bansal classes - doubtion
TRANSCRIPT
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]
B B ansal ansal C C llaasssseess Problems fo Problems for JEE-20r JEE-2007 07 [3] [3]
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]
B B ansal ansal C C llaasssseess Problems fo Problems for JEE-20r JEE-2007 07 [4] [4]
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]
B ansal C lasses Problems for JEE-2007 [6]
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]
B ansal C lasses Problems for JEE-2007 [7]
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]
B ansal C lasses Problems for JEE-2007 [8]
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
B ansal C lasses Problems for JEE-2007 [9]
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
B ansal C lasses Problems for JEE-2007 [10]
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.
B ansal C lasses Problems for JEE-2007 [11]
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
B ansal C lasses Problems for JEE-2007 [12]
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 ]
B ansal C lasses Problems for JEE-2007 [13]
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
B ansal C lasses Problems for JEE-2007 [14]
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. ]
B ansal C lasses Problems for JEE-2007 [15]
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
B ansal C lasses Problems for JEE-2007 [16]
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
B ansal C lasses Problems for JEE-2007 [17]
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
B ansal C lasses Problems for JEE-2007 [18]
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
B ansal C lasses Problems for JEE-2007 [19]
=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
B ansal C lasses Problems for JEE-2007 [20]
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
B ansal C lasses Problems for JEE-2007 [21]
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. ]
B ansal C lasses Problems for JEE-2007 [22]
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. ]
B ansal C lasses Problems for JEE-2007 [23]
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. ]
B ansal C lasses Problems for JEE-2007 [24]
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 ]
B ansal C lasses Problems for JEE-2007 [25]
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}
B ansal C lasses Problems for JEE-2007 [26]
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)
B ansal C lasses Problems for JEE-2007 [27]
(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
B ansal C lasses Problems for JEE-2007 [28]
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
B ansal C lasses Problems for JEE-2007 [29]
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. ]
B ansal C lasses Problems for JEE-2007 [30]
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. ]
B ansal C lasses Problems for JEE-2007 [31]
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
B ansal C lasses Problems for JEE-2007 [32]
=)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° ]
B ansal C lasses Problems for JEE-2007 [33]
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
B ansal C lasses Problems for JEE-2007 [34]
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. ]
B ansal C lasses Problems for JEE-2007 [35]
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
B ansal C lasses Problems for JEE-2007 [36]
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(
B ansal C lasses Problems for JEE-2007 [37]
= 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
B ansal C lasses Problems for JEE-2007 [38]
)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. ]
B ansal C lasses Problems for JEE-2007 [39]
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 ]
B ansal C lasses Problems for JEE-2007 [40]
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. ]
B ansal C lasses Problems for JEE-2007 [41]
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