Title Subject Final Math - C- Solution.pdf ✅

1 Varsity Subject Final Math [Set-C (Solution)] c~Y©gvb: 140 †b‡MwUf gvK©: 0.25 mgq: 1 NÈv 30 wgwbU 1. hw` GKwU eM© g ̈vwUa· A Ggb nq †h, A 2 – 3AI2 + I2 = 0 Z‡e, A –1 = ? [If a square matrix A is such that A2 – 3AI2 + I2 = 0 then A–1 = ?] A – 3I A 2 + 3I 3I – A 3I – A 2 DËi: 3I – A e ̈vL ̈v: A 2 – 3AI2 + I2 = 0  A 2 – 3AI + I = 0  A – 3I + A–1 = 0 [A–1 Øviv ̧Y K‡i]  A –1 = 3I – A 2. Awae„‡Ëi Dc‡K›`a؇qi ga ̈eZx© `~iZ¡ 12 GKK Ges e = 2 n‡j, mgxKiY †KvbwU? [If the distance between the foci of a hyperbola is 12 units and its eccentricity is 2, what is its equation?] y 2 – x 2 = 18 x 2 – y 2 = 18 2x2 – y 2 = 4 x 2 4 – y 2 2 = 1 DËi: x 2 – y 2 = 18 e ̈vL ̈v: awi, Awae„Ë x 2 a 2 – y 2 b 2 = 1 Dc‡K›`a؇qi `~iZ¡ 2ae = 12  ae = 6  a = 6 2  a = 2  2  3 2  a = 3 2  a 2 = 18 Avevi, e = 1 + b 2 a 2  1 + b 2 18 = 2  1 + b 2 18 = 2  b 2 18 = 1  b 2 = 18 Awae„‡Ëi mgxKiY: x 2 18 – y 2 18 = 1  x 2 – y 2 = 18 3. (x + 1)3 + y2 + 6y + 1 = x3 + 1 n‡j, KwYKwUi †K‡›`ai ̄’vbv1⁄4 KZ? [If (x + 1)3 + y2 + 6y + 1 = x3 + 1, what is the coordinate of the center of the conic?]     – 3  – 1 2     – 1 2  – 3     1 2  3     3  1 2 DËi:     – 1 2  – 3 e ̈vL ̈v: x 3 + 3x2 + 3x + 1 + y2 + 6y + 1 = x3 + 1  3x2 + 3x + y2 + 6y + 1 = 0  3(x2 + x) + (y2 + 6y + 1) = 0  3       x 2 + 2.x. 1 2 +     1 2 2 + (y2 + 2.y.3 + 32 ) = 8 + 3 4  3     x + 1 2 2 + (y + 3)2 = 35 4  12 35    x + 1 2 2 + (y + 3)2  4 35 = 1     4 35 Øviv fvM K‡i      x + 1 2 2 35 12 + (y + 3) 2 35 4 = 1  †K›`a     – 1 2  – 3 4. (2, 3) we›`y‡Z 3x2 – 4y2 = 12 Awae„‡Ëi ̄úk©‡Ki Xvj KZ? [What is the slope of the tangent to the hyperbola 3x2 – 4y2 = 12 at point (2, 3)?] 0.5 0.7538 0.9019 2.375 DËi: 0.5 e ̈vL ̈v: 3x2 – 4y2 = 12 6x – 8y dy dx = 0 dy dx = 6x 8y = 3x 4y = 3  2 4  3 = 1 2 Shortcut: †h‡nZz calculator use Kiv hv‡e bv †m‡nZz L, M, N n‡e bv| 5. sinP . sin(P + 30) + cosP . sin(P + 120) Gi gvb KZ? [What is the value of sinP . sin(P + 30) + cosP . sin(P + 120)?] 1 2 3 2 1 2 1 DËi: 3 2 e ̈vL ̈v: sinP . sin(P + 30) + cosP . sin(P + 120) = sinP . sin(P + 30) + cosP . sin(90 + 30 + P) = sinP . sin(P + 30) + cosP . cos(30 + P)
2 = cos(30 + P – P) = cos30 = 3 2 Shortcut: P = 0 a‡i, sin0.sin(0 + 30) + cos0sin(0 + 120) = sin120 = sin(180 – 60) = sin60 = 3 2 6. †Kv‡bv Zi‡1⁄2i ch©vqKvj 1sec Ges Ae ̄’vb 5sint + 4cost 1sec ci Zi‡1⁄2i Z¡iY KZ n‡e? [A wave has period of 1 sec and position of 5sint + 4cost what will be the acceleration after 1 sec?] 16 2 36 2 – 16 2 – 36 2 DËi: – 16 2 e ̈vL ̈v: awi, x = 5sint + 4cost  dx dt = v = 5cost – 4sint  dv dt = a = –  2 5sint –  2 4cost = –   5sint + 4cost = – (2) 2  (0 + 4) t = 1sec Ges  = 2rads–1 = – 4 2  4 = – 16 2 7. tany = 1 x 2 – 1 n‡j, dy dx = ? [If tany = 1 x 2 – 1, then dy dx = ?] 1 x 2 – 1 1 1 – x 2 – 1 1 – x 2 – 1 x 2 – 1 DËi: – 1 1 – x 2 e ̈vL ̈v: y = tan–1 1 – x 2 x = cos–1 x  dy dx = – 1 1 – x 2 1 x 1 – x 2 8. 2sin  32 = ? 