IIT-JEE 2006 Mathematics Paper (solved) - Class 12 PDF Download

 Page 1


IIT-JEE 2006-MA-1
           Solutions  to  IITJEE–2006  
M
M
a
a
t
t
h
h
e
e
m
m
a
a
t
t
i
i
c
c
s
s
Time: 2 hours 
Note: Question number 1 to 12 carries (3, -1) marks each, 13 to 20 carries (5, -1) marks each, 21 to 32 
carries (5, -2) marks each and 33 to 40 carries (6, 0) marks each. 
Section – A (Single Option Correct) 
1. For x > 0, 
( )
1/ x sin x
x0
lim (sin x) (1/ x)
?
+ is  
(A)0 (B) - 1 
(C) 1 (D) 2 
Sol. (C)
sin x
1/ x
x0
1
lim (sin x)
x
?
??
??
?+ ?
??
??
??
??
0 + 
x0
1
lim sin x ln
x
e1
?
??
??
??
= (using L’ Hospital’s rule). 
2. 
2
34 2
x1
dx
x2x 2x 1
-
-+
?
 is equal to  
(A) 
42
2
2x 2x 1
x
-+
 + c (B) 
42
3
2x 2x 1
x
- +
 + c 
(C) 
42
2x 2x 1
x
-+
 + c (D) 
42
2
2x 2x 1
2x
- +
 + c 
Sol. (D) 
35
24
11
dx
xx
21
2
xx
??
-
??
??
-+
?
Let 
24
21
2z
xx
-+ = ?  
1dz
4 z
?
?
1
zc
2
×+  ?  
24
12 1
2c
2 xx
-+ + . 
3. Given an isosceles triangle, whose one angle is 120 ° and radius of its incircle = 3 . Then the area of the triangle in sq.
units is
(A) 7 + 12 3 (B)12 - 7 3
(C) 12 + 7 3 (D)4 p 
Sol. (C)
2
3
b
4
?= …(1) 
Page 2


IIT-JEE 2006-MA-1
           Solutions  to  IITJEE–2006  
M
M
a
a
t
t
h
h
e
e
m
m
a
a
t
t
i
i
c
c
s
s
Time: 2 hours 
Note: Question number 1 to 12 carries (3, -1) marks each, 13 to 20 carries (5, -1) marks each, 21 to 32 
carries (5, -2) marks each and 33 to 40 carries (6, 0) marks each. 
Section – A (Single Option Correct) 
1. For x > 0, 
( )
1/ x sin x
x0
lim (sin x) (1/ x)
?
+ is  
(A)0 (B) - 1 
(C) 1 (D) 2 
Sol. (C)
sin x
1/ x
x0
1
lim (sin x)
x
?
??
??
?+ ?
??
??
??
??
0 + 
x0
1
lim sin x ln
x
e1
?
??
??
??
= (using L’ Hospital’s rule). 
2. 
2
34 2
x1
dx
x2x 2x 1
-
-+
?
 is equal to  
(A) 
42
2
2x 2x 1
x
-+
 + c (B) 
42
3
2x 2x 1
x
- +
 + c 
(C) 
42
2x 2x 1
x
-+
 + c (D) 
42
2
2x 2x 1
2x
- +
 + c 
Sol. (D) 
35
24
11
dx
xx
21
2
xx
??
-
??
??
-+
?
Let 
24
21
2z
xx
-+ = ?  
1dz
4 z
?
?
1
zc
2
×+  ?  
24
12 1
2c
2 xx
-+ + . 
3. Given an isosceles triangle, whose one angle is 120 ° and radius of its incircle = 3 . Then the area of the triangle in sq.
units is
(A) 7 + 12 3 (B)12 - 7 3
(C) 12 + 7 3 (D)4 p 
Sol. (C)
2
3
b
4
?= …(1) 
IIT-JEE 2006-MA-2
Also 
sin120 sin30
ab
°°
= ?a3b =
and 3s ?=  and  
1
s(a2b)
2
=+
?
3
(a 2b)
2
?= + …(2)
From (1) and (2), we get 
( )
12 7 3 ?= + .
4. If 0 < ? < 2 p, then the intervals of values of ? for which 2 sin
2
? - 5 sin ? + 2 > 0, is
(A) 
5
0, , 2
66
pp ?? ? ?
?p
?? ? ?
?? ? ?
(B)
5
,
86
p p ??
??
??
(C) 
5
0, ,
866
pp p ?? ? ?
?
?? ? ?
?? ? ?
(D)
41
,
48
p ??
p
??
??
Sol. (A) 
2sin
2
? - 5sin ? + 2 > 0 
? (sin ? - 2) (2sin ? - 1) > 0
? sin ? < 
1
2
? ? ? 
5
0, , 2
66
pp ?? ? ?
?p
?? ? ?
?? ? ?
. 
5. If w = a + i ß, where ß ? 0 and z ? 1, satisfies the condition that 
wwz
1z
- ??
??
-
??
 is purely real, then the set of values of z is 
(A) {z :  |z| = 1} (B) {z :  z = z } 
(C) {z :  z ? 1} (D) {z :  |z| = 1, z ? 1} 
Sol. (D)
wwz wwz
1z 1 z
--
=
--
? (zz 1)(w w) 0 -- =
? zz 1 =  ?  
2
z1 =  ?  z1 = . 
6. Let a, b, c be the sides of a triangle. No two of them are equal and ? ? R. If the roots of the equation x
2
 + 2(a + b+ c) x
+ 3 ?  (ab + bc + ca) = 0 are real, then
(A) 
4
3
?< (B)
5
3
? >
(C) 
15
,
33
??
??
??
??
(D) ? ? 
45
,
33
??
??
??
Sol. (A) 
D = 0 
? 4(a + b + c)
2
 - 12 ? (ab + bc + ca) = 0
? 
22 2
ab c 2
3(ab bc ca) 3
++
?= +
++
 Since |a - b| < c ?  a
2
 + b
2
 - 2ab < c
2
 …(1) 
 |b - c| < a  ?  b
2
 + c
2
 - 2bc < a
2
 …(2)
 |c - a| < b  ?  c
2
 + a
2
 - 2ac < b
2
 …(3)
From (1), (2) and (3), we get
22 2
ab c
2
ab bc ca
++
<
++
. 
Hence 
22
33
?< +  ?  ? < 
4
3
. 
Page 3


