BANSAL CLASSES - Doubtion

FINFINAAL L PRPRAACTCTICE ICE PRPROBOBLLEMEMS FOS FORR

IIIT IT JJEEE-E-20200707(With Hin(With Hints and Solutions at ts and Solutions at the End)the End)

ALL ALL THE THE BEST BEST FOR FOR JEE JEE -2007-2007

Advise :Advise : Do not spend more than 10 minutes for each problem and then read Do not spend more than 10 minutes for each problem and then read the solution and then do it.the solution and then do it.

XII & XIIIXII & XIII

M M AA T HT H EEM M AA T IT IC SC S

BANSAL CLASSESBANSAL CLASSESTTARGET ARGET IIT IIT JEE JEE 2007 2007

Q.1 to Q.1 to Q.29 Q.29 are are of 6 of 6 Marks Marks ProblemsProblems Q.30 to Q.30 to Q.66 Q.66 are are of 8 of 8 Marks Marks ProblemsProblems Q.67 to Q.67 to Q.82 Q.82 are of are of 10 Marks 10 Marks ProblemsProblems Q.83 to Q.1Q.83 to Q.100 are Objective type problems.00 are Objective type problems.

B B ansal ansal C C llaasssseess Problems fo Problems for JEE-20r JEE-2007 07 [2] [2]

SUBJECTIVE:SUBJECTIVE:

Q.1Q.1 If the If the susum m of of ththe e roroots ots of of ththe e eqequauatitionon 11222222 22xx22222211xx11111122xx333333 is expressed in the is expressed in the formform22

11

SS

SS find find

SS11 + + SS22, where, where22

11

SS

SS is in i is in its lowest ts lowest form.form. [6][6]

Q.2Q.2 Let K iLet K is a posis a positivtive ine integer steger such uch that 36 + Kthat 36 + K, 300 + K, 300 + K, 596 + K a, 596 + K are the sre the squarquares of es of threthree conse consecuecutivtiveeterterms of an ms of an arithmetic proarithmetic progression. Find K.gression. Find K. [6][6]

Q.3Q.3 FiFind thnd the nue numbmber of er of 4 di4 digigit numt numbers sbers startintarting wig with 1 ath 1 and hnd haviaving eng exactlxactly twy two ideo identintical cal didigitsgits.. [6][6]

QQ.4.4 A A cchhoorrd od of f ththe pe paarraabboolla ya y22 = 4ax touches the parabo = 4ax touches the parabola yla y22 = 4bx. Show that the tangents at t = 4bx. Show that the tangents at the extremihe extremitiestiesof the chord meet on the paraboof the chord meet on the parabola byla by22 = = 4a4a22x.x. [6][6]

Q.5Q.5 ConsConsiider a cider a circlrcle S wie S with centh centre at the oritre at the origin gin and raand radiudius 4. Fs 4. Four cirour circlecles As A, B, C a, B, C and D end D each wiach with radth radiiususunity and centres (–3, 0), unity and centres (–3, 0), (–1, 0)(–1, 0), (1, 0, (1, 0) and () and (3, 0) respectively are drawn. 3, 0) respectively are drawn. A chA chord PQ ord PQ of the circlof the circle Se Stouctouches the circle B and passes through thes the circle B and passes through the centre of the he centre of the circle C. If the length of this chord can becircle C. If the length of this chord can be

expressed asexpressed as xx , find x., find x. [6][6]

QQ..66 IInntteeggrraattee xx

xx

77

22 5511(( ))dxdx [6][6]

QQ..77 IIff

22

0022

dxdx))xx22sisinn11((

xx22sisinn11 = =

b b

aa where a, b ar where a, b are relate relatively ively prime find a + b + ab.prime find a + b + ab. [6][6]

Q.8Q.8 A A bus bus contraccontractor agrees tor agrees to run sto run specipecial al busbuses fes for the eor the empmployloyees ees of of ABABC C Co. LCo. Ltd. He atd. He agrees grees to run thto run thee buses buses if if atleatleast ast 200 200 persons persons travel travel by by hihis s busesbuses. . The The fare fare per per person person is is to be to be Rs. Rs. 10 p10 per er day day if if 200 200 traveltraveland will and will be decreased for evbe decreased for everybody by erybody by 2 paise per perso2 paise per person over 200 that n over 200 that travels. How many passengerstravels. How many passengerswill will give the contgive the contractractor maximum dailor maximum daily revenue?y revenue? [6][6]

Q.9Q.9 If If ththe pe poioint nt P(P(a, a, b) b) lliies es on on ththe ce cururve ve 9y9y22 = = xx33 such that the normal to the curve at such that the normal to the curve at P mP makes equal interceptsakes equal interceptswith the axes. Find the value of (a + 3with the axes. Find the value of (a + 3b).b). [6][6]

Q.10Q.10 Let xLet x(t) be (t) be the cthe concentrationcentration of on of gluglucose per cose per uniunit volt volumume of e of blblood at timood at time t,e t, being the amount of glucose being the amount of glucose being being injinjected per uniected per unit volumt volume per unie per unit time. t time. If If the glucose ithe glucose is dis disappearing fsappearing from the rom the blood at a ratblood at a ratee proport proportionional al to to the the conconcencentratitration on of of glglucosucose e (K (K bebeining g the the conconstanstant t of of proportproportionionalalityity), ), fifind nd x(t)x(t). . AAllso so ffinind d the ultimthe ultimate coate concentration of gluncentration of glucose as tcose as t .. [6][6]

QQ.1.111 FFiinnd td thhe ve vaallueue(s) of the parameter 'a' (a > 0) (s) of the parameter 'a' (a > 0) for each of which the area of the fifor each of which the area of the figure bounded by thegure bounded by the

strastraight line, y =ight line, y =aa aa xx

aa

22

4411

& the parabola y & the parabola y ==

xx aa xx aa

aa

22 22

44

22 33

11

is is the greatest.the greatest. [6][6]


Q.12Q.12 MrMr. A . A is is a coma compulpulsisive lve liaiarr. He l. He lieiess 5522 of the time. However a clue to of the time. However a clue to his valhis validity iidity is that his ears dros that his ears droopop

3322 of t of the time when he is tellhe time when he is telling a lie. They onling a lie. They only drooy droopp 101011 of the t of the time when hime when he is telline is telling the tg the truth.ruth.

Mr. A tells hiMr. A tells his friend Ms friend Mr. B r. B that "that "certain event has occured" certain event has occured" and hiand his ears wes ears were droppre dropping ing as noticed byas noticed by

Mr. B. Find the probability that Mr. A was telling the truth.Mr. B. Find the probability that Mr. A was telling the truth. [6][6]

Q.13Q.13 FiFive persons entered the ve persons entered the lilift cabin on the ft cabin on the ground floor of ground floor of an eighan eight t flfloor oor house. house. Suppose Suppose that that eacheachof them of them , , independentindependently ly & & with equal with equal proprobabilibability ty can can leave the leave the cabin cabin at any at any flfloooor r beginning beginning witwith h thethefirst, find first, find out tout the probability he probability of all of all 5 5 persopersons leavins leaving at diffng at different floorerent floors.s. [6][6]

QQ..1144 LLeett vvand and uu

be non zero vectors on a plane or in 3-space. Show that the vector be non zero vectors on a plane or in 3-space. Show that the vector uu||vv||vv||uu||ww

bi bisects sects the the anglangle e betweenbetween vvand and uu

.. [6][6]

Q.15Q.15 FiFind nd the the didistanstance ce frfrom om the the llinine e x x = = 2 2 + + t t , , y y = = 1 1 + + t t , , z z == tt22

11

22

11 to t to the plane x + 2y + 6z = 10.he plane x + 2y + 6z = 10.

[6][6]

QQ..1166 IIff is the angle between the lines iis the angle between the lines in which the pln which the planes 3x – 7y – 5z = 1 and 5x – 13y + 3z + 2 = 0 cutsanes 3x – 7y – 5z = 1 and 5x – 13y + 3z + 2 = 0 cutsthe plane 8the plane 8x – 11y + 2z = 0, x – 11y + 2z = 0, find sinfind sin.. [6][6]

Q.17Q.17 Suppose Suppose u, v u, v and and w w are are twictwice e difdifferenferentiabtiable le funfunctionctions of s of x x that sthat satisfatisfy y the the relrelations ations au + bv + au + bv + cw = 0cw = 0

where a, b and c are constants , not where a, b and c are constants , not all all zero. Show tzero. Show thathat''''ww''''vv''''uu''ww''vv''uu

wwvvuu

= = 0.0. [6][6]

Q.18Q.18 In aIn any ny triatriangngle le ABABC, pC, prove rove that, that, cos cos A A · sin· sin22

22

AA + cos B · + cos B · sisinn22

22

BB + cos C · + cos C · sisinn22

22

CC

88

33.. [6][6]

Q.19Q.19 If If ththe ne normormalals to ths to the ce cururve ve y y = x= x22 at t at the points Phe points P, Q and R , Q and R pass through the pass through the pointpoint

22

33,,00 , find the radius, find the radius

of the of the circle circumcircle circumscribing the trscribing the triangliangle PQR.e PQR. [6][6]

QQ.2.200 LLeet A t A = = {{aa R | the equation (1 + 2 R | the equation (1 + 2ii)x)x33 – 2(3 + – 2(3 + ii)x)x22 + (5 – 4i)x + 2a + (5 – 4i)x + 2a22 = = 0}0}

has at least one real roohas at least one real root. Fint. Find td the value ofhe value of AAaa

22aa .. [6][6]

Q.21Q.21 FiFind the equatind the equation of a lion of a line passne passining through (– 4, –2) havig through (– 4, –2) having equang equal il intercepts on the coordintercepts on the coordinate axnate axeses..[6][6]

Q.2Q.222 LeLet S bt S be e the the seset of t of alall l x sx sucuch h thathat t xx44 – 10x – 10x22 + + 99 0. Find t 0. Find the maximum value ofhe maximum value of f f (x) = x (x) = x33 – 3x on S. – 3x on S.[6][6]

