Title Polynomial Varsity Practice Sheet Solution.pdf ✅

eûc`x I eûc`x mgxKiY  Varsity Practice Sheet 1 04 eûc`x I eûc`x mgxKiY Polynomial and Polynomial Equation 1. wb‡Pi †KvbwU eûc`x bq? ax 2 + 2hxy + by2 2x2 + 3xy + y2 x 2 + y2 + 2gx + 2fg + c 2x2 + 3y x + y2 DËi: 2x2 + 3y x + y2 e ̈vL ̈v: KviY x Gi NvZ Negative 2. †KvbwU eûc`x? 2x2 – 5 x + 1 x 3 – 3 x 2 + 4x + 1 x 3 + 2x2 – 3x + x–1 2x2 – x + 1 DËi: 2x2 – x + 1 e ̈vL ̈v: eûc`x n‡Z n‡j Pj‡Ki NvZ c~Y©msL ̈v n‡Z n‡e, fMœvsk ev FYvZ¥K n‡j n‡e bv| 3. 4x3 + 2x2 + 3x – 6 †K x – 1 Øviv fvM Ki‡j fvM‡kl KZ n‡e? 1 3 –11 0 DËi: 3 e ̈vL ̈v: awi, f(x) = 4x3 + 2x2 + 3x – 6 f(1) = 4 + 2 + 3 – 6 = 3 4. x 3 – 9x2 + 24x – 15 †K x – 2 Øviv fvM Ki‡j fvM‡k‡lÑ 4 5 6 7 DËi: 5 e ̈vL ̈v: f(x) = x3 – 9x2 + 24x – 15  fvM‡kl f(2) = 23 – 9 (22 ) + 24  2 – 15 = 5 5. 2x3 + bx2 – 9x – 26 eûc`xi GKwU Drcv`K x – 2 n‡j, b Gi gvbÑ 1 3 5 7 DËi: 7 e ̈vL ̈v: f(x) = 2x3 + bx2 – 9x – 26  f(2) = 0  2.23 + b.22 – 9.2 – 26 = 0  b = 7 6. y = x 2 – 12x + 40 GB mgxKiY m¤ú‡K© †KvbwU mZ ̈? x Aÿ‡K GKUv we›`y‡Z †Q` K‡i x Aÿ‡K `yBUv we›`y‡Z †Q` K‡i x Aÿ‡K wZbUv we›`y‡Z †Q` K‡i x Aÿ‡K †Q`B K‡i bv DËi: x Aÿ‡K †Q`B K‡i bv e ̈vL ̈v: D = 144 – 4.40 < 0 ev ̄Íe g~j bvB|  x Aÿ‡K †Q`B K‡i bv 7. y = x 3 mgxKiYwU x Aÿ‡K KqwU we›`y‡Z †Q` K‡i? 0 1 2 3 DËi: 1 e ̈vL ̈v: x 3 – 1 = 0  x 2 + x + 1 = 0 Gi ev, (x – 1) (x2 + x + 1) = 0 †Kv‡bv ev ̄Íe mgvavb †bB x 3 – 1 = 0 Gi ev ̄Íe mgvavb 1wU  x Aÿ‡K 1wU we›`y‡Z †Q` Ki‡e| 8. y = x 4 + x3 – x 2 + 1 mgxKiYwU y Aÿ‡K KZevi †Q` Ki‡e? 0 1 2 4 DËi: 1 e ̈vL ̈v: y A‡ÿ, x = 0 y = 0 + 0 – 0 + 1 = 1  y Aÿ‡K (0, 1) we›`y‡Z GKevi †Q` K‡i| 9. y = x 2 – 3x – ix + 3i †iLvwU x Aÿ‡K †Kvb we›`y‡Z †Q` Ki‡e? 0, 3 3, 0 Awb‡Y©q †Q` Ki‡e bv DËi: 3, 0 e ̈vL ̈v: x A‡ÿ y = 0  x 2 – 3x – ix + 3i = 0  x (x – 3) – i (x – 3) = 0  x = 3, i  x A‡ÿi †Q`we›`y (3, 0) 10. 3 x+5 = 3x+3 + 8 3 n‡j x = ? [RU 17-18] 1 –2 – 4 9 DËi: – 4 e ̈vL ̈v: 3 x .35 – 3 x .33 = 8 3  3 x+3(32 – 1) = 8 3  3 x+3 = 3–1  x = – 4 11. 5 2x – 24.5x – 25 = 0 GB mgxKi‡Yi g~j KqwU? 1wU 2wU 3wU 4wU DËi: 1wU
