(80)-高数专项练题3_08.2026考研数学高途王喆全程班_赠送2025课程_25考研数学（三）全年智达班_{2}--资料

第二章 一元函数微分学例2.1 函数 f (x)在(−1,1)有定义，且lim f (x) = 0，则（ C ）． x→0 f ( x) fN 9 Y . （A）若lim = 0，则 f (x)在 x = 0可导； R (31 : = x→0 x f ( x) （B）若lim = 0，则 f (x)在 x = 0可导； S2(1 fix [ = = x2 x→0 f ( x) （C）若 f (x)在 x = 0可导，则lim = 0； x→0 x f ( x) （D）若 f (x)在 x = 0可导，则lim = 0. x2 x→0 (c) (D) fix Strc]E = ONES = fio = Me to fix = 0 . fix-fro M fix My = = #0 X j -finf = fix I = M c => = m => ↓ ↓ O C AFO O #Fas例2.2 设奇函数 f (x)在 x = 0处可导，则函数  f ( x)( x + sin x)  , x  0 F(x) =  x 在 x = 0处( D ).  0, x = 0  (A) 不连续 (B) 不可导 (C) 可导且F(0) = 0 (D) 可导且F(0) = 2 f (0) flol f : o : = f(x Mix SmX) fix = IMEN + = : + X # f(0) F(d) FINE L =2 = 0 = . X =f. XSM is Fio Fi-Fol fix M 8 (xism - = Um O = X2 Y fix f fil-f M b m = + 2 = * X fir 2 =f (e2x − 1) − 1 例2.3 设 f (x)一阶可导且 f (0)  0，且 f (0) = 1，则lim ln f ( x) x→0 __________. . = M (n [1 f(x) 1) + - fil-fi fiel M fid M = = . . X - o 2 fix 2 = = . fixax2 + bx + c, x  0 例2.4 已知 f ( x) =  在 x = 0处有二阶导数, 试确定参数 ln(1 + x), x  0  a,b,c的值. 4 fix * To : = fix E Of I - fio fill fio - = ( = 0 = o , : C = 0 fIN0(x = Bit · = fin To LES :↑ fix #J 2axth X0 = , x0AJ fix i , = , fi Mfn fix = fixb 1 , = = "fix 92ax + 1 x = , - . 2a = -1 : a 1 X = 0 = & X > ItX fix-fo s f1(d A Hm 2ax + 1 - 1 = = 2a = X - 8 x)0 X I fix-fo find run i um - I * m = 1+X - = = 1 x+jt = * X ( +x) X -=例2.5 设 f (x)为周期为 5 的连续函数,它在 x = 0的某个邻域内满足: f (1+ sin x) − 3 f (1− sin x) = 8x + o(x) 其中o(x)是当 x → 0时比 x 高阶的无穷小量, 且 f (x)在 x = 1处可导, 求 曲线 y = f (x)在点(6, f (6))处的切线方程. : FIN E :hm IfCHSmX) 3 fU-smx)) My EXTON] - = Xtr f() fill -3 0 => = fill => = rfitsmx-fi-sIfcsmx-fu 20 M mus + 0(X) => #0 ShX ShX fl flitsmx- fli-smx-fl Mr M pur 8X +ox) => 3 + = # 0 & SmX SmX - f(6) fill => fil + 3 fix = 8 : = = o f(6) fill fill = = 2 4 S => = f(6) (6 IT SEE ~ fill . => 2 = * fix A B 5 : y - 0 = 2(X - 6) fix A 5 y . 2X 12 : = - .例2.6 设函数 f (x), g(x)在(−,+)上有定义，且满足 f (x + y) = f (x)g( y) + f ( y)g(x)， f (0) = 0, g(0) = 1, f (0) = 1, g(0) = 0. 