BANSAL CLASSES TARGET IIT JEE 2007
MATHEMATICS XII & XIII
FINAL PRACTICE PROBLEMS FOR IIT JEE-2007 (With Hints and Solutions at the End)
Q.1 to Q.29 are of 6 Marks Problems Q.30 to Q.66 are of 8 Marks Problems Q.67 to Q.82 are of 10 Marks Problems Q.83 to Q.100 are Objective type problems.
Advise : Do not spend more than 10 minutes for each problem and then read the solution and then do it.
ALL THE BEST FOR JEE -2007
SUBJECTIVE: Q.1
If the sum of the roots of the equation 2
333 x 2
111x 1
2
2
222 x 2
S1 1 is expressed in the form S find 2
S1 S1 + S2, where S is in its lowest form. 2
[6]
Q.2
Let K is a positive integer such that 36 + K, 300 + K, 596 + K are the squares of three consecutive terms of an arithmetic progression. Find K. [6]
Q.3
Find the number of 4 digit numbers starting with 1 and having exactly two identical digits.
Q.4
A chord of the parabola y2 = 4ax touches the parabola y2 = 4bx. Show that the tangents at the extremities of the chord meet on the parabola by2 = 4a2x. [6]
Q.5
Consider a circle S with centre at the origin and radius 4. Four circles A, B, C and D each with radius unity and centres (–3, 0), (–1, 0), (1, 0) and (3, 0) respectively are drawn. A chord PQ of the circle S touches the circle B and passes through the centre of the circle C. If the length of this chord can be expressed as
Q.6
Integrate
2
Q.7
If
0
[6]
x , find x.
[6]
x7 dx (1 x 2 ) 5
[6]
1 sin 2 x dx = a where a, b are relatively prime find a + b + ab. 2 (1 sin 2 x ) b
[6]
Q.8
A bus contractor agrees to run special buses for the employees of ABC Co. Ltd. He agrees to run the buses if atleast 200 persons travel by his buses. The fare per person is to be Rs. 10 per day if 200 travel and will be decreased for everybody by 2 paise per person over 200 that travels. How many passengers will give the contractor maximum daily revenue? [6]
Q.9
If the point P(a, b) lies on the curve 9y2 = x3 such that the normal to the curve at P makes equal intercepts with the axes. Find the value of (a + 3b). [6]
Q.10 Let x(t) be the concentration of glucose per unit volume of blood at time t, being the amount of glucose being injected per unit volume per unit time. If the glucose is disappearing from the blood at a rate proportional to the concentration of glucose (K being the constant of proportionality), find x(t). Also find the ultimate concentration of glucose as t . [6] Q.11
Find the value (s) of the parameter 'a' (a > 0) for each of which the area of the figure bounded by the straight line, y =
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a2 a x x2 2 a x 3a 2 & the parabola y = is the greatest. 1 a4 1 a4
Problems for JEE-2007
[6]
[2]
Q.12 Mr. A is a compulsive liar. He lies 2 5 of the time. However a clue to his validity is that his ears droop
2 3 of the time when he is telling a lie. They only droop 1 10 of the time when he is telling the truth. Mr. A tells his friend Mr. B that "certain event has occured" and his ears were dropping as noticed by Mr. B. Find the probability that Mr. A was telling the truth. [6] Q.13 Five persons entered the lift cabin on the ground floor of an eight floor house. Suppose that each of them , independently & with equal probability can leave the cabin at any floor beginning with the first, find out the probability of all 5 persons leaving at different floors. [6] Q.14 Let u and v be non zero vectors on a plane or in 3-space. Show that the vector w | u | v | v | u bisects the angle between u and v . [6]
Q.15 Find the distance from the line x = 2 + t , y = 1 + t , z =
1 1 t to the plane x + 2y + 6z = 10. 2 2 [6]
Q.16 If is the angle between the lines in which the planes 3x – 7y – 5z = 1 and 5x – 13y + 3z + 2 = 0 cuts the plane 8x – 11y + 2z = 0, find sin. [6] Q.17 Suppose u, v and w are twice differentiable functions of x that satisfy the relations au + bv + cw = 0
u v w u ' v' w ' = 0. where a, b and c are constants , not all zero. Show that u ' ' v' ' w ' '
[6]
A B C 3 Q.18 In any triangle ABC, prove that, cos A · sin2 + cos B · sin2 + cos C · sin2 . [6] 2 2 2 8 3 Q.19 If the normals to the curve y = x2 at the points P, Q and R pass through the point 0, , find the radius 2 of the circle circumscribing the triangle PQR. [6]
Q.20 Let A = {a R | the equation (1 + 2i)x3 – 2(3 + i)x2 + (5 – 4i)x + 2a2 = 0} has at least one real root. Find the value of
a2 .
[6]
aA
Q.21 Find the equation of a line passing through (– 4, –2) having equal intercepts on the coordinate axes. [6] Q.22 Let S be the set of all x such that x4 – 10x2 + 9 0. Find the maximum value of f (x) = x3 – 3x on S. [6] Q.23 Solve the differential equation, (x4y2 – y)dx + (x2y4 – x)dy = 0
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Problems for JEE-2007
(y(1) = 1)
[6]
[3]
Q.24 All the face cards from a pack of 52 playing cards are removed. From the remaining pack half of the cards are randomly removed without looking at them and then randomly drawn two cards simultaneously from the remaining. If the probability that two cards drawn are both aces is
p( 38C 20 ) 40
C 20 · 20C 2
, find p. [6]
x 2 y2 Q.25 A circle intersects an ellipse 2 2 = 1 precisely at three points A, a b B, C as shown in the figure. AB is a diameter of the circle and is perpendicular to the major axis of the ellipse. If the eccentricity of the ellipse is 4/5, find the length of the diameter AB in terms of a. [6] Q.26 Suppose R is set of reals and C is the set of complex numbers and a function is defined as f : R C, f (t) =
1 ti where t R, prove that f is injective. 1 ti
[6]
Q.27 Circles A and B are externally tangent to each other and to line t. The sum of the radii of the two circles is 12 and the radius of circle A is 3 times that of circle B. The area in between the two circles and its b external tangent is a 3 – then find the value of a + b. 2
[6]
0 0 1 Q.28 Define a matrix A = 3 0 . Find a vertical vector V such that (A8 + A6 + A4 + A2 + I) V = 11 where I is a unit matrix of order 2. [6] Q.29 A circle is inscribed in a triangle with sides of lengths 3, 4 and 5. A second circle, interior to the triangle, is tangent to the first circle and to both sides of the larger acute angle of the triangle. If the radius of teh second circle can be expressed in the form (0, 90°), find the value of k + w.
sin k where k and w are in degrees and lie in the interval cos w [6]
ax 2 24x b Q.30 If the equation = x, has exactly two distinct real solutions and their sum is 12 then find x2 1 the value of (a – b). [8] Q.31 If a, b, c and d are positive integers and a < b < c < d such that a, b, c are in A.P. and b, c, d are in G.P. and d – a = 30. Find the four numbers. [8] Q.32 Let the set A = {a, b, c, d, e} and P and Q are two non empty subsets of A. Find the number of ways in which P and Q can be selected so that P Q has at least one common element. [8] Q.33 If the normals drawn to the curve y = x2 x + 1 at the points A, B & C on the curve are concurrent at the point P (7/2, 9/2) then compute the sum of the slopes of the three normals. Also find their equations and the co-ordinates of the feet of the normals onto the curve. [8]
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Problems for JEE-2007
[4]
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 the equation of the conic and also the equation of a circle touching the conic at A(1, 4) and passing through its 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 + b 2 where a, b N, find the value of a3 + b3. [8]
Q.37 For a 2, if the value of the definite integral
0
dx 1 a x x
2
equals
2
. Find the value of a. 5050
[8]
4
Q.38 If
2 ( 4) tan = ln k – , find the value of (kw), where k, w N. d 1 tan w 4
[8]
1
Q.39 Given a function g, continuous everywhere such that g(1) = 5 and g (t) dt = 2. 0
If f(x) =
1 2
x
(x t)2 g (t) dt , then compute the value of f (1) f (1).
[8]
0
1
Q.40 Let f : [0, 1] R is a continuous function such that
f (x )dx = 0. Prove that there is c (0, 1) such 0
c
that f ( x )dx = f (c).
[8]
0
Q.41 Consider the equation in x, x3 – ax + b = 0 in which a and b are constants. Show that the equation has only 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 salt are 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 the amount 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-III contains 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]
Bansal Classes
Problems for JEE-2007
[5]
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 probability that 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 that the 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 hidden in 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 animal with 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 the quadrilateral 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 x 1 y3 z2 x2 y4 z 1 = = and = = 2 1 4 3 2 2 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 to the line x = 1 + t, y = 1 – t and z = 2t and also intersects this line. [8] Q.51 Suppose that r1 r2 and r1r2 = 2 (r1 , r2 need not be real). If r1 and r2 are the roots of the biquadratic x4 – x3 + ax2 – 8x – 8 = 0 find r1, r2 and a. [8]
Q.52 Express
x 2 y2 a 2 2ax xy 2ay x 2
2ax xy 2ay x 2 a 2 2x 2 2ax xy 2 2ax xy x y 2 a 2
as a product of two polynomial.
