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PAPER-B
IIT–JEE
(2012)
(Algebra + Straight Lines)
Solutions
“TOWARDS IIT- JEE IS NOT A JOURNEY,
IT’S A BATTLE, ONLY THE TOUGHEST WILL SURVIVE”
TIME: 60 MINS
MAX. MARKS: 75
MARKING SCHEME
For each question in Section I : you will be awarded 5 marks if you have darkened only the
1.
bubble corresponding to the correct answer and zero mark if no bubbles are darkened. In all other cases,
−2) marks will be awarded.
minus two (−
2.
For each question in Section II : you will be awarded 3 marks if you darken the bubble
corresponding to the correct answer and zero mark if no bubble is darkened. No negative marks will be
awarded for incorrect answers in this Section.
For each question in Section III : you will be awarded 3 marks if you darken only the bubble
3.
corresponding to the correct answer and zero mark if no bubbles are darkened. In all other cases, minus
−1) mark will be awarded.
one (−
For each question in Section IV : you will be awarded 2 marks for each row in which you have
4.
darkened the bubble (s) corresponding to the correct answer. Thus, each question in this section carries a
maximum of 8 marks. There is no negative marks awarded for incorrect answer (s) in this Section.
NAME OF THE CANDIDATE
PHONE NUMBER
L.K. Gupta (Mathematics Classes)
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Section I
This Section contains 6 multiple choice questions. Each question has 4 choices A), B), C)
and D) out of which ONLY ONE is correct.
If a < 1 and b < 1 , then the sum of the series 1 + (1 + a)b + (1 + a + a2)b2 +
1.
(1 + a + a2 + a3)b3 + …. is
1
1
1
1
(a)
(b)
(c)
(d)
(1 − a)(1 − b)
(1 − a)(1 − ab)
(1 − b)(1 − ab)
(1 − a)(1 − b)(1 − ab)
Sol. (c)
We have, 1 + (1 + a)b + (1 + a + a2 )b2 + (1 + a + a2 + a3 )b3 + ...∞
∞
= ∑(1 + a + a2 + ... + a n−1 )bn−1
n =1
 1 − a n  n−1
=∑ 
b
n =1  1 − a 
∞
bn−1 ∞ a n bn−1
=∑
−∑
n =1 1 − a
n =1 1 − a
∞
1 ∞ n−1
a ∞
=
b −
(ab)n−1
∑
∑
1 − a n=1
1 − a n=1
1
a
=
[1 + b + b2 + ... ∞]−
[1 + ab + (ab)2 + ... ∞]
1−a
1−a
1
1
a
1
=
×
−
=
1 − a 1 − b (1 − a)(1 − ab) (1 − ab)(1 − b)
2.
2F(n) + 1
for n = 1, 2, 3, ……and F(1) = 2. Then, F(101) equals
2
(b) 52
(c) 54
(d) none of these
Suppose that F(n + 1) =
(a) 50
Sol. (b)
Given,
F(n + 1) =
2F(n) + 1
2
1
2
Hence, the given series is an A.P. with common difference ½ and first term being 2. F (101)
is 101st term A.P. given by 2 + (101 − 1)(1 / 2) = 52
⇒ F(n + 1) − F(n) =
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1
If t n = (n + 2)(n + 3) for n = 1, 2, 3, ……., then
4
1 1 1
1
+
+ + ..... +
=
t1 t2 t3
t 2003
3.
4006
4003
(b)
3006
3007
Sol.(d)
1
Let t n = (n + 2)(n + 3) . Then
4
1 1 1
1
+ + + ... +
t1 t2 t3
t 2003
(a)
(c)
4006
3008
(d)
4006
3009
1
1
1
 1

=4 
+
+
+ ... +
2005 × 2006 
 3× 4 4 × 5 5× 6
1 
1
=4  −
 3 2006 
2003
4006
=4×
=
3(2006) 3009
If a , β be the roots of the equation 2x 2 − 35x + 2 = 0 , then the value of
4.
