Tiêu đề Integration CQ & MCQ Practice Sheet Sulution (HSC 26).pdf ✅

†hvMRxKiY  CQ & MCQ Practice Sheet Solution (HSC 26) 3 (M) GLv‡b, mij‡iLv, y = 4x ......(i) eμ‡iLv, y = x – x 2 .....(2) (i) bs I (ii) bs n‡Z cvB,  x 2 + 4x – x = 0  x 2 + 3x = 0  x(x + 3) = 0  x = 0, – 3 y = 4x X C X Y Y O y = x– x 2 x = 0 n‡j, (i) bs n‡Z cvB, y = 4  0  y = 0 x = – 3 n‡j, (i) bs n‡Z cvB, y = 4  (– 3)  y = – 12  (i) bs I (ii) bs ci ̄úi (0, 0) Ges (– 3, – 12) we›`y‡Z †Q` K‡i| GLb, (i) bs n‡Z, y = 4x = y1 (awi) (ii) bs n‡Z, y = x – x 2 = y2 (awi) wP‡Îi QvqvK...Z As‡ki †ÿÎdj, =  –3 0 (y1 – y2) dx =  –3 0 (4x – x + x2 )dx =  –3 0 (3x + x2 ) dx = 3     x 2 2 –3 0 +     x 3 3 –3 0 = 3 2  9 + 1 3 (– 27) = 27 2 – 9 = 27 – 18 2 = 9 2 eM© GKK (Ans.) 3| `„k ̈Kí-1: f(x) = 1 (4 + x2 ) 3 2 `„k ̈Kí-2: x 2 + y2 = 64 ; y = 5 [h‡kvi †evW©- Õ23] (K)  sin9x sin11x dx wbY©q Ki| (L) `„k ̈Kí-1 e ̈envi K‡i  4 0 f(x) dx wbY©q Ki| (M) `„k ̈Kí-2 Gi e„Ë I mij‡iLv Øviv Ave× ÿz`aZi As‡ki †ÿÎdj wbY©q Ki| mgvavb: (K)  sin9x sin11x dx = 1 2  2 sin11x sin9x dx = 1 2  {cos(11x – 9x) – cos(11x + 9x)} dx = 1 2  (cos2x – cos20x) dx = 1 2     sin2x 2 – sin20x 20 + c = sin2x 4 – sin20x 40 + c (Ans.) (L) †`Iqv Av‡Q, f(x)= 1 (4 + x2 ) 3 2   4 0 f(x) dx =  4 0 1 (4 + x2 ) 3 2 dx =  tan–1 2 0 2 sec2  (4 + 4tan2 ) 3 2 d awi, x = 2 tan  dx = 2 sec2  d x 0 4  0 tan–1 2 =  tan–1 2 0 2 sec2 d (4sec2 ) 3 2 =  tan–1 2 0 2 sec2  8 sec3  d =  tan–1 2 0 1 4sec d = 1 4  tan–1 2 0 cos d = 1 4 [sin] tan–1 2 0 = 1 4 [sin(tan–1 2) – 0] [⸪ x = 2 tan   = tan–1 2] = 1 4 sin    sin–1 2 5 = 1 4  2 5 = 1 2 5 (Ans.) 1 2 2 + 12 1 = 5 tan–1 2 2 (M) cÖ`Ë e„Ë, x 2 + y2 = 64  x 2 + y2 = 82 .....(i) Ges cÖ`Ë mij‡iLv, y = 5 ......(ii) (0, – 8) X X Y Y (– 8, 0) O (8, 0) (0, 5) (0, 8) y = 5 (i) bs I (ii) bs Øviv Ave× ÿz`aZi As‡ki †ÿÎdj = 2  8 5 xdy = 2  8 5 64 – y 2 dy = 2   2 sin–15 8 64 – 64sin2  8cos d awi, y = 8sin  dy = 8 cos d y 5 8  sin–15 8  2 = 2  8  8   2 sin–1 5 8 1 – sin2  . cos d = 128  2 sin–1 5 8 cos2  .cos d = 128   2 sin–1 5 8 cos2  d
