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SOLUTION OF IIT JEE – 2010
BY SUHAAG SIR
&
HIS STUDENTS OF CLASS MOVING FROM 11TH TO 12TH
S.No.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Student’s Name
Syd. Almas Ali
Anmol Rehani
Devashish Saxena
Mujahid Mohd. Khan
Shahrukh Ahmed
Sparsh Mehta
Geet Soni
Amit Sarathe
Ranjeet Singh
Rahul Jharwade
Anamika Singh
Shailja Aouthanere
Yamini Jain
Kirti Chopariya
School
All Saints’ School
Vikram Hr. Sec.
St. Mary’s Sr. Sec.
All Saints’ School
People’s Public School
People’s Public School
Jawahar Lal Nehru
K.V. – 3
Peragatisheel School
People’s Public School
K.V. – 2
People’s Public School
Chavara Vidya Bhawan Mandideep
Bourbon School
Results of year 2009 -- 15 IIT & 37 AIEEE selections out of 70 Fresh students
International Maths Olympiad
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Question sequence as per
MATHEMATICS PAPER – 2
SECTION – 1
Paper CODE - 1
Single Correct Choice Type
This section contains 8 multiple choice question. Each question has 4 choices A), B), C) and D) for its
answer, out of which ONLY ONE is correct.

20.
Two adjacent sides of a parallelogram ABCD are given by AB 2iˆ 10jˆ 11kˆ and

AD ˆi 2jˆ 2kˆ The side AD is rotated by an acute angle in plane of the parallelogram so
21.
22.
23.
that AD becomes AD' . If AD' makes a right angle with the side AB, then the cosine of the
angle is given by
8
1
17
4 5
(a)
(b)
(c) .
(d)
9
9
9
9
4
1
A sinnal which can be green or red with probability
and
respectively, is received by
5
5
station A then transmitted to station B. The probability of each station receiving the signal
received at station B is green, then the probability that the original signal was green is
3
6
20
9
(a)
(b)
(c)
(d)
5
7
23
20
If the distance of the point P 1, 2,1 from the plane x 2y 2z
, where
0, is 5, then
the foot of the perpendicular from P to the plane is
8 4 7
4 4 1
1 2 10
2 1 5
(a)
(b)
(c)
(d)
, ,
,
,
, ,
, ,
3 3 3
3 3 3
3 3 3
3 3 2
Let f be a real – valued function defined on the interval on the interval
1,1 such that
x
x
e f (x)
t 4 1dt, for all x
2
1,1 , and let f
1
be the inverse function of f. Then
0
f
1
'(2) is equal to
(a) 1
24.
For r
1
1
1
(c)
(d)
3
2
e
0,1......10, let A r , Br and C r denote, respectively, the coefficient of x r in the
expansions of 1 x
(b)
10
,1 x
20
and 1 x
30
10
. Then
Ar B10 Br C10 Ar is equal to
r 1
(a) B10 C10
2
C10 A10
(b) A10 B10
(c) 0
(d) C10
B10
1,2,3,4 . The total number of unordered pairs of disjoint subsets of S is equal to
(B) 34
(c) 42
(d) 41
SECTION – II
(Integer Type)
This Section contains 5 questions. The answer to each question is a single-digit integer, ranging from
0 to 9. The correct digit below the question no. in the ORS is to be bubbled.
26.
Let k be a positive real number and let
25.
Let S
(a) 25
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2k 1 2 k
A
27.
28.
29.
30.
2 k
2 k
1
2k
2 k
2k
1
and B
0
2k 1
k
1 2k
0
2 k .
k
2 k
0
If det (adj A) + det(adj B) 106 , then [k] is equal to
[NOTE : adj M denotes the adjoint of a square matrix M and[k] denotes the largest integer less
then or equal to k].
Two parallel chords of a circle of radius 2 are at a distance 3 1 apart. If the chords subtend at
2