(2a − 35)3 (2 β − 35)3 is equal to
(a) 8
(b) 1
(c) 64
(d) none of these
Sol. (c)
Since, α , β are the roots of the equation 2x 2 − 35x + 2 = 0 , therefore,
2α2 − 35α = −2 or 2α − 35 =
−2
α
and 2β2 − 35β = −2 or 2β − 35 =
−2
β
Now,
3
3
 −2   −2 
(2a − 35) (2β − 35) =    
 α   β 
8 × 8 64
= 3 3=
= 64
(∵ αβ = 1)
1
αβ
3
3
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If a and β are the roots of the equation x2 − ax + b = 0 and A n = a n + βn , then which of
5.
the following is true?
(a) A n +1 = aA n + bA n −1
(b) A n +1 = bA n + aA n −1
(c) A n +1 = aA n − bA n −1 (d) A n +1 = bA n + aA n −1
Sol.(c)
A n +1 = α n+1 + βb+1
= α n+1 + α nβ + βn +1 + αβn − α nβ − αβn
= α n (α + β) + βn (β + α) − αβ(α n −1 + βn −1 )
= α n (α + β) + βn (β + α) − αβ(α n −1 + βn −1 )
= ( α + β)(α n + βn ) − αβ(α n −1 + βn −1 )
= aA n − bA n −1
2
2
1
2m
If
+
+
=
, then orthocenter of the triangle having sides
1!9 ! 3 ! 7 ! 5 ! 5 ! n !
6.
x − y + 1 = 0, x + y + 3 = 0 and 2x + 5y − 2 = 0 is
(a) (2m − 2n, m − n)
(b) (2m − 2n, n − m)
(c) (2m − n, m + n)
Sol.
2
2
1
2m
∵
+
+
=
1!9 ! 3 ! 7 ! 5 ! 5 ! n !
(d) (2m − n, m − n)
1  2 × 10! 2 × 10! 10!  2m
+
+

=
10!  1!9!
3!7! 5!5!  n !
1 10
2m
10
10
2 C1 + 2 C3 + C5 =
⇒
10!
n!
(
1
10!
(
10
)
)
C1 +10 C3 +10 C5 +10 C7 +10 C9 =
2m
n!
1
2m
10−1
⇒
( 2) =
10!
n!
∴ m = 9 and n = 10
Hence x – y + 1 = 0 and x +y +3 = 0 are perpendicular to each other, then orthocenter is the
point of intersection which is (–2, –1).
∴ –2 = 2m – 2n
and –1 = m – n
∴ Point is (2m – 2n, m–n).
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SECTION – II (Integer Type)
This section contains 5 questions. The answer to each question is a single-digit integer,
ranging from 0 to 9. Correct digit below the question no. in the answer sheet is to be
bubbled.
7. In a ∆ ABC, A ≡ (α, β), B ≡ (1, 2), C ≡ (2,3) and point A lies on the line y = 2x + 3,
whereα,β ∈ I . If the area of ∆ ABC be such that [ ∆ ] = 2, where [.] denotes the greatest
integer function, then the number of all possible coordinates of A must be
Sol. 4
∵ ( α , β) lies on y = 2x + 3
Then, β = 2α + 3
Thus, the coordinates of A are (α , 2α + 3) ---------(i)
∆=
2
3 
1  α 2α + 3 1 2
+
+


2
2 3
α 2α + 3 
2 1
1
2α − (2α + 3) + 3 − 4 + 4α + 6 − 3α
2
1
= α +2
2
but [∆ ] = 2
1

 2 a + 2  = 2
a +2
⇒2≤
<3
2
⇒4≤ a +2 <6
⇒ 4 ≤ α + 2 < 6 and − 6 < α + 2 ≤ − 4
⇒ 2 ≤ α < 4 and − 8 < α ≤ − 6
∵a ∈I
∴ a = 2, 3, − 7, − 6
Hence, possible coordinates of A are
(2, 7), (3, 9), (−7, 11) and (−6, − 9).
Number of all possible coordinates of A are 4.
=
2007
8. If x1, x2, x3, ………x 2008 are in HP and
∑x
i
x i +1 = λ x1 x2008 , then λ − 2000 must be equal
t =1
to :
Sol. 7
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Given x1, x2, x3,……x2008 are in HP
1 1 1
1
1
∴ , , ,.......
