Tiêu đề Differentiation Engineering Practice Sheet Solution.pdf ✅

AšÍixKiY  Engineering Practice Sheet Solution 1 09 AšÍixKiY Differentiation WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1. x Gi mv‡c‡ÿ AšÍiR wbY©q Ki: y = tan–1 4x 1 – 4x2 [BUET 23-24] mgvavb: y = tan–1 4x 1 – 4x2  dy dx = 1 1 +     4x 1 – 4x2 2  1 – 4x2  4 – 4x  1 2 1 – 4x2  (– 8x) (1 – 4x2 ) = (1 – 4x2 ) 1 – 4x2 + 16x2  4 1 – 4x2 + 16x2 1 – 4x2 (1 – 4x2 ) = 1 1 + 12x2  4(1 – 4x2 ) + 16x2 1 – 4x2 = 4 (1 + 12x2 ) 1 – 4x2 (Ans.) 2. GKwU †MvjvKvi cvwbi †duvUvi AvqZb e„w×i nvi 13 cm3 /s Ges Gi e ̈vm 34 cm n‡j, c„ôZ‡ji †ÿÎdj e„w×i nvi KZ? [BUET 23-24] mgvavb: GLv‡b, dV dt = 13 cm3 /s, 2r = 34 cm GLv‡b, V = 4 3 r 3  dV dt = 4 3 r 2  dr dt  13 = 4r 2  dr dt  dr dt = 13 4(17) 2 = 3.58  10–3 cm/s GLb, A = 4r 2  dA dt = 4  2r  dr dt = 4  34  3.58  10–3 = 1.53 cm2 /s (Ans.) 3. a Gi gvb wbY©q Ki hLb lim x  0 2ex – 2e–5x + ax x 2 Gi gvb we` ̈gvb| ‘a’ Gi mxgvi gvbI wbY©q Ki| [BUET 22-23] mgvavb: lim x  0 2ex – 2e–5x + ax x 2     0 0 form = lim x  0 2e x + 10e–5x + a 2x [L Hôpital’s Rule] = lim x  0 2ex – 50e–5x 2 [L Hôpital’s Rule] = 2 – 50 2 = – 24 (Ans.) [‘a’ Gi Limiting value wbY©q m¤¢e bq| GLv‡b g~jZ dvsk‡bi Limiting value eySv‡bv n‡q‡Q] GLv‡b, 2e0 + 10e0 + a = 0  a = – 12 (Ans.) 4. hw` y = (x + 1 + x ) 2 20 nq, Z‡e   d 2 y dx2 x = 0 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  d 2 y dx2 = 400 (Ans.) 5. 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
2  Higher Math 1st Paper Chapter-9 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 (Ans.) 6. 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 jNygvb 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 + 2sin2 3 = 2 3 + 3 (Ans.)  F     4 3 = – 2sin    4 3 = (– 2)      – 3 2 = 3 > 0  jNygvb = F     4 3 = 4 3 + 2sin4 3 = 4 3 – 3 (Ans.) GLb Inflection point G, F(x) = 0  – 2 sin x = 0  sinx = 0  x = 0, , 2 wKš‘ [0, 2] e ̈ewa‡Z x = 0, 2 n‡jv F(x) dvsk‡bi mgvwß we›`y|  x =  Gi Rb ̈ Inflection Point cvIqv hv‡e| GLb, F() =  – 2sin2 =   Inflection Point (, ) †R‡b iv‡Lv: Inflection Point n‡jv †mB we›`y †hLv‡b dvsk‡bi AeZj‡Z¡i cwieZ©b nq| cÖ`Ë F(x) = x + 2sinx; [0, 2] Gi MÖvd A1⁄4b Ki‡j, (0, 0) mgvwß we›`y Y X (, ) (2, 2) mgvwß we›`y †h‡nZz, (0, 0) Ges (2, 2) we›`y‡Z dvsk‡bi AeZj‡Z¡i cwieZ©b ̄úó bq, ZvB GB `ywU Point of Inflection n‡e bv| 7. f(x) = aex + be–x Gi †ÿ‡Î a > b > 0 kZ©v‡ivwcZ n‡j †`LvI †h, Gi jNygvb, ̧iægvb A‡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 Note: cÖ`Ë dvsk‡bi †Kv‡bv ̧iægvb we` ̈gvb †bB 8. 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 + x2 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 + e0  0 1 = 1 (Ans.)
AšÍixKiY  Engineering Practice Sheet Solution 3 9. 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 Hôpital’s Rule] = 0 (Ans.) 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 7x0 (1 + 7x) 1 7x 21 = e21 (Ans.) 11. tany = 2t 1 – t 2 Ges sinx = 2t 1 + t2 n‡j, dy dx gvb wbY©q Ki| [BUET 18-19; KUET 04-05] mgvavb: y = tan–1 2t 1 – t 2 = 2tan–1 t x = sin–1 2t 1 + t2 = 2tan–1 t = y  x = y  dy dx = 1 (Ans.) 12. †`LvI †h, x + y = a eμ‡iLvi †h‡Kv‡bv ̄úk©K Øviv Aÿ `yBwU †_‡K KwZ©Z Ask؇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| (Showed) 13. 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  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) (Proved) (ii) bs †K x Gi mv‡c‡ÿ cybivq AšÍixKiY K‡i cvB, (1 + x2 )y3 + y2 .2x + xy2 + 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.)
4  Higher Math 1st Paper Chapter-9 14. 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 form = lim x0 e x 2  2x + sinx 2x [L Hôpital’s Rule] = lim x0    e  x 2 2 + 2x.ex 2 2x + cosx 2 [L Hôpital’s Rule] = 2 + 0 + 1 2 = 3 2 (Ans.) 15. 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) – sin3x 3h = Lt h0 sin(3x + 9h) – sin3x 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 = 3cos3x (Ans.) 16. 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) 17. 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 = DE 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  x = r 2 n‡j wmwjÛv‡ii eμZj e„nËg n‡e| (Showed) 18. 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 (x2 – 1)y2 + m2 xy1 – y = 0 (Ans.) 19. 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; KUET 06-07] 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