Title Polynomial CQ & MCQ Practice Sheet Solution.pdf ✅

eûc`x I eûc`x mgxKiY  Higher Math Academic Batch 1 04 eûc`x I eûc`x mgxKiY Polynomial and Polynomial Equation Board Questions Analysis m„Rbkxj cÖkœ †evW© mvj XvKv gqgbwmsn ivRkvnx Kzwgjøv h‡kvi PÆMÖvg ewikvj wm‡jU w`bvRcyi 2023 2 2 2 2 1 2 2 2 2 2022 2 2 2 2 2 2 2 2 2 eûwbe©vPwb cÖkœ †evW© mvj XvKv gqgbwmsn ivRkvnx Kzwgjøv h‡kvi PÆMÖvg ewikvj wm‡jU w`bvRcyi 2023 4 5 6 4 4 4 5 5 4 2022 5 4 3 4 4 5 4 4 5 weMZ mv‡j †ev‡W© Avmv m„Rbkxj cÖkœ 1| wØNvZ mgxKiY ax2 + bx + b = 0 ; [a  0] [XvKv †evW©- Õ23] (K) 2x2 – 2(p + q)x + (p2 + q2 ) = 0 mgxKi‡Yi g~jØq ev ̄Íe I mgvb n‡j, cÖgvY Ki †h, p = q| (L) DÏxc‡Ki g~j؇qi AbycvZ m : 3n n‡j, cÖgvY Ki †h, m n + 3 n m + 3b a = 0 (M) a = 1, b = – 4 Ges DÏxc‡Ki mgxKi‡Yi g~jØq  I  n‡j, ( + ) I ( – ) g~jwewkó mgxKiYwU wbY©q Ki| mgvavb: (K) †`Iqv Av‡Q, 2x2 – 2(p + q)x + (p2 + q2 ) = 0 mgxKiYwUi g~jØq ev ̄Íe I mgvb n‡j, wbðvqK k~b ̈ n‡e| A_©vr mgxKiYwUi wbðvqK, D = {– 2(p + q)}2 – 4.2.(p2 + q2 ) = 0  4(p + q)2 – 8(p2 + q2 ) = 0  4(p2 + 2pq + q2 – 2p2 – 2q2 ) = 0  2pq – p 2 – q 2 = 0  p 2 – 2pq + q2 = 0  (p – q)2 = 0  p – q = 0  p = q (Proved) (L) cÖ`Ë mgxKiY, ax2 + bx + b = 0 ; a  0 Ges mgxKiYwUi g~j؇qi AbycvZ m : 3n g‡b Kwi, g~jØq m I 3n  m + 3n = – b a  m + 3n = – b a Ges m  3n = b a  mn = b 3a 2 evgcÿ = m n + 3 n m + 3b a = m n + 3 n m + 3b a = ( m) 2 + 3( n) 2 mn + 3b a = m + 3n mn + 3b a = – b a b 3a 2 + 3b a = – b a b 3a + 3b a = – b a  3a b + 3b a = – b  3 a + 3b a = – 3b a + 3b a = 0 = Wvbcÿ|  m n + 3 n m + 3b a = 0 (Proved) (M) DÏxc‡Ki mgxKiYwU, ax2 + bx + b = 0 ; [a  0] a = 1, b = – 4 n‡j mgxKiYwU n‡e, x 2 – 4x – 4 = 0 mgxKiYwUi g~jØq  I    +  = – – 4 1 = 4  = – 4 1 = – 4 GLb, ( + ) I ( – ) g~jwewkó mgxKiY wbY©q Ki‡Z n‡e|


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