Tiêu đề Binomial Expansions Engineering Practice Sheet Solution.pdf ✅

wØc`x we ̄Í...wZ  Engineering Practice Sheet Solution (HSC 26) 1 Rhombus Publications 05 wØc`x we ̄Í...wZ Binomial Expansions WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1| y = 3x + 6x2 + 10x3 + ...  n‡j, †`LvI †h, x = 1 3 y – 1 3 2 . 4 2! y 2 + 1 . 4 . 7 3 3 . 3! y 3 – ...... [BUET 19-20] mgvavb: y = 3x + 6x2 + 10x3 + ...  1 + y = 1 + 3x + 6x2 + 10x3 + ...  1 + y = (1 – x)–3  (1 + y) – 1 3 = (1 – x) – 3  ( ) – 1 3  1 – x = (1 + y)– 1 3 1 – x = 1 – 1 3 y + ( ) – 1 3 ( ) – 1 3 – 1 2! y 2 + ( ) – 1 3 ( ) – 1 3 – 1 ( ) – 1 3 – 2 3! y 3 + ...  x = 1 3 y – 1 3 2  4 2! y 2 + 1  4  7 3 3  3! y 3 – ...... (Showed) 2| y Gi Nv‡Zi DaŸ©μg Abymv‡i (2y + 1)10 Gi we ̄Í...wZ‡Z y r–1 Gi mnM Cr Ges Cr+2 = 4Cr n‡j, r Gi gvb wbY©q Ki| [BUET 19-20] mgvavb: GLv‡b, (2y + 1)10 y Gi Nv‡Zi DaŸ©μg Abymv‡i mvRv‡j (2y + 1)10 †K †jLv hvq (1 + 2y)10 GLb, Tr + 1 = 10Cr 2 r y r  y r–1 Gi mnM = 10Cr–1 2 r–1 cÖkœg‡Z, Cr = 10Cr-1 2 r–1  Cr+2 = 10Cr+2–1 2 r+2–1 = 10Cr+1 2 r+1 Avevi kZ©vbymv‡i, Cr+2 = 4Cr  10Cr+1 2 r+1 = 4 . 10Cr–1 2 r–1  10Cr+1 = 10Cr–1  4  2 r–1–r–1  10Cr+1 = 10Cr–1  4  1 2 2  r + 1 + r – 1 = 10 [ nCx = nCy n‡j, x + y = n]  r = 5 (Ans.) 3| Amxg avivwUi †hvMdj wbY©q Ki: 1 3 + 1  3 3  6 + 1.3.5 3.6.9 + 1.3.5.7 3.6.9.12 + 1.3.5.7.9 3.6.9.12.15 + ...  [BUET 18-19] mgvavb: †`Iqv Av‡Q, avivwU, 1 3 + 1  3 3  6 + 1.3.5 3.6.9 + 1.3.5.7 3.6.9.12 + 1.3.5.7.9 3.6.9.12.15 + ...  ........ (i) Avgiv Rvwb, (1 + x)n = 1 + nx + n(n – 1) 2 x 2 + ...   (1 + x)n – 1 = nx + n(n – 1) 2 x 2 + ...  ......... (ii) (i) bs †K (ii) bs ivwki mv‡_ Zzjbv K‡i cvB, nx = 1 3  n 2 x 2 = 1 9 ...... (iii) Ges n(n – 1) 2 x 2 = 1 6 ...... (iv) (iv)  (iii) n‡Z cvB, n(n – 1) 2 x 2  1 n 2 x 2 = 3 2  n – 1 n = 3  2n = – 1  n = – 1 2 x = – 2 3 [(iii) †_‡K]  Amxg avivwUi †hvMdj = (1 + x)n – 1 =    1 –  2 3 – 1 2 – 1 = 3 – 1 (Ans.) 4| n  Ges |x| < 1 n‡j, †`LvI †h, (1 + x) n 1 – x Gi we ̄Í...wZ‡Z x n Gi mnM 2 n | [BUET 18-19] mgvavb: (1 + x)n (1 – x)–1 = (1 + x)n (1 + x + x2 + x3 + ... + xn + ...) = (1 + nC1x + nC2x 2 + ... + nCnx n + ...)(1 + x + x2 + x3 + ... + xn + ...) x n Gi mnM = 1 + nC1 + nC2 + nC3 + ... + nCn GLb, (1 + x)n = 1 + nC1x + nC2x 2 + nC3x 3 + ... + nCnx n Gi we ̄Í...wZ‡Z x = 1 n‡j, 2 n = 1 + nC1 + nC2 + nC3 + ... + nCn = x n Gi mnM  x n Gi mnM = 2n (Showed)


