7 x2,当-≤x≤时 则f(-x) (-x)2,当-5≤-x≤时 故f(-x)≠f(x),f(-x)≠-f(x),于是此函数是非奇非偶函数 2.试证两个偶函数的乘积是偶函数,两个奇函数的乘积是奇函数,一个奇函数与一个偶函数的乘积是奇函数 证明:设f1(x),f2(x)为定义在(-a,a)(a>0)内的偶函数,g1(x),92(x)为定义在(-a,a)(a>0)内的奇函 数,F1(x)=f1(x)f2(x),F2(x)=g1(x)92(x),F3(x)=f1(x)f2(x) 则f1(-x)=f1(x),f2(-x)=f2(x),g1(x)=-g1(x),92(-x)=-92(x),于是 F1(-x)=f1(-x)f2(-x)=f1(x)(x)=F1(x) F2(-x)=91(-)92(-x)=(-91(x)(-92(x)=91(x)92(x)=F2(x) F3(-x)=f1(-x)91(-x)=f1(x)(-91(x)=-f1(x)g1(x)=-F3(x) 从而F1(x)是偶函数;F2(x)是偶函数;F3(x)是奇函数 23.设f(x)为定义在(-∞,+∞)内的任何函数,证明F1(x)≡=f(x)+f(-x)是偶函数,F2(x)≡f(x)-f(-x)是奇 函数写出对应于下列函数的F1(x),F2(x) (2)y=(1+x) 证明:因F1(-x)=f(-x)+f(x)=F1(x),则F1(x)=∫(x)+f(-x)是偶函数 又F2(-x)=f(-x)-f(x)=-F2(x),则F2(x)=f(x)-f(-x)是奇函数 (1)F1(x)=f(x)+f(-x)=a2+a-x,F2(x)=f(x)-f(-x)=a2 (2)F1(x)=f(x)+f(-x)=(1+x)2+(1-x)n,F2(x)=f(x)-f(-x)=(1+x)-(1-x)n 4.说明下列函数哪些是周期函数,并求最小周期: (1)y=sina (2)y=sinr (3)y=sin r +o sin 2r (5)y=lsin r|+Icos rl (6)y=√tanx (7)y=x-[x (8)y=sinner (1)因=出n2s1os2x,则r= 22 (2)假设y=sinx2为一周期函数且T=>0 据周期函数的定义,对任何x∈(-∞,+∞),有sin(x+u)2=sinx2,特别对x=0也应该成立 则sinu2=0,于是u2=kr,u=Vkr(k∈z+) 又对x=√2u=√2也成立,故sin(√②u+u)2=sin2=0,则(√2+1)2kr=n丌(n∈z+),于 是(√2+1)2=(k,n∈z+) 又(√2+1)2=3+2V2∈Q,而∈Q+,则假设不成立,即函数y=sinx2不是周期函数 (3)因n=smz的T=27:=2sin2的T=,则y=inx+2sm2的T=27 (4)T=元=87 (6) œy = f(x) =    2 x2 ,  1 2 < x < +∞û sin x 2 ,  − 1 2 6 x 6 1 2 û 1 2 x 2 ,  − ∞ < x < − 1 2 û ß Kf(−x) =    2 (−x) 2 ,  1 2 < −x < +∞û sin(−x) 2 ,  − 1 2 6 −x 6 1 2 û 1 2 (−x) 2 ,  − ∞ < −x < − 1 2 û =    1 2 x 2 ,  1 2 < x < +∞û sin x 2 ,  − 1 2 6 x 6 1 2 û 2 x2 ,  − ∞ < x < − 1 2 û ß f(−x) 6= f(x), f(−x) 6= −f(x)ßu¥dºÍ¥ö¤öÛºÍ. 22. £y¸áۺͶ»¥ÛºÍ߸᤺Ͷ»¥¤ºÍßò᤺ÍÜòáۺͶ»¥¤ºÍ. y²µf1(x), f2(x)转3(−a, a)(a > 0)SÛºÍßg1(x), g2(x)转3(−a, a)(a > 0)S¤º ÍßF1(x) = f1(x)f2(x), F2(x) = g1(x)g2(x), F3(x) = f1(x)f2(x) Kf1(−x) = f1(x), f2(−x) = f2(x), g1(x) = −g1(x), g2(−x) = −g2(x)ßu¥ F1(−x) = f1(−x)f2(−x) = f1(x)f2(x) = F1(x) F2(−x) = g1(−x)g2(−x) = (−g1(x))(−g2(x)) = g1(x)g2(x) = F2(x) F3(−x) = f1(−x)g1(−x) = f1(x)(−g1(x)) = −f1(x)g1(x) = −F3(x) l F1(x)¥ÛºÍ¶F2(x)¥ÛºÍ¶F3(x)¥¤ºÍ. 23. f(x)转3(−∞, +∞)S?¤ºÍßy²F1(x) ≡ f(x) + f(−x)¥ÛºÍßF2(x) ≡ f(x) − f(−x)¥¤ ºÍ.—ÈAueºÍF1(x), F2(x)µ (1) y = a x (2) y = (1 + x) n y²µœF1(−x) = f(−x) + f(x) = F1(x)ßKF1(x) = f(x) + f(−x)¥ÛºÍ qF2(−x) = f(−x) − f(x) = −F2(x)ßKF2(x) = f(x) − f(−x)¥¤ºÍ. (1) F1(x) = f(x) + f(−x) = a x + a −x , F2(x) = f(x) − f(−x) = a x − a −x (2) F1(x) = f(x) + f(−x) = (1 + x) n + (1 − x) n , F2(x) = f(x) − f(−x) = (1 + x) n − (1 − x) n 24. `²eºÍ= ¥±œºÍßø¶Å±œµ (1) y = sin2 x (2) y = sin x 2 (3) y = sin x + 1 2 sin 2x (4) y = cos π 4 x (5) y = | sin x| + | cos x| (6) y = √ tan x (7) y = x − [x] (8) y = sin nπx )µ (1) œy = sin2 x = 1 2 − 1 2 cos 2xßKT = 2π 2 = π (2) by = sin x 2èò±œºÍÖT = ω > 0 ‚±œºÍ½¬ßÈ?¤x ∈ (−∞, +∞)ßksin(x + ω) 2 = sin x 2ßAOÈx = 0èAT§·ß Ksin ω 2 = 0ßu¥ω 2 = kπ, ω = √ kπ(k ∈ Z +) qÈx = √ 2ω = √ 2kπ觷ßsin(√ 2ω + ω) 2 = sin ω 2 = 0ßK( √ 2 + 1)2 kπ = nπ(n ∈ Z +)ßu ¥( √ 2 + 1)2 = k n (k, n ∈ Z + ) q( √ 2 + 1)2 = 3 + 2√ 2 ∈ Q −ß k n ∈ Q +ßKbÿ§·ß=ºÍy = sin x 2ÿ¥±œºÍ. (3) œy1 = sin xT = 2π¶y2 = 1 2 sin 2xT = πßKy = sin x + 1 2 sin 2xT = 2π. (4) T = 2π π 4 = 8