[8]
1 2 2 Q.53 Given the matrices A = 2 2 3 ; C = 1 1 3 Solve the matrix equation Ax = b. Q.54 Prove that
2 1 1 2 2 1 and D = 1 1 1
10 13 and that Cb = D. 9
a b c 3 + + for a, b, c > 0. bc ca ab 2
Q.55 Given x, y R,
x2 + y2 > 0. If the maximum and minimum value of the expression
M and m, and A denotes the average value of M and m, compute (2007)A.
[8] [8] x 2 y2 are x 2 xy 4 y 2 [8]
Q.56 Prove that the triangle ABC will be a right angled triangle if cos
A B C A B C 1 cos cos – sin sin sin = 2 2 2 2 2 2 2
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Problems for JEE-2007
[8]
[6]
Q.57 A point P is situated inside an angle of measure 60° at a distance x and y from its sides. Find the distance of the point P from the vertex of the given angle in terms of x and y. [8] Q.58 In ABC, 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 of n. [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 any combination of 1's and 2's. Find the total number of ways in which the person can go up the stairs. [8] b
Q.61 Let a and b be two positive real numbers. Prove that
ex a eb x x dx = 0 a
[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 the distance from A to l is 3. Two tangents intersect line l 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 possible area of the triangle. [8] R = 3 + 1, where R is the radius of the r circumcircle and r is the radius of the incircle. Find C in degrees. [8]
Q.65 In triangle ABC, max {A, B} = C + 30° and
Q.66 The parabola P : y = ax2 where 'a' is a positive real constant, is touched by the line L: y = mx – b (where m 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 of
1 . A
[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 point A 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 the DAB. (v) Equation of the circle circumscribing the DAB and also the intercepts made by this circle on the coordinate axes. [10] 7
Q.68 If
i 2 x i 1 and i 1
7
7
(i 1) 2 x i 12 and
(i 2) 2 x i
i 1
i 1
123 ,
7
then find the value of
(i 3)2 x i .
[10]
i 1
Bansal Classes
Problems for JEE-2007
[7]
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 lies along 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 = 0 respectively. Let m be the smallest positive value of 'a' for which the line y = ax contains the centre of a p circle that is externally tangent to W2 and internally tangent to W1. Given that m2 = where p and q are q relatively prime integers, find (p + q). [10]
Q.72 If
3 b 3 dx = a – where a, b, c N and b, c are relatively prime, find the value of 2 c 5 6 (1 sin x )
a + b + c + abc. 1
Q.73 If
0
[10]
dx = 1 x 1 x 2
a b
where a,b,c N, find the value a2 + b2 + c2. c
[10]
Q.74 Suppose f (x) and g (x) are differentiable functions such that x g f ( x ) f ' g (x )g ' (x ) = f g (x )g ' f (x ) f '( x ) a
for all real x. Moreover, f (x) is nonnegative and g (x) is positive. Furthermore,
f g( x ) dx 1
0
for all reals a. Given that g f (0) = 1. If the value of g f (4) = e–k where k N, find k.
e 2a 2
[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 n
of Lim
n
f (k) .
[10]
k 1
x f (x) Q.76 Let f be a differentiable function satisfying the condition f = (y 0, f(y) 0) V x, y R and y f (y)
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. Find
1
a b
.
[10]
i i
Q.78(i) Let Cr's denotes the combinatorial coefficients in the expansion of (1 + x)n, n N. If the integers an = C0 + C3 + C6 + C9 + ........ bn = C1 + C4 + C7 + C10 + ........ and cn = C2 + C5 + C8 + C11 + ........, then prove that (a) a 3n b3n c3n – 3anbncn = 2n, (ii) Prove the identity:
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(b) (an – bn)2 + (bn – cn)2 + (cn – an)2 = 2.
[10]
(C0 – C2 + C4 – C6 + .....)2 + (C1 – C3 + C5 – C7 + .......)2 = 2n Problems for JEE-2007
[8]
5 1 3 Q.79 Given the matrix A = 1 3 5 and X be the solution set of the equation Ax = A, 1 3 5 where x N – {1}. Evaluate
x3 1 x 3 1 where the continued product extends x X.
[10]
c Q.80 If a, b, c are the sides of triangle ABC satisfying log 1 + log a – log b = log 2. Also a 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
1 For x (0, /2) and sin x = , if 3
sin( nx ) ab b = then find the value of (a + b + c), n 3 c n 0
where a, b, c are positive integers. (You may Use the fact that sin x =
eix e ix ) 2i
[10]
Q.82 Two distinct numbers a and b are chosen randomly from the set {2, 22, 23, 24, ......, 225}. Find the probability that logab is an integer. [10] O BJEC T IV E 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? ("The wood medium red square" is such a block) (A) 29 (B) 39 (C) 48 (D) 56 49
Q.84 The sum
n!
n
(1) k 299k where r r!(n r)! equals
k 0
(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 of (A) 4
(B) – 4
(C) 2
u c equals A
(D) – 2
Q.86 Consider the polynomial equation x4 – 2x3 + 3x2 – 4x + 1 = 0. Which one of the following statements describes 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
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(C) 7
Problems for JEE-2007
(D) 9
[9]
Q.88 The polynomial f (x) = x4 + ax3 + bx2 + cx + d has real coefficients and f (2i) = f (z + i) = 0. The value of (a + b + c + d) equals (A) 1 (B) 4 (C) 9 (D) 10
Q.89 If the sum
( k 2) k 1
equals (A) 6
1 = k k k2
(B) 8
a b where a, b, c N and lie in [1, 15] then a + b + c c (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. cm is (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 the coefficient 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 balls in 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 toy helicopter 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 4 5 Q.93 The value of tan arc sin arc cos is equal to 5 13 (A)
25 63
(B) –
3 7
(C) –
33 56
(D)
16 63
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 byone the A.P. becomes a G.P. In original A.P. if the largest number is increased by two, the A.P. also becomes a G.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 B intersect each other at the point G. If a third normal to the parabola through G meets the parabola at C then 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.
Bansal Classes
Problems for JEE-2007
[10]
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, ±
1 2
(C) (x) attains its local maxima at 0.
Q.98
(B) (x) increases in 1
(D) (x) decreases in 1
2 , 0 1
2,0 1
2,
2,
2 2 sin 3 sin x x = If tan where 0 < x < , then the value of x is 2 3 cos cos x 3 (A)
12
(B)
5 12
(C)
7 12
(D)
MATCH THE COLUMN: Q.99 Column-I (A) The smallest positive integeral value of n for which the complex
(B)
(C)
(D)
Q.100 (A)
11 12
Column-II (P) 4
number 1 3 i n 2 is real, is Let z be a complex number of constant non zero modulus such that z2 is purely imaginary, then the number of possible values of z is 3 whole numbers are randomly selected. Two events A and B are defined as A : units place in their product is 5. B : their product is divisible by 5. If p1 and p2 are the probabilities of the events A and B such that p2 = kp1 then 'k' equals
(Q)
6
(R)
8
(S)
9
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
Column-I For real a and b if the solutions to the equation Z9 – 1 = 0 are written in the form of a + ib then the number of distinct ordered pairs (a, b) such that a and b are positive, is
Column-II (P) 0 (Q)
1
(R)
2
(S)
e
x
(B)
x
(C) (D)
Lim e
x
1 e 1 x
Let A, B be two events with P(B) > 0. If B A then P(A/B) equals A real number x is chosen at random such that 0 x 100. 1 a is , where a and b are relatively 3 b primes and [x] denotes the greatest integer then (b – a) equals
The probability that x – [x]
Bansal Classes
Problems for JEE-2007
[11]
HINTS AND SOLUTIONS 1.
Let 2111x = y so that log2y = 111 x equation becomes
x=
log 2 y 111
y3 + 2y = 4y2 + 1 4 y3 – 16y2 + 8y – 4 = 0 sum of the roots of the given equation is
x1 + x2 + x3 = 2.
log 2 y1 log2 y 2 log 2 y3 log 2 ( y1y 2 y 3 ) log 2 4 2 = = = S1 + S2 = 113 Ans.] 111 111 111 111
Let the 3 consecutive terms are a – d, a, a + d d>0 hence a2 – 2ad + d2 = 36 + K ....(1) 2 a = 300 + K ....(2) 2 2 a + 2ad + d = 596 + K ....(3) now (2) – (1) gives d(2a – d) = 264 ....(4) (3) – (2) gives d(2a + d) = 296 ....(5) (5) – (4) gives 2d2 = 32 d2 = 16 Hence from (4) 4(2a – 4) = 264 2a – 4 = 66 2 K = 35 – 300 = 1225 – 300 = 925 Ans.]
d = 4 (d = – 4 rejected)
2a = 70
a = 35
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 2 2at1t 2 y=t t x +t t 1 2 1 2 2a 2h x + a ....(1) k k Since (1) touches y2 = 4bx , using the condition of tangency
y=
2ah bk k 2a Locus is by2 = 4a2x ]
Bansal Classes
Problems for JEE-2007
[12]
5.