,
are in AP.
x1 x 2 x3
x2007 x 2008
1 1
1
1
Then, −
=
−
x 2 x 1 x3 x 2
1
1
= .... =
=−
=D
x 2008
x 2007
x − x 2 x 2 − x3 x 2007 − x2008
or 1
=
=
=D
x1x2
x 2x 3
x 2007 x 2008
(x − x 2 ) + (x 2 − x3 ) + .... + (x2007 − x 2008 )
= 1
=D
x1x 2 + x2x3 + .... + x2007 x 2008
(by law of proportion)
x − x2008
or x1x 2 + x 2x3 + ....... + x2007 x 2008 = 1
D
2007
x − x2008
------(i)
or ∑ x i x i +1 = 1
D
i =1
1
1
=
+ (2008 − 1) D
Now,
x2008 x1
x −x
or 1 2008 = 2007 x1x2008 --------(ii)
D
∴ From Eqs. (i) and (ii),we get
2007
∑ xx
i i +1
= 2007 x1x 2008
i =1
∴ λ = 2007 ⇒ λ − 2000 = 7
9. The number of zeros in the end in the product of 56 × 67 × 78 × 89 × ……..× 5051 is
λ then λ − 280 =
Sol. 5
Since, one pair of 2 and 5 makes 10 (i. e., 2 × 5 = 10) which gives one zero at the end.
Here, 56, 1011, 1516, 2021, 2526, 3031, 3536, 4041 , 4546, 5051
∴ Total numbers of fives
= 6 + 11 + 16 + 21 + 26 + 31 + 36 + 41 + 46 + 51
10
=
(6 + 51) = 5 × 57 = 285
2
Clearly number of 2’s is greater than that of 5. Thus, number of zeros at the end = 285.
λ − 280 = 5
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λ
10.
If 12 Pr = 11880, then
∑
λ
C i − 120 must be (where λ = r + 3)
i =1
Sol. 7
∵ 12pr = 12 × 11 × 10 × 9 = 12p4
∴r =4
then , λ = r + 3 = 7
7
∴ ∑ 7 Ci = 27 − 1 = 127
i=1
⇒
λ
∑
λ
C i − 120 = 7
i =1
11. The number of words can be formed with the letters of the word PATALIPUTRA
without changing the relative positions of vowels and consonants is the λ , then λ − 3600 =
Sol. 0
The word ‘PATALIPUTRA’ has eleven letters, in which 2P’s, 3A’s, 2T’s 1L, 1U, 1R, 1I. Vowels
are AAIUA.
These consonants con be PTLPTR these consonants con be arranged themselves in
6!
= 180 ways.
2!2!
The consonants are PTLPTR these consonants can be
6!
arranged themselves in
= 180 ways.
2 !2 !
∴ Required number of words = 20 × 180 = 3600 ways.
⇒ λ − 3600 = 0
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 questions has four choices A), B), C) and D)
out of which Only One is correct.
Paragraph for questions 12 − 14
PASSAGE 5
Let A (x1, y1), B (x2, y2) and ( x3 , y3 ) be three points such that AC = AB and ∠ CAB = θ .
Let z1 = x1 + iy1, z2 = x2 + iy2 and z3 = x3 + iy 3 , where i = − 1 . Therefore, AB = z2 − z1 ,
AC = z3 − z1
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Then, AC will be obtained by rotating AB through an angle θ in anticlockwise sense and
z −z 
therefore AC = AB ei θ or (z3 − z1 ) = (z2 − z1 )eiθ or  3 1  = eiθ
 z 2 − z1 
On the basis of above information, answer the following questions:
