13.y = 4cosx - 3sin²x + 2
= 4cosx - 3(1-cos²x)+2
= 3cos²x + 4cosx - 1
= 3(cosx + 2/3)² - 13/9
因为 cosx∈[-1,1],所以当cosx + 2/3 = 0 时 y 取到最小值-13/19,
当cosx = 1时 y 取到最大值 3(1+2/3)² - 13/9 = 62/9.
14.由 a(n+1) - an = 2n - 1 得:
a2 - a1 = 1
a3 - a2 = 3
a4 - a3 = 5
.
an - a(n-1) = 2(n-1) - 1
等式两端分别相加得:an - a1 = 1+3+5+.+2(n-1)-1 = 2(n-1)(n-1)/2 = (n-1)²
所以 an= (n-1)² + a1 = (n-1)² +2.
15.当n=1时,左式 = 1+ 3/1 = 4 = 2² = (1+1)² = 右式.
设n = k时原式成立,即 ( 1+ 3/1)( 1+5/4).[1+(2k+1)/k²] = (k + 1)².
当n = k+1 时,
( 1+ 3/1)( 1+5/4).[1+(2k+1)/k][1+(2(k+1)+1)/(k+1)²]
= (k + 1)² [1+(2(k+1)+1)/(k+1)²]
= (k + 1)² + 2(k + 1) +1
= (k + 2)² = [(k + 1) + 1]².
所以 n = k+1时原式成立.
所以原式当n为正整数时成立.
11.令 a = x^(1/2),则 1/a = x^(-1/2).
由已知得 a + 1/a = 3.
两边平方得:a^2 + 1/a^2 + 2 = 9.
所以有:a^2 + 1/a^2 = 7.
两边再平方:a^4 + 1/a^4 + 2 = 49.
得:a^4 + 1/a^4 = 47.
即有 x^2 + x^(-2) = 47.
在 a + 1/a = 3 两边取3次方,得
27 = (a + 1/a)^3 = a^3 + 3a + 3/a + 1/a^3
= a^3 + 3(a + 1/a) + 1/a^3
= a^3 + 1/a^3 + 9
所以 a^3 + 1/a^3 = 18.
即有 x^(3/2) + x^(-3/2) = 18.
所以有 [ x^(3/2) + x^(-3/2) - 3] / [ x^2 + x^(-2) -2]
= (18-3)/(47-2) = 15/45 = 1/3.