2  Higher Math 2nd Paper Chapter-4 e ̈vL ̈v: 5 x = t ; t 2 – 24t – 25 = 0  t 2 – 25t + t – 25 = 0  (t – 25) (t + 1) = 0  t = 25 ; 5 x = – 1 [MÖnY‡hvM ̈ bv]  5 x = 52  x = 2  g~j  1wU 12. x 2 – 10x + 34 = 0 mgxKi‡Yi g~j ̧‡jv njÑ [JU 14-15] 3  i 5  i 5  2i 5  3i DËi: 5  3i e ̈vL ̈v: x 2 – 10x + 34 = 0  x = 10  100 – 4  34 2 = 10  – 36 2  x = 5 + 3i, 5 – 3i 13. 9x2 – 12x + 4 = 0 wØNvZ mgxKi‡Yi g~jØq ,  n‡j   = ? [JnU 14-15] 4 : 9 3 : 2 1 : 1 4 : 3 DËi: 1 : 1 e ̈vL ̈v: 9x2 – 12x + 4 = 0  (3x) 2 – 2.3x.2 + 22 = 0  (3x – 2)2 = 0  x = 2 3  = 2 3 ;  = 2 3    = 1 14. 4x5 + 1 = 0 mgxKiYwUi g~jÑ 1 wU 2 wU 5 wU 4 wU DËi: 5 wU e ̈vL ̈v: x Gi m‡e©v”P NvZ hZ n‡e g~‡ji msL ̈v ZZwU n‡e| 15. x 2 – 7x + 12 = 0 mgxKi‡Yi g~jØqÑ 3, 4 –3, 4 3, – 4 – 3, – 4 DËi: 3, 4 e ̈vL ̈v: x 2 – 7x + 12 = 0  x 2 – 3x – 4x + 12 = 0  x(x – 3) – 4(x – 3) = 0  x = 3, 4 16. f(x) = x3 – 2x2 – 5x + 6 eûc`xi `yBwU Drcv`K (x – 1) I (x + 2) n‡j, f(x) = 0 mgxKi‡Yi g~jÎq n‡eÑ 1, – 2, – 3 1, – 2, 3 1, 2, – 3 –1, 2, 3 DËi: 1, – 2, 3 e ̈vL ̈v: (x – 1) I (x + 2) `yBwU Drcv`K nIqvq, `ywU g~j 1 I – 2 GLb f(3) = 33 – 2  3 2 – 5  3 + 6 = 0  Aci g~j 3 17. x 2 + px + q = 0 mgxKi‡Yi GKwU g~j 3 + i n‡j p I q Gi gvb KZ? – 6, – 10 – 6, 10 6, – 10 6, 10 DËi: – 6, 10 e ̈vL ̈v: Aci g~j 3 – i 3 + i + 3 – i = – p  p = – 6 Avevi, 3 2 + 1 = q  q = 10 18. x 2 = 0 mgxKi‡Yi c„_vqK KZ? – 4 0 1 4 DËi: 0 e ̈vL ̈v: 1.x2 + 0.x + 0 = 0  D = 02 – 4.1.0 = 0 19. 2x2 – 2 6x + 3 = 0 GB mgxKi‡Yi g~‡ji cÖK...wZÑ ev ̄Íe I mgvb Aev ̄Íe I mgvb ev ̄Íe I Amgvb Aev ̄Íe I Amgvb DËi: ev ̄Íe I mgvb e ̈vL ̈v: D = 24 – 4  2  3 = 0  ev ̄Íe I mgvb 20. k Gi †Kvb gv‡bi Rb ̈ (k – 1)x2 – (k + 2)x + 4 ivwkwU c~Y©eM© n‡e? –10, 2 10, – 2 2, 10 –2, – 10 DËi: 2, 10 e ̈vL ̈v: ivwk c~Y©eM© n‡Z n‡j, D = 0 (k + 2)2 – 4.(k – 1).4 = 0  k 2 + 4k + 4 – 16k + 16 = 0  k 2 – 12k + 20 = 0  k 2 – 10k – 2k + 20 = 0  k = 10, 2 21. (k + 2) x2 + 3x – (1 – 3k) = 0 mgxKi‡Yi g~jØq ci ̄úi ̧YvZ¥K wecixZ n‡j k = ? 3 3 2 – 3 2 – 3 DËi: 3 2 e ̈vL ̈v: g~jØq ci ̄úi ̧YvZ¥K wecixZ n‡j, g~j؇qi ̧Ydj = 1 x 2 Gi mnM = aaæeK k + 2 = 3k – 1  k = 3 2