12. The line joining the points A (1, 0) and B (1 + 3 ,1) is rotated about A by an angleθ.
If B goes to C (2, 3), then the angle θ is
(a) 150
(b) 300
Sol. (b)
Here, z1 = 1 + i.0,z2 = (1 + 3) + i
(c) 450
(d) 900
and z3 = 2 + i 3
then
⇒
(2 + i 3) − (1 + i . 0)
= e iθ
[(1 + 3) + i] − (1 + i . 0)
(1 + i 3) iθ
=e
( 3 + i)
(1 + i 3)( 3 − i)
= e iθ
4
2 3 + 2i
⇒
= e iθ
4
3 i
⇒
+ = eiθ
2 2
⇒ eiπ/6 = eiθ
π
∴ θ = = 300
6
⇒
13. The extremities of the diagonal of a square are (1, 1), (−2, − 1), then the equation of
the other diagonal is
(a) 8x + 3y + 4 = 0
(b) 8x − 3y + 4 = 0
(c) 6x + 4y + 3 = 0
(d) 6x − 4y + 3 = 0
Sol. (c)
∵ A ≡ (1,1)
∴ zA = 1 + i
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and C ≡ ( −2, − 1)
∴ zc = − 2 − i
 1 
The centre of the square E ≡  − , 0 
 2 
1
zE = −
2
Now in ∆ AEB (EA = EB)
zB − zE
= ei π/2 = i
z A − zE
1
1

⇒ zB + = i  1 + i + 
2
2

3
= i−1
2
3 3
or zB = − + i
2 2
 3 3
∴ B ≡ − , 
 2 2
⇒ Equation of other diagonal is
3
−0
1

2
y −0=
x+ 

3 1
2
− + 
2 2
3
3
⇒ y= − x −
2
4
⇒ 4y = − 6x − 3
⇒ 6x + 4y + 3 = 0
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14. A point (2, 2) undergoes reflection in the x-axis and then the coordinate are rotated
through an angle of π / 4 in anticlockwise direction. The final position of the point in the
new coordinate system is
(a) (0,2 2)
(b) (0, − 2 2)
(c) (2 2,0)
(d) ( − 2 2, 0)
Sol.
Image of the point (2,2) in the x-axis is (2, − 2). If (x, y)be the coordinates of any point and
(X, Y) in its new coordinates then X = x cos θ + y sin θ and Y = − x sin θ+ y cos θ where θ is
the angle through which the axes have been rotated.
Here, θ = π /4 ,x = 2, y = − 2
2
2
∴ X = 0, Y = −
−
= −2 2
2
2
∴ (X, Y) ≡ (0, − 2 2)
Paragraph for questions 15 − 17
Consider the inequation x2 + x + a − 9 < 0.
15. The values of the real parameter ‘a’ so that the given inequation has at least one
positive solution :
(a) ( −∞ , 37 / 4)
(b) ( −∞ , ∞ )
(c)(3, ∞ )
(d) ( −∞ ,9)
Sol.
Let f(X) = = x 2 + x + a − 9
x2 + x + a − 9 < 0 has at least one positive solution, then either both the roots of equation
x2 + x + a − 9 = 0 are non-negative or 0 lies between the roots.
For case I, sum or roots is −1 / 2 > 0 , which is not possible.
For case II. f(0) < 0 ⇒ a − 9 < 0 ⇒ a < 9
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16.
The values of the real parameter ‘a’ so that the given inequation has at least one
negative solution :
(a) ( −∞ ,9)
(b) (37 / 4, ∞ )
(c) [9, 37 / 4)
(d) none of these
Sol.
If x2 + x + a − 9 < 0 has at least one negative solution, then either both the roots of equation
x2 + x + a − 9 = 0 are non-positive or 0 lies between the roots.
For case I, sum of roots is
−1
< 0 . Product of roots is a − 9 > 0
2
⇒ a ≥ 9 and
D > 0 ⇒ 1 − 4(a − 9) > 0 ⇒ a <
Hence, 9 ≤ a <
17.
37
4
37
.
4
The values of the real parameter ‘a’ so the given inequation is true ∀ x ∈ ( − 1, 3) :
(a) ( −∞ , − 3)
(b) ( − 3, ∞ )
(c) [9, ∞ )
(d) ( −∞ , 37 / 4)
Sol.
x 2 + x + a − 9 < 0 is true ∀ x∈( −1, 3) , then f( −1) < 0 and f(3) < 0.
∴ 1 − 1 + a − 9 < 0 and 9 + 3 + a − 9 < 0
⇒ a < 9 and a < −3
⇒ a < −3
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SECTION − IV (Matrix type)
Each question contains statements given in two columns, which have to be matched.
Statements a, b, c, d in columns I have to be matched with statements p, q, r, s in columns II.
If the correct matched are a → p, a → s, b → q, b → r, c → p, c → q and d → s, then the
correctly bubbled 4 × 4 matrix should be as follows:
18. Column I contains rational algebraic expression and column II contains possible
integers, which lie in their range.
