solutions http://www.genesismentors.com 1.Let n be total days then number of days with no rain = n-11 = days with no rain in aft & morn of the 11 let b days have rain in morn. Therefore b days have clear after also 11-b days will have rain in aft. Therefore 11-b days have clear mornings thus(n-11)+b = 12 => (n-11)+(11-b) = 9 Solve to get n =16 or n-11 = 5 2.Just number the boxes as #0, #1, #2 #9 and now take 0, 1, 2 9 balls respectively from each box in that order. Now weigh all the 45 balls together in one weighing. Total weightwillbe(45a)kg.acaneasilybecalculated fromtheweightobtained.Nowboxwithlighterballsis 10a. Hence only one weighing is sufficient. 3.Besttechniqueforsuchquestionsistogobyoptions. Takeoption(a):P(1)=0;option(b):P(-1)=-12; option (c): P(2) = 0; Option (d): P(3) = 20. 4.(x + a) (x + 1991) + 1 = 0; (x + a) (x + 1991) = -1 Since the factors are integral, we have two cases I.(x + a) = 1 and(x + 1991) = -1 x = -1992 or a = 1-x = 1993 II.(x + a) = -1 and(x + 1991) = 1 x = -1990 or a = -1-x = 1989. 5.Put x = -1. We get 2f(1) =-3 => f(1) = -3/2 6.S = ..... 670 10 670 10 + + + +S S + + = 670 10 Squaring both the side we get S S + = 670 102 From option put the value. S=6 satisfies the equation 7.I.graphofcirclecenteredattheoriginwithequation 162 2= + y xsymmetricaboutthey-axiseven function II. ) ( ) ( x f x f = symmetricabouttheoriginodd function III.) ( ) ( x f x f = axisofsymmetryatx=4neither even nor odd IV. ) ( ) ( x f x f = not an even function 8.Descartes rule of signs: # of sign changes for positive x = 4 -> 4, 2 or 0 possible positive real roots #ofsignchangesfornegativex=2->2or0possible negative real roots 8 roots total -> possible combination could be 4 positive, 0negative,4non-realrootsor0positive,0negative,8 non-real. 9.Brijeshsequation:18 3 ) 6 )( 3 (2 = + x x x x ; Manishsequation:10 9 ) 10 )( 1 (2 = + x x x x Correct equation: 5 , 2 0 10 32 = = x x x 10. Apply AM-GM concept. Ans 72. 11. Givenfraction 80 5 40 20 1015 + +=5 4 5 10 2 5 2 1015 + += 5 105 10.5 1055 3 10 315++== = ) 1 2 ( 5 5 10 + = + 12.) 2 12 17 ( + 2 2 3 ] 2 2 . 3 . 2 ) 2 2 ( 3 [2 2+ = + + =1 2 ) 2 2 3 (4) 2 12 17 ( + = + = + 13.0 = + + b a a c c b .Henceonerootis1.Alsoasroots are equal, other root will also be equal to 1. Also c bb a= | o. c bb a= 1 . 1 c b b a = c a b + = 2 14.If R is the number right then the score is R (30 R)/4 = 5(R 6)/4. Thus the answer is a multiple of 5/4. Note 8.75 = 7(5/4). 15.Since0 0 0 ) 0 ( < = + + = c c f Rootswillbeofopposite sign,ve b = = + | o(b > 0) It is given that| o < . So,ve = + | ois possible only when| o > | | | o | o > > < | | , 0 , 0| | 0 o | o < < < 16.f(x) = x3 4x + pf(0) = p and f(1) = p 3; p 3 < p As p and p 3 are of opposite signs, p 3 < 0 and p > 0. p < 3 and p > 0; 0 < p < 3 17.0 2 32> + x xor0 ) 2 )( 1 ( > x x) , 2 ( ) 1 , ( e x ..(i) Again 0 4 32s x xor0 ) 1 )( 4 ( s + x x] 4 , 1 [ e x..(ii) Fromeq.