IIT-JEE 2006-MA-1
           Solutions  to  IITJEE–2006  
M
M
a
a
t
t
h
h
e
e
m
m
a
a
t
t
i
i
c
c
s
s
Time: 2 hours 
Note: Question number 1 to 12 carries (3, -1) marks each, 13 to 20 carries (5, -1) marks each, 21 to 32 
carries (5, -2) marks each and 33 to 40 carries (6, 0) marks each. 
Section – A (Single Option Correct) 
1. For x > 0, 
( )
1/ x sin x
x0
lim (sin x) (1/ x)
?
+ is  
(A)0 (B) - 1 
(C) 1 (D) 2 
Sol. (C)
sin x
1/ x
x0
1
lim (sin x)
x
?
??
??
?+ ?
??
??
??
??
0 + 
x0
1
lim sin x ln
x
e1
?
??
??
??
= (using L’ Hospital’s rule). 
2. 
2
34 2
x1
dx
x2x 2x 1
-
-+
?
 is equal to  
(A) 
42
2
2x 2x 1
x
-+
 + c (B) 
42
3
2x 2x 1
x
- +
 + c 
(C) 
42
2x 2x 1
x
-+
 + c (D) 
42
2
2x 2x 1
2x
- +
 + c 
Sol. (D) 
35
24
11
dx
xx
21
2
xx
??
-
??
??
-+
?
Let 
24
21
2z
xx
-+ = ?  
1dz
4 z
?
?
1
zc
2
×+  ?  
24
12 1
2c
2 xx
-+ + . 
3. Given an isosceles triangle, whose one angle is 120 ° and radius of its incircle = 3 . Then the area of the triangle in sq.
units is
(A) 7 + 12 3 (B)12 - 7 3
(C) 12 + 7 3 (D)4 p 
Sol. (C)
2
3
b
4
?= …(1) 
IIT-JEE 2006-MA-2
Also 
sin120 sin30
ab
°°
= ?a3b =
and 3s ?=  and  
1
s(a2b)
2
=+
?
3
(a 2b)
2
?= + …(2)
From (1) and (2), we get 
( )
12 7 3 ?= + .
4. If 0 < ? < 2 p, then the intervals of values of ? for which 2 sin
2
? - 5 sin ? + 2 > 0, is
(A) 
5
0, , 2
66
pp ?? ? ?
?p
?? ? ?
?? ? ?
(B)
5
,
86
p p ??
??
??
(C) 
5
0, ,
866
pp p ?? ? ?
?
?? ? ?
?? ? ?
(D)
41
,
48
p ??
p
??
??
Sol. (A) 
2sin
2
? - 5sin ? + 2 > 0 
? (sin ? - 2) (2sin ? - 1) > 0
? sin ? < 
1
2
? ? ? 
5
0, , 2
66
pp ?? ? ?
?p
?? ? ?
?? ? ?
. 
5. If w = a + i ß, where ß ? 0 and z ? 1, satisfies the condition that 
wwz
1z
- ??
??
-
??
 is purely real, then the set of values of z is 
(A) {z :  |z| = 1} (B) {z :  z = z } 
(C) {z :  z ? 1} (D) {z :  |z| = 1, z ? 1} 
Sol. (D)
wwz wwz
1z 1 z
--
=
--
? (zz 1)(w w) 0 -- =
? zz 1 =  ?  
2
z1 =  ?  z1 = . 
6. Let a, b, c be the sides of a triangle. No two of them are equal and ? ? R. If the roots of the equation x
2
 + 2(a + b+ c) x
+ 3 ?  (ab + bc + ca) = 0 are real, then
(A) 
4
3
?< (B)
5
3
? >
(C) 
15
,
33
??
??
??
??
(D) ? ? 
45
,
33
??
??
??
Sol. (A) 
D = 0 
? 4(a + b + c)
2
 - 12 ? (ab + bc + ca) = 0
? 
22 2
ab c 2
3(ab bc ca) 3
++
?= +
++
 Since |a - b| < c ?  a
2
 + b
2
 - 2ab < c
2
 …(1) 
 |b - c| < a  ?  b
2
 + c
2
 - 2bc < a
2
 …(2)
 |c - a| < b  ?  c
2
 + a
2
 - 2ac < b
2
 …(3)
From (1), (2) and (3), we get
22 2
ab c
2
ab bc ca
++
<
++
. 
Hence 
22
33
?< +  ?  ? < 
4
3
. 
IIT-JEE 2006-MA-3
7. If f ?(x) = - f(x) and g(x) = f '(x) and F(x) = 
22
xx
fg
22
?? ? ? ?? ??
+
?? ? ? ?? ??
?? ?? ?? ? ?
 and given that F(5) = 5, then F(10) is equal to  
(A) 5 (B) 10 
(C) 0 (D) 15 
Sol. (A)
f''(x) = -f(x) and f '(x) = g(x)
? f''(x) . f '(x) + f(x) . f '(x) = 0
? f(x)
2
 + (f '(x))
2
 = c  ?  (f(x)
2
 + (g(x))
2
 = c
? F(x) = c   ?  F(10) = 5.
8. If r, s, t are prime numbers and p, q are the positive integers such that the LCM of  p, q  is r
2
t
4
s
2
, then the number of
ordered pair (p, q) is
(A) 252 (B) 254 
(C) 225 (D) 224 
Sol. (C)
Required number of ordered pair (p, q) is (2 × 3 - 1) (2 × 5 -1) (2 × 3 - 1) = 225.
9. Let  ? ? 0,
4
p ??
??
??
 and t
1
 = (tan ?)
tan ?
,  t
2
 = (tan ?)
cot ?
, t
3
 = (cot ?)
tan ?
 and t
4
 = (cot ?)
cot ?
, then 
(A) t
1
 > t
2
 > t
3
 > t
4
(B) t
4
 > t
3
 > t
1
 > t
2
 
