Tiêu đề Binomial Expansions Board CQ & MCQ Practice Sheet Solution.pdf ✅

2  Higher Math 2nd Paper Chapter-5 mgvavb: K (a + x)4 Gi we ̄Í...wZ = a4 + 4C1 a 4–1 x + 4C2 a 4–2 x 2 + 4C3 a 4–3 x 3 +.... = a4 + 4C1 a 3 x + 4C2 a 2 x 2 + 4C3 ax3 + ....  x 3 Gi mnM = 4C3 a = 4a GLb, 4a = 16  a = 4 (Ans.) L †`Iqv Av‡Q, Z = 2x + 3y  Z = 2x + 3   –  1 x 2     y = –  1 x 2  Z =    2x –  3 x 2  Z 12 =    2x –  3 x 2 12 awi,    2x –  3 x 2 12 Gi we ̄Í...wZ‡Z (r + 1) Zg c`wU x ewR©Z|  Tr + 1 = 12Cr (2x)12 – r    –  3 x 2 r = 12Cr 2 12 – r x 12 – r (– 3)r x –2r = 12Cr 2 12 – r (– 3)r x 12 – 3r (r + 1) Zg c` x ewR©Z n‡j, x 12–3r = 1 = x0  12 – 3r = 0  r = 4  (4 + 1) Zg A_©vr 5 Zg c`wU x ewR©Z Ges c`wUi gvb = 12C4  2 12–4  (– 3)4 = 10264320 (Ans.) M Awfó dvskb: Z = 2x + 3y kZ©mg~n: x + 2y  8 x + y  6 x, y  0 AmgZv ̧wj‡K mgxKi‡Y cÖKvk K‡i cvB, x + 2y = 8  x 8 + y 4 = 1......(i) Ges x + y = 6  x 6 + y 6 = 1......(ii) x = 0......(iii) y = 0......(iv) MÖvd KvM‡R x, y Aÿ I g~jwe›`y O wPwýZ Kwi| † ̄‹j: Dfq A‡ÿi w`‡K ÿz`a 2 e‡M©i evûi •`N© ̈ = 1 GKK a‡i A(8, 0), B(0, 4), C(6, 0), D(0, 6) we›`y ̧wj ̄’vcb Kwi| MÖvd KvMR n‡Z mgvavb GjvKv OCEB wbY©q Kwi| Y (ii) (i) X O C(6,0) Y E(4,2) A(8,0) X D(0,6) B(0,4)  O(0, 0) we›`y‡Z, Z = 2  0 + 3  0 = 0 C(6, 0) we›`y‡Z, Z = 2  6 + 3  0 = 12 E(4, 2) we›`y‡Z, Z = 2  4 + 3  2 = 14 B(0, 4) we›`y‡Z, Z = 2  0 + 3  4 = 12  Z Gi m‡e©v”P gvb = 14 (Ans.) cÖkœ3 DÏxcK: h(x) = – 8x 1 – x 2 GKwU fMœvsk Ges  n = 1  n! n (n – 1)!3n n‡jv GKwU avivi mgwó| (K) x = i n‡j h(x) Gi eM©g~j †ei Ki| [i GKwU KvíwbK msL ̈v] [Kz. †ev. 19] (L) DÏxc‡Ki avivwUi AwfmvwiZv hvPvB Ki| [Kz. †ev. 19] (M) h(x) Gi we ̄Í...wZ‡Z x r Gi mnM wbY©q Ki| [Kz. †ev. 19] mgvavb: K †`Iqv Av‡Q, h(x) = – 8x 1 – x 2 x = i n‡j, h(x) = – 8i 1 – i 2 = – 8i 1 + 1 [∵ i 2= –1] = – 4i = 2(– 2i) = 2(1 – 2i – 1) = 2(1 – 2i  i 2 ) = 2(1 – i) 2 [∵ (a – b)2 = a2 – 2ab  b 2 ]  wb‡Y©q eM©g~j =  2(1 – i) 2 =  2 (1 – i) (Ans.) L cÖ`Ë avivi mgwó =  n = 1  n!n (n – 1)! 3n =  n = 1  n.(n – 1)!n (n – 1)!3n =  n = 1  n 2 3 n