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💡 Solution & Explanation

2 2 2 3 3 a P b kt P a 1 b kt P on differentiating 2a dP b k dt P dP bk P dt 2a         AITS-CRT-III (Paper-1)-PCM (Sol.)-JEE(Advanced)/18 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 10 Mathematics PART – III SECTION – A (One Options Correct Type) This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE option is correct. 37. (D) EFM ~ AFD    2 area EFM MF area AFD FD          FD = 2MF …(1)  BF = 2FD Let BD = a then area 2 1 a 3 a EFD a 2 3 2 4 3      Area 2 1 4a 3 2a AFC 2a. 2 3 2 3            So area FED 1 area AFC 8    ….(2) Now a max z z a(1 3)      a min z z a 3 1     2 a max a min z z 2 3 z z             …..(3) from (2) & (3) k = 24. C D F M B A E Locus of z 38. (B)         n n n r=1 r=1 r=1 sinθ 1 sin(r+1)θ-rθ 1 S= tan(r+1 θ-tanrθ) 2cos(r+1)θcosrθsinθ 2sin cos(r+1)θcosrθ 2sin 1 sin tan(n+1 θ-tanθ 2sinθ cos n+1 θ.sinn                                     n 39. (C) Any point on the line through P(a,2)is(a+rcosθ,2+rsinθ) and meets the ellipse 2 2 x y + =1 9 4 at two distinct points A and D 2 2 2 2 2 2 2 2 2 (a +rcosθ) (2+rsinθ) + =1 9 4 (4cos θ +9sin θ)r +4(2acosθ+9sinθ)r+4a =0 4a PA.PD= (4cos θ+9sin θ)    AITS-CRT-III (Paper-1)-PCM (Sol.)-JEE(Advanced)/18 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 11 also line meets the axes at B and C 2a PB.PC= sinθ.cosθ  since PA, PB, PC and PD are in G.P. 2 2 2 2 4a 2a = sinθ.cosθ 4cos θ+9sin θ 2asin2θ+5cos2θ=13 13 -1 1 a 6 or a 6 4a +25        40. (D) n 2 k=1 n k=1 As [(k+3) -(k+2)](k+2)! + 3.3! = 2003.2003! [(k+3)(k+3)!-(k+2)(k+2)!] + 3.3! = 2003.2003! (n+3)(n+3)! = 2003.2003! n 2000       41. (B) 1/2 x 1 2 3/2 2 2 0 1 1 x e sin x dx (1 x ) 1 x 1 x                             x 1 1/ 2 0 2 1 e sin x 1 x           1/2 2 e 1(0 1) 6 3            42. (B), (C) Applying Leibnitz rule, we get f(x) = x – x2 f(x) 2 x f(x) 1 x  1 1 2 2 1 1 2 2 x x dx dx 1 x 1 x        1 8 ln 2 5  43. (A),(B), (C), (D) 2 1 4 tan ,tan tan 2 3 2 2 2 7     B C A AITS-CRT-III (Paper-1)-PCM (Sol.)-JEE(Advanced)/18 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 12 tan .tan 2 3 42 2 2 B C s a s a s      Perimeter=42  2 . . 84 r s cm   tan , tan , tan 2 2 2 A B C  all are less than 1. All angles are acute. 44. (A), (B), (C) (B) (a + c - b)12 + 2c (-1) + (b + c - a) = 0 -1 is root of the equation which is rational. 2nd must also be rational (C) 2 x c  Satisfies given equation which is rational 45. (B), (C) 2 2 dt I 1 e 1 e t 1 e            (t = tanx/2) if 0 < e < 1, 1 e 0 1 e    , So, (B) is correct If e > 1, 1 e 0 1 e    So, (C) is correct. 46. (A), (B), 1 1 2 1 0 4 3 5 5 5 x y z        47. (A), (C) ,(D) dy Y y (X x) dx    y dy A x , 0 , B(x, 0) dx              A B O y dy x 2x, y x dy dx dx     dy dx y x  AITS-CRT-III (Paper-1)-PCM (Sol.)-JEE(Advanced)/18 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 13 LN |Y| = LN|X| + C CURVE IS PASSING FROM (1, 1), C = 0 XY = 1 48. (A), (B), (D) Multiply 1 2 3 R , R , R respectively by a, b, c and take out a, b, c common from 1 2 3 C ,C ,C respectively. 