, where k>0, then the value of [k] is
the center, angles of and
k
k
[NOTE : [k] denotes the largest integer less than or equal to k]
Consider a triangle ABC and let a,b and c denote the lengths of the sides opposite to vertices
A,B and C respectively. Suppose a = 6, b = 10 and the area of the triangle is 15 3. If ACB is
obtuse and if r denotes the radius of the incircle of the triangle, then r 2 is equal to
Let f be a function defined on R (the set of all real numbers) such that
2
3
4
f (x) 2010 x 2009 x 2010 x 2011 x 2012 , for all x R.
If g is a function defined on R with values in the interval 0,
such that f (x) n g(x) , for
all x R. then the number of points in R at which g has a local maximum is
Let a1 ,a 2 ,a 3 ,....,a11 be real numbers satisfying a1 15, 27 2a 2 0 and a k 2a k 1 a k 2 for
k
3, 4,....11. If
a12
2
a 22 .... a11
11
90, then the value fo
a1 a 2 ... a11
is equal to
11
SECTION – III
Paragraph Type
This section contains 2 paragraphs. Based upon each of the paragraphs 3 multiple choice questions
have to be answered. Each of these question has four choices A), B), C) and D) out of which ONLY
ONE is correct.
Paragraph for questions 31 to 33.
x2 y2
1 touching the ellipse at
Tangents are drawn from the point P(3, 4) to the ellipse
9
4
points A and B.
31.
The coordinates of A and B are
8 2 161
9 8
,
and
,
A) (3,0) and (0, 2)
B)
5
15
5 5
C)
32.
33.
8 2 161
,
and 0, 2
5
15
D) 3, 0 and
9 8
,
8 5
The orthocenter of the triangle PAB is
8
7 25
11 8
8 7
A) 5,
B)
C)
D)
,
,
,
5 8
5 5
25 5
7
The equation of the locus of the point whose distances from the point P and the line AB are
equal, is
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2
2
2
2
A) 9 x y 6 xy 54 x 62 y 241 0
B) x 9 y 6 xy 54 x 62 y 241 0
2
2
C) 9 x 9 y 6 xy 54 x 62 y 241 0
D) x 2 y 2 2 xy 27 x 31 y 120 0
Paragraph for question 34 to 36.
Consider the polynomial f x 1 2x 3x2 4x3. Let s be the sum of all distinct real roots of
34.
35.
36.
f x and let t s .
The real number of s lies in the interval
1
3
3 1
1
A)
B) 11,
C)
D) 0,
,0
,
4
4
4 2
4
The area bounded by the curve y f x and the lines x 0, y 0 and x t , lies in the
interval
3
21 11
21
A)
B)
C) 9, 10
D) 0,
,3
,
64
4
64 16
The function f ' x is
1
1
and decreasing in
,t
4
4
1
1
B) decreasing in t ,
and increasing in
,t
4
4
C) increasing in t , t
D) decreasing in t , t
SECTION – IV
(Matrix Type)
This Section contains 2 question. Each question has four statements (A, B, C and D) given in
Column I and five statements (p, q, r, s and t) in Column II. Any given statement in Column I
can have correct matching with one or more statement(s) given in Column II. For example, if
for a given question, statement B matches with the statements given in q and r, then for that
particular question, against statement B, darken the bubbles corresponding to q and r in the
ORS.
Match the statements in Column – I with the values in Column - II
A) increasing in
37.
t,
Column I
A. A line from the origin meets the lines
8
x
x 2 y 1 z 1
3 y 3 z 1 at P and Q respectively. If
and
1
2
1
2
1
1
2
length PQ = d, then d is
3
B. The values of x satisfying tan 1 x 3 tan 1 x 3 sin 1
are
5
   
  

C. Non-zero
vectors a,b,c satisfy a.b 0, b a . b c 0 and
   
 