2 + 2 + 2 + 2 2 + 2 – 2 + 2 2 – 2 + 2 – 2 2 – 2 + 2 + 2 DËi: 2 – 2 + 2 + 2 e ̈vL ̈v: Shortcut: 2sin  2 n = 2 – 2 + 2 + ... (n – 1) msL ̈K  2sin  2 5 = 2 – 2 + 2 + 2 9. cosx + cosy = 5 2 I sinx + siny = 3 2 n‡j, tan x + y 2 Gi gvb KZ? [If cosx + cosy = 5 2 and sinx + siny = 3 2 then what is the value of tanx + y 2 ?] 1 5 2 3 3 5 5 2 DËi: 3 5 e ̈vL ̈v: cosx + cosy = 5 2  2cos x + y 2 cos x – y 2 = 5 2 ............ (i) [⸪ cosC + cosD = 2cos C + D 2 cos C – D 2 ] sinx + siny = 3 2  2sinx + y 2 cos x – y 2 = 3 2 ............ (ii) [⸪ sinC + sinD = 2sin C + D 2 cos C – D 2 ] (ii)  (i) n‡Z- tan x + y 2 = 3 2  2 5 = 3 5 10. y = (ex ) x 3 n‡j, dy dx = KZ? [If y = (ex ) x 3 , then dy dx =?] e x 4 4ex 4 4ex 4 x 3 4x3 DËi: 4ex 4 x 3 e ̈vL ̈v: y = (ex ) x 3 = e x 4 dy dx = ex 4 . d dx(x4 ) = 4ex 4 .x 3 11. y 2 = 12ax cive„ËwU (3, – 2) we›`yMvgx n‡j, cive„ËwUi Dc‡Kw›`aK j‡¤^i ˆ`N© ̈ KZ GKK? [If the parabola y2 = 12ax passes through the point (3, 2), what is the length of its latus rectum?] 9 4 4 3 4 9 2 3 DËi: 4 3 e ̈vL ̈v: y 2 = 12ax cive„ËwU (3, – 2) we›`yMvgx|  (– 2)2 = 12a  3  36a = 4  a = 1 9  y 2 = 12  1 9  x
3 = 4  1 3  x  Dc‡Kw›`aK j‡¤^i ˆ`N ̈© = |4a| = 4  1 3 = 4 3 GKK 12. lim x   sin(ln1 + aloga x ) x = ? ln1 + a 0  1 DËi: 1 e ̈vL ̈v: lim x  0 sin(0 + x) x = lim x  0 sinx x = 1 [ ln1 = 0 a loga x = x →j‡Mi ag©] 13. x = 2 y †iLvi †Kvb we›`y‡Z ̄úk©‡Ki Xvj x A‡ÿi mgvšÍivj n‡e? [At what point on the line x = 2 y will the slope of the tangent be parallel to the x axis?] (2, 1) (1, 2) (, 0) (0, 0) DËi: (, 0) e ̈vL ̈v: x = 2 y ........(i)  xy = 2 kZ©g‡Z, m = 0  xy1 + y = 0  y1 = – y x  0 = – y x y = 0 (i) bs n‡Z x =   wb‡Y©q we›`ywU (, 0) 14. lim x       x + 5 x + 7 x = ? e 2 e –2 e e –3 DËi: e –2 e ̈vL ̈v: lim x       x + 5 x + 7 x = lim x   x x          x      1 + 5 x x     1 + 7 x x = lim x                 1 + 1 x 5 x 5 5               1 + 1 x 7 x 7 7    lim  x       1 + 1 x x = 0 = e 5 e 7 = e5–7 = e–2 kU©KvU: lim x        1 +  a x 1 + b x x = ea–b 15. Dce„‡Ëi e„nr A‡ÿi ˆ`N© ̈ ÿz`a A‡ÿi mgvb n‡j, Gi Dr‡Kw›`aKZv KZ? [If the length of the major axis of the ellipse is equal to the minor axis, then what is the eccentricity?] 1 2 2 1 2 2 DËi: 2 e ̈vL ̈v: awi, Dce„ËwU x 2 a 2 + y 2 b 2 = 1 [a > b]  e„nr A‡ÿi ˆ`N© ̈ = 2a GKK ÿz`a A‡ÿi ˆ`N© ̈ = 2b GKK cÖkœg‡Z, 2a = 2b  a = b AvqZvKvi Awae„‡Ëi Dr‡Kw›`aKZv, e = 1 + b 2 a 2 = 1 + a 2 a 2 = 2  Awae„ËwU AvqZvKvi Ges Dr‡Kw›`aKZv = 2 16. P Gi gvb KZ n‡j, 4x2 + py2 = 80 Dce„ËwU (0,  4) we›`y w`‡q AwZμg Ki‡e? [For what value of P, will the ellipse 4x2 + py2 = 80 pass through the point (0,  4)?] 8 6 5 4 DËi: 5 e ̈vL ̈v: Dce„‡Ëi mgxKiY, 4x2 + py2 = 80 mgxKiYwU (0,  4) we›`yMvgx n‡j, 4(0)2 + P( 4)2 = 80  16P = 80 P = 5 17. R S P Q PQ Ges RS Gi mgxKiY †KvbwU? [ R S P Q What is the equation of PQ and RS?] x = 0 y =  be x =  ae None of these DËi: y =  be e ̈vL ̈v: RS, PQ Dc‡Kw›`aK j¤^ y =  be [b > a] 18. tan2 x + sec2 x = 3 n‡j, x = ?