Note that triangles BCM and OCN are similar now let ON = p. N will be mid point of chord PQ
p 1 = 1 2
now
R = 2 r 2 p 2 for large circle
p=
= 2 16 (1 4) = 63 Equation of large circle as x2 + y2 = 16
Alternatively:
now
C = (1, 0) with slope PQ = –
equation of PQ :
x7 dx = (1 x 2 ) 5
=
x 3
x
1 =– 2
7.
2
1
5
I=
0
1 2
x7 1 x 2 1 x
5
result
dx
]
Taking x2 out of the bracket
10
Put x –2 – 1 = t =
2 dx = dt x3
4 1 1 dt 1 t 1 1 x8 = – = + C = = 4 8 1 2 4 8 t4 t5 8 1 x2 2 1 x
Using sin 2x =
2
dx
1 (think !) 3
3y+x=1
P (from origin) =
6.
1 2
4
+C]
2 tan x 1 tan 2 x
2 tan x 1 tan 2 x dx = 2 tan x 1 1 tan 2 x 1
put y = tan x
2
0
(1 tan x ) 2 ·(1 tan 2 x ) dx = 4 (1 tan x )
2
0
(1 tan x ) 2 ·sec 2 x dx 4 (1 tan x )
dy = sec2x dx
(1 y) 2 dy I= 4 ( 1 y ) 0
now put 1 + y = z
dy = dz
3z 2 6 z 4 (2 z ) 2 dz = I= 3z 3 z4 1 2
Alternatively:
I=
0
Bansal Classes
= 1
1 3
a = 1, b = 3 1 + 3 + 3 = 7 Ans. ]
(cos x sin x ) 2 dx (cos x sin x ) 4
Problems for JEE-2007
[13]
1 I=– 3
2
d
1
dx x sin x ) · (cos dx (cos x sin x )3 0
I
II
integrating by parts 1 (cos x sin x ) =– 3 (cos x sin x )3
= 8.
2
0
2
0
0
2 (sin x cos x ) 1 dx dx ( 1 ) ( 1 ) = – 1 sin 2x 3 (cos x sin x )3 0
2 tan x
using sin 2x = 2 1 = – 3 3
2
1 tan 2 x
sec 2 x 2 1 dx = – 2 3 3 (1 tan x )
1 2 1 – = 3 3 3
dt
t2
=
1
2 1 t + t 3 3
0
=
2 1 + [(0) – (1) 3 3
a = 1, b = 3 1 + 3 + 3 = 7 Ans. ]
Let the number of passengers be x ( x > 200) Fair changed per person = 10 – (x – 100)
2 100
2x 2 2x 2 4x ( x 200) = 10 x Total revenue = x . 10 ( x 200) = 10 x 100 100 100
f (x) = 14x – f (x) = 14 – f (x) < 0 9.
2x 2 100 4x =0 x = 350 100 x = 350 gives maxima]
Given 9y2 = x3 Let the point on the curve be x = t2 and y = dx = 2t ; dt
t3 3
dy = t2 dt
dy dy dt t 2 t2 = × = = slope of the normal = – dx dt dx 2 t 2t normal makes equal intercept
hence –
2 =–1 t
t=2
8 ) 3
a + 3b = 4 + 3 ·
Hence P = (4,
Bansal Classes
8 = 4 + 8 = 12 Ans. ] 3
Problems for JEE-2007
[14]
10.
Amount of glucose in blood at time t is x (t)
hence
dx = – K x dt
dx dt K x 1 ln ( – K x) = t + C K ln ( – K x) = – Kt + C – K x = e– K t + C e K t C x= K Lim x ( t ) ] t K
–
x2
11.
A=
(a 2 a x) (x 2 2 a x 3 a 2 ) 1 a4
x1
dx
where x1 & x2 are the roots of , x2 + 2 a x + 3 a2 = a2 a x x = a or x = 2 a A=
12.
dA a3 = 0 gives a = 31/4 Ans. ] 4 d a 6(1 a )
A : ears of Mr A formed to be drooping B1 : Mr A was telling a truth P(B1) = 3/5 B2 : Mr B was telling a false P(B2) = 2/5 P(A/B1) = 1/10 P(A/B2) = 2/3 3 1 · 9 3 5 10 P(B1/A) = = = Ans. ] 40 49 3 1 2 2 3 · · 3 5 10 5 3
13.
E : all the 5 persons leave at different floors n(S) = 85 n(A) = 8C5 · 5! 8
105 C 5 5! = ans. ] 5 512 8 u ·w u ·(| u | v | v | u ) ( u · v) | u | | v | | u |2 = cos = | w = || u | | u || w | | u || w | ( u · v ) | v | | u | cos = ....(1) |w| v ·w v ·(| u | v | v | u ) | v |2 | u | ( v ·u ) | v | = cos = | w = || v | | w || v | | w || v|
14.
P(E) =
Bansal Classes
Problems for JEE-2007
[15]
| v | | u | ( v ·u ) cos = ....(2) |w| from (1) and (2) cos = cos = ] 15.
1 z x 2 y 1 2 t The line is 1 1 1 2
....(1)
1 1 line passes through 2ˆi ˆj kˆ and is parallel to the vector V ˆi ˆj kˆ 2 2 ˆ ˆ ˆ vector normal to the plane x + 2y + 6z = 10, is n i 2 j 6k line (1) is | | to the plane V .n = 1 + 2 – 3
16.
d=
2 2 3 10 = 1 4 36
9 Ans 41
]
Vector v1 along the line of intersection of 3x – 7y – 5z = 1 and 8x – 11y + 2z = 0 is given by
ˆi ˆj kˆ v1 n1 n 2 = 3 7 5 = – 23( 3ˆi 2ˆj kˆ ) 8 11 2 |||ly vector v 2 along the line of intersection of the planes 5x – 13y + 3z = 0 and 8x – 11y + 2z = 0 is
ˆi ˆj v 2 n 3 n 4 = 5 13 8 11
kˆ 3 = 7 ( ˆi 2ˆj 7 kˆ ) 2
now v1 · v 2 = 0 angle is 90° sin90° = 1
17.
Given and
]
au + bv + cw = 0 ....(1) au + bv + cw = 0 ....(2) au + bv + cw = 0 ....(3)
u v w u For non trivial solution (non zero) solution of a, b and c . We must have ' v' w ' = 0 ] u ' ' v' ' w ' ' 18.
Let
A B C y = cos A · sin2 + cos B · sin2 + cos C · sin2 2 2 2
=
1 [cosA (1 – cosA) + cosB (1 – cosB) + cos C (1 – cos C)] 2
=
1 [(cosA – cos2A) + (cosB – cos2B) + (cosC – cos2C)] 2
2 2 2 1 1 1 1 1 1 1 cos A cos B cos C = 2 2 4 2 4 2 4
Bansal Classes
Problems for JEE-2007
[16]
1 y= 2
2 2 2 3 1 1 1 cos A cos B cos C 2 2 2 4
now y will be maximum if cosA = cos B = cos C =
1 2
hence ymax = 3/8 ] 19.
y = x2; x = t; y = t2 dy = 2x = 2t dx
slope of normal m = –
1 2t
equation of normal y – t2 = – if
1 (x – t) 2t
x = 0; y =
2t(y – t2) = – x + t
or
3 2
3 2 2t t = t t = 0 2 2 or 3 – 2t = 1 t = 1 or – 1 hence one of the point is origin and the other two are (–1, 1) and (1, 1) PQR is a right triangle radius of the circle is 1 its equation is x2 + (y – 1)2 = 1 x2 + y2 – 2y = 0 ]
20.
Let x be a real root. Equating real and imaginary part x3 – 6x2 + 5x + 2a2 = 0 .....(1) and 2x3 – 2x2 – 4x = 0 .....(2) 2 2x(x – x – 2) = 0 2x(x – 2)(x + 1) = 0 the given x = 0, 2 or – 1 if x=0 a=0 x=–1
a2 = 6
a=±
6
a2 = 3
a=±
3
x=2
a { 0, 6 , 6 , 3 , 3 } S = 0 + 6 + 6 + 3 + 3 = 18 Ans. ] 21.
For non zero intercepts slope = – 1 y=–x+c point (– 4, – 2) –2=4+c
Bansal Classes
c=–6 Problems for JEE-2007
[17]
lines is y = – x – 6 x+y+6=0 for 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) 0 hence – 3 x – 1 or 1x3 3 now f (x) = x – 3x f ' (x) = 3x2 – 3 = 0 x=±1 maximum occurs when x = 3 f (3) = 18 ]
23.
x4y2dx + x2y4dy = xdy + ydx x2y2(x2dx + y2dy) = xdy + ydx x2dx + y2dy =
d (xy ) ( xy ) 2 2 x dx +
Integrating,
2 y dy =
d (xy )
( xy) 2
1 x3 y3 + =– +C xy 3 3 3 = C; now if x = 1; y = 1 C = 5, xy hence x3 + y3 + 3(xy)–1 = 5 Ans. ] (x3 + y3) +
24.
card removed 52 face 40 20 drawn randomly 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 /\ 4
=
4 aces 36 other
C 0 · 36 C 20 40
36
=
Bansal Classes
C 20
4
4
C2
· 20
C2
+
C1 · 36 C19 40
C 20
3
C2
· 20
C2
4
+
C 2 · 36 C18 40
C 20
2
C · 20 2 C2
C 20 · 4C 2 4C1 · 36 C19 · 3C 2 4C 2 · 36 C18 · 2C 2 40
C 20 · 20 C 2
Problems for JEE-2007
[18]
=
=
25.
e=
6 · 36C 20 12 · 36C19 6 · 36C18 40
C 20 · 20C 2
6( 37 C 20 37 C19 ) 40
C 20 · 20 C 2
=
=
6 [36 C 20 36C19 36C19 36C18 ]
6( 38C 20 ) 40
C 20 · 20 C 2
40
C 20 · 20C 2
p = 6 Ans. ]
4 5
16 9 b 3 b2 = ....(1) 2 = 1 – 25 = 25 ; a 5 a 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 ) 1 cos 9 = 1 cos 25 25 – 25 cos = 9 + 9 cos 16 = 34 cos
cos =
26.
sin =
AB = 2b sin = 2 ·
Let
a, b R, such that f (a) = f (b)
27.