Q.23Q.23 SolSolve ve the the didiffffereerentintial al equequatiation, on, (x(x44yy22 – y)dx + (x – y)dx + (x22yy44 – – xx))ddy y = = 00 ((yy((11) ) = = 11)) [6][6]


Q.24Q.24 AlAll the fl the face cards from a ace cards from a pack of 52 playpack of 52 playining cards are remg cards are removed. From oved. From the remthe remainainining pack halg pack half of thef of thecards are cards are randomly remrandomly removed withouoved without lookt looking at them and then randomly ing at them and then randomly drawn two cadrawn two cards simultaneouslyrds simultaneously

from the remainifrom the remaining. If the prong. If the probabilbability that two ity that two cards dracards drawn are botwn are both aces ish aces is22

20202020

40402020

3838

CC··CC

))CC(( p p, find p., find p. [6][6]

Q.25Q.25 A A cicirclrcle ine intertersesects acts an eln elllipipsese 22

22

22

22

b b

yy

aa

xx = 1 pr = 1 preciselecisely at ty at three points A,hree points A,

B, C as shown in the figure. AB is a diameter of the circle and isB, C as shown in the figure. AB is a diameter of the circle and is perpen perpendicdiculaular to r to the the majmajor axor axis is of of the the ellellipsipse. e. If If the the ecceneccentricitricity ty of of thetheellipse is 4/5, find the length ellipse is 4/5, find the length of the diameteof the diameter AB ir AB in terms on terms of a.f a. [6][6]

Q.26Q.26 Suppose R iSuppose R is set of reals set of reals and C is and C is the set of comps the set of complex nlex numbumbers and a fers and a functiunction is defon is definined as f : Red as f : RC,C,

f (t) =f (t) =ii

ii

tt11

tt11

where t where t R, prove that R, prove that f f is injective. is injective. [6][6]

Q.27Q.27 CirCirclecles s A A and B and B are exare externalternally ly tangentangent to each other t to each other and to land to linineet t . The sum of the radii of the t. The sum of the radii of the two circleswo circlesis 12 and tis 12 and the radius of circle he radius of circle A iA is 3 ts 3 timimes that es that of circle B. The area in between the of circle B. The area in between the two two circles ancircles and itsd its

external tangent isexternal tangent is 33aa ––22

b b then find the value of a + then find the value of a + b.b. [6][6]

Q.Q.2828 DeDeffiine ne a ma matatririx x A A ==

00331100

. Find a vertical vector. Find a vertical vector VV

such that (A such that (A88 + + AAAA66 + + AAAA44 + + AA22 + I) + I)VV

= =

111100

where I is a unit matrix of order where I is a unit matrix of order 2.2. [6][6]

Q.29Q.29 A A circlcircle is ie is inscribnscribed in a trianed in a triangle wigle with sides of lth sides of lengths 3, 4 and 5. engths 3, 4 and 5. A A second circlsecond circle, interie, interior to or to tthe trianglhe triangle,e,is tangent to is tangent to the first circlthe first circle and to e and to both sides of the larger acutboth sides of the larger acute angle of the triangle. If the re angle of the triangle. If the radius of tehadius of teh

second circle can be expresecond circle can be expressed in the formssed in the formwwcoscos

k k sinsin where k and w ar where k and w are in degrees and lie ie in degrees and lie in the intervaln the interval

(0, 9(0, 90°), find the value o0°), find the value of k + wf k + w.. [6][6]

QQ.3.300 If If ththe ee eququatatiionon11xx

b bxx2424axax22

22

= x, has exactly two distinct = x, has exactly two distinct real solutreal solutions and their sum is 12 then find ions and their sum is 12 then find

the valthe value of ue of (a – b).(a – b). [8][8]

Q.31Q.31 If a, b, c anIf a, b, c and d are positive id d are positive integers and a < b < ntegers and a < b < c < d such that a, b, c are in c < d such that a, b, c are in A.PA.P. and b, c, d a. and b, c, d are in re in GG.P.P..and d – a and d – a = 30. Find the four numbers.= 30. Find the four numbers. [8][8]

Q.32Q.32 Let the set A Let the set A = {a, b, c, d, e} and P = {a, b, c, d, e} and P and Q are two non empand Q are two non empty subsets of ty subsets of A. FiA. Find the numnd the number of wayber of ways s ininwhiwhich P and Q can ch P and Q can be selected so that be selected so that PP Q has at least o Q has at least one common elemne common element.ent. [8][8]

Q.33Q.33 If If the the nornormmalals ds drawrawn to thn to the cue curve rve y = y = xx22 x + 1 x + 1 at tat the points he points A, B & C on tA, B & C on the curve are concurrent athe curve are concurrent atthe point Pthe point P(7/2, 9(7/2, 9/2) then compute /2) then compute the sum of tthe sum of the slopes of the he slopes of the three normals. Athree normals. Also filso find their equationsnd their equationsand the co-ordand the co-ordininates oates of the feef the feet of the normals onto the curve.t of the normals onto the curve. [8][8]

B ansal C lasses Problems for JEE-2007 [5]

Q.34 A conic passing through the point A(1, 4) is such that the segment joining a point P(x, y) on the conic and the point of intersection of the normal at P with the abscissa axis is bisected by the y- axis. Find theequation of the conic and also the equation of a circle touching the conic at A(1, 4) and passing throughits focus. [8]

Q.35 A hyperbola has one focus at the origin and its eccentricity = 2 and one of its directrix is x + y + 1 = 0.

Find the equation to its asymptotes. [8]

Q.36 Let A, B, C be real numbers such that(i) (sin A, cos B) lies on a unit circle centred at origin.(ii) tan C and cot C are defined.

If the minimum value of (tan C – sin A)2 – (cot C – cos B)2 is a + 2 b where a, b N, find the valueof a3 + b3. [8]

Q.37 For a 2, if the value of the definite integral

0

22

x

1xa

dx equals5050

. Find the value of a.

[8]

Q.38 If

4

4

d tan1

tan)4( = ln k –

w

2, find the value of (kw), where k, w N. [8]

Q.39 Given a function g, continuous everywhere such that g(1) = 5 and0

1

g (t) dt = 2.

If f(x) =1

2 0

x

(x t)2 g(t) dt , then compute the value of f (1) f (1). [8]

Q.40 Let f : [0, 1] R is a continuous function such that 1

0

dx)x(f = 0. Prove that there is c (0, 1) such

that c

0

dx)x(f = f (c). [8]

Q.41 Consider the equation in x, x3 – ax + b = 0 in which a and b are constants. Show that the equation hasonly one solution for x if a 0, for a = 3, find the values of b for which the equation has three solutions.

[8]

Q.42 A tank consists of 50 litres of fresh water. Two litres of brine each litre containing 5 gms of dissolved saltare run into tank per minute; the mixture is kept uniform by stirring, and runs out at the rate of one litre per minute. If 'm' grams of salt are present in the tank after t minute, express 'm' in terms of t and find theamount of salt present after 10 minutes. [8]

Q.43 Urn-I contains 3 red balls and 9 black balls. Urn-II contains 8 red balls and 4 black balls. Urn-IIIcontains 10 red balls and 2 black balls. A card is drawn from a well shuffled back of 52 playing cards. If a face card is drawn, a ball is selected from Urn-I. If an ace is drawn, a ball is selected from Urn-II. If any other card is drawn, a ball is selected from Urn-III. Find

(a) the probability that a red ball is selected.(b) the conditional probability that Urn-I was one from which a ball was selected, given that the ball selected

was red. [8]


Q.44 The digits of a number are 1, 2 , 3, 4 , 5, 6 , 7 , 8 & 9 written at random in any order. Find the probabilitythat the order is divisible by 11. [8]

Q.45 A number is chosen randomly from one of the two sets, A = {1801, 1802,.....,1899, 1900} &B = {1901, 1902,.....,1999, 2000}. If the number chosen represents a calender year. Find the probability

that it has 53 Sundays. [8]

Q.46 A box contains 2 fifty paise coins, 5 twenty five paise coins & a certain fixed number N ( 2) of ten & five paise coins. Five coins are taken out of the box at random. Find the probability thatthe total value of these five coins is less than Re. 1 & 50 paise. [8]

Q.47 A hunter knows that a deer is hidden in one of the two near by bushes, the probability of its being hiddenin bushI being 4/5. The hunter having a rifle containing 10 bullets decides to fire them all at

bushI or II . It is known that each shot may hit one of the two bushes , independently of the other with probability 1/2. How many bullets must he fire on each of the two bushes to hit the animalwith maximum probability. (Assume that the bullet hitting the bush also hits the animal). [8]

Q.48 ABCD is a tetrahedron with A( 5, 22, 5); B(1, 2, 3); C(4, 3, 2); D( 1, 2, 3). Find

AB

( )BC BD

. What can you say about the values of ( )AB BC

BD

and ( )AB BD

BC

.

Calculate the volume of the tetrahedron ABCD and the vector area of the triangle AEF where thequadrilateral ABDE and quadrilateral ABCF are parallelograms. [8]

Q.49 Find the equation of the line passing through the point (1, 4, 3) which is perpendicular to both of the lines

2

1x =

1

3y =

4

2z and

3

2x =

2

4y =

2

1z

Also find all points on this line the square of whose distance from (1, 4, 3) is 357. [8]

Q.50 Find the parametric equation for the line which passes through the point (0, 1, 2) and is perpendicular tothe line x = 1 + t, y = 1 – t and z = 2t and also intersects this line. [8]

Q.51 Suppose that r 1 r 2 and r 1r 2 = 2 (r 1 , r 2 need not be real). If r 1 and r 2 are the roots of the biquadraticx4 – x3 + ax2 – 8x – 8 = 0 find r 1, r 2 and a. [8]

Q.52 Express2222

22

2222

ayxxyax2xay2

xyax2x2axyax2

xay2xyax2ayx

as a product of two polynomial. [8]

Q.53 Given the matrices A =

311322221

; C =

111122112

and D =

91310

and that Cb = D.

