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

2  Higher Math 2nd Paper Chapter-4  g~j؇qi †hvMdj = ( + ) + ( – ) = 4 + ( – ) 2 = 4 + ( + ) 2 – 4 = 4 + 4 2 – 4(– 4) = 4 + 32 = 4  4 2 = 4(1  2) Ges g~j؇qi ̧Ydj = ( + ) ( – ) = 4 ( – ) 2 = 4 ( + ) 2 – 4 = 4 4 2 – 4(– 4) = 4 32 = 4( 4 2) =  16 2 wb‡Y©q mgxKiY, x 2 – (g~j؇qi †hvMdj)x + g~j؇qi ̧Ydj = 0  x 2 – 4(1  2) x +( 16 2) = 0  x 2 – 4(1  2) x  16 2 = 0 2| DÏxcK-1: 2mx2 + nx + 1 = 0 Ges nx2 + 2mx + 1 = 0 DÏxcK-2: x 3 + px2 + qx + r = 0 [XvKv †evW©- Õ23] (K) x 2 + (p2 – 3)x – (p + 2) = 0 mgxKi‡Yi GKwU g~j – 1 + ip n‡j, mgxKiYwU mgvavb Ki| (L) DÏxcK-1 Gi mgxKiY `yBwUi GKwUgvÎ mvaviY g~j _vK‡j, cÖgvY Ki †h, 2m + n + 1 = 0| (M) DÏxcK-2 Gi mgxKiYwUi g~jÎq , ,  n‡j, ( – ) 2 Gi gvb wbY©q Ki| mgvavb: (K) †`Iqv Av‡Q, x 3 + (p2 – 3)x – (p + 2) = 0 mgxKi‡Yi GKwU g~j – 1 + ip Avgiv Rvwb, RwUj g~j ̧‡jv AbyeÜx hyMjiƒ‡c _v‡K|  mgxKiYwUi Aci GKwU g~j = – 1 – ip awi, mgxKiYwU Aci g~j  cÖ`Ë mgxKiY n‡Z cvB, x 3 + (p2 – 3)x – (p + 2) = 0  (– 1 + ip) + (– 1 – ip) +  = 0  – 1 + ip – 1 – ip +  = 0   – 2 = 0   = 2  wb‡Y©q mgvavb, x = 2, – 1 + ip, – 1 – ip (L) †`Iqv Av‡Q, 2mx2 + nx + 1 = 0 Ges nx2 + 2mx + 1 = 0 g‡b Kwi, mgxKiY `yBwUi mvaviY g~j  hv Dfq mgxKiY‡K wm× K‡i|  2m 2 + n + 1 = 0 ......(i) Ges n 2 + 2m + 1 = 0 .....(ii) (i) I (ii) bs mgxKiY n‡Z eRa ̧Yb m~Îvbymv‡i cvB,  2 n – 2m =  n – 2m = 1 4m2 – n 2   n – 2m = 1 4m2 – n 2   n – 2m = 1 (2m) 2 – n 2   n – 2m = 1 (2m + n) (2m – n)   = – 1 2m + n Avevi,  2 n – 2m =  n – 2m   = 1  – 1 2m + n = 1  2m + n + 1 = 0 (Proved) (M) †`Iqv Av‡Q, x 3 + px2 + qx + r = 0 mgxKi‡Yi g~jÎq , ,    +  +  = – p 1 = – p  +  +  = q 1 = q  = – r 1 = – r GLb, ( – ) 2 = ( – ) 2 + ( – ) 2 + ( – ) 2 = ( 2 – 2 +  2 ) + ( 2 –  +  2 ) + ( 2 – 2 +  2 ) = 2( 2 +  2 +  2 ) – 2( +  + ) = 2{( +  + ) 2 – 2( +  + )} – 2( +  + ) = 2( +  + ) 2 – 6( +  + ) = 2(– p)2 – 6q = 2p2 – 6q = 2(p2 – 3q)  wb‡Y©q gvb: 2(p2 – 3q) 3| f(x) = 3x2 – 4x + 1 Ges P(x) = x3 – 7x2 +8x + 10 [ivRkvnx †evW©- Õ23] (K) f(x) = 0 mgxKi‡Yi g~‡ji cÖK...wZ wbY©q Ki| (L) f(x) = 0 mgxKi‡Yi g~jØq ,  n‡j, | – | Ges  2 +  2 g~jwewkó mgxKiY wbY©q Ki| (M) P(x) = 0 mgxKi‡Yi GKwU g~j 5 n‡j, Aci g~j ̧‡jv wbY©q Ki| mgvavb: (K) †`Iqv Av‡Q, f(x) = 3x2 – 4x + 1 Ges f(x) = 0  3x2 – 4x + 1 = 0 mgxKiYwUi wbðvqK, D = (– 4)2 – 4.3.1 = 16 – 12 = 4 > 0 †h‡nZzD > 0 AZGe, mgxKiYwUi g~jØq ev ̄Íe I Amgvb|

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