4  Higher Math 2nd Paper Chapter-5 Rhombus Publications weMZ mv‡j KUET-G Avmv cÖkœvejx 15| (1 – x)–3 Gi we ̄Í...wZ‡Z e„nËg c‡`i gvb KZ, hLb x = 3 5 ? [KUET 19-20] mgvavb: (1 – x)–3 Gi we ̄Í...wZ‡Z e„nËg c‡`i Rb ̈, Tr+1 Tr = n–r+1 r  x 1 = – 1 [hLb n < 0]  –3–r+1 r  3 5 = –1 [n, x Gi gvb ewm‡q]  – 6 – 3r = –5r  2r = 6  r = 3; hv c~Y©msL ̈v myZivs, Tr I Tr+1 A_©vr T3 I T4 `yBwUi mvsL ̈gvb e„nËg c`| e„nËg c‡`i gvb, T2+1 = (– 3)(– 3 – 1) 2!     3 5 2 = 54 25 A_ev, T3+1 = (– 3)(– 3 – 1)(– 3 – 2) 3!     3 5 3 = 54 25 (Ans.) 16| hw` (1 + x) (a – bx)12 Gi we ̄Í...wZ‡Z x 8 Gi mnM 0 nq, Zvn‡j a b Abycv‡Zi gvb wbY©q Ki| [KUET 13-14, 05-06; CUET 14-15] mgvavb: GLv‡b, (1 + x) (a – bx)12 = (1 + x) {a12 – 12C1 a 11 bx + ...... – 12C7 a 5 b 7 x 7 + 12C8 a 4 b 8 x 8 ...... + b12 x 12}  x 8 Gi mnM = 12C8 a 4 b 8 – 12C7 a 5 b 7 kZ©g‡Z, 12C8 a 4 b 8 – 12C7 a 5 b 7 = 0  12! 8! 4! a 4 b 8 = 12! 7! 5! a 5 b 7  b 8 = a 5  a b = 5 8 (Ans.) 17|    2x  2 + k x 3 10 Gi we ̄Í...wZ‡Z x 5 Ges x 15 Gi mnMØq mgvb n‡j k Gi abvZ¥K gvb wbY©q Ki| [KUET 10-11; BUTex 01-02; BUET 00-01; SUST 09-10;] mgvavb:    2x  2 + k x 3 10 Gi we ̄Í...wZ‡Z (r + 1)Zg c`, = 10Cr (2x2 ) 10–r     k x 3 r = 10Cr 2 10–r k r x 20–5r GB (r + 1)Zg c‡` x 5 _vK‡j 20 – 5r = 5  r = 3 GB (r + 1)Zg c‡` x 15 _vK‡j 20 – 5r = 15  r = 1 cÖkœg‡Z, 10C3 2 10–3 k 3 = 10C1 2 10–1 k  120  128k3 = 5120 k  3k3 = k  k = 0,  1 3  a abvZ¥K gvb wb‡j, k = 1 3 (Ans.) 18|     x y + y x 10 Gi we ̄Í...wZ‡Z ga ̈c`wU wbY©q Ki| [KUET 04-05] mgvavb: c`msL ̈v 11; ga ̈c` n 2 + 1 = 10 2 + 1 = 6 Zg c` T6 = T5+1 = 10C5     x y 10–5     y x 5 = 10C5 (Ans.) weMZ mv‡j RUET-G Avmv cÖkœvejx 19| (1 + x)p , p > 0 Gi x 5 Gi mnM x 4 Gi mn‡Mi wØ ̧Y n‡j p Gi gvb wbY©q Ki| x 6 I x 4 Gi mn‡Mi g‡a ̈ m¤úK© wK? [RUET 17-18] mgvavb: Tr+1 = PCr x r x 5 Gi mnM = PC5 Ges x 4 Gi mnM = PC4 cÖkœg‡Z, PC5 = 2  PC4  p! 5!(p – 5)! = 2  p! 4!(p – 4)!  1 5  4!  (p – 5)! = 2 4!(p – 4)(p – 5)!  p – 4 = 10  p = 14 (Ans.) GLb, x 6Gi mnM x 4Gi mnM = 14C6 14C4 = 3  x 6 Gi mnM = 3  x 4 Gi mnM| (Ans.) weMZ mv‡j CUET-G Avmv cÖkœvejx 20| hw` (1 + x) (a – bx)12 Gi we ̄Í...wZ‡Z x 8 Gi mnM 0 nq, Zvn‡j a b Abycv‡Zi gvb wbY©q Ki| [CUET 14-15; KUET 13-14, 05-06] mgvavb: GLv‡b, (1 + x) (a – bx)12 = (1 + x) {a12 – 12C1 a 11 bx + ...... – 12C7 a 5 b 7 x 7 + 12C8 a 4 b 8 x 8 ...... + b12 x 12}  x 8 Gi mnM = 12C8 a 4 b 8 – 12C7 a 5 b 7 kZ©g‡Z, 12C8 a 4 b 8 – 12C7 a 5 b 7 = 0  12! 8! 4! a 4 b 8 = 12! 7! 5! a 5 b 7  b 8 = a 5  a b = 5 8 (Ans.)