Solve the matrix equation Ax = b. [8]

Q.54 Prove thatc b

a

+

ac

b

+

ba

c

2

3 for a, b, c > 0. [8]

Q.55 Given x, y R, x2 + y2 > 0. If the maximum and minimum value of the expression 22

22

y4xyx

yx

are

M and m, and A denotes the average value of M and m, compute (2007)A. [8]

Q.56 Prove that the triangle ABC will be a right angled triangle if

cos2

A cos

2

Bcos

2

C – sin

2

A sin

2

Bsin

2

C =

2

1[8]


Q.57 A point P is situated inside an angle of measure 60° at a distance x and y from its sides. Find the distanceof the point P from the vertex of the given angle in terms of x and y. [8]

Q.58 InABC, a = 4 ; b = 3 ; medians AD and BE are mutually perpendicular. Find ‘c’ and ‘’. [8]

Q.59 The lengths of the sides of a triangle are log1012, log1075 and log10n, where n N. Find the number of possible values ofn. [8]

Q.60 A flight of stairs has 10 steps. A person can go up the steps one at a time, two at a time, or anycombination of 1's and 2's. Find the total number of ways in which the person can go up the stairs.

[8]

Q.61 Let a and b be two positive real numbers. Prove that b

a

x bax

dxx

ee= 0 [8]

Q.62 Let f (x) = 2kx + 9 where k is a real number. If 3 f (3) = f (6), then the value of f (9) – f (3) is equal to N, where N is a natural number. Find all the composite divisors of N. [8]

Q.63 Line l is a tangent to a unit circle S at a point P. Point A and the circle S are on the same side of l, and thedistance from A to l is 3. Two tangents intersect linel at the point B and C respectively. Find the value of (PB)(PC). [8]

Q.64 A triangle has one side equal to 8 cm the other two sides are in the ratio 5 : 3. What is the largest possiblearea of the triangle. [8]

Q.65 In triangle ABC, max {A, B} = C + 30° andr

R = 3 + 1, where R is the radius of the

circumcircle and r is the radius of the incircle. FindC in degrees. [8]

Q.66 The parabola P : y = ax2 where 'a' is a positive real constant, is touched by the line L: y = mx – b (wherem is a positive constant and b is real) at the point T.Let Q be the point of intersection of the line L and the y-axis is such that TQ = 1. If A denotes the

maximum value of the region surrounded by P, L and the y-axis, find the value ofA

1. [8]

Q.67 A point moving around circle (x + 4)2 + (y + 2)2 = 25 with centre C broke away from it either at the pointA or point B on the circle and moved along a tangent to the circle passing through the point D (3, – 3).Find the following.

(i) Equation of the tangents at A and B.(ii) Coordinates of the points A and B.(iii) Angle ADB and the maximum and minimum distances of the point D from the circle.(iv) Area of quadrilateral ADBC and theDAB.(v) Equation of the circle circumscribing theDAB and also the intercepts made by this circle on the

coordinate axes. [10]

Q.68 If 1x7

1

2 i

ii and 12x)1(

7

1

2 i

ii and 123x)2(

7

1

2 i

ii ,

then find the value of

7

1

2 x)3(i

ii . [10]


Q.69 The normals to the parabola y2 = 4x at the points P, Q & R are concurrent at the point (15, 12). Find (a) the equation of the circle circumscribing the triangle PQR (b) the co-ordinates of the centroid of the triangle PQR. [10]

Q.70 The triangle ABC, right angled at C, has median AD, BE and CF. AD lies along the line y = x + 3, BE liesalong the line y = 2x + 4. If the length of the hypotenuse is 60, find the area of the triangle ABC.

[10]

Q.71 Let W1 and W2 denote the circles x2 + y2 + 10x – 24y – 87 = 0 and x2 + y2 – 10x – 24y + 153 = 0respectively. Letm be the smallest positive value of 'a' for which the line y = ax contains the centre of a

circle that is externally tangent to W2 and internally tangent to W1. Given that m2 =q

p where p andq are

relatively prime integers, find ( p+ q). [10]

Q.72 If

652

dx)xsin1(

3= a –

c

3 b where a, b, c N and b, c are relatively prime, find the value of

a + b + c + abc. [10]

Q.73 If

1

02x1x1

dx =

c ba

where a,b,c N, find the value a2 + b2 + c2. [10]

Q.74 Suppose f (x) and g (x) are differentiable functions such that

x g )x( f )x(')x(' gg f = )x(')x(')x( f f gg f

for all real x. Moreover, f (x) is nonnegative andg (x) is positive. Furthermore, 2

e1dx)x(g

a2a

0

f

for all reals a. Given that )0( f g = 1. If the value of )4( f g = e –k where k N, find k. [10]

Q.75 Let f (x) be a differentiable function such that f ' (x) + f (x) = 4xe –x · sin 2x and f (0) = 0. Find the value

of

n

1k n

)k (f Lim . [10]

Q.76 Let f be a differentiable function satisfying the condition f x

y

=

f

f

(x)

(y)(y 0, f(y) 0) V x, y R and

f (1) = 2, then find the area enclosed by y = f(x), x2 + y2 = 2 and x – axis. [10]

Q.77 The equation Z10 + (13 Z – 1)10 = 0 has 5 pairs of complex roots a1, b1, a2, b2, a3, b3, a4, b4, a5, b5.

Each pair ai, bi are complex conjugate. Findii ba

1. [10]

Q.78(i)Let Cr's denotes the combinatorial coefficients in the expansion of (1 + x)n, n N. If the integersan = C0 + C3 + C6 + C9 + ........

bn = C1 + C4 + C7 + C10 + ........and cn = C2 + C5 + C8 + C11 + ........, then

prove that (a) 3n

3n

3n c ba – 3an bncn = 2n, (b) (an – bn)2 + (bn – cn)2 + (cn – an)2 = 2. [10]

(ii) Prove the identity: (C0 – C2 + C4 – C6 + .....)2 + (C1 – C3 + C5 – C7 + .......)2 = 2n


Q.79 Given the matrix A =

531531

531

and X be the solution set of the equation AAx = A,

where x N – {1}. Evaluate

1x

1x3

3

where the continued product extends x X. [10]

Q.80 If a, b, c are the sides of triangle ABC satisfying log

a

c1 + log a – log b = log 2. Also

a(1 – x2) + 2bx + c(1 + x2) = 0 has two equal roots. Find the value of sin A + sin B + sin C. [10]

Q.81 For x (0, /2) and sin x =3

1, if

0nn3

)nxsin(=

c

b ba then find the value of (a + b + c),

where a, b, c are positive integers.

(You may Use the fact that sin x =i2

ee ixix ) [10]

Q.82 Two distinct numbersa and b are chosen randomly from the set {2, 22, 23, 24, ......, 225}. Find the probability that loga b is an integer. [10]

OBJECTIVE

Select the correct alternative. (Only one is correct):

Q.83 A child has a set of 96 distinct blocks. Each block is one of two material (plastic, wood), 3 sizes (small,medium, large), 4 colours (blue, green, red, yellow), and 4 shapes (circle, hexagon, square, triangle).How many blocks in the set are different from "Plastic medium red circle" in exactly two ways? ("Thewood medium red square" is such a block)(A) 29 (B) 39 (C) 48 (D) 56

Q.84 The sum

49

0k

k

k 299)1( where

)!r n(!r

!nr n

equals

(A) – 298 (B) 298 (C) – 249 (D) 249

Q.85 If A > 0, c, d , u, v are non-zero constants, and the graphs of f (x) = | Ax + c | + d and

g (x) = – | Ax + u | + v intersect exactly at 2 points (1, 4) and (3, 1) then the value ofA

cu equals

(A) 4 (B) – 4 (C) 2 (D) – 2

Q.86 Consider the polynomial equation x4 – 2x3 + 3x2 – 4x + 1 = 0. Which one of the following statementsdescribes correctly the solution set of this equation?(A) four non real complex zeroes. (B) four positive zeroes(C) two positive and two negative zeroes. (D) two real and two non real complex zeroes.

Q.87 The units digit of 31001 · 71002 · 131003 is(A) 1 (B) 3 (C) 7 (D) 9


Q.88 The polynomial f (x) = x4 + ax3 + bx2 + cx + d has real coefficients and f (2i) = f (z + i) = 0. The valueof (a + b + c + d) equals(A) 1 (B) 4 (C) 9 (D) 10

Q.89 If the sum

1k 2k k k )2k (

1 =

c

ba where a, b, c N and lie in [1, 15] then a + b + c

equals(A) 6 (B) 8 (C) 10 (D) 11

Q.90 Triangle ABC is isosceles with AB = AC and BC = 65 cm. P is a point on BC such that the perpendicular distances from P and AB and AC are 24 cm and 36 cm respectively. The area of triangle ABC in sq. cmis(A) 1254 (B) 1950 (C) 2535 (D) 5070

Q.91 The polynomial function f (x) satisfies the equation f (x) – f (x – 2) = (2x – 1)2 for all x. If p and q are thecoefficient of x2 and x respectively in f (x), then p + q is equal to(A0 0 (B) 5/6 (C) 4/3 (D) 1

Q.92 Three bxes are labelled A, B and C and each box contains four balls numbered 1, 2, 3 and 4. The ballsin each box are well mixed. A child chooses one ball at random from each of the three boxes. If a, b, and c are the numbers on the balls chosen from the boxes A, B and C respectively, the child wins a toyhelicopter when a = b + c. The odds in favour of the child to receive the toy helicopter are(A) 3 : 32 (B) 3 : 29 (C) 1 : 15 (D) 5 : 59

Q.93 The value of tan

13

5cosarc

5

4sinarc is equal to

(A)63

25(B) –

7

3(C) –

56

33(D)

63

16

Select the correct alternatives. (More than one are correct):

Q.94 Three positive integers form the first three terms of an A.P. If the smallest number is increased by one theA.P. becomes a G.P. In original A.P. if the largest number is increased by two, the A.P. also becomes aG.P. The statements which does not hold good?(A) first term of A.P. is equal to 3 times its common difference.(B) Sn = n(n + 11)(C) Smallest term of the A.P. is 8(D) The sum of the first three terms of an A.P. is 36.