2 b c b a . If a
b 4c, then the possible values of are
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Column II
P. 4
Q. 0
R. 0
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D.
Let
f
f (x) sin
38.
be
the
9x
2
sin
function
x
for x
2
on
given
,
0. The value of
2
by
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f 0 9 and S. 5
f (x)dx is
T. 6
Match the statements in Column-I with those in Column-II. [Note : Here z takes values in the
complex plane and Im z and Re z denote, respectively, the imaginary part and the real part of z.]
Column I
Column II
4
A. The set of points z satisfying z i z z i z
P. an ellipse with eccentricity
5
is contained in or equal to
B. The set of points z satisfying Q. the st of points z satisfying Im z = 0
z 4 z 4 10 is contained in or equal to
2, then the set of points z
C. If w
w
1
w
R. the set of points z satisfying Imz
1
w
1
w
S. the set of points z satisfying Rez
2
is contained in or equal to
1. then the set of points z
D. If w
is contained in or equal to
T. the set of points z satisfying z
3
SOLUTION – IIT JEE 2010 (PAPER - 2)
20.
(B)
cos
so cos
21.
2 20 22
1 4 4 4 100 121
cos
=
=
cos
8
9
8
9
17
9
(C) Event G = original signal is green , E1 A receives the signal correct, E 2 B receives
the signal correct, E = signal received by B is green, P(signal received by B is green)
46
40 / 5 16 20
.
, P(G / E)
5 16
46 / 5 16 23
5
1 4 2
5 3
5
(A) Let PN be normal to plane. PN
, 5
2
2
1
3
1 2 2
5 15
10 or 15 (cancle) plane x 2y 2z 10 Now check all option only A
= P( GE1E2 )+ P(GE1E2 ) P GE1E2 P(E)
22.
will satisfy the plane.
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x
x
4
0 1 (form equation. 1)
23.
Diff. Eq e f '(x) e f(x)
x 1 Put x = 0 1.f '(0) 1.f(0)
1
1
if g(x)
is f (x), so g (x) is f(x) , g 1(x) f(x),x g(f(x))
f '(0) 2 1,f '(0) 3
, 1 g'(f(x)).f '(x)
24.
25.
1
f '(x)
g'(f(x)),x
0
1
f '(0)
2 …..(1)
f(0)
g'(f(0))
1
3
g'(2).
(D) Here we are applying INDUCTION CONCEPT GIVEN BY SUHAAG SIR so here we
can’t print it.
Total No. of subsets = 2 n , Here n 4 therefore total sets = 16
1 , 2 , 3 , 4 , 1, 2 , 2,3 , 3, 4 , 4,1 , 2, 4 , 1,3
1, 2,3 , 2,3, 4 , 3, 4,1 , 2, 4,1 , 1, 2,3, 4
Set
26.
Set
1
its disjoints
8
2,4
its disjoints
1
2
7
1,3
2
3
1,2,3
6
1
4
2,3,4
5
1
1, 2
3,4,1
3
1
2,3
2,4,1
2
1
3,4
1,2,3,4
1
1
4,1
1
0
Therefore Ans = Total No. of disjoint sets i.e. = 41
(5)
A (2k 1)3, B 0 (since B is a skew – symmetric matrix of order 3)
A
27.
n 1
(2)
2k 1
2cos
2cos2
1
2
2
106
2cos
1 cos
2 3 1
4
28.
2k
3
2
k
2k 1 10
3 1, cos
3 1
,cos
2
2
2 2 3 3
 t
,
4
2
2k
9, k
cos
2k
t,2t 2
3 1
2
k
0,cos
k
1 4 3
0 ,t
3
,
2 2
2
Let
1
3 3
2
t
1,1 ,cos
4.
3 1
2
2
3
4
k
6
cos
3.
(3)
1
ab sinC
2
62 102
29.
det(adj A)
(1)
2
ab
sinC
2 6 10 cos120o
2 15 3
6 10
14
r
3
2
s
C 120o
225 3
r2
6 10 14
f(x) ln g(x) ,g(x) ef(x),f '(x) ef(x).f '(x),g'(x) 0
2
3
2010(x 2009) (x 2011) (x 2012)
4
a2
c
2
f '(x) 0
b2
2abcosC
3.
as
ef (x)
0
0 so there is only one point of local maxima.
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30.
This series is A.P. so a = 15 and 2nd
Ans (0)
27 2(a d)
31.
32.
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term is a + d , given
0,27 2(15 d)
0,27 30 2d 0, 3 2d 0, 3
2d.,
3
2
d. so acceding to
condition of question d must be 3 so now check answer will be 0.
x.x1 y. y1
x
1
y 1 , x 3y
(D) Coordinate of point P 3, 4 equation of tangency
9
4
3
9 8
points (3, 0) and
satisfies the above equation hence the answer is (D)
Ans (D)
,
5 5
OR II METHOD
(D) By using Suhaag Short Trick 1. They satisfy the given equation
according to the graph (3,4)
Ans (D)
8
0
1
5
(C) Slope
9
3 3
5
….. (1)
y 4 3x 9 3x y 5 0
y 4 3 x 3
3 Because the
…… (2)
y
2x 6
y 0 2 x 3
3x y 5
2x y 6
11
By putting the value of x in eq. (1)
5x 11 x
5
8
11
11 8
y
Ans
Ans (C)
3
y 5
,
5
5
5 5
OR II METHOD (C) By using Suhaag Short Trick According to the figure given above
11 8
Ans (C)
,
2.2, 1.6
5 5
33.
(A)
x 3
34.
35.
Locus is parabola Equation of AB is
2
y 4
x 3y 3
2
3x
9
4y
4
1
x
3
y 1
x 4y 3
0
2
,
10
2
2
10x 90 60x 10y 160 80y x 2 9y 2 9 6xy 6x 18y
9x 2 v 2 6xy 54x 62y 241 0 .
1
3
3 1
.f
0
,
.
(C) Since f
S lie in
2
4
4 2
3
4
(A)
1 1
,
2 2
s
x4
x3
x2
1
,
16
1
8
1
4
x
1
2
1/ 2
0
t
3
,
4
area
area
81
256
1/ 2
3/4
4x
3
3x
2
2x 1 dx
0
x4
27
64
4x 3
area
3x 2
2x 1 dx
0
x3
9
16
x2
x
3/4
0
3 15
,
4 16
area
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525
.
256
Page 7 of 7
36.
37.
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(B)
f '(x) 2 12x 3
1/ 4.
x y z
(A) - (T) Let the line be
intersects the
a b c
a : b : c :: 5r : 5r : 2r
3a b 5c 0
(B) - (P&R) tan 1(x 3) tan 1(x 3) sin 1(3 / 5)
x 3
x 3
3
6
3
tan 1
tan 1
x2
2
2
4
x 8 4
1 x 9