4  Higher Math 1st Paper Chapter-10 = 64   2 sin–1 5 8 2cos2  d = 64   2 sin–1 5 8 (1 + cos2) d = 64      + sin2 2  2 sin–1 5 8 = 64      + 2sin.cos 2  2 sin–1 5 8 = 64 [ + sin . cos]  2 sin–1 5 8 = 64          2 + sin 2 . cos  2 –     sin–15 8 + sin sin–15 8 .cos sin–15 8 = 64      2 + 0 – sin–15 8 – 5 8 . cos cos–1 39 8 sin–1 5 8 8 2 – 5 2 = 39 8 5 = 64      2 – sin–15 8 – 5 8  39 8 =     32 – 64 sin–15 8 – 5 39 eM© GKK (Ans.) 4| f(x) = x3 – 6x2 + 9x + 5 [h‡kvi †evW©- Õ23] g(x) = x + 2 h(x) = (1 – x) (x2 + 4) (K)  x lnx dx wbY©q Ki| (L) f(x) dvskbwUi gvb †h mKj e ̈ewa‡Z e„w× ev n«vm cvq Zv wbY©q Ki| (M)  g(x) h(x) dx wbY©q Ki| mgvavb: (K)  x lnx dx = lnx  x dx –       d  dx(lnx)  x dx dx = x 2 2 lnx –      1 x  x 2 2 dx = x 2 2 lnx – 1 2  x 2 2 + c = x 2 2 lnx – x 2 4 + c (Ans.) (L) †`Iqv Av‡Q, f(x) = x3 – 6x2 + 9x + 5  f (x) = 3x2 – 12x + 9 dvskbwU μgea©gvb n‡e hw` f (x) > 0 nq  3x2 – 12x + 9 > 0  3(x2 – 4x + 3) > 0  x 2 – 4x + 3 > 0  x 2 – 3x – x + 3 > 0  x(x – 3) – 1(x – 3) > 0  (x – 3) (x – 1) > 0 .....(i) (i) bs AmgZvwU mZ ̈ n‡e hw` Ges †Kej hw` (x – 3) I (x – 1) DfqB abvZ¥K A_ev DfqB FYvZ¥K nq| mgvavb (x – 3) Gi wPý (x – 1) Gi wPý x < 1 – – 1 < x < 3 – + x > 3 + +  (i) bs AmgZvwU mZ ̈ n‡e hw` x < 1 A_ev x > 3 nq| A_©vr, x < 1 A_ev x > 3 n‡j dvskbwU μgea©gvb n‡e| (Ans.) Avevi, dvskbwU μgn«vmgvb n‡e hw`,  f(x) < 0 nq  3x2 – 12x + 9 < 0  (x – 3) (x – 1) < 0 .....(ii)  1 < x < 3 A_©vr,1 < x < 3 n‡j dvskbwU μgn«vmgvb n‡e| (Ans.) (M) †`Iqv Av‡Q, g(x) = x + 2 Ges h(x) = (1 – x) (x2 + 4)   g(x) h(x) dx =  x + 2 (1 – x) (x 2 + 4) dx awi, x + 2 (1 – x) (x 2 + 4) = A 1 – x + Bx + C x 2 + 4 ....(i) (i) bs Gi Dfqcÿ‡K (1 – x) (x2 + 4) Øviv ̧Y K‡i cvB, x + 2 = A(x2 + 4) + (Bx + C) (1 – x) .......(ii) (ii) bs G x = 1 ewm‡q cvB, 3 = 5A  A = 3 5 (ii) bs n‡Z x 2 Gi mnM mgxK„Z K‡i cvB, A – B = 0  3 5 – B = 0  B = 3 5 (ii) bs n‡Z x Gi mnM mgxK„Z K‡i cvB, 1 = B – C  C = B – 1 = 3 5 – 1 = – 2 5 (i) bs G A, B I C Gi gvb ewm‡q cvB,  x + 2 (1 – x) (x 2 + 4) = 3 5(1 – x) + 3 5 x – 2 5 x 2 + 4 = 3 5(1 – x) + 3x – 2 5(x 2 + 4)   g(x) h(x) dx =  x + 2 (1 – x)(x 2 + 4) dx =  3 5(1 – x) dx +  3x – 2 5(x 2 + 4) dx = 3 5  1 1 – x dx + 1 5      3x x 2 + 4 dx –  2 x 2 + 4 dx