(C) t
3
 > t
1
 > t
2
 > t
4 
(D) t
2
 > t
3
 > t
1
 > t
4
 
Sol. (B)
Given ? ? 0,
4
p ??
??
??
, then tan ? < 1  and  cot ? > 1. 
Let tan ?  = 1 - ?
1
  and  cot ?  = 1 + ?
2
  where ?
1
 and ?
2
 are very small and positive.
then  
12
11
11 2 1
t(1 ) ,t (1 )
-? + ?
=-? =-?
12
11
32 4 2
t(1 ) andt (1 )
-? + ?
=+? = +?
 Hence t
4
 > t
3
 > t
1
 > t
2
. 
10. The axis of a parabola is along the line y = x and the distance of its vertex from origin is 2 and that from its focus is
2 2 . If vertex and focus both lie in the first quadrant, then the equation of the parabola is
(A)  (x + y)
2
 = (x - y - 2)  (B) (x - y)
2
 = (x + y - 2)
(C) (x - y)
2
 = 4 (x + y - 2) (D) (x - y)
2
 = 8 (x + y - 2)
Sol. (D)
Equation of directrix is x + y = 0.
Hence equation of the parabola is
22
xy
(x 2) (y 2)
2
+
=- + -
Hence equation of parabola is  
 (x - y)
2
  =  8(x + y - 2). 
11. A plane passes through (1, - 2, 1) and is perpendicular to two planes 2x - 2y + z = 0 and x - y + 2z = 4. The distance
of the plane from the point (1, 2, 2) is
(A) 0 (B) 1 
(C) 2 (D)2 2
Sol. (D)
The plane is a(x - 1) + b(y + 2) + c(z - 1) = 0
where 2a - 2b + c = 0 and a - b + 2c = 0
?
ab c
11 0
==
So, the equation of plane is x + y + 1 = 0