(i)and(ii),] 4 2 ( ) 1 , 1 [ e x 1 1 < s x or 4 2 s < x 18.We have,0 1 ) )( ( = b x a x0 1 ) )( ( > = b x a x0 ) )( ( > b x a x[b > a] [ , ] [ , ] + e b a x,i.e.) , ( a and (b, ). 19.Sum of the reciprocals= -(-450)/27 = 50/3 20.Take a = 1 and b = 2 and match the options. Option d 21.Let'scallthefirsttermoftheoriginalgeometricseries a and the common ratio r, so 2005 = a+ar+ar2+.. Using thesumformulaforinfinitegeometricseries,we have a/(1-r)=2005. Thenwe formanew series, a2+a2 r2+ a4 r4.We know this series has sum 20050 = a2/(1-r2). Dividing both this equation,wegeta/(1+r)=10.Then a=2005-2005r and a=10+10rso2005-2005r=10+10r,2015r= 1995 and finally r = 1995/2015 22.y0=y0(x2 + 2x + 9) = x2 2x + 9 x2(y0 1) + 2x(y0 + 1) + 9(y0 1) = 0 The equation has a real solution if discriminant 0 4(y0 + 1)2 4(y0 1)2 . 9 0 (y0 + 1)2 9(y0 1)2 0 9 29 222+ ++ x xx x21 ++ b a ++ solutions http://www.genesismentors.com (y0 + 1 + 3y0 3)(y0 + 1 3y0 + 3) 0 (4y0 2)(-2y0 + 4) 0 (y0 2) 0; 1/2 y0 2Range = 23. 16 ) 5 ( 9 25 ) 5 ( 9 10 ) (2 2 2 2 2 4 = + = + = x x x x x f. Minimum value = -16 24.Rewriteequationasy(5y+2x)=80,soyisadivisorof 80.Moreover,ymustbeeven,forifywereoddthen both terms in the product would be odd, which is clearly impossible. Additionally, y cannot be divisible by 16, since this would force x to be non integral; but all other choices ofyyieldintegralvaluesforx.Hencey{2,4,8,10, 20, 40, -2, -4, -8, -10, -20, -40} 25.It is given to us that the sum of all the numbers of the AP is 4500 and the APs have the same middle term. We can seethatthesumofanAPwithnterms,wherenisodd will be n times the middle term (e.g. a 1 + a + a + 1 = 3a) Since there are 9 terms in the APs, hence the sum of each AP will be 9X. Also, as there are 5 APs, the sum of all these will be 5 9X = 45X = 4500 Thus, X = 100 26. 27.From ca=b2a we get c = b2. From 2c=2.4a => 2c=22a+1 or c =2a+1.Combiningresultswegeta=(b2-1)/2and therefore (b2-1)/2 +b +b2 = 16 or 3b2+2B33=0.Thus(3b+11)(b-3)=0.Butwhenb=3,c=9,anda=4, abc>0 Therefore b = -11/3, c = 121/9, a = 56/9 and 9a -6b +9c = 199. 28.Note that 0 = (x+y +z)2 = x2 +y2 +z2 +2xy +2yz +2zx. Rearranging,wegetthatxy+yz+zx=-1/2*(x2+y2 +z2), so that in fact the quantity is always equal to -1/2. 29. 30.Required area = (1/2)x2x2 sq. units = 2 sq. units 31.Put x = 1 we geta0 + a1 + a1 + + a15 = 45 = 1024, Put x = -1 , we get a0 - a1 + a2 - - a15 = 0 Adding both the equation 2(a0 + a2 + a4 + + a14) =1024(a0 + a2 + a4 + + a14) =1024/2 = 512. 32.Put x2 = 1 in the expression we get150 -2.125x + 1 = 2-2x 33.Let P = (1 + x) (1 + x2) (1 + x4) (1 + x8) (1 + x16) (1 + x32) (1 + x64). Multiplying (1-x) both the side we get(1-x)P = (1-x)(1 + x) (1 + x2) (1 + x4) (1 + x8) (1 + x16) (1 + x32) (1 + x64) = (1-x2) (1 + x2) (1 + x4) (1 + x8) (1 + x16) (1 + x32) (1 + x64) = (1 - x4) (1 + x4) (1 + x8) (1 + x16) (1 + x32) (1 + x64) and so on we get(1-x)P = (1 x128) or P = (1 x128)/ (1-x) 34.a2 + b2 + c2 + d2 = (a + b + c + d)2 2(ab + ac + ad + bc + bd + cd) = 40. For d to be greatest, a, b and c must be least and equal. So let a = b = c = x => x = 83d and 3283d | | |\ .+ d2 = 40 => d2 4d 14 = 0 => dmax= 2 + 32 35.f(n) = 24 4 12 1 2 1n nn n+ + +

