close

標題:

F.4 a.maths....

發問:

1a) Prove that (cosx-sinx)^2 = 1 - sin2x.b) Find the maximum and minimum values of cosx-sinx.Hence find the range of the values of 1/(cox-sinx-2) for all possible values of x.2 Solve the following equations for 00 and... 顯示更多 1a) Prove that (cosx-sinx)^2 = 1 - sin2x. b) Find the maximum and minimum values of cosx-sinx. Hence find the range of the values of 1/(cox-sinx-2) for all possible values of x. 2 Solve the following equations for 00 and n>=4. In the expansion of (x-2/x)^2 (1+mx)^n in ascending powers of x, the constant term is 20 and there is no x term. Find the values of m and n. 更新: 1b) cosx-sinx既MAX.係√2 , MIN.係 -√2

 

此文章來自奇摩知識+如有不便請留言告知

最佳解答:

但(b)中係二次方後先係1-sin2x 所以應該係0-sin2x>-1 2>1-sin2x>0 01/(cosx-sinx-2)>1/(-2+sqrt2) 2 Solve the following equations for 00 and n>=4. In the expansion of (x-2/x)^2 (1+mx)^n inascending powers of x, the constant term is 20 and there is no x term.Find the values of m and n. (x-2/x)^2(1+mx)^n =(x^2-2(x)(2/x)+(2/x)^2)(1+nmx+n(n-1)(mx)^2/2!+...) =(x^2-4+4/x^2)(1+(nm)x+n(n-1)m^2x^2/2!+n(n-1)(n-2)(m^3)x^3/3!...) constant term=-4+4n(n-1)m^2/2!=20 4n(n-1)m^2/2=24 4n(n-1)m^2=48 n(n-1)m^2=12----(1) x term=-4nm+4n(n-1)(n-2)m^3/6=0 4n(n-1)(n-2)m^3/6=4nm n(n-1)(n-2)m^3/6=nm n(n-1)(n-2)m^3=6nm (n-1)(n-2)m^2=6----(2) (1)/(2) n(n-1)/(n-1)(n-2)=2 n/(n-2)=2 n=2n-4 4=n when n=4 4(4-1)m^2=12 4x3xm^2=12 m^2=1 m=1

其他解答:

1a) Prove that (cosx-sinx)^2 = 1 - sin2x. L.H.S. = (cosx-sinx)^2 = cos^2x - 2cosxsinx + sin^2x = 1 - 2cosxsinx = 1-sin2x = R.H.S. 2010-07-14 12:23:50 補充： b)Given (cosx-sinx)^2 = 1 - sin2x., For Max: sin2x = -1, (cosx-sinx)^2 = 1+1 = 2 For Min: sin2x = 1, (cosx-sinx)^2 = 1-1 = 0 Max for cosx-sinx = 2^(1/2) Min for cosx - sinx = 0 0=1/(cosx - sinx-2)>=1/[2^(1/2)-2]E2A5F59BAA12C031

arrow
arrow
    文章標籤
    没有合适的结果
    全站熱搜
    創作者介紹
    創作者 lgzrelv 的頭像
    lgzrelv

    lgzrelv的部落格

    lgzrelv 發表在 痞客邦 留言(0) 人氣()

    E-mail轉寄