Title 1st-Paper-Viva-Exam-Written-Solution.pdf ✅

 1st Paper Viva Exam (Written) 1 Higher Math 1 st Paper Viva Exam 1| x = e2y + 1 , x = e– y + 1 , y = – 2 Ges y = 1 Øviv Ave× †ÿ‡Îi †ÿÎdj wbY©q Ki| mgvavb: X (0, 0) (e, 0) x = e2y + 1 y = 1 y = – 2 x = e–y + 1 Y  wb‡Y©q †ÿÎdj =  0 – 2 (e–y + 1 – e 2y + 1 )dy +  1 0 (e2y + 1 – e –y + 1)dy =     – e 1–y – 1 2 e 2y + 1 0 – 2 +     1 2 e 2y + 1 + e1 – y 0 1 = 22.998 eM© GKK (Ans.) 2| wbY©vqKwUi gvb wbY©q Ki:       1 mC1 mC2 1 m+1C1 m+1C2 1 m+2C1 m+2C2 mgvavb:         1 m m(m – 1) 2 1 m+1 (m + 1)m 2 1 m + 2 (m + 2)( m + 1) 2 =         1 m m(m – 1) 2 1 m + 1 (m 2 + m) 2 1 m + 2 (m 2 + 3m + 2) 2 =         1 m m(m – 1) 2 0 1 m 0 1 m + 1     c3 = c3 – c2 c2 = c2 – c1 = m + 1 – m = 1 (Ans.) 3| GKwU mgevû wÎfz‡Ri GKwU kxl© (4, 5) Ges GKwU evûi mgxKiY x + y = 6| Aci `ywU evûi mgxKiY wbY©q Ki| mgvavb: Aci evûi Xvj = m tan60 =  m – (– 1) 1 + m(– 1) =  m + 1 l – m  m = 2 + 3 , 2 – 3 Aci evû `yBwUi mgxKiY, y – 5 = (2 + 3) (x – 4) A(4, 5) B C x + y = 6 (m = – 1) 60 60 Ges y – 5 = (2 – 3) (x – 4) (Ans.) 4| x 2 + y2 = 4x e„‡Ëi (1 3) we›`y‡Z Aw1⁄4Z ̄úk©K, Awfj¤^ I x Aÿ Øviv Ave× wÎfz‡Ri †ÿÎdj wbY©q Ki| mgvavb: x 2 + y2 – 4x = 0 e„‡Ëi †K›`a C = (2, 0) e ̈vmva© r = (– 2) 2 + 0 – 0 = 2 GKK ̄úk©KwUi mgxKiY, x.(1) + y.( 3) – 2.(x + 1) = 0  x – 3y + 2 = 0 ..... (i) (Ans.) Bnv x Aÿ‡K B (x, 0) we›`y‡Z †Q` Ki‡j, x + 2 = 0  x = – 2  B Gi ̄’vbv1⁄4 (– 2, 0) C (2,0) y ● x A(1 3) r = 2 B (–2,0) ● Awfj¤^wUi mgxKiY, 3x + y + k = 0 ..... (ii) hv (2, 0) we›`yMvgx|  3  2 + 0 + k = 0  k = – 2 3 (ii) n‡Z, 3x + y – 2 3 = 0 (Ans.) Avevi, ̄úk©K I Awfj‡¤^i †Q`we›`y A (1 3)  †ÿÎdj = 1 2  AB  AC = 1 2  2  2 3 [AB = (1 + 2) ] 2 + ( 3 – 0) 2 = 12 = 2 3 GKK = 2 3 eM© GKK (Ans.)