Q.95 If the line 2x + 9y + k = 0 is normal to the hyperbola 3x2 – y2 = 23 then the value of k is(A) 31 (B) 24 (C) – 31 (D) – 24

Q.96 The line 2x – y = 1 intersect the parabola y2 = 4x at the points A and B and the normals at A and Bintersect each other at the point G. If a third normal to the parabola through G meets the parabola at Cthen which of the following statement(s) is/are correct.(A) sum of the abscissa and ordinate of the point C is – 1.(B) the normal at C passes through the lower end of the latus rectum of the parabola.(C) centroid of the triangle ABC lies at the focus of the parabola.(D) normal at C has the gradient – 1.


Q.97 If (x) = f (x2) + f (1 – x2) and f '' (x) > 0 for x R then which of the following are correct?

(A) (x) attains its extrema at 0, ±2

1(B) (x) increases in ,210,21

(C) (x) attains its local maxima at 0. (D) (x) decreases in ,210,21

Q.98 If tan

x3

2 =

xcos3

2cos

xsin3

2sin

where 0 < x < , then the value of x is

(A)12

(B)

12

5(C)

12

7(D)

12

11

MATCH THE COLUMN:

Q.99 Column-I Column-II

(A) The smallest positive integeral value of n for which the complex (P) 4

number 2n31 i is real, is

(B) Let z be a complex number of constant non zero modulus (Q) 6such that z2 is purely imaginary, then the number of possiblevalues of z is

(C) 3 whole numbers are randomly selected. Two events A and B are (R) 8defined as

A : units place in their product is 5.B : their product is divisible by 5. (S) 9

If p1 and p2 are the probabilities of the events A and B such that p2 = kp1 then 'k' equals

(D) For positive integers x and k, let the gradient of the line connecting(1, 1) and (x, x3) be k. Number of values of k less than 31, is

Q.100 Column-I Column-II

(A) For real a and b if the solutions to the equation Z9 – 1 = 0 (P) 0are written in the form of a + ib then the number of distinctordered pairs (a, b) such thata and b are positive, is (Q) 1

(B) x

e1x

x

x

1eLim

(R) 2

(C) Let A, B be two events with P(B) > 0. If B A then P(A/B) equals (S) e(D) A real number x is chosen at random such that 0 x 100.

The probability that x – [x]3

1is

b

a, where a and b are relatively

primes and [x] denotes the greatest integer then (b – a) equals


HINTS AND SOLUTIONS

1. Let 2111x = yso that log2y = 111 x x =

111

ylog2

equation becomes

4

y3

+ 2y = 4y2 + 1

y3 – 16y2 + 8y – 4 = 0sum of the roots of the given equation is

x1 + x2 + x3 =111

ylogylogylog 322212 =

111

)yyy(log 3212 =111

4log2 =111

2 S1 + S2 = 113 Ans.]

2. Let the 3 consecutive terms area – d, a, a + d d > 0

hence a2 – 2ad + d 2 = 36 + K ....(1)a2 = 300 + K ....(2)

a2 + 2ad + d 2 = 596 + K ....(3)now (2) – (1) gives

d(2a – d) = 264 ....(4)(3) – (2) gives

d(2a + d) = 296 ....(5)(5) – (4) gives

2d 2 = 32 d 2 = 16 d = 4 (d = – 4 rejected)Hence from (4)

4(2a – 4) = 264 2a – 4 = 66 2a = 70 a = 35 K = 352 – 300 = 1225 – 300 = 925 Ans.]

3. Case-I : When the two identical digits are both unity as shown. any one place out of 3 block for unity can be taken in 3 ways and the remaining two

blocks can be filled in 9 · 8 ways.Total ways in this case = 3 · 9 · 8 = 216

Case-II : When the two identical digit are other than unity.

; ;

two x's can be taken in 9 ways and filled in three ways and y can be taken in 8 ways.Total ways in this case = 9 · 3 · 8 = 216

Total of both case = 432 Ans. ]

4. h = a(t1 t2)k = a(t1 + t2)

Equation to the variable chord 2x – (t1 + t2)y + 2at1 t2 = 0

y = xtt

2

21 +

21

21

tt

tat2

y = xk

a2 + a

k

h2....(1)

Since (1) touches y2 = 4bx , using the condition of tangency

a2

bk

k

ah2

Locus is by2 = 4a2x ]


5. Note that triangles BCM and OCN are similar now let ON = p. N will be mid point of chord PQ

1

p =

2

1 p =

2

1

now R = 22 pr 2 for large circle

= )41(162 = 63

Alternatively: Equation of large circle as x2 + y2 = 16

now C = (1, 0) with slope PQ = –3

1(think !)

equation of PQ : 3 y + x = 1

P (from origin) =2

1 result ]

6. x

xdx

7

2 51( ) = x

xx

dx7

10

2

51

1

Taking x2 out of the bracket

= x

xdx

3

25

1 Put x –2 – 1 = t =2

3x dx = dt

= – 1

2 dt

t 5 = – 1

2

t

4

4=

1

8

14

t

+ C =1

8

1

11

2

4

x

=

x

x

8

2 48 1

+ C ]

7. Using sin 2x =xtan1

xtan22

I =

2

02

2dx

xtan1

xtan21

xtan1

xtan21

=

2

0

24

2

dx)xtan1(·)xtan1(

)xtan1( =

2

0

24

2

dxxsec·)xtan1(

)xtan1(

put y = tan x dy = sec2x dx

I =

04

2

dy)y1(

)y1(

now put 1 + y = z dy = dz

I =

14

2

dzz

)z2( =

1

3

2

z3

4z6z3 =

3

1 a = 1, b = 3 1 + 3 + 3 = 7 Ans. ]

Alternatively: I =

2

04

2

dx)xsinx(cos

)xsinx(cos


I = –

2

0

II

3

I

dx)xsinx(cos

1

dx

d ·)xsinx(cos

3

1

integrating by parts

= –

2

03

2

03

dx)xsinx(cos

)xcosx(sin

)xsinx(cos

)xsinx(cos

3

1 = –

2

0x2sin1

dx)1()1(

3

1

using sin 2x =xtan1

xtan22

=3

2 –

3

1

2

02

2

dx)xtan1(

xsec=

3

2 –

3

1

12t

dt=

3

2 +

3

1 0tt =3

2 +

3

1[(0) – (1)

=3

2 –

3

1 =

3

1 a = 1, b = 3 1 + 3 + 3 = 7 Ans. ]

8. Let the number of passengers be x ( x > 200)

Fair changed per person = 10 – (x – 100)100

2

Total revenue = x .

100

2)200x(10 = )200x(

100

x2x10 = x4

100

x2x10

2

f (x) = 14x –100

x2 2

f (x) = 14 –100

x4 = 0 x = 350

f (x) < 0 x = 350 gives maxima]

9. Given 9y2 = x3

Let the point on the curve be x = t2 and y =3

t3

dt

dx= 2t ;

dt

dy = t2

dx

dy =

dt

dy×

dx

dt =

t2

t2

=2

t slope of the normal = –

t

2

normal makes equal intercept

hence –t

2 = – 1 t = 2

Hence P = (4,3

8) a + 3b = 4 + 3 ·

3

8 = 4 + 8 = 12 Ans. ]


10. Amount of glucose in blood at time t is x (t) hencedt

dx = – K x

dtxK

dx

–K

1ln ( – K x) = t + C

ln ( – K x) = – Kt + C – K x = e – K t + C

x =K

e CtK

K )t(xLim

t

]

11. A =( ) ( )a a x x a x a

ax

x 2 2 2

4

2 3

11

2 dx

where x1 & x2 are the roots of ,x2 + 2 a x + 3 a2 = a2 a xx = a or x = 2 a

A =)a1(6

a4

3

dA

da = 0 gives a = 31/4 Ans. ]

12. A : ears of Mr A formed to be droopingB1 : Mr A was telling a truth P(B1) = 3/5B2 : Mr B was telling a false P(B2) = 2/5P(A/B1) = 1/10P(A/B2) = 2/3

P(B1/A) =

3

2·

5

2

10

1·

5

310

1·

5

3

=

3

403

3

=

49

9 Ans. ]

13. E : all the 5 persons leave at different floorsn(S) = 85

n(A) = 8C5 · 5!

P(E) =8

5

5

5!