8 ; 17
15 17
3a 15 18 · = a Ans. ] 5 17 17
1 ai 1 b i = 1 a i 1 bi 1 – bi + ai + ba = 1 + bi – ai + ab 2ai = 2bi a=b f is injective. ]
Let r be the radius of circle A and R be the radius of circle B r + R = 12 and r = 3R 4R = 12; R = 3 and r = 9 Area of trapezium ABCD =
1 (3 + 9) (12) 2 6 2 2
= 6 108 = 36 3 Area of arc ADC =
Bansal Classes
1 27 81 = 2 3 2
Problems for JEE-2007
[19]
Area of arc BCE =
27 33 3 = 36 3 – required area = 36 3 – 2 2 a = 36, b = 33 a + b = 69 Ans. ]
28.
1 2 9 = 3 2 3
0 1 0 1 3 0 A2 = 3 0 3 0 = 0 3 = 3I 4 6 A = 9I; A = 27; A8 = 81I 8 6 4 2 (A + A + A + A + I) = 121 I 1 0 0 hence 121 0 1 V = 11 ;
121 0 a 0 0 121 b = 11
1 121 a 0 121 b = 11 a = 0, b = 11 ;
29.
Radius of the first circle = sin
0 V 1 ] 11
6 = =1 S 6
C 1 r = ....(1) (r < 1) 2 1 r
also
sin C =
now
2sin2 sin2
4 5
C 3 2 = 1 – cos C = 1 – = 2 5 5
C 1 = 2 5 2
1 1 r = 5 1 r
30.
5 – 1 = ( 5 1 )r
5(1 – r)2 = (1 + r)2
r=
5 (1 r ) = 1 + r
5 1 sin 18 = 5 1 cos 36
k + w = 54° Ans. ]
Cross multiplication and rearranging gives the cubic. x3 – ax2 + 23x – b = 0 2 + = a 2 + 2 = 23 and 2 = Also given + = 12 from (2) and (4) 2 + 2(12 – ) = 23 2 + 24 – 22 = 23
Bansal Classes
....(1) ....(2) ....(3) ....(4)
Problems for JEE-2007
[20]
and 31.
2 – 24 + 23 = 0 = 1 (rejected) since x ± 1 = 23; = – 11 a = 35 from (4) b = 2 = 529 × – 11 b = – 5819 a – b = 35 – (–5819) = 5854 Ans. ]
Let the numbers be .P . G (A D) 2 A D, A, A D, A ) ( b ) (c ) (d ) (a A .P.
Given d – a = 30
( A D) 2 – (A – D) = 30 A D2 + 3AD = 30 A D2 = 3A(10 – D)
(A + D)2 – A(A – D) = 30A
D2 ....(1) 3(10 D) 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
A=
3 (rejected) 7 D=6 A = 3 (rejected) D=9 A = 27 Numbers are 18, 27, 36, 48 Ans.
D=3
32.
A=
]
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(24 – 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. ]
Bansal Classes
Problems for JEE-2007
[21]
33.
Slope of the normal m =
1 3 m2 1 m1 x1 = y1 = ; 2 x1 1 2m 4 m2
equation of the normal in terms of slope of the normal is y = mx +
5 m2 2 m 3 1 . 4 m2
It passes through (7/2, 9/2) 12 m3 13 m2 + 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 = 0 Point A (0, 1) ; B ( 1, 3) ; C (5/2, 19/4) ] 34.
Equation of normal, Yy=
1 (X x) m
X = 0 gives Y = x2 +
Y = 0 gives X = x + m y and
x my m
x x my =0 2
Hence
y2 = C ; passes through (1, 4) 2
conic is
1 x2 y2 = 1 with e = 9 18 2
2x + y
dy =0 dx
C=9
focii are (0, 3) & (0, 3)
Equation of the circles are ; (x 1)2 + (y 4)2 + (x + 2 y 9) = 0 where x + 2 y 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 x y 1 x 2 y2 2 2 x2 + y2 = (x + y + 1)2 2xy + 2x + 2y + 1 = 0 Let 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 are xy + 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 PQ now d2PQ = (Q – OP)2 2 2 d2PQ = (tan C cot C) 1
2
2 d2PQ = (tan C cot C) 2 1
d2min =
2
2
2 1 = 3 – 2 2 a = 3; b = 2 a3 + b3 = 27 – 8 = 19 Ans. ]
Bansal Classes
Problems for JEE-2007
[22]
37.
I=
dx
= 1 2 0 x ( a 2 ) x2 2
x 2 dx
x 4 ( a 2 2) x 2 1
(a2 – 2 = k 0)
0
x 2 dx
1 ( x 2 1) ( x 2 1) dx = 4 = 2 2 4 2 x kx 1 x kx 1 0 0
1 1 (1 x 2 ) 1 1 (1 x 2 ) dx dx + 2 = 2 2 0 x (1 x 2 ) k 2 0 x (1 x 2 ) k I2
I1
now proceed, I1 =
I
38.
=
Let
2a
and I2 = 0 2a
;
x 4
= 2a 5050
a = 2525 Ans. ]
d = dx
or
4 = + 4x
x (1 tan x ) (4x ) tan x 0 4 dx = = – 4 1 tan x dx = – 4 1 tan x 2 1 tan x 2 1 1 tan x 4 0
0
– 4 = – 4x
0
x (1 tan x ) (1 tan x ) · dx 1 tan x ( 2 ) tan x 2
0
x (1 tan x ) x dx = 2 x dx =2 tan x tan x 2 2 I=
0
0 x2 2
I=–
+
2 +2 4
x dx tan x 2
2
0
t dt tan t
x=–t
2
now
I1 =
t dt = t ln sin t t cot II 0 I
I1 = 0 +
2 0
–
ln sin t dt 0
ln 2 2
2 2 Hence 2 · ln 2 – = ln 2 – 2 4 4
Bansal Classes
2
k = 2, w = 4
Problems for JEE-2007
kw = 8 Ans. ]
[23]
1
39.
g(1) = 5 and
g (t) dt = 2 0
x
2f (x) =
x
(x2
– 2xt
+ t2) g(t) dt
0
x
x
2 2 = x g( t ) dt 2 x t g( t ) dt t g ( t ) dt 0
0
0
Differentiating x x 2 f '(x) = x2 g(x) + g ( t )dt ·2 x – 2 x 2g ( x ) t g( t ) dt + x2g(x) 0 0 x
x
2 f '(x) = 2x g( t )dt – 2 t g( t ) dt 0
0 x
x
f " (x) = x g (x) + g( t )dt – x g (x) = g ( t )dt 0
0
1
hence f " (1) = g ( t ) dt = 2 0
also
f ''' (x) = g (x) f ''' (1) = g (1) = 5 f ''' (1) – f ''(1) = 5 – 2 = 3 Ans. ] x
40.
Consider a function
g (x) = e–x f ( t )dt in [0, 1] 0
obvious continuous and derivable g (0) = 0 and g (1) = 0 (given) hence some c (0, 1) such that g ' (c) = 0 x
now
g ' (x) =
e–x
f (x) –
e–x
f ( t )dt 0 c
g ' (c) =
e–c
f (c) –
e–c
f ( t )dt 0
41.
x
=0
f (t )dt
= f (c) ]
0
Consider f (x) = x3– ax + b f '(x) = 3x2 – a if a 0 then f ' (a) 0 for all x hence f is strictly increasing hence f (x) = 0 has exactly one root for a = 3 f ' (x) = 3x2 – 3 = 0 x = 1 or – 1 in order that f (x) may have 3 roots f (x1) · f (x2) 0 where x1 and x2 and the roots of f ' (x) = 0 hence (1 – a + b)(– 1 + a + b) 0 put a = 3 (b – 2)(b + 2) 0 or –2b2 ]
Bansal Classes
Problems for JEE-2007
[24]
42.
Let m gms of salt is present at time t differential equation of the process is dm m(1) = 10 – dt 50 t
dm 1 m = 10; + dt 50 t dt
(50 t ) 2 I.F = e = 50 + t; m(50 + t) = (50 t )dt = 10 +C 2 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
50t
m = 5(50 + t)2 –
5(50) 2 50 t
m(t = 10) = 5 · 60 – m= 43.
