Tiêu đề Polynomial CQ & MCQ Practice Sheet Solution.pdf ✅

eûc`x I eûc`x mgxKiY  Higher Math Academic Batch 3 (L) †`Iqv Av‡Q, f(x) = 3x2 – 4x + 1 Ges f(x) = 0  3x2 – 4x + 1 = 0 mgxKiYwUi g~jØq  I    +  = – – 4 3 = 4 3  = 1 3 wb‡Y©q mgxKi‡Yi g~jØq | – | Ges  2 +  2 g~j؇qi †hvMdj = | – | + ( 2 +  2 ) = | ( – ) | 2 + ( 2 +  2 ) – 2 = | ( + ) | 2 – 4 +     4 3 2 – 2  1 3 =         4 3 2 – 4  1 3 + 16 9 – 2 3 =    16  9 – 4 3 + 16 – 6 9 =    16 – 12 9 + 10 9 =    4 9 + 10 9 = 2 3 + 10 9 = 6 + 10 9 = 16 9 g~j؇qi ̧Ydj = | – |  ( 2 +  2 ) = | ( – ) | 2  {( + ) 2 – 2} = | ( + ) | 2 – 4 {( + ) 2 – 2} =         4 3 2 – 4  1 3            4 3 2 – 2  1 3 =    16  9 – 4 3      16 9 – 2 3 =    4 9  10 9 = 2 3  10 9 = 20 27 wb‡Y©q mgxKiY, x 2 – (g~j؇qi †hvMdj)x + g~j؇qi ̧Ydj = 0  x 2 – 16 9 x + 20 27 = 0  27x2 – 48x + 20 = 0 (M) †`Iqv Av‡Q, P(x) = x3 – 7x2 + 8x + 10 Avevi, P(x) = 0  x 3 – 7x2 + 8x + 10 = 0 mgxKiYwUi GKwU g~j 5  (x – 5), x3 – 7x2 + 8x + 10 ivwki GKwU Drcv`K| GLb, x 3 – 7x2 + 8x + 10 = 0  x 3 – 5x2 – 2x2 + 10x – 2x + 10 = 0  x 2 (x – 5) – 2x(x – 5) – 2(x – 5) = 0  (x – 5) (x2 – 2x – 2) = 0 nq, x – 5 = 0  x = 5 A_ev, x 2 – 2x – 2 = 0 x = – (– 2)  (– 2) 2 – 4  1  (– 2) 2  1 = 2  4 + 8 2 = 1  3  mgxKi‡Yi g~j ̧‡jv: 5, 1 + 3 , 1 – 3 (Ans.) 4| q(x) = lx 2 + mx + n [ivRkvnx †evW©- Õ23] r(x) = nx2 + mx + l Ges z – = x + iy (K) †`LvI †h, p = q bv n‡j 2x2 – 2(p + q)x + (p2 + q2 ) = 0 mgxKi‡Yi g~j ̧‡jv ev ̄Íe n‡Z cv‡i bv| (L) |z + 3| + |z – – 3| = 10 Øviv wb‡`©wkZ mÂvic‡_i mgxKi‡Yi kxl©we›`yi ̄’vbvsK wbY©q Ki| (M) r(x) = 0 mgxKi‡Yi GKwU g~j q(x) = 0 mgxKi‡Yi GKwU g~‡ji wØ ̧Y n‡j, †`LvI †h, l = 2n A_ev 2m2 = (l + 2n)2 | mgvavb: (K) cÖ`Ë mgxKiY, 2x2 – 2(p + q)x + (p2 + q2 ) = 0 mgxKiYwUi g~j ̧‡jv ev ̄Íe n‡e hw` Gi wbðvq‡Ki gvb k~b ̈ A_ev abvZ¥K nq| mgxKiYwUi wbðvqK, = {– 2(p + q)}2 – 4.2.(p2 + q2 ) = 4(p2 + 2pq + q2 – 2p2 – 2q2 ) = 4(– p 2 – q 2 + 2pq) = – 4(p2 – 2pq + q2 ) = – 4(p – q)2  0 wKš‘ wbðvq‡Ki gvb FYvZ¥K n‡j, g~j ̧‡jv ev ̄Íe n‡Z cv‡i bv| Kv‡RB mgxKiYwUi g~j ̧‡jv ev ̄Íe n‡e hw` wbðvq‡Ki gvb k~b ̈ nq| A_©vr, (p – q)2 = 0  p – q = 0  p = q  p = q n‡j mgxKi‡Yi g~j ̧‡jv ev ̄Íe n‡e| AZGe, p = q bv n‡j mgxKiYwUi g~j ̧‡jv ev ̄Íe n‡Z cv‡i bv| (Showed) (L) †`Iqv Av‡Q, z – = x + iy  z = x – iy GLv‡b, |z + 3| + |z – – 3| = 10  |x – iy + 3| + |x + iy – 3| = 10  |(x + 3) – iy| + |(x – 3) + iy| = 10  (x + 3) 2 + (– y) 2 + (x – 3) 2 + y2 = 10  (x + 3) 2 + y2 + (x – 3) 2 + y2 = 10  (x + 3) 2 + y2 = 10 – (x – 3) 2 + y2  ( (x + 3) ) 2 + y2 2 = (10 – (x – 3) ) 2 + y2 2 [eM© K‡i]  (x + 3)2 + y2 = 100 – 20 (x – 3) 2 + y2 + (x – 3)2 + y2