8

C =

105

512 ans. ]

14. cos = |u||w|

w·u

= |w||u|

)u|v|v|u(|·u

=|w||u|

|u||v||u|)v·u( 2

cos =|w|

|u||v|)v·u(

....(1)

cos = |v||w|

w·v

= |v||w|

)u|v|v|u(|·v

=|v||w|

|v|)u·v(|u||v| 2


cos =|w|

)u·v(|u||v|

....(2)

from (1) and (2) cos = cos = ]

15. The line is t

2

12

1z

1

1y

1

2x

....(1)

line passes through k ˆ2

1 ji2 and is parallel to the vector k ˆ

2

1 jiV

vector normal to the plane x + 2y + 6z = 10, is k ˆ6 j2in

n.V

= 1 + 2 – 3 line (1) is | | to the plane

d =3641

10322

=

41

9Ans ]

16. Vector 1v

along the line of intersection of 3x – 7y – 5z = 1 and 8x – 11y + 2z = 0 is given by

211 nnv

=2118573

k ˆ ji

= – 23( k ˆ j2i3 )

|||ly vector 2v

along the line of intersection of the planes 5x – 13y + 3z = 0 and 8x – 11y + 2z = 0 is

432 nnv

=21183135k ˆ ji

= 7 ( k ˆ7 j2i )

now 21 v·v

= 0 angle is 90° sin90° = 1 ]

17. Given au + bv + cw = 0 ....(1)au + bv + cw = 0 ....(2)

and au + bv + cw = 0 ....(3)

For non trivial solution (non zero) solution of a, b and c . We must have''w''v''u'w'v'u

wvu

= 0 ]

18. Let y = cos A · sin2

2

A + cos B · sin2

2

B + cos C · sin2

2

C

=2

1[cosA (1 – cosA) + cosB (1 – cosB) + cos C (1 – cos C)]

=2

1[(cosA – cos2A) + (cosB – cos2B) + (cosC – cos2C)]

=2

1

4

1

2

1Ccos

4

1

2

1Bcos

4

1

2

1Acos

222


y =2

1

222

2

1Ccos

2

1Bcos

2

1Acos

4

3

now y will be maximum if cosA = cos B = cos C =2

1

hence ymax = 3/8 ]

19. y = x2; x = t; y = t2

dx

dy = 2x = 2t

slope of normal m = –t2

1

equation of normal

y – t2 = –t2

1(x – t) or 2t(y – t2) = – x + t

if x = 0; y =2

3

2t

2t2

3 = t t = 0

or 3 – 2t2 = 1 t = 1 or – 1hence one of the point is origin and theother two are (–1, 1) and (1, 1) PQR is a right triangle radius of the circle is 1its equation is x2 + (y – 1)2 = 1 x2 + y2 – 2y = 0 ]

20. Let x be a real root. Equating real and imaginary partx3 – 6x2 + 5x + 2a2 = 0 .....(1)

and 2x3 – 2x2 – 4x = 0 .....(2)2x(x2 – x – 2) = 02x(x – 2)(x + 1) = 0

the given x = 0, 2 or – 1if x = 0 a = 0

x = – 1 a2 = 6 a = ± 6

x = 2 a2 = 3 a = ± 3

a }{ 3,3,6,6,0

S = 0 + 6 + 6 + 3 + 3 = 18 Ans. ]

21. For non zero interceptsslope = – 1

y = – x + c point (– 4, – 2) – 2 = 4 + c c = – 6


lines is y = – x – 6 x + y + 6 = 0for zero intercept

line is y = mx – 2 = m(– 4) m = 1/2

2y = x lines are 2y = x and x + y + 6 = 0 ]

22. x4 – 10x2 + 9 0(x2 – 9)(x2 – 1) 0hence – 3 x – 1 or 1 x 3now f (x) = x3 – 3x

f ' (x) = 3x2 – 3 = 0x = ± 1

maximum occurs when x = 3f (3) = 18 ]

23. x4y2dx + x2y4dy = xdy + ydxx2y2(x2dx + y2dy) = xdy + ydx

x2dx + y2dy = 2)xy(

)xy(d

Integrating, dxx2 + dyy2 = 2)xy(

)xy(d

3

x3

+3

y3

= –xy

1 + C

(x3 + y3) +xy

3= C; now if x = 1; y = 1 C = 5,

hence x3 + y3 + 3(xy) –1 = 5 Ans. ]

24. 52 removed card face 40 randomlydrawn20

Let E0 : 20 cards randomly removed has no aces.E1 : 20 cards randomly removed has exactly one ace.E2 : 20 cards randomly removed has exactly 2 aces.E : event that 2 drawn from the remaining 20 cards has both the aces.P(E) = P(E E0) + P(E E1) + P(E E2)

= P(E0) · P(E / E0) + P(E1) · P(E / E1) + P(E2) · P(E / E2)

= 40 \/

other 36

aces4

=2

202

4

2040

2036

04

C

C·

C

C·C +

220

23

2040

1936

14

C

C·

C

C·C +

220

22

2040

1836

24

C

C·

C

C·C

=2

2020

402

218

362

42

319

361

42

420

36

C·C

C·C·CC·C·CC·C


=2

2020

4018

3619

3620

36

C·C

C·6C·12C·6 =

220

2040

1836

1936

1936

2036

C·C

]CCCC[6

=2

2020

4019

3720

37

C·C

)CC(6 =

220

2040

2038

C·C

)C(6 p = 6 Ans. ]

25. e =5

4

2

2

a

b = 1 –

25

16 =

25

9;

a

b =

5

3....(1)

now radius of the circle r = a –(where, 0 is the centre of the circle)

also r = AC = b sin a – = b sin where = a cos a(1 – cos ) = b sin

a2(1 – cos )2 = b2(1 – cos )(1 + cos ) a2(1 – cos ) = b2(1 + cos )

cos1

cos1 =

25

9

25 – 25 cos = 9 + 9 cos 16 = 34 cos

cos =17

8; sin =

17

15

AB = 2b sin = 2 ·5

a3 ·

17

15 =

17

18a Ans. ]

26. Let a, b R, such thatf (a) = f (b)

i

i

a1

a1

=i

i

b1

b1

1 – bi + ai + ba = 1 + bi – ai + ab2ai = 2bi a = b

f is injective. ]

27. Let r be the radius of circle Aand R be the radius of circle B r + R = 12 and r = 3R 4R = 12; R = 3 and r = 9

Area of trapezium ABCD =2

1(3 + 9) 22 6)12(

= 6 108 = 336

Area of arc ADC =3

812

1 =

2

27


Area of arc BCE =3

29

2

1 = 3

required area = 336 –

32

27 = 336 –

2

33

a = 36, b = 33 a + b = 69 Ans. ]

28. A2 =

0310

0310

=

3003

= 3I

A4 = 9I; A6 = 27; A8 = 81I (A8 + A6 + A4 + A2 + I) = 121 I

hence 121

1001

V

=

110

;

12100121

ba

=

110

b121a121

=

110

a = 0, b =11

1;

11

10

V

]

29. Radius of the first circle =S

=

6

6= 1

sin2

C =

r 1

r 1

....(1) (r < 1)

also sin C =5

4

now 2sin2

2

C = 1 – cos C = 1 –

5

3 =

5

2

sin2

2

C =

5

1

2

r 1

r 1

=5

1 5(1 – r)2 = (1 + r)2 )r 1(5 = 1 + r

5 – 1 = ( 15 )r r =15

15

=

36cos

18sin k + w = 54° Ans. ]

30. Cross multiplication and rearranging gives the cubic.

x3 – ax2 + 23x – b = 0

2 + = a ....(1)2 + 2 = 23 ....(2)

and 2 = ....(3)Also given + = 12 ....(4)from (2) and (4)

2 + 2(12 – ) = 232 + 24 – 22 = 23


2 – 24 + 23 = 0 = 1 (rejected) since x ± 1

= 23; = – 11 a = 35 from (4)and b = 2 = 529 × – 11 b = – 5819 a – b = 35 – (–5819) = 5854 Ans. ]

31. Let the numbers be

)d ()c() b()a(A

)DA(,DA,A,DA

.P.A

2

.P.G

Given d – a = 30

A

)DA( 2 – (A – D) = 30 (A + D)2 – A(A – D) = 30A

D2 + 3AD = 30 AD2 = 3A(10 – D)

A =)D10(3

D2

....(1)

since 'A' is a + ve integer 0 < D < 10 ....(2)Also since '3' is prime and A is an integer D2 must be divisible 3 D must be of the form of 3K possible values of D are 3, 6, 9

D = 3 A =7

3(rejected)

D = 6 A = 3 (rejected)D = 9 A = 27

Numbers are 18, 27, 36, 48 Ans. ]

32. Total number of ways in which P and Q can be chosen simultaneously= (25 – 1)(25 – 1)= 45 – 26 + 1

number of ways when P and Q have no common element= 5C1(2

4 – 1) + 5C2 (23 – 1) + 5C3(22 – 1) + 5C4(21 – 1) + 5C5(20 – 1)= 5C1 · 24 + 5C2 · 23 + 5C3 · 22 + 5C4 · 2 + 5C5 – (5C1 + 5C2 + 5C3 + 5C4 + 5C5)= (5C0 · 25 + 5C1 · 24 + 5C2 · 23 + 5C3 · 22 + 5C4 · 2 + 5C5 – 25) – (25 – 1)= (35 – 25) – (25 – 1)= 35 – 26 + 1

Hence P and Q have atleast one common element = (45 – 26 + 1) – (35 – 26 + 1)= 45 – 35 Ans. ]


33. Slope of the normal m =1

2 11x x1 =

m

m

1

2 y1 =

3 1

4

2

2

m

m

;

equation of the normal in terms of slope of the normal is y = mx +5 2 1

4

2 3

2

m m

m

.

It passes through (7/2, 9/2) 12 m3 13m2 + 1 = 0 sum = 13/12.Also (m 1) (3m 1) (4m + 1) = 0 m1 = 1 ; m2 = 1/3 ; m3 = 1/4 the normals are x y + 1 = 0 ; x 3y + 10 = 0 & 2x + 8y 43 = 0Point A (0, 1) ; B ( 1, 3) ; C (5/2, 19/4) ]

34. Equation of normal,

Y y = 1

m (X x) Y = 0 gives X = x + my and

X = 0 gives Y =x my

m

Hence

x x my 2

= 0 2 x + ydy

dx = 0

x2 +y2

2= C ; passes through (1, 4) C = 9

conic isx y2 2

9 18 = 1 with e =

1

2 focii are (0, 3) & (0, 3)

Equation of the circles are ;(x 1)2 + (y 4)2 + (x + 2 y 9) = 0 where x + 2y 9 = 0 is the tangent to the ellipse at (1, 4)]

35. Equation to the hyperbola where S = (0, 0) ; directrix is x + y + 1 = 0 and e = 2 is

2

1yx2yx 22

x2 + y2 = (x + y + 1)2

2xy + 2x + 2y + 1 = 0Let the combined equation of the asymptotes is

2xy + 2x + 2y + c = 0 put D = 0 to get c = 2

hence combined equation of the asymptotes arexy + x + y + 1 = 0(x + 1)(y + 1) = 0 x + 1 = 0 and y + 1 = 0 ]