5(50) 2 60
250 25 11 2 11 = 91 = 50 6 = 50 · 60 3 3 6
]
A : red ball is selected B1 : Face card is drawn B2 : ace card is drawn B3 : neither face nor ace is drawn P(A) =
12 3 4 8 36 10 107 · + · + · = Ans. 52 12 52 12 52 12 156
12 3 156 9 P(B1/A) = · · = Ans. ] 52 12 107 107
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 , 9 4 3 1 9 5 2 1 ; If we use five digits then Hence p =
7 6 3 1 8 6 2 1 8 5 3 1 ; 7 5 4 1 ; 6 5 4 2 7 5 3 2 8 4 3 2 7, 1, 2, 3, 4 6, 5, 3, 2, 1
4! 5! 9 5! 4! 2 9!
( 9 cases )
( 2 cases ) =
11 5 ! 4 ! 9!
=
11 126
[ odd in favour 11 : 115 ] 45.
A = {1801, 1802,.....,1899, 1900} B = {1901, 1902,.....,1999, 2000}
Bansal Classes
Problems for JEE-2007
[25]
E : randomly chosen year has 53 sundays P (E) = P (E L) + P (E O) = P (L). P(E/L) + P (O). P(E/O) 1 24 2 76 1 1 25 2 75 1 . . . . = + 2 100 7 100 7 2 100 7 100 7
=
249 Ans.] 1400
46.
P(E) = 1 P (value of 5 coins is more than 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 cu. units 3
Vector area of triangle AEF = 49.
1 1 AF AE = BC BD = 3 i 10 j k ] 2 2
Equation of the line passing through (1, 4, 3) x 1 y 4 z 3 ....(1) a b c
since (1) is perpendicular to
x 1 y 3 z 2 x 2 y 4 z 1 = = and = = 2 1 4 3 2 2
hence 2a + b + 4c = 0 and 3a + 2b – 2c = 0
a b c 2 8 12 4 4 3
a b c 10 16 1
x 1 y 4 z 3 ....(2) Ans. 10 16 1 now any point P on (2) can be taken as 1 – 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)
hence the equation of the lines is
50.
Ans.]
Equation of the line through (0, 1, 2) x 0 y 1 z 2 ....(1) a b c
now given line
Bansal Classes
x 1 y 1 z 0 =t 1 1 2
....(2)
Problems for JEE-2007
[26]
(2) is along the vector V ˆi ˆj 2kˆ a – b + 2c = 0 ....(3) since (1) and (2) intersect; hence must be coplanar
hence
1 0 2 1 1 2 = 0 a b c
2a + 4b + c = 0 solving (3) and (4),
x y 1 z 2 = t Ans. ] 3 1 2
required equation is 51.
....(4) a:b:c=–3:1:2
Since r1r2 = 2,
x2 + px + 2 = 0
r1 r2
r1 r2 r3 r4 = – 8
and
r3 r4 = – 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 = – 2 on comparing coefficient of x2; a = – 4 p=1 x2 + x + 2 = 0
x 52.
1 i 7 Ans. ] 2
r1, 2 =
y a
a x x x a y
x a
x
x a y x a = y x a x y a x
x a y
2
= [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
]
a1 b = a 2 a 3
2 1 1 a1 10 2 2 1 a 2 = 13 1 1 1 a 9 3
i.e.
2a1 2a 1 a1 a1 = 1
Bansal Classes
a2 2a 2 a2
a 3 10 a 3 = 13 a 3 9
; a2 = 3 ; a3 = 5 Problems for JEE-2007
[27]
1 2 2 x1 1 2 2 3 x 2 = 3 1 1 3 x 3 5 x1 2 x 2 2x 2x 2 1 x x 2 1
54.
2 x 3 1 3x 3 = 3 3x 3 5
i.e.
x1 = 1 ; x2 = – 1 ; x3 = 1
TPT
9 abc abc abc + + 2 b c c a a b x1
x2
Ans.
]
....(1)
x3
Consider AM between the numbers x1, x2, x3 a bc 1 1 1 3 b c c a a b now HM between the numbers x1, x2, x3
=
3(a b c) 3 3 = = bc ca ab 2(a b c) 2 a bc a bc abc AM HM =
a bc 1 1 1 3 3 2 b c c a a b
1 1 9 1 (a + b + c) b c c a a b 2
55.
Let x = r cos and y = r sin y r2 = x2 + y2; tan = x N=
r2 r 2 [cos 2 sin cos 4 sin 2 ]
Nmax = Nmax = A= 56.
2 5 10 2 5 10
=
2 5 10 = M 15
=
2 5 10 = m 15
2 ·10 2 Mm = 15 ·2 = 2 3
Hence proved ]
(0, /2) =
r2 2 = 5 sin 2 3 cos 2 (1 cos 2) sin 2 4(1 cos 2)
2007 ×
2 = 1338 Ans. ] 3
Transposing 2 on RHS using 2 cos A · cos B relation, cos
A 2
Bansal Classes
BC B C A cos 2 cos 2 – sin 2
BC B C cos 2 cos 2 = 1
Problems for JEE-2007
[28]
or cos
A BC A A A BC A A BC sin ) sin + cos cos – sin cos + sin2 – 1 = 0 ( cos 2 2 2 2 2 2 2 2 2
BC A A A A A cos cos sin + cos sin – cos2 = 0 2 2 2 2 2 2 cos
BC A A A cos sin – cos 2 2 2 2
A A cos 2 sin 2 = 0
A A BC A cos 2 sin 2 cos 2 cos 2 = 0
if
A A – sin = 0 2 2
tan
A =1 2
BC A = cos 2 2 B – C =A B = C +A B– C= –A B + A = C = 90° hence triangle must be right angled. ] OAMB is a cyclic quadrilateral using sine law in OBM and OAM
if
57.
cos cos
x d = sin( 60 ) sin 90 and
A = 90°
d y = sin 90 sin
(1) and (2)
B = 90°
.....(1) ....(2)
x y = sin( 60 ) sin
x sin( 60 ) 1 3 = = cot – y sin 2 2
2x +1= y
3 cot
2x y = cot 3y
d2
from (2) d = y cosec d2
=
y2(1 + cot2)
d2 =
3y 2 4x 2 y 2 4xy 3
d=
2 x 2 y 2 xy Ans. ] 3
Bansal Classes
=
y2 1
( 2 x y) 2 3y 2
d2 =
d2
=
y2
(2 x y) 2 + 3
4x 2 4 y 2 4 xy 3
Problems for JEE-2007
[29]
58.
Let G be the centroid : AD = x ; BE = y
AG =
2x x 2y y ; GD = ; BG = ; GE = 3 3 3 3
4x 2 y 2 9 or 16x2 + 4y2 = 81 .....(1) 9 9 4
In AGE :
x 2 4y 2 4 or In BGD : 9 9
(i) – (ii) , 15x2 = 45
x2 + 4y2 = 36 .....(ii) x=
3
9 4 3 5 9 16 c 2 In ADC, cosC = 2(2) (3) 6 2(4) (3)
20 = 25 – c2
c=
5
2 1 1 5 ab sinC = (3) (4) 1 11 sq. units ] 2 2 6
= 59.
or
From triangle inequality log1012 + log1075 > log10n log10900 > log10n also log1012 + log10n > log1075 log1012n > log1075 12n > 75 n>
75 12
or
n>
n < 900
....(1)
25 4
Hence no. of values = 900 – 7 = 893 Ans. ] 60.
x + 2y = 10 where x is the number of times he takes single steps and y is the number of times he takes two steps Cases Total number of ways I: x = 0 and y = 5
5! = 1 (2 2 2 2 2) 5!
II: x = 2 and y = 4
6! 2!·4! = 15 (1 1 2 2 2 2)
III: x = 4 and y = 3
7! 4!·3! = 35 (1 1 1 1 2 2 2)
IV: x = 6 and y = 2
8! 2!·6! = 28 (1 1 1 1 1 1 2 2)
9C = 9 (1 1 1 1 1 1 1 1 2) V: x = 8 and y = 1 1 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. ]
Bansal Classes
Problems for JEE-2007
[30]
b
61.
I=
e x a eb x x dx a
x = at
let
ba
=a
1
dx = a dt
e t e b at dt at
e t e t dt I= t 1
put
t=
y
(where b/a = )
....(1)
dt = –
y2
dy
1
( e y e y ) y · 2 dy I=– y
1
62.
63.
