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