试证明对一切 x ， f (x)均可导，且 f (x) = g(x). fix Un fixTox-fit) fix.glx flax 9-f # Men + = - 8X+0 -X 6X flox fix [910x) 1 + 9 me . - . = -Xo -X 910x-90 flox - fix he Me 9 = . + . -X f(x = 90) 9141 fi0) 9(x) . + · =例2.7 设函数 y = y(x)是由方程 y3 + xy + y + x2 = 0满足 y(0) = 0确定的连 x  y(t)dt 续二阶可导的函数，求lim 0 . x3 x→0 In Hat YYYY M v X3 0x5X y x y 44) + xy + y + = = , => 3y y + y + x - y + y + 2x = 00X =0 . y = 0(x = yi) = 0 . #IFE XcF* Y YIN %1 Y = = , , Jy(y'p 342 %" 2y1 %" y" => + . + + x . + + 2 = 02X = 0 . 4 =0 , Yir) = oft = Y T -5 -2 : = ..↑ 例2.8 设 f (x) = (1+ 2x)ln(1+ 2x)，计算 f (2022)(0). T ER : 35- (022) (2021) 2022) In Go2 In f(x) (2: C + 2x) . C + 2x) + · 2 . (1 +2x) = 22 2 + 2x)] (H [In (1 + = = 2 +2x) (t2X 2x)]" ? 2x32 2x)]" 3 > [m (1 + = 2 (1) · (1 + [In ( + = 2 · ( 1) · (-2) · C +2x) " mC - 2x) 2 ? () .. . . ( - (n - 11) (H2X) 2 ! ()* (n +) ! ~ + = = ( 2x)" + (2022) ? 2202 (1) ! 2021 fix (H 2x) . 22021 : + · ! - = + 2022 2 . 2020 . . ( 2x(2022 + [ (2021 +2x? for 2202 (1) ! 2011 · 2021 (H2x) · 2 ! + 2022 2 . 2020 . . ( 2x(2022 + [ (2021 +2x flor 22022021 2202-2020 ! - + 2022 ! . 22022 2021 2020 ! + 2022 2202? 2020 ! =- . . 22022 ! 2020 => , 88 10例2.8 设 f (x) = (1+ 2x)ln(1+ 2x)，计算 f (2022)(0). Effor 33 = fix fla fir 0 (222 = + . X + + ! 2022 f(x) = (1 + 2x) - (n(1 + 2x) = (x 5(2x (2x % = (H + 2x) . - + +. ) ... * x =..... 2022 (x) - ... fo 2022 2 -' =.... 240?X 2022 = +... 2022 ! 2021 x2022 I flor 2022020 22022 2022 ~ y = - ... ! 2021 X 2012例2.9 设 f (x) = (x − 1)5e−x ,求 f (10)(1) * *) 35 - fin Ci (x-1(* ) (16 ! (x1 51x-11 = . + . . (x-11" ** 1 8 ? *) ( 20 ( ( 60 (x-11 (E-X( + . . + · 1 * -> 20120 (x+1 · ( -X( + ( : 120 (eX)51 . f ! ( ( -eY) 10 x et) : 10 ( = . T . . - x= 1 5 !. ! 10 + et - = e 30240 · = - . 5!例2.9 设 f (x) = (x − 1)5e−x ,求 f (10)(1) (5 f()( = X 1 5 (( t+ ) Et -t =: + e g(t) . = . = t 90 9 (t) 9 o = , % + + +. + + tet + - +) + t(1 9(t) ( t) o(t) = e = e + + + + ett +" +... +... = - gib g . e -Get fi -Fet : = = I - : 10& 例2.10 设 y = f (x)是区间[0,1]上的任一非负连续函数. ( ) (1)试证存在 x (0,1),使得在区间0, x 上以 f x 为高的矩形面积,等   0 0 0 于在x ,1上以 y = f (x)为曲边的梯形面积.   