( e y e y )dy I= y
( e t e t )dt I=– t 1
or
from (1) and (2)
2I = 0
....(2)
I = 0 Ans. ]
f (3) 1 23k 9 = 6k = ; f (9) – f (3) = (29k + 9) – (23k + 9) = 29k – 23k f (6) 2 9 3 3(23k + 9) = 26k + 9 26k – 3(23k) – 18 = 0 23k = y y2 – 3y – 18 = 0 (y – 6)(y + 3) = 0 y = 6; y = – 3 (rejected) 23k = 6 now f (9) – f (3) = 29k – 23k { from (1) } = (23k)3 – 23k = 63 – 6 = 210 hence N = 210 = 2 · 3 · 5 · 7 Total number of divisor = 2 · 2 · 2 · 2 = 16 number of divisors which are composite = 16 – (1, 2, 3, 5, 7) = 11 Ans. ]
....(1)
Radius of the circle is 1 r B tan = = 2 PB s(s b) PB = |||ly
r ·s(s b) s = · ·(s b) = (s – b); s PC = (s – c)
(PB)(PC) = (s – b)(s – c) =
Bansal Classes
s(s a )(s b)(s c) s(s a ) Problems for JEE-2007
[31]
=
· =r· s(s a ) (s a )
=
= (s a ) a 3 3a = = 3 Ans. ] 3 2 3a 2 a 2
=
64.
r 1 s s
5x + 3x > 8 x>1 5x + 8 > 3x x>–4 and 3x + 8 > 5x x<4 Hence, x (1, 4). Now perimeter of the triangle = 8(x + 1) s = 4x + 4 2 A (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=
17 2
17 17 15 15 = – 16 1 16 = 16 × × = (2 × 15)2 max 2 2 2 2 (Area)max = 30 sq. units ]
A 2 (t)
65.
From the identity r = 4R · sin
A B C · sin · sin 2 2 2
or
r = 4 3 1 r · sin
let
A B A – C = 30°
then
B C A · sin · sin or 2 2 2
1 A C B = 2 sin ·sin ·sin 2 3 1 2 2 2
AC AC B 3 1 cos sin = cos 2 2 2 4
3 1 6 2 sin B sin B = 4 2 2 4 Let
sin
B = x yields 2
x2 –
6 2 x+ 4
3 1 = 0, 4
B B 6 2 2 and x = . It follows that = 15° or = 45°. The second 2 2 4 2 solution is not acceptable, because A B. Hence B = 30°, A = 90° and C = 60° ]
whose solutions are x =
Bansal Classes
Problems for JEE-2007
[32]
66.
y = ax2
dy dt T = 2ax0 = m hence line is y = (2ax0)x – b
.....(1)
(x0, a x 20 ) lies on parabola and the line (1)
a x 02 = 2a x 20 – b b = a x 20 . Hence Q = (0, – b) = (0, – a x 20 ) now using (TQ)2 = 1 x 20 + 4a2 x 04 = 1
a2 =
(1 x 20 )
.....(2)
4 x 40 x0
now
ax 3 mx 2 bx A = (ax mx b)dx = 3 2 0
=
A2 =
let
A2
ax 30 3
ax 30
a 2 x 60 9
ax 30
max
=
=
= 0
ax 30
3
mx 20 2
bx 0
ax 30 3
x 60 1 x 02 x 2 (1 x 2 ) 0 0 = 9 4 = 4 x 36 0
= f (x0) =
x 02 (1 x 20 ) 36
This is maximum when x 02 = A2
x0
2
1 1 1 1 · · = ; 2 2 36 144
1 2
Amax =
1 12
1 = 12 Ans. ] A
67.
(i)
Equation of tangent from point (3, –3) to the given circle is y + 3 = m(x – 3) mx – 3m – y – 3 = 0
Bansal Classes
Problems for JEE-2007
[33]
4m 3m 2 3
and also
=5 1 m2 (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
4 (x – 3) and 3 3y + 9 = – 4x + 12 4x + 3y = 3 Equation of normals to these 2 tangents are
y+3=–
(ii)
3 (x + 4) 4 4y + 8 = 3x + 12 3(3x – 4y + 4 = 0) 9x – 12y = – 12 16x + 12y = 12 —————— x = 0; y = 1
4 (x + 4) 3 3y + 6 = – 4x – 16 4(4x + 3y= – 22) 16x + 12y = – 88 9x – 12y = 63 —————— 25x = – 25 x = – 1; y = – 6 points A and B are (0, 1) and (–1, – 6) Ans.
y+ 2=
(iii)
and
y+2=–
angle between the 2 tangents = 90° ADB = 90° | A D |max = CD + radius CD =
(iv)
3 (x – 3) 4 4y + 12 = 3x – 9 3x – 4y = 21
y+ 3=
50
| AD |max = 5 2 + 5 | AD |min = 5 2 – 5 Area of quadrilateral ADBC = AC × AD AD =
(iv)
7 2 12 25 = 25 = 5 area of quadrilateral ABCD = 5 × 5 = 25 sq. units.
area of triangle DAB =
1 25 = 12.5 sq. units. 2
Circle circumscribing DAB will have points A and B as its diametrical extremities x2 + y2 – x(–1) – y(–5) – 6 = 0 x2 + y2 + x + 5y – 6 = 0 Ans. x-intercept = 2 g 2 c = 2 (1 4) 6 = 5
Ans.
y-intercept = 2 f 2 c = 2 (25 4) 6 = 7 Ans. ]
Bansal Classes
Problems for JEE-2007
[34]
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 2 hence 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 = 12 9a + 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 = 0 Solving with x = at2 , y = 2at a2t4 + 4a2t2 + 2gat2 + 4aft + c = 0 t1 + t2 + t3 + t4 = = 0 ....(1) 3 N : y + tx = 2at + at passing through (h, k) at3 + t(2a – h) – k = 0 ....(2) t1 + t2 + t3 = 0 ....(3) from (1) and (3) t4 = 0 hence circle passes through the origin c = 0 equation of the circle after cancelling –at at3 + 4at + 2gt + 4f =0 at3 + 2(2a + g)t + 4f = 0 ....(3) Now (2) and (3) must be represent the same equation 2(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 = xa = =– =
a [(t + t + t )2 – 2 t1 t 2 ] 3 1 2 3 2a ( 2a h ) 2a 2 t1 t 2 = – . = – (2a – h) 3 a 3 3
26 (a = 1 ; h = 15 ) 3
70.
a ( t12 t 22 t 32 ) 2a ( t1 t 2 t 3 ) , 3 3
26 C : ,0 ] 3
1 ab ; also a2 + b2 = 3600 2 AD : y = x + 3 solve to get G (1, 2) BE : y = 2x + 4 acute angle between the medians is given by
Area =
m1 m 2 2 1 1 1 tan = 1 m m = = tan = 1 2 3 3 1 2
Bansal Classes
Problems for JEE-2007
[35]
In quadrilateral GDCE, we have (180 – ) + 90° + + = 360° = + – 90° cot = – tan( + )
tan tan –3= 1 tan tan 9ab = 2 × 3600 Area = 400 sq. units ]
71.
or
2 b 2a a b –3= 2 b 2a 1 · a b
2( a 2 b 2 ) 9= ab
1 ab = 400 2
W1: C1 = (–5, 12) W2: C2 = (5, 12) r1 = 16 r2 = 4 now, CC2 = r + 4 CC1 = 16 – r let 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 subtraction 20h = 240 – 40r h = 12 – 2r 12r = 72 – 6h ...(1) By addition 2[h2 + 25 + a2h2 – 24ah + 144] = 272 – 24r + 2r2 12 h + – 24ah + 169 = 136 – 12r + = 136 + (6h – 72) + 2 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 2 2 (96a) – 4 · 4 · 69 (3 + 4a ) 0 576a2 – 69.3 – 276a2 0 69 13 300a2 207 a2 ; hence m (smallest) = 100 10 69 m2 = ; p + q = 169 Ans. ] 100
h2(1
So,
a2)
2
r2
[using (1)]
72.
I=3
2 4 2 4 ( 1 sin x ) sec x dx ( 1 2 sin x sin x ) sec x dx = 3 5 6 5 6
2 2 2 2 2 = 3 sec x (1 tan x )dx 2 sec x tan x sec x dx sec x (tan x )dx 5 6 5 6 5 6
2 2 2 = 3 (1 2 tan x ) sec x dx 2 (sec x tan x sec x ) dx 5 6 5 6
Bansal Classes
Problems for JEE-2007
[36]
1 0 2 2 = 3 (1 2 t ) dt 2 t dt = 1 3 2 3
2 t 3 0 t 3 1
3
2 3 t 3
3
1
1
8 1 2 1 2 (1) = 3 (0) 3 3 3 3 3 3 3 11 6 3 16 1 2 2 8 11 2 3 3 8 = 3 = 3 1 = 9 3 3 9 3 3 3 3 9 3 3 3 3
6 3 5 5 3 b 3 =2– =a– 3 3 9 c 1
73.
dx 1 x 1 x 2
I=
0
put x = cos 2 4
I=2
0
4
I=
0
4
=
0
dx = – 2 sin 2 d
sin 2 d = 2 cos 2 sin 2
sin 2 d = cos 1 4
4
0
0
4
sin 2 d = cos sin 2
4
0
4
0
2
0
sin 2 d 2 cos 2 4
1 2 sin 2 d 1 cos 4
(1 cos ) d – 2 (1 cos ) d sin 2 0 4
4 2
2 sin
cot cosec ) d – 2 (1 cos ) d = cot cosec 0
2 1 Lim 0
= 2 2 1 74.
4
(cosec
2
1 d – 2 (1 cos ) d = 1 cos 0
0
=
4
cos 2 d = cos 1
4
=
a = 2, b = 5, c = 9 a + b + c + abc = 106 Ans.]