2  Higher Math 1st Paper 5| hw` GKwU wÎfz‡Ri j¤^‡K›`a A(– 3, 5) I fi‡K›`a B(3, 3) Ges C cwi‡K›`a nq, Z‡e AC †K e ̈vm a‡i Aw1⁄4Z e„‡Ëi e ̈vmva© wbY©q Ki| mgvavb: Q R P A B C Avgiv Rvwb, GKwU wÎfz‡Ri j¤^‡K›`a, fi‡K›`a I cwi‡K›`a 2:1 Abycv‡Z _v‡K| A_©vr, AB : BC = 2 : 1  BC = 1 2 AB GLb, AC = AB + BC = AB + 1 2 AB =     1 + 1 2 (– 3 – 3) 2 + (5 – 3) 2 = 3 2  40 = 3 2  2 10  wb‡Y©q e„‡Ëi e ̈vmva© = AC 2 = 3 10 2 GKK (Ans.) 6| logcosx (cotx) + 4 logsinx (tanx) = 1 n‡j sinx = ? mgvavb: ln(cotx) ln(cosx) + 4 ln(tanx) ln(sinx) = 1  ln(cosx) – ln(sinx) ln(cosx) + 4ln(sinx) – 4ln(cosx) ln(sinx) = 1  ln(sinx) ln(cosx) – {ln(sinx)} 2 + 4ln(sinx) ln(cosx) – 4{ln(cosx)} 2 ln(sinx) ln(cosx) = 1  5ln(sinx) ln(cosx) – {ln(sinx)}2 – 4{ln(cosx)}2 = ln(sinx) ln(cosx)  {ln(sinx)}2 – 4ln(sinx) ln(cosx) + {2ln(cosx)}2 = 0  {ln(sinx) – 2ln(cosx)}2 = 0  ln(sinx) = 2ln(cosx)  ln(sinx) = ln(cos2 x)  sinx = 1 – sin2 x  sin2 x + sinx – 1 = 0  sinx = – 1  1 2 – 4  1  (– 1) 2  1  sinx = – 1  5 2 (Ans.) 7| ABC G 3 sinA + 4 cosB = 6 Ges 4 sinB + 3 cosA = 1 n‡j C = ? mgvavb: 3 sinA + 4 cosB = 6 ...............(i) 4 sinB + 3 cosA = 1 ................(ii) (i)2 + (ii)2 K‡i cvB, 9(sin2A + cos2A) + 16(sin2B + cos2B) + 24 sinA cosB + 24 sinB cosA = 36 + 1  24 sin(A + B) = 12  sin(A + B) = 1 2  sin(180 – C) = 1 2  sinC = 1 2  C = 30, 150 wKš‘ C = 150 n‡j (i) mgxKiY wm× n‡e bv|  C = 30 (Ans.) 8| a evûwewkó eM© Gi Pvi‡KvYv †_‡K 4wU eM©‡ÿÎ †K‡U GKgyL‡Lvjv Nbe ̄‘ •Zwi Kiv n‡jv| D”PZv KZ n‡j Nbe ̄‘wUi AvqZb m‡e©v”P n‡e? mgvavb: x x x a a x v = (a – 2x)2 x  v = a2 x – 4ax2 + 4x3  dv dx = a2 – 8ax + 12x2  d 2 v dx2 = – 8a + 24x  Pigwe›`y‡Z, dv dx = a2 – 8ax + 12x2 = 0  12x2 – 6ax – 2ax + a2 = 0  6x(2x – a) – a(2x – a) = 0  (2x – a) (6x – a) = 0  x = a 6     x = a 2 MÖnY‡hvM ̈ bq x = a 6 n‡j d 2 v dx2 = – 4a < 0  x = a 6 (Ans.)
 1st Paper Viva Exam (Written) 3 9| lim x   (1 + x) 1 x – e x = ? mgvavb: lim x   (1 + x) 1 x – e x     0 0 AvKvi KviY lim x   (1 + x) 1 x = e = lim x   (1 + x) 1 x      1  x(1 + x) – ln(1 + x) x 2 – 0 1 [L' Hôpital’s Rule] awi, y = (1 + x) 1 x  lny = 1 x ln(1 + x)  1 y . dy dx = 1 x . 1 1 + x + ln(1 + x)    –  1 x 2  dy dx = (1 + x) 1 x     1 x(1 + x) – ln(1 + x) x 2 = lim x   (1 + x) 1 x . lim x   x – (1 + x) ln(1 + x) x 2 (1 + x) = e lim x   1 – 1 – ln(1 + x) 3x2 + 2x [L' Hôpital’s Rule] = e lim x   – ln(1 + x) 3x2 + 2x = e lim x   – 1 (1 + x) 6x + 2 [L' Hôpital’s Rule] = – e 2 (Ans.) 10| gvb wbY©q Ki:  tanx dx mgvavb:  tanx dx =  z .2z dz 1 + z4 =  (z 2 + 1) + (z 2 – 1) 1 + z4 dz awi, tanx = z2  sec2 x dx = 2z dz  dx = 2z dz 1 + z4 =  1 + 1 z 2 z 2 + 1 z 2 dz +  1 – 1 z 2 z 2 + 1 z 2 dz =  d     z – 1 z     z – 1 z 2 + ( 2) 2 +  d     z + 1 z     z + 1 z 2 – ( 2) 2 = 1 2 tan–1       z – 1 z 2 + 1 2 2 ln      z +  1 z – 2 z + 1 z + 2 + c [ †hLv‡b z = tanx] (Ans.)