4 [If tan2 x + sec2 x = 3 then, x = ?] 2n ±  3 n ±  4 K I L DfqB †Kv‡bvwUB bq DËi: †Kv‡bvwUB bq e ̈vL ̈v: tan2 x + sec2 x = 3  sec2 x – 1 + sec2 x = 3  sec2 x = 2  cos2 x = 1 2  cosx = ± 1 2 = ± cos  4  x = 2n ±  4 19. x = a + ib a – ib n‡j, (a2 + b2 )x2 + (a2 + b2 ) = ? [If x = a + ib a – ib then (a2 + b2 )x2 + (a2 + b2 ) = ?] (a2 – b 2 )x – 2(a2 – b 2 )x 2(a2 – b 2 x) – 2(b2 – a 2 )x DËi: – 2(b2 – a 2 )x e ̈vL ̈v: x + 1 x – 1 = a ib [†hvRb-we‡qvRb K‡i]  (x + 1) 2 (x – 1) 2 = a 2 – b 2  a 2 (x2 – 2x + 1) = – b 2 (x2 + 2x + 1)  (a2 + b2 )x2 + (a2 + b2 ) = – 2(b2 – a 2 )x 20. (2 –  + 2 2 )(2 + 2 –  2 ) = ? 10 9 8 7 DËi: 9 e ̈vL ̈v: (2 + 2 2 – )(2 + 2 –  2 ) = (– 2 – )(– 2 2 –  2 ) = (– 3 )(– 3 2 ) = 9 21. cotx – tanx = 2 n‡j, x Gi gvb KZ? [If cotx – tanx = 2, what is the value of x?] (4n + 1) 4 2n +  8 (4n + 1) 8 me ̧‡jv DËi: (4n + 1) 8 e ̈vL ̈v: cotx – tanx  2  cosx sinx – sinx cosx  2  cos2 x – sin2 x sinx.cosx  2  cos2x  2sinx.cosx  cos2x  sin2x  tan2x  1  2x  (4n  1)  4  x  (4n  1)  8 ; n  z 22.       0 a 2 b a 2 c ab2 0 cb2 ac2 bc2 0 = †KvbwU? [       0 a 2 b a 2 c ab2 0 cb2 ac2 bc2 0 = ?] abc 0 2a3 b 3 c 3 1 DËi: 2a3 b 3 c 3 e ̈vL ̈v: a 2 b 2 c 2       0 b c a 0 c a b 0 = a2 b 2 c 2       0 b c 0 – b c a b 0 [c2 = c2 – c3] = a2 b 2 c 2  a     b c – b c = a3 b 2 c 2 (bc + bc) = 2a3 b 3 c 3 23. Lvov Dc‡ii w`‡K wμqviZ 12N GKwU e‡ji `yBwU Ask‡Ki g‡a ̈ GKwU Avbyf~wgK w`‡K 5N n‡j AciwU KZ? [What is the force of a force of 12N acting vertically upward on one of the two components if the other is 5N horizontally?] 3N 17N 13N 7N DËi: 13N e ̈vL ̈v: P 12 5  P tan90 = Psin 5 + Pcos 5 + Pcos = 0  Pcos = – 5  cos = – 5 P  R 2 = P2 + 52 + 2  P  5 cos  P 2 + 25 + 10Pcos  144 = P2 + 25  10P  – 5 P )  144 = P2 + 25 – 50  P 2 = 144 + 25  P 2 = 169  P = 13 Shortcut: P 2 = 52 + 122 = 25 + 144 = 169  P = 13 24. sec = sec n‡j,  = ? [If sec = sec then,  = ?] n +  n + (– 1)n