Tiêu đề Differentiation Engineering Practice Sheet With Solution.pdf ✅

AšÍixKiY  Engineering Question Bank Solution 1 AšÍixKiY Differentiation beg Aa ̈vq WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1| y = (x + 1 + x ) 2 20 n‡j x = 0 we›`y‡Z y2 Gi gvb wbY©q Ki| [BUET 22-23] mgvavb: y = (x + 1 + x ) 2 20  y1 = 20 (x + 1 + x ) 2 19     1 + x 1 + x2  y1 1 + x2 = 20 (x + 1 + x ) 2 20  y2 1 + x2 + y1 x 1 + x2 = 400 (x + 1 + x ) 2 19     1 + x 1 + x2 x = 0 we›`y‡Z,  y2 × 1 + y1 × 0 = 400 × 1  y2 = 400 (Ans.) 2| lim x  0 e 2x – e –5x + ax x 2 we` ̈gvb _vK‡j a Gi gvb KZ Ges Gi mxgv wbY©q Ki| [BUET 22-23] mgvavb: lim x  0 e 2x – e –5x + ax x 2 [ 0 0 form] = lim x  0 2e 2x + 5e–5x + a 2x [L Hospital Rule] = lim x  0 4e2x – 25e–5x 2 [L Hospital Rule] = 4 – 25 2 = – 21 2 (Ans.) GLv‡b, 2e0 + 5e 0 + a = 0  a = – 7 (Ans.) 3| lim x   3x2 – sin2x x 2 + 5 wbY©q Ki| [BUET 21-22] mgvavb: Avgiv Rvwb, – 1  – sin2x  1  3x2 – 1  3x2 – sin2x  3x2 + 1  3x2 – 1 x 2 + 5  3x2 – sin2x x 2 + 5  3x2 + 1 x 2 + 5  lim x   3x2 – 1 x 2 + 5  lim x   3x2 – sin2x x 2 + 5  lim x   3x2 – 1 x 2 + 5 GLb, lim x   3x2 – 1 x 2 + 5 = lim x   x 2     3 – 1 x 2 x 2     1 + 5 x 2 = lim x   3 – 1 x 2 1 + 5 x 2 = 3 Avevi, lim x   3x2 + 1 x 2 + 5 = lim x   x 2     3 + 1 x 2 x 2     1 + 5 x 2 = lim x   3 + 1 x 2 1 + 5 x 2 = 3  m ̈vÛDBP Dccv` ̈ Abymv‡i cvB, lim x   3x2 – sin2x x 2 + 5 = 3 4| F(x) = x + 2sinx dvskbwUi [0, 2] e ̈ewa‡Z jNygvb/ ̧iægvb wbY©q Ki Ges Gi Inflection Point wbY©q Ki| [BUET 21-22] mgvavb: †`Iqv Av‡Q, F(x) = x + 2sinx  F(x) = 1 + 2cosx Ges F(x) = – 2 sinx jNy Ges ̧iægv‡bi Rb ̈, F(x) = 0  1 + 2cosx = 0  cosx = – 1 2  cosx = – cos  3  cosx = cos      –  3  cosx = cos 2 3  x = 2n  2 3  [0, 2] e ̈ewa‡Z x = 2 3 , 4 3  F     2 3 = – 2sin     2 3 = – 2  3 2 = – 3 < 0  ̧iægvb = F     2 3 = 2 3 + 2 sin 2 3 = 2 3 + 3  F     4 3 = – 2sin     4 3 = (–2)      – 3 2 = 3 > 0  jNygvb = F     4 3 = 4 3 + 2 sin 4 3 = 4 3 – 3 GLb lnflection point G, F(x) = 0  – 2 sin x = 0  sinx = 0  x = 0, , 2 wb‡Y©q ̧iægvb = 2 3 + 