高等数学C答案(2016-2017学年第一学期期末A卷) 1.(1) dxe2+1'dr2(e2+1)3 1(3)x+ln2 14 001 (7)p<4 2.设f(x)= r arctan r-ln(1+x2),则f(x)= arctan r 1+x2 2x2 f"(r)=(1+a2)2.由f(0)=0及x>0时f(a)>0得到x>0时 f(x)>0.,又由f(0)=0得到x>0时f(x)> 3ex+1 3.(1)f(x) f"(x) 驻点为r1=lm 3+√5 由f"(x1)<0,f"(x2)>0得到f在x1取到极大值,在 取到极小值 (2)由lim(f(x)-x)=-1,lim(f(x)-x)=0得到f(x)有渐近线 =x-1,y=x.又有垂直渐近线x=0 4.(1)根据Roll定理,由f(1)=f(2)=f(3)=f(4)=0得到存 在51∈(1,2),52∈(2,3),53∈(3,4),使得∫(5)=0(i=1,2,3);存在 n∈(51,2),m2∈(52,53),使得∫"(m)=0(=1,2);存在(∈(mh,m2)使得 (2)在行列式的第j列提出(x-j)(j=1,2,3,4),得到 ∫(x)=(x-1)(x-2)(x-3)(x-4) 1 x2+2x+22 x2+3x+32 x2+4x+4 x3+x2+x+1x3+2x2+22x+23x3+3x2+32x+3x3+4x2+42x+43 第三行乘以(-x)加到第四行上,第二行乘以(-x)加到第三行上,第一行pêÆ C ‰Y (2016-2017 Æc1ÆÏÏ" A ò) 1. (1) dy dx = e t e t + 1 , d 2 y dx 2 = 1 (et + 1)3 (2) a = 1, b = −1 (3) π 4 + 1 2 ln 2 (4) π 2 2 (5) 3 2 √3 2 (6)   1 −1 4 0 0 1 −1 2 −7   (7) p < 4 2.  f(x) = x arctan x − ln(1 + x 2 ), K f 0 (x) = arctan x − x 1 + x 2 , f 00(x) = 2x 2 (1 + x 2 ) 2 . d f 0 (0) = 0 9 x > 0 ž f 00(x) > 0  x > 0 ž f 0 (x) > 0, qd f(0) = 0  x > 0 ž f(x) > 0. 3. (1) f 0 (x) = e 2x − 3ex + 1 (ex − 1)2 , f 00(x) = e x (ex + 1) (ex − 1)3 . 7: x1 = ln 3 − √ 5 2 , x2 = ln 3 + √ 5 2 . d f 00(x1) < 0, f 00(x2) > 0  f 3 x1 4ŒŠ, 3 x2 4Š. (2) d lim x→−∞ (f(x) − x) = −1, lim x→+∞ (f(x) − x) = 0  f(x) kìC‚ y = x − 1, y = x. qkR†ìC‚ x = 0. 4. (1) Šâ Rolle ½n, d f(1) = f(2) = f(3) = f(4) = 0  3 ξ1 ∈ (1, 2), ξ2 ∈ (2, 3), ξ3 ∈ (3, 4), ¦ f 0 (ξi) = 0 (i = 1, 2, 3); 3 η1 ∈ (ξ1, ξ2), η2 ∈ (ξ2, ξ3), ¦ f 00(ηi) = 0 (i = 1, 2); 3 ζ ∈ (η1, η2) ¦ f 000(ζ) = 0. (2) 31ª1 j JÑ (x − j) (j = 1, 2, 3, 4),  f(x) = (x − 1)(x − 2)(x − 3)(x − 4)· 0 BBB@ 1 1 1 1 x + 1 x + 2 x + 3 x + 4 x 2 + x + 1 x 2 + 2x + 22 x 2 + 3x + 32 x 2 + 4x + 42 x 3 + x 2 + x + 1 x 3 + 2x 2 + 22x + 23 x 3 + 3x 2 + 32x + 33 x 3 + 4x 2 + 42x + 43 1 CCCA . 1n1¦± (−x) \1o1þ, 11¦± (−x) \1n1þ, 11 1