= ( ) ( ) ( ) ( )( ) ( )2 22 1 2 1 2 1 2 12 1 2 1n n n nn n+ + + + + + = ( ) ( )( ) ( )3 32 22 1 2 12 1 2 1n nn n+ + =( ) ( )3 32 1 2 12n n + So f(1) + f(2) + f(3) + + f(40)=( ) ( )3 32 40 1 2 1 12 + = 729 12=364 36.Let 1x yx+ =and 331x zx+ =,sotheexpression becomes 6 233y zy zy z= + = 3y = 31xx| |+ |\ .. By AM-GM inequalityweknowthat 12 xx| |+ > |\ ..Henceminimum value of the expression is 6. 37.For x = y = 0, we have F(0)[F(0) 1] = 0 => F(0) = 0 or 1. For x = 2010 and y = 0, we have F(0)[F(2010) 1] = 2010 => F(0)=0and F(0)=1hence F(2010)=2010+1= 2011 38.a2 4(b + 2) = a2 + b2 (b2 + 4b + 8) > 0. So a2 + b2 > b2 + 4b + 8 => a2 + b2 > (b + 2)2 + 4 => a2 + b2 > 4. Hence the required minimum value is 4 |.|

\|210y((

2 ,21 solutions http://www.genesismentors.com 39. P(x)=(x19200x18)(x18200x17)+(x17 200x16) - ... (x2 200x) + x. So P(201) = 200 40.By given two equation we find that a + b + c = abc = 1 Now M =1 1 11 1 1 a ab b bc c ca+ ++ + + + + +(1)

=> M = 1 1 11 (1/ ) 1 (1/ ) 1 (1/ ) a c b a c b+ ++ + + + + + => M = 1 1 1c a bc ac a ab b bc+ ++ + + + + + ..(2) => M =1 (1/ ) 1 (1/ ) 1 (1/ )c a bc b a c b a+ ++ + + + + +=> M = 1 1 1bc ca abb bc c ca a ab+ ++ + + + + + ..(3) Adding (1), (2) and (3), we get3M = 3 => M = 1 41.Let S = 1 1 2 3 5 8 13 21...4 8 16 32 64 128 256 512+ + + + + + + + So S = 1 1 2 3 5 8 13 21...8 16 32 64 128 256 512 1024+ + + + + + + + Subtracting the two we get, S = 1 1 1 2 3 5 8...4 16 32 64 128 256 512+ + + + + + + =1 14 4S +Hence S = 1 42.Total number of houses = 39 + 66 1 = 104 43.Total numbers should a number of the form 4k + 1 near 80. So most probable such number is 81. Sum of first 81 naturalnumbersis3321.Sumofgiven80numbersis 40.75*80 = 3260, so number 61 was removed. 44.Firstfindthecriticalnumbersofthefunction. 3 2( ) 19 f x x x x = + + ,sothatmeansthat 2'( ) 3 2 1 f x x x = + . '( ) (3 1)( 1) f x x x = +,thecritical numbers are 1, 13x = . To determine which is the relative minandwhichistherelativemax,usethosevaluesin the second derivative test''( ) 6 2 f x x = +. Testing1 x = , we get -4. This is less than 0, and therefore must be the relativemaximum.Testingat 13x =,weget4,whichis greaterthan0.Sothisisourrelativeminimum.Sonow we know that 13A = and1 B = . So3BA = . 45.There are 73 students between 7 and 81; therefore, there must also be 73 students between 81 and 7.Six of these arestudents1through6,whiletheremaining67are students 82 through 148. 46.Using the formulas for the sum and product of the roots, we have 10/3 c =8/3c, which gives c = {4/3, 2}. Since c is an integer, c = 2, and ab = 8/3x2 = 4/3. 47.Outof128only1winnerhence127teamwillloss therefore number of matches to be played = 127 48.Sincethisiscubicfunction,takingdifferenceof consecutive terms 3 times will result in equal difference. -1 -5 -5 5 31f(6) -40 1026f(6)-31 410 16 f(6)-57 6 6f(6)-73 Thus f(6)-73 = 6 or f(6) = 79 49. 