36. Note that (tan C – sin A)2 + (cot C – cos B)2 denotes the square of the distance PQnow d 2

PQ = (Q – OP)2

d 2PQ =2

22 1)CcotC(tan

d 2PQ =2

2 12)CcotC(tan

d 2min = 212 = 3 – 22

a = 3; b = 2 a3 + b3 = 27 – 8 = 19 Ans. ]


37. I =

02

22 )2a(

x

1x

dx =

0224

2

1x)2a(x

dxx(a2 – 2 = k 0)

=

024

2

1kxx

dxx =

024

22

dx1kxx

)1x()1x(

2

1

=

1I

022

2

dxk )x1(x

)x1(1

2

1

+

2I

022

2

dxk )x1(x

)x1(1

2

1

now proceed, I1 =a2

and I2 = 0

a2

I

;a2

=

5050

a = 2525 Ans. ]

38. Let = x4

d = dx or 4 = + 4x – 4 = – 4x

=

0

2

dx

x4

tan1

x4

tan)x4(

= – 4

0

2

dx

xtan1

xtan11

xtan1

)xtan1(x

= – 4

0

2

dxxtan)2(

)xtan1(·

xtan1

)xtan1(x

= 2

0

2

dxxtan

)xtan1(x = 2

0

2

dxxxtan

x

I = 0

2

2x

+

0

2

dxxtan

x

I = –4

2 + 2

2

0

dtttan

tx = – t

now I1 =

2

0 III

dttcott = 2

0tsinnt

l –

2

0

dttsinnl

I1 = 0 +2

ln 2

Hence 2 ·2

ln 2 –

4

2 = ln 2 –

4

2 k = 2, w = 4 kw = 8 Ans. ]


39. g(1) = 5 and0

1

g (t) dt = 2

2f (x) = x

0(x2 – 2xt + t2) g(t) dt =

x

0

2x

0

x

0

2 dt)t(gtdt)t(gtx2dt)t(gx

Differentiating

2 f '(x) = x2 g(x) + x2·dt)t(gx

0 –

x

0

2 dt)t(gt)x(gx2 + x2g(x)

2 f '(x) = 2x x

0

dt)t(g – x

0

dt)t(gt2

f " (x) = x g (x) + x

0

dt)t(g – x g (x) = x

0

dt)t(g

hence f " (1) = 1

0

dt)t(g = 2

also f ''' (x) = g (x) f ''' (1) = g (1) = 5 f ''' (1) – f ''(1) = 5 – 2 = 3 Ans. ]

40. Consider a function g (x) = e –x x

0

dt)t(f in [0, 1]

obvious continuous and derivableg (0) = 0 and g (1) = 0 (given)

hence some c (0, 1) such that g ' (c) = 0

now g ' (x) = e –x f (x) – e –x x

0

dt)t(f

g ' (c) = e –c f (c) – e –c c

0

dt)t(f = 0 x

0

dt)t(f = f (c)]

41. Consider f (x) = x3 – ax + bf '(x) = 3x2 – a

if a 0 then f ' (a) 0 for all x hence f is strictly increasinghence f (x) = 0 has exactly one root

for a = 3f ' (x) = 3x2 – 3 = 0

x = 1 or – 1in order that f (x) may have 3 roots

f (x1) · f (x2) 0where x1 and x2 and the roots of f ' (x) = 0hence (1 – a + b)(– 1 + a + b) 0

put a = 3(b – 2)(b + 2) 0

or – 2 b 2 ]


42. Let m gms of salt is present at time t differential equation of the process is

dt

dm = 10 –

t50

)1(m

dt

dm + m

t50

1

= 10;

I.F = t50

dt

e = 50 + t; m(50 + t) = dt)t50( =2

)t50(10

2 + C

m(50 + t) = 5(50 + t)2 + C; t = 0; m = 0, C = – 5.(50)2

m(50 + t) = 5(50 + t)2 – 5 (50)2

m = 5(50 + t)2 –t50

)50(5 2

m(t = 10) = 5 · 60 –60

)50(5 2

m =3

1125 =

3

291 = 50

60

2506 = 50 ·

6

11]

43. A : red ball is selected B1 : Face card is drawnB2 : ace card is drawnB3 : neither face nor ace is drawn

P(A) =12

3·

52

12 +

12

8·

52

4 +

12

10·

52

36 =

156

107 Ans.Ans.

P(B1/A) =

12

3·

52

12 ·

107

156 =

107

9 Ans. ]

44. 1, 2, 3, 4, 5, 6, 7, 8, 9

x + y = 45 ; x y = 11 x = 28 ; y = 17 Now to realise a sum 17 using 4 digits we can have different cases ,

12591349

;

234813581268

;

235714571367

; 6 5 4 2 ( 9 cases )

If we use five digits then 7 , 1 , 2 , 3 , 4 ( 2 cases )6 , 5 , 3 , 2 , 1

Hence p =4 5 9 5 4 2

9

! ! ! !

!

=

11 5 4

9

! !

!=

11

126[ odd in favour 11 : 115 ]

45. A = {1801, 1802,.....,1899, 1900}B = {1901, 1902,.....,1999, 2000}


E : randomly chosen year has 53 sundaysP (E) = P (E L) + P (E O)

= P (L). P(E/L) + P (O). P(E/O)

=

7

1.

100

76

7

2.

100

24

2

1 +

7

1.

100

75

7

2.

100

25

2

1

=1400

249 Ans.]

46. P(E) = 1 P (value of 5 coins is morethan or equal to Rs. 1.50)

= 1 P(A A B B B or A A B B C or A B B B B) ]

47 6 on bush-I & 4 on bush-II

48. AB

( )BC BD

= 0 ;( )AB BC

BD

= 0 ;( )AB BD

BC

= 0 ;

Note that AB

;BC

;BD

are mutually perpendicular Þ BC

×

BD is collinear with

AB and so on

Volume =1

6[AB

, BC

, BD

] =

220

3 cu. units

Vector area of triangle AEF =1

2AF

AE

=

1

2BC

BD

= 3 10 i j k ]

49. Equation of the line passing through (1, 4, 3)

c

3z

b

4y

a

1x

....(1)

since (1) is perpendicular to2

1x =

1

3y =

4

2z and

3

2x =

2

4y =

2

1z

hence 2a + b + 4c = 0and 3a + 2b – 2c = 0

34

c

412

b

82

a

1

c

16

b

10

a

hence the equation of the lines is1

3z

16

4y

10

1x

....(2) Ans.

now any point P on (2) can be taken as1 – 10 ; 16 + 4 ; + 3

distance of P from Q (1, 4, 3)(10)2 + (16)2 + 2 = 357(100 + 256 + 1)2 = 357 = 1 or – 1 Hence Q is (–9, 20, 4) or (11, – 12, 2) Ans.]

50. Equation of the line through (0, 1, 2)

c

2z

b

1y

a

0x

....(1)

now given line2

0z

1

1y

1

1x

= t ....(2)


(2) is along the vector k ˆ2 jiV

a – b + 2c = 0 ....(3)since (1) and (2) intersect; hence must be coplanar

hencec ba211201

= 0

2a + 4b + c = 0 ....(4)solving (3) and (4), a : b : c = – 3 : 1 : 2

required equation is2

2z

1

1y

3

x

= t Ans. ]

51. Since r 1r 2 = 2,

x2 + px + 2 = 02

1

r

r and r 1r 2r 3r 4 = – 8 r 3r 4 = – 4

x4 – x3 + ax2 – 8x – 8 = (x2 + px + 2)(x2 + qx – 4)compare coefficient of x3 and x

p + q = – 1 .....(1)and 2q – 4p = – 8 q – 2p = – 4 ....(2) p = 1 and q = – 2on comparing coefficient of x2; a = – 4

p = 1 x2 + x + 2 = 0

r 1, 2 =2

7i1 Ans. ]

52.

yax

xxa

ayx

yxa

axy

xax

=

2

yxa

axy

xax

= [x (xy – ax) – a(y2 – a2) + x (xy – ax) ]2

= [2x2 (y – a) – a (y – a) (y + a) ]2

= (y – a)2 [2x2 – a(y + a)]2

Hence D = (y2 + a2 – 2ay) (2x2 – ay – a2)2 ]

53. Let b =

3

2

1

aaa

111122112

3

2

1

a

aa

=

91310

321

321

321

aaa

aa2a2aaa2

=

91310

i.e. a1 = 1 ; a2 = 3 ; a3 = 5


311322221

3

2

1

xxx

=

531

321

321

321

x3xxx3x2x2x2x2x

=

531

i.e. x1 = 1 ; x2 = – 1 ; x3 = 1 Ans. ]

54. TPT

1x

c b

c ba

+

2x

ac

c ba

+

3x

ba

c ba

2

9....(1)

Consider AM between the numbers x1, x2, x3

=

ba

1

ac

1

c b

1

3

c ba

now HM between the numbers x1, x2, x3

=

c ba

ba

c ba

ac

c ba

c b3

=

)c ba(2

)c ba(3

=2

3

AM HM

ba

1

ac

1

c b

1

3

c ba

2

3

(a + b + c)

ba

1

ac

1

c b

1

2

9Hence proved ]

55. Let x = r cos and y = r sin

r 2 = x2 + y2; tan =x

y (0, /2)

N =]sin4cossin[cosr

r 222

2

=

)2cos1(42sin)2cos1(

r 2

=

2cos32sin5

2

Nmax = 105

2

= 105

15

2 = M

Nmax = 105

2

= 105

15

2 = m

A =2

mM = 2·15

10·2 =

3

2 2007 ×

3

2 = 1338 Ans. ]

56. Transposing 2 on RHS using 2 cos A · cos B relation,

cos2

A

2

CBcos

2

CBcos – sin

2

A

2

CBcos

2

CBcos = 1


or cos2

A sin

2

A +

2

CBcos

2

Acos

–

2

CBcos

2

Asin

+ sin2

2

A– 1 = 0 (

2

Asin

2

CBcos

)