3 ; jNygvb = 4 3 – 3 Ges lnflection Point x = 0  2 (Ans.) 5| f(x) = aex + be–x Gi †ÿ‡Î a > b > 0 kZ©v‡ivwcZ n‡j †`LvI Gi jNygvb, ̧iægvb Av‡cÿv e„nËi| [BUET 20-21] mgvavb: f(x) = aex + be–x  f(x) = aex – be–x GLb, aex – be–x = 0  aex = be–x  aex = b e x  (ex ) 2 = b a  e x = b a [ex Gi gvb FYvZ¥K n‡Z cv‡i bv]  x = ln    b a GLb, f(x) = aex + be–x f     ln b a = a  b a + b  a b = ab + ab = 2 ab > 0 [jNygvb we` ̈gvb]  jNygvb = a  b a + b  a b = 2 ab [we.`a.: cÖ`Ë dvsk‡bi †Kv‡bv ̧iægvb we` ̈gvb †bB]
2  Higher Math 1st Paper Chapter-9 6| lim x0 e x + e–x – 2 x Gi gvb wbY©q Ki| [BUET 19-20] mgvavb: lim x0 e x + e–x – 2 x [ 0 0 form] = lim x0 e x – e –x 1 [L Hospital Rule] = 0 (Ans.) 7| hw` tan (lny) = x nq, Z‡e y2(0) Gi gvb wbY©q Ki| [BUET 19-20] mgvavb: tan (lny) = x  lny = tan–1 x  y = etan–1x y1 = e tan–1x 1 + x 2 y2 = 1 (1 + x2 ) 2  e tan–1x + etan–1x  (–2x) (1 + x2 ) 2 x = 0 we›`y‡Z y2 (0) = 1 1  e 0  0 1 = 1 (Ans.) 8| lim x0 e x 2 – cosx x 2 Gi gvb wbY©q Ki| [BUET 17-18] mgvavb: lim x0 e x 2 – cosx x 2     0 0 AvKvi = lim x0 e x 2  2x + sinx 2x [L' Hospital Rule] = lim x0      e  x 2 2 + 2x.ex 2 2x + cosx 2 [L' Hospital Rule] = 2 + 0 + 1 2 = 3 2 (Ans.) 9| †`LvI †h, x + y = a eμ‡iLvi †h‡Kv‡bv ̄úk©K Øviv Aÿ `yBwU †_‡K KwZ©Z Asÿ؇qi †hvMdj GKwU aaæeK| [BUET 18-19] mgvavb: x + y = a  1 2 x + 1 2 y dy dx = 0  dy dx = – y x (x1, y1) we›`y‡Z ̄úk©K, y – y1 = – y1 x1 (x – x1) x Aÿ‡K †Q` Ki‡j, y = 0  y1 = y1 x1 (x – x1)  x1y1 = x – x1  x = x1 + x1y1 ... ... ... (i) y Aÿ‡K †Q` Ki‡j, x = 0  y = y1 + x1y1 ... ... ... (ii) mgxKiY (i) + (ii) n‡Z cvB, x + y = x1 + y1 + 2 x1y1 = ( x1 + y1) 2 = a 2 = a hv GKwU aaæeK| [Showed] weKí mgvavb: x + y = a ... ... ... (i) (i) †K x-Gi mv‡c‡ÿ AšÍixKiY K‡i cvB, 1 2 x + 1 2 y dy dx = 0  dy dx = – y x eμ‡iLvi Dci (x1, y1) †h‡Kv‡bv we›`y‡Z, x1 + y1 = a ........ (i) Ges dy dx = – y1 x1  (x1, y1) we›`y‡Z ̄úk©‡Ki mgxKiY, y – y1 = y1 x1 (x – x1)  y x1 – x1y1 = – x y1 + x1 y1  x y1 + y x1 = x1y1 + x1 y1  x y1 + y x1 = x1 y1 + ( y1 + x1)  x y1 + y x1 = x1 y1 a  x a x1 + y a y1 = 1  Aÿ `yBwU †_‡K KwZ©Z As‡ki †hvMdj = a x1 + a y1 = a ( y1 + x1) = a a = a  †h‡Kv‡bv ̄úk©‡Ki †ÿ‡Î KwZ©Z As‡ki †hvMdj = a, hv GKwU aaæeK| 10| lim x0 (1 + 7x) 5x + 3 x Gi gvb wbY©q Ki| [BUET 18-19] mgvavb: lim x0 (1 + 7x) 5x + 3 x  lim x0 (1 + 7x)5 + 3 x  lim x0 (1 + 7x)5  lim x0 (1 + 7x) 3 x  1  lim x0 (1 + 7x) 1 7x  7  3 { } lim x0 (1 + 7x) 1 7x 21 = e21 (Ans.) 11| tan y = 2t 1 – t 2 Ges sin x = 2t 1 + t2 n‡j, dy dx gvb wbY©q Ki| [BUET 18-19] mgvavb: y = tan–1     2t 1 – t 2 = 2 tan–1 t x = sin–1     2t 1 + t2 = 2 tan–1 t GLb, dy dx = dy dt dx dt = d dt (2 tan–1 t) d dt (2 tan–1 t) = 1 1 + t2 1 1 + t2 = 1 (Ans.) 12| y = (x + 1 + x ) 2 m n‡j cÖgvY Ki †h, (1 + x2 ) d 2 y dx2 + x dy dx – m 2 y = 0| AZtci x = 0 we›`y‡Z d 3 y dx3 Gi gvb †ei Ki| [BUET 17-18] mgvavb: †`Iqv Av‡Q, y = (x + 1 + x ) 2 m  y1 = m (x + 1 + x ) 2 m (x + 1 + x ) 2       1 + 2x 2 1 + x2  y1 = m (x + 1 + x ) 2 m (x + 1 + x ) 2      x + 1 + x  2 1 + x2
AšÍixKiY  Engineering Question Bank Solution 3  1 + x2 y1 = m (x + 1 + x ) 2 m ... ... ... (i)  (1 + x2 )  y1 2 = m2 y 2 [eM© K‡i]  (1 + x2 ) 2 y1 y2 + y1 2 2x = m2 2y y1  (1 + x2 ) y2 + x y1 – m 2 y = 0 ... ... ... (ii) (Porved) (ii) bs †K x Gi mv‡c‡ÿ cybivq AšÍixKiY K‡i cvB, (1 + x2 ) y3 + y2 2x + x y2 + y1 – m 2 y1 = 0 ... (iii) x = 0 we›`y‡Z y = 1, y1 = m Ges y2 = m2 [(i) I (ii) bs n‡Z gvb ewm‡q] (iii) bs mgxKi‡Y y, y1 Ges y2 Gi gvb ewm‡q cvB, (1+ 0) y3 + m2  2  0 + 0  m 2 + m – m 2  m = 0  y3 = m3 – m (Ans.) 13| GKwU mge„Ëf~wgK †KvY‡Ki g‡a ̈ GKwU Lvov e„ËvKvi wmwjÛvi ̄’vcb Kiv Av‡Q| wmwjÛv‡ii eμZj e„nËg n‡Z n‡j †`LvI †h, wmwjÛv‡ii e ̈vmva© †KvY‡Ki f~wgi e ̈vmv‡a©i A‡a©K| [BUET 16-17] mgvavb: awi, wmwjÛv‡ii e ̈vmva© BE = x; †KvY‡Ki f~wgi e ̈vmva© BC = r Ges D”PZv AB = h| GLb, ABC I DEC m`„k|  AB BC = DE EC = DE BC – BE  h r = DF r – x  DE = h r (r – x) A D B C E wmwjÛv‡ii eμZ‡ji †ÿÎdj, A = 2xDE = 2x  h r (r – x) = 2xh – 2hx2 r A e„nËg n‡j, dA dx = 0  2h – 4h r x = 0  1 – 2x r = 0  2x r = 1  x = r 2 Avevi, d 2A dx2 = – 4h r < 0 myZivs, x = r 2 