2


(C) - (Q,S) As a b 4c
b
4b.c and b

   
b.c
2b c b a
solve and eliminating
2
2 2 10 b
0
0 and 5.
(D) - (R)
/2
8
0
16
2
dx
sin x / 2
/2 I
x
x
sin9 x / 2
2
I
0
sin9 x / 2
2
sin x / 2
0
sin9
d
sin
sin9
8
cos6
cos2
/2
8
0
2
z
z
(A) - (Q)
1
0
4
z
z
1,z
z
z
1
z 4
10
bisector of I and i
(B) - (P)
z 4
length of major axis is 10
(C)
0,
- (P,T)
w
8
1d
3
cos
2
5
i sin
2
sin5
2d ,x
sin3
we
0,
sin 6
6
A is real number
1z
sin
sin
d
sin
sin 4
4
sin 2
2
Im(z) 0.
z lies on an ellipse whose focus are (4,0) and ( 4,0) and
8 and 2a 10
w
2 cos
x3
3/2
isin
y2
2
5/2
2
1,e2
4 / 5 Re(z)
e
,x iy
1
(D)
2 cos
9/4
25 / 4
1
5.
isin
9
25
16
25
1
cos isin
2
4
e
.
5
w
1
get
0
sin
16 sin8
8
d
2
a
eliminating
dx
Again, as
0
z
is unimodular complex number and lies on perpendicular
z
z
2ae
2
4.
sin
/2
0
0
38.
and
0 and
a 3b 5c
2  

4a.c and b b.c d.c
sin
cos 4
8 or x
sin7
0
8
8
S.D 0
dx,x / 2
sin7
/2
cos8
lines
x iy cos isin
cos
isin ,x iy
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2cos Re(z)
(Q,T)
1, Im(z)
0.
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