2 2 2 2 2 2 2 2 2 a 1 a a D b b 1 b c c c 1      Apply 1 2 3 R R R   and take 2 1 a  common   2 2 2 2 2 2 2 1 1 1 D 1 a b b 1 b c c c 1      Make two zeros by 2 1 3 1 C C , C C     2 2 2 D a b c 1 1 given      2 2 2 a b c 0      c  49. (A), (B), (C),(D) GIVEN   1 2 3 1 2 3 1 a b c aP +bP +cP =K y= + + 2 P P P  IS MINIMUM. WHEN   1 2 3 1 2 3 1 a b c y = aP +bP +cP + + 2K P P P       IS MINIMUM. BUT , 2 2 2 3 3 1 2 2 1 2 1 3 2 3 1 P P P P P P 1 y= a +b +c +ab + +bc + +ab + 2K P P P P P P                             2 2 2 1 a +b +c +2ab+2bc+2ac 2K    2 3 3 1 2 2 1 2 1 3 2 3 1 a+b+c P P P P P P y when = = = = = 2K P P P P P P   =1 1 2 3 i.e, when P =P =P  P is incentre of ΔABC  SECTION – C 50. (1) Let AD, BE & CF are medians M1, M2 & M3 respectively Clearly 2 2 2 1 1 M AD 2(b c ) a 2     2 2 2 2 1 M 2(a c ) b 2    2 2 2 3 1 M 2(a b ) c 2    AITS-CRT-III (Paper-1)-PCM (Sol.)-JEE(Advanced)/18 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 14    2 2 2 2 2 2 1 2 3 3 a b c M M M 4      Now apply AM > GM on 2 2 2 1 2 3 M ,M ,M  2 2 2 2/3 1 2 3 3 (a b c ) 4 (M M M ) 3    since a > b > c  1 2 3 M M M    1 2 3 M ,M ,M are in G.P.  1 2 3 M M M = (M2)3  M2 = 1 2 so          2 3 2 2 2 3 a b c 1 4 2     2 2 2 a b c 1 51. (5) The given expression can be re written as       2 2 ,s ,r,q , p x x 1 x 2     where 0  x  1 consider the function f(x) = 2x2 + ( 1-x)2 , 0  x  1 So f (x) = 4x – 2 ( 1 – x) For critical points f(x) = 0 or x = 1/3 Thus set of critical points is   1, 3 1 ,0 Now f(0) = 1 , f       3 1 = 3 2 , f(1) = 2 Thus maximum value of f(x) is 2 and minimum value of f(x) is 2/3. Therefore maximum value of given expression is 4  2 = 8 and minimum value is 4 ( 2/3) = 3 8 52. (9) Let x1, x2, x3, x4,x5 are the no of objects obtained by different person Then x1+ x2 + x3 + x4 + x5 = 21 1 x1 < x2 < x3 <x4 < x5 Let y1 = x1 y2 = x2 – x1 y3 = x3 – x2 y4 = x4 – x3 y5 = x5 – x4  5y1 + 4y2 + 3y3 + 2y4 + y5 = 21 5t5 + 4t4 + 3t3 + 2t2 + t1 = 6 (ti = yi – 1, 2,3,4,5) t1, t2, t3, t4, t5  0 the no. of solution of above equation is Coeff. of x6 in ( 1+ x5 + x10 + …) (1+ x4 + x8 + ….) (1+x3 + x6 + …) ( 1+ x2 + x4 + …..) ( 1+ x+ x2 + ….) =Coeff. of x6 in (1+x5) (1+x4) ( 1+x3 + x6) ( 1+ x2 + x4 + x6) ( 1+ x+ x2+ …..+ x6) AITS-CRT-III (Paper-1)-PCM (Sol.)-JEE(Advanced)/18 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 15 =Coeff , of x6 in (1+x4 + x5) ( 1+ x3 + x6) (1+ x2 + x4 + x6) ( 1+x+ x2+ …. +x6) = Coeff. of x6 in ( 1+ x3 + x4 + x5 + x6) ( 1+ x2 + x4 + x6) (1+x + x2 + … + x6) = Coeff. pf x6 in ( 1+ x2 + x3 + 2x4 + 2x5+ 3x6) (1+x+ x2+ … + x6) = 9 Now the variables x1, x2, …., x5 can permute in 5! ways  Total no of ways = 9 5! = 1080 53. (1) 3 3 2 2 2 1 - αx 1 - αx 1 - α 1 1 1 x x x α α α Roots of the equation ax +x-1-a = 0 are 1,α,β 1+a α + β = -1 and αβ = a (1 + a)x - x - a (x-1)[(1 + a) x + ax + a] [aαβx + ax + a] lim = lim lim (e -1)(x - 1) (e -1)(x - 1) (e       x 1 - αx 1 1 x x α α -1) a(1 - αx)(1 - βx) a(α - β) lim lima(1 - βx) = α (e -1)     54. (4) , ,  are the roots of equation x3 + px2 + qx – q = 0 …..(i) then equation whose roots are 1 1 1 , ,  will be qx3 – qx2 – px – 1 = 0 1 after replacing x by x       …..(ii) roots of this equation , ,  are in A.P. 2 =  +  also  +  +  = 1 3 = 1   +   +   p q   1 ' 3  1 ' ' ' q  Put 1 x 3  in equation (ii) 9p + 2q = 27 Using 2 2 2 2 1 1 1 1 1 1 1 1 1 2                              Now 1 1 1 1 1 1 also . 3         So 2 2 2 1 1 1 1 1 1 1 1 2                        , 2 2 2 2 1 1 1 2 1 1 2                Now let 1 1 d 3    , 1 1 d 3    , 2 2 2 1 1 1 2 1 1 1 2 d d 9 3 3                          2 2 2 2 1 1 1 3 1 2 d 9               , 2 2 2 2 1 1 1 1 2d 0 3         , so 2 2 2 1 1 1 1 3      