1 cos 1 = cos 2 4
2
8 1
4 0
0
2 1
2 2
a = 8, b = 1, c = 4 a2 + b2 + c2 = 81 Ans. ] 4
x · g f ( x ) f ' g (x )g ' (x ) = f g (x )g ' f (x ) f '( x ) x ·g f (x)
d d f g ( x ) = f g (x ) g f (x ) dx dx
d d f g (x ) g f (x) x · dx dx f g (x ) g f (x)
Bansal Classes
Problems for JEE-2007
[37]
x·
d d ln f g ( x ) ln g f ( x ) dx dx a
now,
f g (x) dx 1
0
e 2a 2
differentiate w.r.t. 'a' f g (a) = e–2a from (1) and (2) we get
75.
f g (x) = e–2x
put
d ln g f (x) dx x = 0, C = 0
g f (x) e x ;
...(1)
– 2x =
2
ln f g (x) = – 2x
....(2)
ln g f (x) = – x2 + C
Hence g f (4) = e–16 k = 16 Ans. ]
Let f (x) = y dy + y = 4xe–x · sin 2x dx I.F. ex
(linear differenial equation)
yex = 4 x sin 2x dx I
II
cos 2 x 1 cos 2 x dx yex = 4 x 2 2
now
x cos 2 x sin 2x yex = 4 +C 2 4 yex = (sin 2x – 2x cos 2x) + C f (0) = 0 C=0 –x y = e (sin 2x – 2x cos 2x) f (k) = e–k (sin 2k – 2k · cos 2k) = e–k (0 – 2k) f (k) = – 2 (k · e–k)
f (k) = – 2 e–
2e–2
kek
k 1
S –3 3e +
S =1· + + ......... + – –2 –3 Se = +e + 2e + ......... + —————————————————— S(1 – e–) = e– + e–2 + e–3 + ...... 1 e S(1 – e–) = = e 1 1 e 1 e S= = (e 1)(1 e ) (e 1) 2
2 e Ans. ] (e 1) 2
Bansal Classes
Problems for JEE-2007
[38]
76.
f (x) = Limit f (x h) f (x) h0 h f (x h) x h f f (x) 1 1 x f (x) = Limit = f(x) · Limit h0 h h 0 h
h f 1 1 f 1 t 1 f (x) x Limit = f(x) · Limit = x t 0 t h h0 x x Now putting x = 1, y = 1 in functional rule f(1) =
f (1) =1 f (1)
f (x) =
f (x) · f (1) x
=
f ' (x) 2 = f (x) x ln (f(x)) = 2lnx + C x = 1; f(1) = 0 C = 0 Now solving y = x2 and x2 + y2 = 2 y2 + y – 2 = 0 (y + 2) (y – 1) = 0 y=1 1
A= 2
2f (x) x
;
f(x) = x2
2 y 2 y dy
0
1 1 2 = 2 2 y dy y dy 0 0 1
= now
2 1 1 y dy y 2 3
0
1
= 0
2 3
1
and
y=
2 y 2 dy
2 sin
0
/ 4
/4
2 cos 2 cos d
0
0
/ 4
1 = + sin 2 2 0
Hence
Bansal Classes
2 2 cos d =
1 2 A= 2 4 2 3
;
/ 4
(1 cos 2) d 0
1 4 2
1 A = sq. units ] 2 3
Problems for JEE-2007
[39]
77.
Z10
+
10
Z10 13
1 Z
=0
10
1 13 Z
13 –
= – 1 = cos + i sin
1 = cos2m 1 i sin 2m 1 10 Z i
=e
( 2 m 1) 10 ( 2 m 1)
i 1 = 13 – e 10 Z substituting m = 0, 1, 2,.......9 we get
i 1 = 13 – e 10 Z1
1 1 and are complex conjugate note Z1 Z10
3
i 1 = 13 – e 10 Z2
5
i 1 = 13 – e 10 Z3
19
i 1 = 13 – e 10 Z10
Let
1 1 1 1 = and = Z1 a1 Z10 b1 and so on
i i 1 = 169 – 13 [ e 10 + e 10 ] + 1 a i bi
= 169 – 13
3 [ e 10
+e
3 10
]+1
i 1 10 = 170 – 26 Re e a i bi
and
3 1 i 10 etc = 170 – 26 Re e a 2b 2
3 5 3 9 1 = 850 – 26 cos cos cos cos cos 10 10 10 10 a i bi 10
= 850 – 26[cos18º + cos54° + cos90° + cos126° + cos162°] = 850 Ans. ]
Bansal Classes
Problems for JEE-2007
[40]
78.(i)
|||ly
an + bn + cn = C0 + C1 + C2 + C3 + C4 + ................ an + bn + cn = 2n ....(1) n (1 + x) = C0 + C1 x + C2 x2 + C3 x3 + ................ x= (1 + )n = C0 + C1 + C2 2 + C3 3 + C4 4 +................ = (C0 + C3 + C6 + .......) + (C1 + C4 + C7 + ........) + 2(C2 + C5 + C8 + ........) n (1 + ) = an + bn + 2cn ....(2) 2 n 2 (1 + ) = an + bn + cn ....(3)
now
a 3n b3n c3n – 3anbncn = (an + bn + cn) (an + bn + 2cn) (an + 2bn + cn)
now put
= 2n(1 + )n (1 + 2)n = 2n(– 2)n (– )n = 2n also
(a n b n ) 2 = 2(an + bn + 2cn) (an + 2bn + cn) (a n b n ) 2 = 2 Ans.
78.(ii) Let and
x = C0 – C2 + C4 – C6 + ..... 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
x 2n y 2n = 2n/2
hence x 2n y 2n = 2n hence proved ] 79.
5 1 3 5 1 3 5 1 3 1 3 5 1 3 5 1 3 5 = A matrix A is idempotent = = 1 3 5 1 3 5 1 3 5 Hence A2 = A3 = A4 = ....... = A x = 2, 3, 4, 5, .......... A2
x3 1 x3 1 x 2 n
now
Lim n
n
Lim n
x 1 x 1 x 2
x2 x 1 x2 x 1 x 2 n
3 4 5 n ( n 1) 3 7 13 n 2 n 1 Lim · · ....... · · ....... n 1 2 3 ( n 1) 7 13 21 n 2 n 1 Lim n
80.
n ( n 1) 3 3 · 2 = 1· 2 n n 1 2
Ans.
]
ac a + log = log 2 Given log a b
ac = log 2 log b
Bansal Classes
Problems for JEE-2007
[41]
also
a + c = 2b .....(1) 2 a – ax + 2bx + c + cx2 = 0 (c – a)x2 + 2bx + (c + a) = 0 has equal roots D=0 4b2 – 4(c2 – a2) = 0 b2 = c2 – a2 ....(2) from (1) and (2) b2 = (c – a)(c + a) b2 = (c – a) 2b 2(c – a) = b ....(3) 2 2 from (2) c = a + b2 triangle is a right at C. C = 90° A + B = 90° from (3) using sine law 2(sin C – sin A) = sin B C = 90° sin C = 1 A + B = 90° B = 90° – A 2(1 – sin A) = sin(90 – A) = cos A squaring both sides 4(1 – sin A)2 = cos2A = (1 – sin2A) 4(1 – sin A) = (1 + sin A) 3 = 5 sin A sin A = 3/5 B = 90 – A sin B = cos A = 4/5 and sin C = 1
sin A + sin B + sin C =
3 4 12 + +1= Ans. ] 5 5 5
81.
sin( nx ) 3n n 0
put
e nix e nix sin (nx) = 2i
sin( nx ) 3n = 21i n 0
ix e n ix e n ix 1 e 3n = 2i n0 3 n 0
1 1 1 = ix 2i e e ix 1 1 3 3
n
ix e n 0 3
n
3 3 = ix 3 e ix 3 e
(3 e ix ) (3 e ix ) 3 2i sin x = ix ix 9 3(e e ) 1 2i 10 6 cos x
=
3 2i
=
3 sin x 1 1 = = 2(5 3 cos x ) 2(5 2 2 ) 2 5 3 1 (1 9)
Bansal Classes
Problems for JEE-2007
[42]
52 2 34 a = 5, b = 2 , c = 37
= 82.