2

Asin

2

Acos

2

CBcos + cos

2

A sin

2

A – cos2

2

A = 0

2

Asin

2

Acos

2

CBcos – cos

2

A

2

Asin

2

Acos = 0

2

Asin

2

Acos

2

Acos

2

CBcos = 0

if cos2

A – sin

2

A = 0 tan

2

A = 1 A = 90°

if cos2

CB = cos

2

A

B – C = A B = C + A B = 90°B – C = – A B + A = C = 90°

hence triangle must be right angled. ]57. OAMB is a cyclic quadrilateral

using sine law in OBM and OAM

90sin

d =

)60sin(

x

.....(1)

and 90sin

d =

sin

y....(2)

(1) and (2) )60sin(

x

=sin

y

y

x =

sin

)60sin( =

2

3cot –

2

1

y

x2 + 1 = cot3

y3

yx2 = cot

from (2)d = y cosec

d 2 = y2(1 + cot2) d 2 = y2

2

2

y3

)yx2(1 d 2 = y2 +

3

)yx2( 2

d 2 =3

xy4yx4y3 222 d 2 =

3

xy4y4x4 22

d = xyyx3

2 22 Ans. ]


58. Let G be the centroid : AD = x ; BE = y

AG =3

x2 ; GD =

3

x ; BG =

3

y2 ; GE =

3

y

In AGE :4

9

9

y

9

x4 22

or 16x2 + 4y2 = 81 .....(1)

In BGD : 49

y4

9

x 22

or x2 + 4y2 = 36 .....(ii)

(i) – (ii) , 15x2 = 45 x = 3

In ADC, cosC =)3()4(2

c169

6

5

)3()2(2

349 2

20 = 25 – c2 or c = 5

=2

1ab sinC =

2

1(3) (4) 11

6

51

2

sq. units ]

59. From triangle inequalitylog1012 + log1075 > log10nlog10900 > log10n n < 900 ....(1)

also log1012 + log10n > log1075log1012n > log107512n > 75

n >12

75or n >

4

25

Hence no. of values = 900 – 7 = 893 Ans. ]

60. x + 2y = 10where x is the number of times he takes single stepsand y is the number of times he takes two stepsCases Total number of ways

I: x = 0 and y = 5!5

!5= 1 (2 2 2 2 2)

II: x = 2 and y = 4 !4·!2

!6 = 15 (1 1 2 2 2 2)

III: x = 4 and y = 3 !3·!4

!7 = 35 (1 1 1 1 2 2 2)

IV: x = 6 and y = 2 !6·!2

!8 = 28 (1 1 1 1 1 1 2 2)

V: x = 8 and y = 1 9C1 = 9 (1 1 1 1 1 1 1 1 2)VI: x = 10 and y = 0 1 (1 1 1 1 1 1 1 1 1 1)hence total number of ways = 1 + 15 + 35 + 28 + 9 +1 = 89 Ans. ]


61. I = b

a

x bax

dxx

ee

let x = at dx = a dt

=

a b

1

at bt

dtat

eea

I =

1

tt

dtt

ee....(1) (where b/a =)

put t =y

dt = –

2y

dy

I = –

1

2

yy

dyy

·y)ee(

I =

1 yy

y

dy)ee(or I = –

1

tt

t

dt)ee(....(2)

from (1) and (2) 2I = 0 I = 0 Ans. ]

62.)6(f

)3(f =

92

92k 6

k 3

=

3

1; f (9) – f (3) = (29k + 9) – (23k + 9) = 29k – 23k ....(1)

3(23k + 9) = 26k + 9 26k – 3(23k ) – 18 = 0

23k = yy2 – 3y – 18 = 0(y – 6)(y + 3) = 0y = 6; y = – 3 (rejected)23k = 6

now f (9) – f (3) = 29k – 23k { from (1) }= (23k )3 – 23k

= 63 – 6 = 210hence N = 210 = 2 · 3 · 5 · 7Total number of divisor = 2 · 2 · 2 · 2 = 16number of divisors which are composite = 16 – (1, 2, 3, 5, 7) = 11 Ans. ]

63. Radius of the circle is 1

tan2

B =

PB

r =

) bs(s

PB = ) bs(s·r

= ) bs(·s

·s

= (s – b);

|||ly PC = (s – c)

(PB)(PC) = (s – b)(s – c) =)as(s

)cs)( bs)(as(s


=)as(s

·

= r ·)as(

=)as(

=

a

s1s

r

=

a

2

a32

a3 =

23

3

= 3 Ans. ]

64. 5x + 3x > 8 x > 15x + 8 > 3x x > – 4

and 3x + 8 > 5x x < 4Hence, x (1, 4). Now perimeter of the triangle = 8(x + 1)

s = 4x + 4A2(x) = ( 4(x + 4)(4 – x)(4x – 4)(x + 4) )

= – 16(x2 – 1)(x2 – 16)A2(t) = – 16(t – 1)(t – 16), where x2 = t, t (1, 16)A2 (t) = – 16[t2 – 17t + 16] = f (t)

f ' (t) = 0 t =2

17

max

2 )t(A = – 16

12

17

162

17= 16 ×

2

15 ×

2

15 = (2 × 15)2

(Area)max = 30 sq. units ]

65. From the identity

r = 4R · sin2

A · sin

2

B · sin

2

C

or r = 134 r · sin2

A · sin

2

B · sin

2

Cor 132

1

= 22

Bsin·

2

Csin·

2

Asin

let A B A – C = 30°

then4

13 =

2

CAcos

2

CAcos sin

2

B

4

13 = 2

Bsin

2

Bsin

4

26

Let sin2

B = x yields x2 –

4

26 x +

4

13 = 0,

whose solutions are x =4

26 and x =

2

2. It follows that

2

B = 15° or

2

B = 45°. The second

solution is not acceptable, because A B. Hence B = 30°, A = 90° and C = 60° ]


66. y = ax2

Tdt

dy = 2ax0 = m

hence line isy = (2ax0)x – b .....(1)

(x0, a 20x ) lies on parabola and the line (1)

a 20x = 2a 2

0x – b

b = a 20x . Hence Q = (0, – b) = (0, – a 2

0x )

now using (TQ)2 = 1

20x + 4a2 4

0x = 1

a2 = 40

20

x4

)x1( .....(2)

now A = 0x

0

2 dx) bmxax( =0x

0

23

bx2

mx

3

ax = 0

20

30 bx

2

mx

3

ax

= 30

30

30 axax

3

ax =

3

ax30

A2 =9

xa 60

2

=

40

20

60

x4

x1

9

x =

36

)x1(x 20

20

let A2 = f (x0) =36

)x1(x 20

20

This is maximum when 20x =

2

1

max

2A =36

1·

2

1·

2

1 =

144

1; Amax = 12

1

A

1 = 12 Ans. ]

67.

(i) Equation of tangent from point (3, –3) to the given circle isy + 3 = m(x – 3)mx – 3m – y – 3 = 0


and also 2m1

32m3m4

= 5

(1 + 7m)2 = 25(1 + m2) 1 + 49m2 + 14m = 25 + 25m2 12m2 + 7m – 12 = 0 (4m – 3)(3m + 4) = 0 m = 3/4 or m = – 4/3 equation of tangent at point A and B are

y + 3 = –3

4(x – 3) and y + 3 =

4

3(x – 3)

3y + 9 = – 4x + 12 4y + 12 = 3x – 94x + 3y = 3 3x – 4y = 21

(ii) Equation of normals to these 2 tangents are

y + 2 =4

3(x + 4) and y + 2 = –

3

4(x + 4)

4y + 8 = 3x + 12 3y + 6 = – 4x – 163(3x – 4y + 4 = 0) 4(4x + 3y= – 22)9x – 12y = – 12 16x + 12y = – 8816x + 12y = 12 9x – 12y = 63

—————— —————— x = 0; y = 1 25x = – 25

x = – 1; y = – 6 points A and B are (0, 1) and (–1, – 6) Ans.

(iii) angle between the 2 tangents = 90° ADB = 90°

| AD |max = CD + radius

CD = 50

| AD |max = 25 + 5

| AD |min = 25 – 5

(iv) Area of quadrilateral ADBC = AC × AD

AD = 2517 22 = 25 = 5

area of quadrilateral ABCD = 5 × 5 = 25 sq. units.

area of triangle DAB = 252

1 = 12.5 sq. units.

(iv) Circle circumscribing DAB will have points A and B as its diametrical extremitiesx2 + y2 – x(–1) – y(–5) – 6 = 0x2 + y2 + x + 5y – 6 = 0 Ans.

x-intercept = cg2 2 = 6)41(2 = 5 Ans.

y-intercept = cf 2 2 = 6)425(2 = 7 Ans. ]


68. Let, f (x) = x2 x1 + (x +1)2x2 + ........ + (x + 6)2x7 [if x = 1, we get 1st relation, and so on]note that degree of f (x) is 2hence f (x) = ax2 + bx + c where f (1) = 1, f (2) = 12 and f (3) = 123 to find f (4) = ?hence a + b + c = 1

4a + 2b + c = 129a + 3b + c = 123

solving a = 50, b = – 139, c = 90 f (4) = 16a + 4b + c = 800 – 556 + 90 = 334 Ans. ]

69. Suppose, circle x2 + y2 + 2gx + 2fy + c = 0Solving with x = at2 , y = 2ata2t4 + 4a2t2 + 2gat2 + 4aft + c = 0

t1 + t2 + t3 + t4 = = 0 ....(1) N : y + tx = 2at + at3

passing through (h, k)at3 + t(2a – h) – k = 0 ....(2)t1 + t2 + t3 = 0 ....(3)from (1) and (3) t4 = 0hence circle passes through the origin c = 0

equation of the circle after cancelling –atat3 + 4at + 2gt + 4f =0at3 + 2(2a + g)t + 4f = 0 ....(3)

Now (2) and (3) must be represent the same equation2(2a + g) = 2a – h 2g = – (2a + h)

and 4f = – k 2f = – k/2 equation of circle is x2 + y2 – (2a + h)x – (k/2)y = 0

x2 + y2 – 17x – 6y = 0 Ans.