4  Higher Math 2nd Paper Chapter-4  (x + 3)2 – (x – 3)2 = 100 – 20 (x – 3) 2 + y2  4.3x = 100 – 20 (x – 3) 2 + y2  3x = 25 – 5 (x – 3) 2 + y2  5 (x – 3) 2 + y2 = 25 – 3x  (5 (x – 3) ) 2 + y2 2 = (25 – 3x)2 [cybivq eM© K‡i]  25(x2 – 6x + 9 + y2 ) = 625 – 150x + 9x2  25x2 – 150x + 25y2 + 225 = 9x2 – 150x + 625  16x2 + 25y2 = 400  x 2 25 + y 2 16 = 1  x 2 5 2 + y 2 4 2 = 1 ; hv GKwU Dce„Ë wb‡`©k K‡i Ges GwU‡K x 2 a 2 + y 2 b 2 = 1 ; a > b Gi mv‡_ Zzjbv K‡i cvB, a = 5, b = 4 ; a > b  Dce„ËwUi kxl©we›`yi ̄’vbvsK  ( a, 0)  ( 5, 0) (M) †`Iqv Av‡Q, q(x) = lx 2 + mx + n Ges q(x) = 0  lx 2 + mx + n = 0 ......(i) Avevi, r(x) = nx2 + mx + l Ges r(x) = 0  nx2 + mx + l = 0 ......(ii) GLv‡b, (ii) bs mgxKi‡Yi GKwU g~j (i) bs mgxKi‡Yi GKwU g~‡ji wØ ̧Y| g‡b Kwi, (i) bs mgxKi‡Yi GKwU g~j   (ii) bs mgxKi‡Yi GKwU g~j 2 (i) bs mgxKiY n‡Z cvB, l 2 + m + n = 0 .....(iii) (ii) bs mgxKiY n‡Z cvB, n(2) 2 + m.2 + l = 0  4n 2 + 2m + l = 0 ......(iv) (iii) bs I (iv) bs mgxKi‡Y eRa ̧Yb m~Î cÖ‡qvM K‡i cvB,   lm – 2mn =  4n2 – l 2 = 1 2lm – 4mn    lm – 2mn = 1 2lm – 4mn    lm – 2mn = 1 2(lm – 2mn)   2 = 1 2   =  1 2 Avevi,   lm – 2mn =  4n2 – l 2   lm – 2mn = 1 (2n) 2 – l 2   1 2 – m(2n – l) = 1 (2n + l) (2n – l)   1 2 (2n + l) (2n – l) = – m(2n – l)   1 2 (2n + l) (2n – l) + m(2n – l) = 0  (2n – l)       m  1 2 (2n + l) = 0 nq, 2n – l = 0  l = 2n A_ev, m  1 2 (2n + l) = 0  2m  (2n + l) = 0  2m =  (2n + l)  2m2 = (l + 2n)2 AZGe, l = 2n A_ev 2m2 = (l + 2n)2 (Proved) 5| (i) mx2 + nx + n = L [h‡kvi †evW©- Õ23] (ii) S = 6x3 – 20x2 + 5 Ges T = 6 – 6x – 9x2 (K) GKwU wØNvZ mgxKiY wbY©q Ki hvi GKwU g~j 1 2 + i3 (L) hw` L = 0 mgxKi‡Yi g~j `ywUi AbycvZ p : q nq Zvn‡j cÖgvY Ki †h, p q + q p + n m = 0 (M) hw` S = T mgxKiYwUi g~j ̧‡jv mgvšÍi cÖMg‡bi †MŠwYK wecixZ cÖMgbfz3 nq Z‡e x Gi gvb wbY©q Ki| mgvavb: (K) GLv‡b, wØNvZ mgxKi‡Yi GKwU g~j 1 2 + i3 = 2 – i3 (2 + i3) (2 – i3) = 2 – i3 2 2 – (i3) 2 = 2 – i3 4 – 9i2 = 2 – i3 4 + 9 [⸪ i2 = – 1] = 2 13 – i 3 13 Avgiv Rvwb, RwUj g~j ̧‡jv hyMjiƒ‡c _v‡K|  Aci g~jwU n‡e 2 13 + i 3 13 wb‡Y©q mgxKiY, x 2 – (g~j؇qi †hvMdj)x + g~j؇qi ̧Ydj = 0  x 2 –     2 13 – i 3 13 + 2 13 + i 3 13 x +     2 13 – i 3 13     2 13 + i 3 13  x 2 – 4 13 x +     2 13 2 –     i 3 13 2 = 0  x 2 – 4 13 x + 4 169 – i 2 9 169 = 0 x 2 – 4 13 x + 4 169 + 9 169 = 0 [⸪ i2 = – 1]  x 2 – 4 13 x + 1 13 = 0  13x2 – 4x + 1 = 0 (L) †`Iqv Av‡Q, mx2 + nx + n = L Avevi, L = 0  mx2 + nx + n = 0 mgxKiYwUi g~j؇qi AbycvZ p : q g‡b Kwi, g~jØq p I q  p + q = – n m  p + q = – n m Ges p  q= n m  pq = n m 2