n‡j wmwjÛv‡ii eμZj e„nËg n‡e| 14| f(x) = sin3x n‡j Lt h0 f(x + 3h) – f(x) 3h Gi gvb wbY©q Ki| [BUET 16-17] mgvavb: Lt h0 f(x + 3h) – f(x) 3h = Lt h0 sin3 (x + 3h) – sinx 3h = Lt h0 sin(3x + 9h) – sin 3x 3h = Lt h0 2cos 6x + 9h 2 sin 9h 2 3h = Lt h0 3sin 9h 2 cos 6x + 9h 2 9h 2 = 3.1.cos 6x + 0 2 = 3 cos 3x (Ans.) 15| hw` y = f(x) Ges x = 1 z nq, Z‡e †`LvI †h, d 2 f dx2 = z 4 d 2 y dz2 + 2z3 dy dz. [BUET 16-17] mgvavb: x = 1 z  1 = – 1 z 2  dz dx [x Gi mv‡c‡ÿ AšÍixKiY K‡i]  dz dx = – z 2 GLb, df dx = dy dz  dz dx = – z 2 dy dz  d 2 f dx2 = – z 2 d dx     dy dz – dy dz  d dx (z2 ) = – z 2 d 2 y dz2  dz dx – 2z dy dz  dz dx  d 2 f dx2 = z 4 d 2 y dz2 + 2z3 dy dz [Showed] 16| y m + y–m = 2x n‡j m 2 (x2 – 1)y2 + m2 xy1 – y = ? [BUET 15-16] mgvavb: y m + y–m = 2x  y 2m + 1 = 2xym  y 2m – 2xym + 1 = 0  y m = 2x  4x2 – 4 2 = x  x 2 – 1  y = (x  x ) 2 – 1 1 m  y1 = 1 m (x  x ) 2 – 1 1 m – 1     1  x x 2 – 1  my1 = (x  x ) 2 – 1 1 m – 1 (x  x ) 2 – 1 x 2 – 1  my1 = x 2 – 1 =  (x  x ) 2 – 1 1 m  my1 x 2 – 1 =  y  m 2 y1 2 (x2 – 1) = y2  2m2 y1y2(x2 – 1) + 2xm2 y1 2 = 2yy1  m 2 y2 (x2 – 1) + m2 xy1 – y = 0 (Ans.) 17| y = 3 mij‡iLvi mgvšÍivj †Kvb †iLv y = (x – 3)2 (x – 2) eμ †iLvi †h mg ̄Í we›`y‡Z ̄úk©K †mB we›`y ̧‡jvi ̄’vbv1⁄4 wbY©q Ki| [BUET 15-16] mgvavb: y = 3 mij‡iLvi mgvšÍivj mij‡iLvi mgxKiY, y = k y = k mij‡iLvi Xvj, dy dx = 0 y = (x – 3)2 (x – 2) dy dx = 2(x – 3) (x – 2) + (x – 3)2  †hme we›`y‡Z mij‡iLvwU eμ‡iLv‡K ̄úk© Ki‡e †mme we›`y‡Z Df‡qi Xvj mgvb n‡e|
4  Higher Math 1st Paper Chapter-9  2(x – 3) (x – 2) + (x – 3)2 = 0  x = 3, 7 3 GLb, x = 7 3 n‡j, y = 4 27 Ges x = 3 n‡j, y = 0 myZivs, wb‡Y©q we›`ymg~n     7 3  4 27 Ges (3, 0) 18| †`LvI †h, f(x) = x 1 x Gi gvb e„nËg n‡e hw` x = e nq| [BUET 12-13] mgvavb: f(x) = x 1 x f(x) = x 1 x       1 x x  dx dx + lnx d dx     1 x = x 1 x     1 x 2 – lnx x 2 ̧iægvb A_ev jNygv‡bi Rb ̈, f(x) = 0  x 1 x     1 x 2 – lnx x 2 = 0  x 1 x – 2 [1 – lnx] = 0 wKš‘ x 1 x – 2  0  1 – lnx = 0  lnx = 1 = lne  x = e f (x) = x 1 x       – 2 x 3 – x 2  1 x – lnx  2x x 4 + [ ] 1 x 2 – lnx x 2 x 1 x     1 x 2 – lnx x 2 GLb x = e we›`y‡Z, f (e) = e 1 e      –1 e 3 hv FYvZ¥K|  x = e n‡j, f(x) e„nËg n‡e| (Showed) 19| 1 wjUvi (1000 Nb †mwg.) Zij aviY ÿgZv m¤úbœ `yB cÖv‡šÍ Ave× GKwU Lvov e„ËvKvi wmwjÛvi cÖ‡qvRb| wmwjÛviwUi D”PZv I e ̈vmva© wKiƒc n‡j me©v‡cÿv Kg †ÿÎdj wewkó wUb w`‡q Zv •Zix Kiv m¤¢e? [BUET 14-15; 06-07] mgvavb: Avgiv Rvwb 1000 cm3 = 1 dm3 awi, wmwjÛv‡ii D”PZv h dm Ges e ̈vmva© r dm  r 2 h = 1  h = 1 r 2 wmwjÛv‡ii †ÿÎdj, A = 2r 2 + 2rh = 2r(r + h) = 2r     r + 1 r 2 = 2r 2 + 2 r  dA dr = 4r – 2 r 2  d 2A dr2 = 4 + 4 r 2 > 0 (me©wb¤œ gvb cvIqv hv‡e) A me©wb¤œ n‡j, 4r – 2 r 2 = 0  4r – 2h = 0  2r – h = 0  2r = h kZ©g‡Z, r 2  2r = 1  2r 3 = 1  r = 3 1 2 = 0.542 dm = 5.42 cm Ges h = 2r = 10.84 cm 20| y = (x + 1 + x ) 2 m + (x + 1 + x ) 2 –m n‡j (1 + x2 )y2 + xy1 – m 2 y Gi gvb wbY©q Ki| [BUET 14-15] mgvavb: y = (x + 1 + x ) 2 m + (x + 1 + x ) 2 –m  y1 = m(x + 1 + x ) 2 m–1      1 + x 1 + x2 – m(x + 1 + x ) 2 m–1      1 + x 1 + x2  y1 = m 1 + x2 (x + 1 + x ) 2 m – m 1 + x2 (x + 1 + x ) 2 –m  (1 + x2 )y1 2 = m 2 {(x + 1 + x ) } 2 m – (x + 1 + x ) 2 –m 2 [eM© K‡i]  (1 + x2 )y1 2 = m2 [{(x + 1 + x ) } ] 2 m + (x + 1 + x ) 2 –m 2 – 4 [∵ (a – b)2 = (a + b)2 – 4ab]  (1 + x2 )y1 2 = m 2 (y2 –4)  (1 + x2 )2y1y2 + 2xy1 2 = m2 2yy1  (1 + x2 )y2 + xy1 – m 2 y = 0 (Ans.) 21| k~b ̈ e ̈ZxZ k Gi Ggb GKwU gvb wbY©q Ki hv D‡jøwLZ dvskb‡K x = 0 we›`y‡Z Awew”Qbœ Ki‡e| †Zvgvi Dˇii †hŠw3KZv e ̈vL ̈v Ki| f(x) =    tan kx x  x < 0 3x + 2k2  x  0 [BUET 14-15] mgvavb: †`Iqv Av‡Q, f(x) =    tan kx x  x < 0 3x + 2k2  x  0 L.H.L = lim x0 – tan kx x = lim x0 – tankx kx  k = k R.H.L = lim x0 + (3x + 2k2 ) = 2k2 Ges f(0) = 3  0 + 2k2 = 0 + 2k2 = 2k2  x = 0 we›`y‡Z f(x) Awew”Qbœ n‡e hw` L.H.L = R.H.L nq| A_©vr, 2k2 = k  2k2 – k = 0  k(2k – 1) = 0  k = 0, 1 2 AZGe, k~b ̈ e ̈ZxZ k = 1 2 n‡j dvskbwU x = 0 we›`y‡Z Awew”Qbœ n‡e| Dˇii †hŠw3KZv: k = 1 2 n‡j, f(x) =     tan x 2 x  x < 0 3x + 1 2  x  0 GLb, lim x0 – tan x 2 x 2  1 2 = 1 2 lim x0 +     3x + 1 2 = 1 2