23x 31 A Bx 3 x 5x 2x 33x 31 A(x 5) B(x 3)Let x = -5 B = -2Let x = 3 A = 5-2 5 = -10+= + ++ + = + + 50. ( ) ( )6 2 5 A B226 2 5 A B6 2 5 A 2 AB B6 A B2 5 2 AB 5 ABA AB B 6 5 11 = = = += + = =+ + = + = 51.Eachrootwillbereducedby5.Hencetherootsofnew equation M(x+5) = {-5, -3, -8} 52.We have {} +{-} = (-3) + (--(-4)) = 4-3 = 1 53.{x} is a proper fraction hence -1 f(-3) = 2x(-3)+5 = -1 => f(-1) = 5 => f(5)= 52 1 = 24. 65.Sum of the coefficients of f(x) is at x= 1. Given f(3x-5) = x4- x3 -3x +7, putting x = 2 we get f(1) = 24- 23 -3x2 +7 = 9. 66.The constant term = f(0) ; putting x = -2 we getf(0) = -13 67.The first common term = 21 Thecommondifference=LCMofcommondifferenceof the two series i.e. 4 and 5= 20.Lettherebentermscommonhencetheseriesof common term has first term a = 21 common difference d = 20. The last term = 21 + (n-1)x20 417 => n = 20. 68.xcannotfractionnotnegativeintegerandcannotbe more than 3. Put x = 1, 2 and 3 and check the equation only x = 3 satisfies. Hence only one solution. 69.Ifallthepolynomialshadrealroots,theirdiscriminants would all be nonnegative: a24bc; b24ca; c24ab Multiplyingtheseinequalitiesgives(abc)264(abc)2,a contradiction.Henceoneofthequadraticshasnoreal roots.Themaximumof4realrootsisattainable:for example,the values(a;b;c)=(1; 5;6)give-2;-3as roots to x2 + 5x + 6 and -1/2; -1/3 roots to 6x2 + 5x + 1. 70.Solution: ( )( ) 2 112F nF n++ = f(n+1) f(n) = 1/2 f(2) f(1) = 1/2 f(3) f(2) = 1/2 . . f(101) f(100) = 1/2 adding we getf(101) - f(1) = 50 = > f(101) = 52. 71.Substitute x with 2/x we get 24 2 0 cx bx a + + =. 72. x2 + 6x + y2= 4 - (x+3)2 + y2 = 13 Also 13 = 4 + 9 => x+3 = 2, y = 3 or x+3 = 3, y = 2. Total 8 solutions 73. x6+x3-2 =0 x3 = 1, -2 or x = 1, -21/3 Hence two points. 74.Cyclicity is observed, take p = q = r = 2 andx = y = z = 1. We get the value of expression = 1. 75.Forax2+bx+c,a>0theminimumvalueisatx=-b/2a.Comparing at putting the value we havea = -p/8 => p = -8a. put in option and check. We get 15a + 2p + + >+ +abb aabb aabb a Given 2a +3b+ 1/6ab = 3 that means AM = GM hence all terms should be equal i.e.2a = 3b = 1/6ab = 1 a = 1/2 , b = 1/3. Putting in a2 + b2 +ab= 19/36 77. q p qpx+=24applyingcomponendoanddividendoweget p qp qp xp x+=+ 344Similarly we get q pq pq xq x+=+ 344 Therefore23 34444=+++=+++q pq pp qp qq xq xp xp x 78.Subtractthetwoequationwegetx45x2+4=0. Solving for x we have x = 1, 2 Forall4valuesofxthecorrespondingvalueofyis positive hence answer is 4. 79.Clearly,since (0,0) isanintercept, e mustbe 0.Butif d was0, x2 woulddividethepolynomial,whichmeansit would have a double root at 0, which is impossible, since all five roots are distinct. 80.wehave a+b+c=0,andsothedesiredanswer is (a+b)3+(b+c)3+(c+a)3=-(c3)-(a3)-(b3).We knowthatifa+b+c=0then,a3+b3+c3=3abc Therefore ( a3 + b3 + c3) = -3abc = -3x(-2008)/8 = 753