0 ( X0fixd 16 i fitidt [T] EPGEEXO = : , fixe 1x fitat = x0 = 0 - Xf(x-l FNxd , F fitnt # 35 : = 0 = - , F10) -1! fieldt F(ll full < = > 0 = fixo 1 * fitidt (X fitat)'(x 35 == EX0 + = 0 = =xo = 0 & / fat FI X Fidlo Fill = = - , , ,2 f ( x) (2)又设 f (x)在区间(0,1)内可导,且 f (x)  − ,证明 x ( 1 ) 中的 x 0 是唯 一的. fixo 1x fitat x0 = 0 - Pl9N & X fin-1 fit at 0 91H = = , 9 fin fix f(x) fix f(x) x = + X + = X + 2 x( 2) f(x) > 2 = 0 - + ↑ 9 : &R-is : 9(o) = 0 Xo ,例2.11 设 f (x)在[a,b]连续，在(a,b)可导，证明在(a,b)内至少存在一点 f () − f (a) 使得 = f (). b − [ FT) = < f 15)-f(al = (b - 5) fiz) < ) f(3)-f(a) + (5 - b) fig) = 0 -f(u) FIs) . F() (x b) (fix) = E = 0 - fix-f(a) f Fix (x b) = + - FN) (fix -(f) F(al F(b) it = = (b) = 0 = 0 . , -↑ x 2 例2.12 设函数 f (x) =  et dt ；（1）证明：存在(1,2)，使得 1 2 f () = (2 −)e . es 2)e ( , et at f(s) (3-2) (5 NET) : + = 0 < + - = 0 . / etat * Fl Fi 35 -i = 0 = + (x-2 e , · / etat , Fill -eco F(2 = > o = . at)' ) =((X 2)) Yet (x / etat 3 F(x (x -2) = = - =3 = 0 , = . , , 175 . F(2) F(l = 0 = - .2 （2）证明：存在(1,2)，使得 f (2) = ln 2e . en fixe f12) < EE) IAFT : = , In2 it = Inx . /2 MI InX EF E E Me (12) EE : = , er fin) f12l-ful f(2) o - = => - t gin) 9121-911 In2 0 - F => er n = .    例2.13 设函数 f (x)在 0, 上连续，且 f (x)dx = 0, f (x)cos xdx = 0, 0 0 试证：在(0,)内至少存在两个不同的点 , ,使 f ( ) = f ( ) = 0. 1 2 1 2 ( fN ax f(i) f(sl 1 : = x . = 0 = =o & fil f(52) losXdX = - losE = 0 - ⑳ / FI =ofInt EPEESE Fil Er 2 = =0 : , / to GEH FIN "fitut : = , = F(0 fittat Fix = 0 0 = ,2) fiN XaX = 0 . in fixcoxaX FA losX 1? F-Saxax = + . ↑ ↓ 1? flat Fo F(x) = 0 =o = , (FM Saxax ! ficoax : = · 7) FIX) . SmXo =0 Y((0 , 2) , F(xd 0 : = : F(0) F(xd) = FIX) = 0 = - u Fil fIsEr FErl fIErI = = =O例2.14 已知 f (x)在[0,1]上连续,在(0,1)内可导,且 f (1) = 1, f (0) = 0,证明 在(0,1)内至少有一点,使得e−1  f () + f ()  = 1. e5 [ f(3) fig)] I #11) : + = e f(x) (et (x E e = fIN GERA : / FIX et #JE EEE (0 1) = . . . . F(1) Fros [F(sl - ecfisi fiel = => eTo + = e 1 - 0 =例2.15 已知函数 f (x)在区间[a,+]上具有 2 阶导数, f (a) = 0, f (x)  0, f (x)  0,设b  a,曲线 y = f (x)在点(b, f (b))处的切线与 x 轴的交点是 f(b)) (b ( ) · x ,0 ,证明a  x  b. . 0 0 i f( b)) +RESTE 4-fIbl f'(b) (x b) LEHA (b : = . - : , S f(b) Yo & y =0 = Xo b = - f'(b) fix fin ↑ fibl f(al F = co : < = 01 78 , b Xo < i · f(b) f(b) Es b b JeiE Xoca - > a => - ac f(n) f(D) - f(n) f(b) f() f(D) ↳ < = b a -E FEIGHE ESE (a b) It . , f(b)-fal f(b) fis) = = b b a -a - fin co fin ↑ : & f(b) f() fin) : > = b -a f(b) b : - a > f'(k) b-flb : .. Xo > a , · f(b)