Title 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|