Centroid of PQR =3

)ttt(a 23

22

21

,3

)ttt(a2 321

xa = 3

a[(t1 + t2 + t3)2 – 2 21 tt ]

= – 21 tt3

a2 = –

3

a2.

a

)ha2( = –

3

2(2a – h)

=3

26(a = 1 ; h = 15 )

C :

0,3

26]

70. Area = ab2

1; also a2 + b2 = 3600

AD : y = x + 3BE : y = 2x + 4

)2,1(Ggettosolve

acute angle between the medians is given by

tan =21

21

mm1

mm

=

21

12

=3

1 tan =

3

1


In quadrilateral GDCE, we have(180 – ) + 90° + + = 360°

= + – 90°cot = – tan( + )

– 3 =

tantan1

tantanor – 3 =

b

a2·

a

b21

b

a2

a

b2

9 =

ab

) ba(2 22

9ab = 2 × 3600 ab2

1 = 400

Area = 400 sq. units ]

71. W1: C1 = (–5, 12) W2: C2 = (5, 12)r 1 = 16 r 2 = 4now, CC2 = r + 4

CC1 = 16 – rlet C(h, k) = c(h, ah)

CC12 = (16 – r)2

(h + 5)2 + (12 – ah)2 = (16 – r)2

CC22 = (4 + r)2

(h – 5)2 + (12 – ah)2 = (4 + r)2

By subtraction20h = 240 – 40r

h = 12 – 2r 12r = 72 – 6h ...(1)By addition

2[h2 + 25 + a2h2 – 24ah + 144] = 272 – 24r + 2r 2

h2(1 + a2) – 24ah + 169 = 136 – 12r + r 2 = 136 + (6h – 72) +

2

2

h12

[using (1)]

4[h2(1 + a2) – 24ah + 169] = 4[64 + 6h] + (12 – h)2 = 256 + 144 + h2

h2(3 + 4a2) – 96ah + 105 · 4 – 36 · 4 = 0 h2(3 + 4a2) – 96ah + 69 · 4 = 0; for 'h' to be real D 0 (96a)2 – 4 · 4 · 69 (3 + 4a2) 0 576a2 – 69.3 – 276a2 0

300a2 207 a2 100

69; hence m (smallest) =

10

13

So, m2 =100

69; p + q = 169 Ans. ]

72. I = 3

65

42 dxxsec)xsin1( = 3

65

42 dxxsec)xsinxsin21(

= 3

65

22

65

2

65

22 dx)x(tanxsecdxxsecxtanxsec2dx)xtan1(xsec

= 3

65

2

65

22 dx)xsecxtanx(sec2dxxsec)xtan21(


= 3

1

32

20

31

2 dtt2dt)t21( =

1

31

3

0

31

3

t3

2

3

t2t

= 3

33

8)1(

3

2

33

1

3

2

3

1)0(

= 3

33

81

3

2

39

2

3

1 =

33

833

3

2

39

11 = 3

39

163611

33

536 = 2 –

9

35 = a –

c

3 b a = 2, b = 5, c = 9 a + b + c + abc = 106 Ans.]

73. I =

1

0 2x1x1

dx

put x = cos 2 dx = – 2 sin 2 d

I = 2

4

02sin2cos2

d 2sin =

4

0 2sincos

d 2sin2 =

4

0 24

cos2

d 2sin2

I =

4

0 14

cos

d 2sin =

4

0

d 1cos

2cos =

4

0

2

d cos1

sin21

=

4

0

d cos1

1 –

4

0

d )cos1(2 =

4

02

d sin

)cos1( –

4

0

d )cos1(2

=

4

0

2 d )coseccot(cosec –

4

0

d )cos1(2 = 4

0

4

0

sin2coseccot

=

2

1

4cos

cos1Lim12

0 = 2

212

=2

122

4

18

a = 8, b = 1, c = 4 a2 + b2 + c2 = 81 Ans. ]

74. x · g )x( f )x(')x(' gg f = )x(')x(')x( f f gg f

)x(dx

d )x(·x g f f g = )x(

dx

d )x( f gg f

x ·

)x(

)x(dx

d

)x(

)x(dx

d

f g

f g

g f

g f


)x(ndx

d )x(n

dx

d ·x f glg f l ...(1)

now, 2

e1dx(x)

a2a

0

g f

differentiate w.r.t. 'a'

(a)g f = e –2a (x)g f = e –2x (x)n g f l = – 2x ....(2)

from (1) and (2) we get

– 2x = (x)ndx

d f gl (x)n f gl = – x2 + C

put x = 0, C = 0

2xe(x) f g ; Hence )4( f g = e –16 k = 16 Ans. ]

75. Let f (x) = y

dx

dy + y = 4xe –x · sin 2x (linear differenial equation)

I.F. ex

yex = 4 dxx2sinx

III

yex = 4

dxx2cos2

1

2

x2cosx

yex = 4

4

x2sin

2

x2cosx + C

yex = (sin 2x – 2x cos 2x) + Cf (0) = 0 C = 0

y = e –x(sin 2x – 2x cos 2x)now f (k ) = e –k (sin 2k – 2k · cos 2k ) = e –k (0 – 2k )

f (k ) = – 2 (k · e –k )

)k (f = – 2

S

1k

k ke

S = 1 · e – + 2e –2 + 3e –3 + ......... + S e – = + e –2 + 2e –3 + ......... +

—————————————————— S(1 – e – ) = e – + e –2 + e –3 + ......

S(1 – e – ) =

e1

e =

1e

1

S =)e1)(1e(

1

= 2)1e(

e

2)1e(

e2

Ans. ]


76. f (x) = Limith0

f h f

h

(x ) (x)

=h

1)x(f

)hx(f )x(f

0hLimit

= f(x) · Limit

h

f x h

x

h

0

1

= f(x) ·

x

hx

1x

h1f

0hLimit

=

f

xLimitt

f t

t

(x)

0

1 1

Now putting x = 1, y = 1 in functional rule

f(1) =f

f

( )

( )

1

1 = 1

f (x) =f

x

(x) · f (1) =

2f

x

(x)

f

f

'(x)

(x)=

2

x

ln (f(x)) = 2lnx + Cx = 1; f(1) = 0 C = 0 ; f(x) = x2

Now solving y = x2 and x2 + y2 = 2y2 + y – 2 = 0(y + 2) (y – 1) = 0y = 1

A = 2 2 2

0

1

y y dy

= 2 2 2

0

1

0

1

y dy y dy

= now y dy y

2

3

1

21

0

1

0

1

=2

3

and 2 2

0

1

y dy y = 2sin

2 20

4

cos cos/

d 2 2

0

4

cos/

d = ( cos )/

1 20

4

d

= +1

22

0

4

sin/

4

1

2

Hence A = 2

3

2

2

1

4; A =

3

1

2 sq. units ]


77. Z10 + Z10

10

Z

113

= 0

10

Z

113

= – 1 = cos + i sin

13 –Z

1 = 101m2sini1m2cos

= 10

)1m2(

e

i

Z

1 = 13 – 10

)1m2(

e

i

substituting m = 0, 1, 2,.......9 we get

1Z

1 = 13 – 10e

i

2Z

1 = 13 – 10

3

e

i

3Z

1 = 13 – 10

5

e

i

conjugatecomplexareZ

1and

Z

1note

101

10Z

1 = 13 – 10

19

e

i

Let1Z

1 =

1a

1 and

10Z

1 =

1 b

1 and so on

ii ba

1 = 169 – 13 [ 10e

i

+ 10e

i

] + 1

= 169 – 13 [ 10

3

e

+ 10

3

e

] + 1

ii ba

1 = 170 – 26 Re 10e

i

and 22 ba

1 = 170 – 26 Re 10

3

e

i

etc

ii ba

1 = 850 – 26

10

9cos

10

3cos

10

5cos

10

3cos

10cos

= 850 – 26[cos18º + cos54° + cos90° + cos126° + cos162°]= 850 Ans. ]


78.(i) an + bn + cn = C0 + C1 + C2 + C3 + C4 + ................ an + bn + cn = 2n ....(1)now (1 + x)n = C0 + C1 x + C2 x2 + C3 x3 + ................

put x =(1 + )n = C0 + C1 + C2

2 + C3 3 + C4

4 +................= (C0 + C3 + C6 + .......) + (C1 + C4 + C7 + ........) + 2(C2 + C5 + C8 + ........)

(1 + )n = an + bn + 2cn ....(2)|||ly (1 + 2)n = an + 2 bn + cn ....(3)

now 3n

3n

3n c ba – 3an bncn = (an + bn + cn) (an + bn + 2cn) (an + 2 bn + cn)

= 2n(1 + )n (1 + 2)n

= 2n(– 2)n (– )n = 2n

also 2nn ) ba( = 2(an + bn + 2cn) (an + 2 bn + cn)

2nn ) ba( = 2 Ans.

78.(ii) Let x = C0 – C2 + C4 – C6 + .....and y = C1 – C3 + C5 – C7 + .......

(1 + i)n = C0 + C1 i + C2 i2 + C3 i3 + C4 i4 + .........equating the real and imaginary part

xn + i yn = (1 + i)n

| xn + iyn | = | 1 + i |n = 2n/2

2n

2n yx = 2n/2

hence 2n

2n yx = 2n hence proved ]

79. A2 =

531531

531

531531

531

=

531531

531

= A matrix A is idempotent

Hence A2 = A3 = A4 = ....... = A x = 2, 3, 4, 5, ..........

now

n

2x3

3

n 1x

1xLim

n

2x2

2n

2xn 1xx

1xx

1x

1xLim

1nn

1nn.......

21

13·

13

7·

7

3

)1n(

)1n(n.......

3

5·

2

4·

1

3Lim

2

2

n

1nn

3·

2·1

)1n(nLim

2n

=

2

3Ans. ]

80. Given log

a

ca + log

b

a = log 2

log

b

ca = log 2

BANSAL CLASSES - Doubtion

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