Tiêu đề Polynomial Practice Sheet Solution HSC FRB 25.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|
eûc`x I eûc`x mgxKiY  Final Revision Batch '25 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