Bansal Classes: Mathematics

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.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
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
Q.7
If
[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 =
Bansal Classes
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.AtellshisfriendMr.Bthat"certaineventhasoccured"andhisearsweredroppingasnoticedby
Mr.B.FindtheprobabilitythatMr.Awastellingthetruth.
[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
Bansal Classes
(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]
Bansal Classes
[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
dx
1
a x
x
equals
. Find the value of a.

5050
[8]
Q.38 If
2
( 4) tan
=
ln
k
, find the value of (kw), where k, w N.

d
1 tan
w
4
[8]
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 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
f (x )dx = 0. Prove that there is c (0, 1) such

0
that f ( x )dx = 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 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
[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}.Ifthenumberchosenrepresentsacalenderyear.Findtheprobability
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
Bansal Classes
[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
(i 1) 2 x i 12 and
(i 2) 2 x i
i 1
i 1
123 ,
then find the value of
(i 3)2 x i .
[10]
i 1
Bansal Classes
[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
[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
for all reals a. Given that g f (0) = 1. If the value of g f (4) = ek where k N, find k.
e 2a
2
[10]
Q.75 Let f (x) be a differentiable function such that f ' (x) + f (x) = 4xex sin 2x and f (0) = 0. Find the value
n
of Lim
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
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:
Bansal Classes
(b) (an bn)2 + (bn cn)2 + (cn an)2 = 2.
[10]
(C0 C2 + C4 C6 + .....)2 + (C1 C3 + C5 C7 + .......)2 = 2n

[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!
(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
Bansal Classes
(C) 7
(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
[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)
(R)
(S)
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)
(R)
(S)
(B)
(C)
(D)
Lim e
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
[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
[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
1
=
2
7.
I=
1
2
x7
1
x 2 1
x
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 =
dx
1
(think !)
3
3y+x=1
P (from origin) =
6.
1
2
+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
(1 tan x ) 2
(1 tan 2 x ) dx =
4
(1 tan x )
(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=
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
[13]
1
I=
3
dx
x sin x )
(cos
dx (cos x sin x )3
II
integrating by parts
1 (cos x sin x )
=
3 (cos x sin x )3
=
8.
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
1 tan 2 x
sec 2 x
2 1
dx =
2
3 3
(1 tan x )
1
2 1
=
3
3 3
dt
t2
2
1 t
+
t
3
3
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
[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
[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
....(1)
1
1
line passes through 2i j k and is parallel to the vector V i j k
2
2

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
i
j
k

v1 n1 n 2 = 3 7 5 = 23( 3i 2j k )
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
i
j

v 2 n 3 n 4 = 5 13
8 11
k
3 = 7 ( i 2j 7 k )
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
[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=
a2 = 3
a=
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
[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
C2
20
C2
C1 36 C19
40
C 20
C2
20
C2
C 2 36 C18
40
C 20
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
[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
27
81 =
2
3
2
[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)
[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
[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
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
d2PQ = (tan C cot C) 2 1
d2min =
2 1 = 3 2 2
a = 3; b = 2
a3 + b3 = 27 8 = 19 Ans. ]
Bansal Classes
[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)
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 =
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
4 = 4x
x (1 tan x ) (1 tan x )
dx
1
tan
x
(
2
)
tan
x
2
x (1 tan x )
x
dx = 2
x dx
=2
tan x
tan x
2
2
I=
0
x2
2
I=
2
+2
4
x
dx
tan
x
2
t
dt
tan t
x=t
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
k = 2, w = 4
kw = 8 Ans. ]
[23]
39.
g(1) = 5 and
g (t) dt = 2
0
2f (x) =
(x2
2xt
+ t2) g(t) dt
2
2
= x g( t ) dt 2 x t g( t ) dt t g ( t ) dt
0
Differentiating
x
2 f '(x) = x2 g(x) + g ( t )dt 2 x 2 x 2g ( x ) t g( t ) dt + x2g(x)
0
0
2 f '(x) = 2x g( t )dt 2 t g( t ) dt
0
0
x
f " (x) = x g (x) + g( t )dt x g (x) = g ( t )dt

0
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) = ex 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) =
ex
f (x)
ex
f ( t )dt
0
c
g ' (c) =
ec
f (c)
ec
f ( t )dt
0
41.
=0
f (t )dt
= f (c) ]
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
[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
[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 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)
[26]

(2) is along the vector V i j 2k
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 a
y x a = y x
a x y
a x
x
a
y
= [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
[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
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
[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
[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=
9 4 3 5 9 16 c 2

In ADC, cosC =
2(2) (3) 6
2(4) (3)
20 = 25 c2
c=
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
[30]
61.
I=
e x a eb x
x dx
a
x = at
let
ba
=a
dx = a dt
e t e b at
dt
at
e t e t
dt
I=
t
1
put
t=
(where b/a = )
....(1)
dt =
y2
dy
( e y e y ) y
2 dy
I=
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 )
[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
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
[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
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
= f (x0) =
x 02 (1 x 20 )
36
This is maximum when x 02 =

A2
x0
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
[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
[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
[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)
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
[36]
1
0
2
2
= 3 (1 2 t ) dt 2 t dt =
1 3
2 3
2 t 3 0
t
3 1
2 3
t
3
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=
put x = cos 2
4
I=2
I=
dx = 2 sin 2 d
sin 2 d
=
2 cos 2 sin 2
sin 2 d
=
cos 1
4
sin 2 d
=
cos sin 2
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.
(cosec
1
d 2 (1 cos ) d =
1 cos
0
cos 2
d =
cos 1
a = 2, b = 5, c = 9 a + b + c + abc = 106 Ans.]
1 cos 1

=
cos
2
4
8 1
4
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
[37]
d
d
ln f g ( x )
ln g f ( x )
dx
dx
a
now,
f g (x) dx 1
e 2a
2
differentiate w.r.t. 'a'
f g (a) = e2a
from (1) and (2) we get
75.
f g (x) = e2x
put
d
ln g f (x)
dx
x = 0, C = 0
g f (x) e x ;
...(1)
2x =
ln f g (x) = 2x
....(2)
ln g f (x) = x2 + C
Hence g f (4) = e16 k = 16 Ans. ]
Let f (x) = y
dy
+ y = 4xex 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) = ek (sin 2k 2k cos 2k) = ek (0 2k)
f (k) = 2 (k ek)
f (k) = 2
e
2e2
kek
S
3
3e +
S
=1
+
+
......... +
2
3
Se =
+e
+ 2e + ......... +
S(1 e) = e + e2 + e3 + ......
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
[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
1
1
2
= 2 2 y dy y dy
0
0
= now
2 1 1
y dy y 2
3
=
0
2
3
and
y=
2 y 2 dy
2 sin
/ 4
/4
2 cos 2 cos d
/ 4
= + sin 2
2
0
Hence
Bansal Classes
2
2 cos d =
1 2
A= 2
4 2 3
/ 4
(1 cos 2) d
0
4 2
1
A = sq. units ]
2 3
[39]
77.
Z10
10
Z10 13
=0
10
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
i
1
= 13 e 10
Z2
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[cos18 + cos54 + cos90 + cos126 + cos162]

= 850 Ans. ]
Bansal Classes
[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
5
1
5
1
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
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
[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
3
3
ix
n 0 3
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
[42]

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BANSAL CLASSES

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