Column I
x 2 − 2x + 4
a. y = 2
, x∈R
x + 2x + 4
x 2 − 3x − 2
b. y =
, x∈R
2x − 3
2x 2 − 2x + 4
c. y = 2
, x∈R
x + 4x + 3
d. x 2 − (a − 3) x + 2 < 0, ∀ x ∈( − 2, 3)
Column II
p. 1
q. 4
r. − 3
s. − 10
Sol.
a → p;b → p,q ,r,s;c → p,q ,s;d → r,s .
x2 − 2x + 4
a. y= 2
x + 2x + 4
2
⇒ x y + 2xy + 4y = x 2 − 2x + 4
⇒ (y − 1)x2 + 2(y + 1)x + 4(y − 1) = 0
D≥0
⇒ 4(y + 1)2 − 16(y − 1)2 ≥ 0
⇒ (y + 1)2 − (2y − 2)2 ≥ 0
⇒ (3y − 1)(3 − y) ≥ 0
1 
⇒ (3y − 1)(y − 3)≤ 0 ⇒ y ∈ ,3
3 
⇒ {1} ⇒ P
x 2 − 3x − 2
b. y =
2x − 3
2
⇒ x − 3x − 2 = 2xy − 3y
⇒ x 2 − (3 + 2y)x + (3y − 2) = 0
D ≥0
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⇒ (3 + 2y)2 − 4(3y − 2) ≥ 0
⇒ 9 + 4y 2 + 12y − 12y + 8 ≥ 0
⇒ 4y 2 + 17 ≥ 0
which is always true. Hence,
y∈ R ⇒ {1,4, −3, −10} ⇒ p,q,r,s
2x2 − 2x + 4
c. y = 2
x − 4x + 3
2
⇒ x y − 4xy + 3y = 2x 2 − 2x + 4
(y − 2)x 2 + 2(1 − 2y)x + 3y − 4 = 0
D≥ 0
4(1 − 2y)2 − 4(y − 2)(3y − 4)≥ 0
⇒ 1 + 4y 2 − 4y − (3y 2 − 10y + 8)≥ 0
⇒ y 2 + 6y − 7 ≥ 0
⇒ (y + 7)(y − 1) ≥ 0
⇒ y ≥ 1 or y ≤ − 7
⇒ {1,4, −10} ⇒ p ,q ,s
d. f(x) = x 2 − (a − 3)x + 2 < 0, ∀x∈ [−2, −1]
⇒ f( −2)< 0 and f( −1)< 0
⇒ 4 + 2 (a − 3) + 2 < 0 and1 + (a − 2) + 2 < 0
⇒ a < 0 and a < −1
⇒ a < −1
⇒ a∈{ −10, −3}
19. Consider the convex polygon, which has 35 diagonals. Then match the following
column.
Column I
Column II
a. Number of triangles joining the vertices of the polygon
p. 210
b. Number of points intersections of diagonal which lies
q. 120
inside the polygon
c. Number of triangles in which exactly one side is
r. 10
common with that of polygon
d. Number of triangles in which exactly two sides are
s. 60
common with that of polygon
Sol.
a. If polygon has n sides, thee number of diagonals is nC2 − n = 35 (given). Solving we get n =
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10. Thus, there are 10 vertices, from which
10
C3 ( = 120) triangles can be formed.
b. Four vertices can be selected in 10 C4 ( = 210) ways. Using these four vertices two
diagonals can be formed which has exactly one point of intersection lying inside the
polygon.
Hence, number of points of intersection of diagonal which lies inside the polygon is
10
C4 × 1 = 210 .
c.
Suppose one of the sides of the triangle is A 1 A 2 . Then third vertex cannot be A3 or A10 .
Thus, for the third vertex six vertices are left. There are six triangles in which side A1 A 2 is
common with that of polygon. Similary, for each of the sides A 2 , A3 , A3 A 4 ,..., A 9 A10 there are
six triangles. Then total number of triangles is 60..
d. Triangles A, A1 A 2A3 , A 2 A3 A 4 , … , A 8 A 9 A10 have two sides common with that of polygon.
Hence, there are 10 such triangles.
PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 320, SECTOR 40–D, CHANDIGARH
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