=原題答案無法選擇

26.ABCD=平行四邊形 => C=?

A=(-1,4), B=(-3,-2), D=(2,-1)

=2M-B

=(1,3)-(-3,-2)

=(1+3,3+2)

=(4,5).......ans=(C)

37.

y1=log<a>x=log(x)/log(a)

y2=log<b>x=log(x)/log(b)

x=1; y1=y2=0

x>1; y1>y2 => log(b)<log(a)

0<b<a=分數<1........ans

43.p=6/7 => q=1-p=1/7

二項式分配: c(5,x)=5!/x!(5-x)!

f(x)=c(5,x)*p^x*q^(5-x)

直接改用商高定理求取:

s^2=h^2+(3πr)^2........繞1.5圈=3π

36(π^2+1)=6^2+(6π)^2

=6^2+(3π2)^2

=> h=6, r=2

A=2πrh

=2π2*6

=24π........ans=(B)

• 麻辣
Lv 7
7 years ago

8.y=x^2+bx+c=0Ax=[-b+√(b^2-4ac)]/2Dx=[-b-√(b^2-4ac)]/2Mx=(Ax+Dx)/2=-2b/4a=-b/2y(0)=0+0+c=c => Cy=By=cBx=2Mx-Cx=-b/a-0=-barea(MBC)=|Bx*Cy/2|=|(-b)*c/2|=|bc/2|.........ans

18.螺旋=最短路線: w=角速度, Q=角位移=wt, v=rw(1) 基本公式:x=r*cosQ=r*cos(wt) => dx=-rw*sin(wt)dty=r*sinQ=r*sin(wt) => dy=rw*cos(wt)dtz=vt=rwt => dz=rwdt(2) 求弧線長s=? (ds)^2=(dx)^2+(dy)^2+(dz)^2=[(rw)^2*(cosQ^2+sinQ^2)+(rw)^2]dt^2=[(rw)^2+(rw)^2]dt^2=2*(rw)^2*dt^2ds=rw√2dts=∫rw√2dt=∫r√2d(wt)=∫r√2dQ........Q=0~3π=r√2*Q=3r√2π=6√(π^2+1)(3) 求半徑r=?兩邊平方: 18r^2*π^2=36(π^2+1)r^2=2(π^2+1)/π^2r=√[2(π^2+1)]/π=√2*a/πwhere a=√(π^2+1)(4) 求柱高h=?: 繞1.5圈 => s=6a繞1圈 => s=4a商高定理:h^2=s^2-(2πr)^2=16a^2-4*2a^2=8a^2h=2√2*aarea(ABCD)=2πrh=2π*(√2*a/π)*2√2*a=8a^2=8(π^2+1)=原題答案無法選擇

26.ABCD=平行四邊形 => C=?A=(-1,4), B=(-3,-2), D=(2,-1)AD中點: M=(A+D)/2=(-1+2,4-1)/2=(1/2,3/2)C=B對稱點 vs AD=2M-B=(1,3)-(-3,-2)=(1+3,3+2)=(4,5).......ans=(C) 37.y1=log<a>x=log(x)/log(a)y2=log<b>x=log(x)/log(b)x=1; y1=y2=0x>1; y1>y2 => log(b)<log(a)0<b<a=分數<1........ans

43.p=6/7 => q=1-p=1/7二項式分配: c(5,x)=5!/x!(5-x)!f(x)=c(5,x)*p^x*q^(5-x)f(3)=(5!/3!2!)*(6/7)^3*(1/7)^2=(5*4/2)*(6*36/7*49)/49=10*216/7*49^2=2160/16807=30*72/72*233.43=30/233.43........ans=(D)

2013-03-20 12:42:54 補充：

第18題修正:

z=v*t 修改為: z=k*v*t

多出未知數k.無法使用積分法求取: r.h.k

直接改用商高定理求取:

s^2=h^2+(3πr)^2........繞1.5圈=3π

36(π^2+1)=6^2+(6π)^2

=6^2+(3π2)^2

=> h=6, r=2

A=2πrh

=2π2*6

=24π........ans=(B)