a + b + c = 5 + 2 + 37 = 41 Ans. ]
log b log a (let b = 2m and a = 2n where m and n denotes the exponents on the base 2 in the given set) logab =
m n hence logab is an integer only if n divides m now total number of ways m and n can be chosen = 25 × 24 = 600 For favourable cases let n=1 hence m can take values 2, 3, 4, 5, 6, ......., 24 if n=2 m = 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24 n=3 m = 6, 9, 12, 15, 18, 21, 24 n=4 m = 8, 12, 16, 20, 24 n=5 m = 10, 15, 20, 25 n=6 m = 12, 18, 24 n=7 m = 14, 21 n=8 m = 16, 24 n = 9, 10, 11, 12 m = 1 for each
=
Hence P =
= 24 = 11 = 7 = 5 = 4 = 3 = 2 = 2 = 4 —— 62
62 31 = Ans. ] 600 300
Select the correct alternative. (Only one is correct): Q.83 A [Sol. There are 4C2 = 6 ways a block can differ from the given block in exactly two ways (1) material and size, (2) material and colour, (3) material and shape, (4) size and colour, (5) size and shape, and (6) colour and shape. Since there is only 1 choice for different material, 2 choices for different size, 3 choices for a different colour, and 3 choices for a different shape, it follows that the number of blocks in each of the above categories is (1 × 2), (1 × 3), (1 × 3), (2 × 3), (2 × 3) and (3 × 3), respectively. The answer is the sum of these six numbers = 29 Ans ] Q.84 C [Sol. Consider the expansion of (1 + x)99 and put x = i and equate the real part to get C] Q.85 B [Sol. f (x) = | Ax + c | + d g (x) = – | Ax + u | + v
figure is parallelogram and diagonals bisect each other
Bansal Classes
Problems for JEE-2007
[43]
u c + = 3 + 1; A A
u c = – 4 Ans. ] A
Q.86 D [Sol. f (x) = x4 – 2x3 + 3x2 – 4x + 1 = 0 f ' (x) = 4x3 – 6x2 + 6x – 4x + 1 = 2(x – 1)(2x2 – x + 1) f ' (x) = 0 only at x = 1 also f '' (x) = 12x2 – 12x + 6 = 6(2x2 – 2x + 1) > 0 x R f is concave up f '' (1) > 0 x = 1 is minima also f (0) > 1; f (1) = – 1 two positive and two non real complex roots (D) ] Q.87 D [Sol. Notice the patterns Power of 3 go like this : 3, 9, 7, 1 Power of 7 go like this : 7, 9, 3, 1 Power of 13 go like this : 3, 9, 7, 1 so knowing that we can see that 31001 3 71002 9 131003 7 so therefore 3 · 9 · 7 = 9 ] Q.88 C [Sol. If a polynomial has real coefficients then roots occur in complex conjugate and roots are 2i, – 2i, 2 + i, 2 – i hence f (x) = (x + 2i)(x – 2i)(x – 2 – i)(x – 2 + i) f (1) = (1 + 2i)(1 – 2i)(1 – 2 – i)(1 – 2 + i) f (1) = 5 × 2 = 10 Also f (1) = 1 + a + b + c + d 1 + a + b + c + d = 10 a + b + c + d = 9 Ans. ] Q.89 D [Sol.
Tk =
1 1 1 ( k 2) k k k 2 (k 2) k k k 2 = = 2 2 2 k k 2 2k (k 2) k ( k 2) k ( k 2)
T1 =
1 1 1 2 1 3
T2 =
1 1 1 2 2 4
T3 =
1 1 1 and so on 2 3 5
as k , sum =
Bansal Classes
1 1 1 2 1 = = 2 2 2 2
1 2 8
a + b + c = 11 Ans. ]
Problems for JEE-2007
[44]
Q.90 C [Sol.
A=
1 2 b sin 2 = b2 sin cos ....(1) 2 x 65 x = 24 36 60 x = 24 · 65 x =26
now
sin =
sin =
again,
b 65 = sin sin 2
from (1)
12 5 and cos = 13 13
A=
b=
65 ·13 132 65 = 2 ·5 = 2 sin 2
134 12 5 · · = 169 · 15 = 2535 Ans. ] 13 13 4
Q.91 B [Hint: Let f (x) = ax3 + px2 + qx + r now use f (x) – f (x – 2) = x2 – 4x + 1 compare the coefficients to get a = 2/3; p = 1; q = – 1/6; hence p + q = 5/6 Ans. ] Q.92 B [Sol. n(S) = 4 × 4 × 4 = 64 n(A) = 211 or 312 or 413 or 431 or 422 6 3 3 = = 64 32 3 29 odds in favour 3 : 29 Ans. ]
P(E) =
Q.93 D [Sol.
1 4 1 5 tan sin cos 5 13 4 5 = – tan sin 1 cos 1 5 13
= – tan( – ) where sin =
4 5 and cos = 5 13
4 12 3 5 tan tan 20 36 16 = = = – =– Ans. ] 4 12 63 63 1 tan tan 1 · 3 5
Bansal Classes
Problems for JEE-2007
[45]
Select the correct alternatives. (More than one are correct): Q.94 A,B [Sol. Let the numbers are a – d, a , a + d (a – d + 1), a, (a + d) in G.P. and (a – d), a, (a + d + 2) also in G.P. (a – d + 1)(a + d) = a2 hence (a2 – d2) + (a + d) = a2 ....(1) and (a – d)(a + d + 2) = a2 (a2 – d2) + 2(a – d) = a2 ....(2) (2) – (1) 2a – 2d – a – d = 0 a = 3d 2nd term of A.P. = 3 times its common difference from (1) d2 – d2 – 3d = 0 d=4 d4 ; A.P. is 8, 12, 16 ]
a 12
Q.95 B,D [Sol. Let the line is normal to the hyperbola at P(x1, y1) hence 2x1 + 4y1 + k = 0 ....(1) 3x1 dy differentiate the curve dx = y P 1
line
y1 slope of normal = – 3x ; 1
slope of line = –
2 9
y1 2 – 3x = – 2x1 = 3y1 9 1 (x1, y1) lies on the hyperbola 3x12 – y12 = 23
x1 = 3 or – 3 hence P is (3, 2) or (–3, –2) k = 24 or – 24 Q.96 A,B,C [Sol. Equation of AB y = 2x – 1 solving it y2 = 4x y2 = 2(y + 1) but y1 + y2 + y3 = 0 but y1 + y2 = 2 y3 = – 2 putting in y2 = 4x x3 = 1
Bansal Classes
3x12 –
4x12 9
= 23
23 x12 = 23 · 9
B, D Ans. ]
y2 – 2y – 2 = 0
Problems for JEE-2007
[46]
hence coordinates of c are (1, – 2) sum of the x and y coordinates of C are – 1 (A) is correct. obviously normal at C passes through the lower end of the latus rectum (B) is correct
x1 x 2 x 3
again centroid of ABC = now solving
with y2
y = 2x – 1 (2x – 1)2 = 4x 4x2 – 8x + 1 = 0
3 = 4x
x1 + x2 = 2;
also x3 = 1
2 1 = (1, 0) 3
centroid of the ABC =
again equation of the normal at C 2 (x – 1) y+2=– 2 y+2=x–1 x–y–3=0 hence gradient of chord at C is 1
(D) is incorrect ]
Q.97 A,B,C [Sol. ' (x) = 2x[f ' (x2) – f ' (1 – x2)] now f ' (x) is an increasing function then f ' (x2) > f ' (1 – x2)
x2 > 1 – x2
so ' > 0
x>
1
1 2
or
2,0 1
x<–
1 2
2 , increasing
] Q.98 B,D
[Sol.
x 2 cos x sin 2 5 3 3 2 x = tan = – cot x = tan x = tan x 3 2 3 3 x 6 2 sin x sin 3 3 2 5 2 + x = n + –x 6 3
x=
n – 2 12
n=
2x = n +
6
5 11 or Ans. ] 12 12
MATCH THE COLUMN: Q.99 Ans. (A) P, (B) Q, (C) R, (D) P [Sol.
(A)
2 cos 3 i sin 3
n 2
is real
n n 2n 2 cos i sin is real 6 6
Bansal Classes
Problems for JEE-2007
[47]
n n = 0; = k; n = 6k 6 6 smallest positive n is 6 Ans. (Q) Let z = x + iy, x, y R and x2 + y2 = 2 (say) z2 is purely imaginary (x + iy)2 is purely imaginary x2 – y2 + 2xyi = 0 + ki k R – {0} 2 2 x = y and 2xy = k [If k = 0 then x = 0 and y = 0] let k > 0 say 2 xy = 1 y = 1/x 4 x =1 x2 = 1 x = 1 or –1 y = 1 or –1 z is 1 + i or – 1 – i if k < 0 say – 2 then xy = – 1; y = – 1/x 4 2 x =1 x =1 x = 1 or –1 y = – 1 or 1 z is 1 – i or –1+i there are four values of z which are ± 1 ± i (P)
hence sin (B)
(D)
x3 1 = x2 + x + 1 < 31 x 1 x2 + x – 30 < 0 (x + 6)(x – 5) < 0 – 6 < x < 5 number of + ve integer is 4.]
k=
Q.100 Ans. (A) R, (B) S, (C) Q, (D) Q [Sol. (A) z = 1 is one solution remaining 8 lie on a circle 2 in each quadrant symmetrically situated on a circle number of solutions are 2 Ans. (B)
Let
l = Lim e x x
1 e x x
ln l = Lim
x
1 e x 1 x
1 e x 1 ex ·ln e x (1 e x ) = Lim [ x ln (1 e x )] x x x x
·ln ( e x 1) = Lim
[1 ln (1 e x )1 x ] = Lim x
(C) (D)
ln l = 1 l = e Ans. BA P(A B) = P(B) P(B) P(A B) P(A / B) = = 1 Ans. ] P(B) P(B) Break the interval into 100 identical cases favourable length
Bansal Classes
2 2 Probability = 3 3
a – b = 1 Ans. ]
Problems for JEE-2007
[48]
