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20 Pages·2008·1.07 MB·English

MATHEMATICS
Directions : Questions number 1 to 5 are Assertion-Reason type questions. Each of these questions
contains two statements : Statement-1 (Assertion) and Statement-2 (Reason). Each of these questions
also has four alternative choices, only one of which is the correct answer. You have to select the correct
choice.
1. Let p be the statement “x is an irrational number”, q be the statement “y is a transcendental
number”, and r be the statement “x is a rational number iff y is a transcendental number”.
Statement-1 :
r is equivalent to either q or p.
Statement-2 :
r is equivalent to ~(p↔~q).
(1) Statement-1 is true, Statement-2 is false
(2) Statement-1 is false, Statement-2 is true
(3) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
(4) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
Sol. (..)
r Statement−I Statement−II
p q ~p ~q ~p↔q q∨p ~(p↔~q)
F F T T F F T
F T T F T T F
T F F T T T F
T T F F F T T
r is not equivalent to either of the statements
2. In a shop there are five types of ice-creams available. A child buys six ice-creams available. A child
buys six ice-creams
Statement-1 :
The number of different ways the child can buy the six ice-creams is 10C .
5
Statement-2 :
The number of different ways the child can buy the six ice-creams is equal to the number of
different ways of arranging 6 A’s and 4 B’s in a row.
(1) Statement-1 is true, Statement-2 is false
(2) Statement-1 is false, Statement-2 is true
(3) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
(4) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
Sol. (2)
a + b + c + d + e = 6
a ice creams of type I, b off type II, . . .
AIEEE 2008 SOLUTIONS MATHS Page 1
∴ total ways = 10C
4
10!
6A′S and 4B′S can be arranged in a row in =10C4 ways
6!4!
3. Statement-1 :
n
∑(r+1)nC =(n+2)2n−1.
r
r=0
Statement-2 :
n
∑(r+1)nC xr =(1+x)n+nx(1+x)n−1.
r
r=0
(1) Statement-1 is true, Statement-2 is false
(2) Statement-1 is false, Statement-2 is true
(3) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
(4) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
Sol. (3)
n n n
∑(r+1)nC =∑r nC +∑nC
r r r
r=0 r=0 r=0
=n×2n−1+2n
=2n−1(n+2)
Statement-1 is true
n n n
∑(r+1)nC xr =∑r nC xr +∑nC xr
r r r
r=0 r=0 r=0
=nx(1+x)n−1+(1+x)n
Statement-2 is true & Statement-2 explains Statement-1
4. Statement-1 :
For every natural number n≥2,
1 1 1
+ +...+ > n.
1 2 n
Statement-2 :
n(n+1)<n+1.
Page 2 AIEEE 2008 SOLUTIONS MATHS
(1) Statement-1 is true, Statement-2 is false
(2) Statement-1 is false, Statement-2 is true
(3) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
(4) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
Sol. (3) n > 2
1
1>
n
1 1
>
2 n
1 1
>
3 n
1 1
>
4 n
1 1
=
n n
Adding them
1 1 1
1+ + +...+ > n
2 3 n
Statement-1 is True
n(n+1)<n+1
⇒ n < n+1 (True)
1 1
>
n n+1
5. Let A be a 2 × 2 matrix with real entries. Let I be the 2 × 2 identity matrix. Denote by tr(A), the sum
of diagonal entries of A. Assume that A = I.
Statement-1 :
If A ≠IandA ≠−I, then det A =−1.
Statement-2 :
If A ≠IandA ≠−I, then tr(A)≠0.
AIEEE 2008 SOLUTIONS MATHS Page 3
(1) Statement-1 is true, Statement-2 is false
(2) Statement-1 is false, Statement-2 is true
(3) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
(4) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
a ba b 1 0
Sol. (1)   = 
c dc d 0 1
a2 +bc b(a+d) 1 0
 = 
 
(a+d)c bc+d2 0 1
a2+bc =1, b(a+d)=0
(a+d)c=0, bc+d2 =1
if b = 0 ⇒a2 = d2 = 1
now if a + d = 0
a = 1, d = –1 or a = –1, d = 1
1 0  −1 0
 or 
c −1  c 1
Then A ≠Ior−I
Then tr(A) = 0, |A|=−1, if a+d≠0,c =0
then A = I or –I
6. The statement p→(q→p) is equivalent to
(1) p→(q↔p) (2) p→(p→q)
(3) p→(p∨q) (4) p→(p∧q)
Sol (3) p q p∨q q→q p→(q→q) p→(p∨q)
F F F T T T
F T T F T T
T F T T T T
T T T T T T
Page 4 AIEEE 2008 SOLUTIONS MATHS
7. The value of cotcosec−15+tan−12 is
 3 3
5 6
(1) (2)
17 17
3 4
(3) (4)
17 17
Sol (2) cottan−13+tan−12
 4 3
 3 2 
+
 
cottan−1 4 3 
 1−3×2
 4 3
cottan−117
 6 
6
=
17
8. The differential equation of the family of circles with fixed radius 5 units and centre on the line
y = 2 is
(1) (x−2)2 y′2 =25−(y−2)2 (2) (x−2)y′2 =25−(y−2)2
(3) (y−2)y′2 =25−(y−2)2 (4) (y−2)2 y′2 =25−(y−2)2
Sol (4) Let centre be (h, 2)
equation of circle becomes
(x−h)2 +(y−2)2 =25
2(x−h)+2(y−2)y′=0
x−h=(2−y)y′
(y−2)2(y′)2+(y−2)2 =25
AIEEE 2008 SOLUTIONS MATHS Page 5
1 1
9. Let I=∫sinxdxandJ=∫cosdx. Then which one of the following is true ?
x x
0 0
2 2
(1) I> andJ<2 (2) I> andJ>2
3 3
2 2
(3) I< andJ<2 (4) I< andJ>2
3 3
1 1
Sol (3) I=∫sinxdx,J=∫cosxdx
x x
0 0
for 0 < x < 1 sin x < x
sinx
⇒ < x
x
1 1
⇒∫sinxdx<∫ x dx
x
0 0
2
⇒I<
3
1 1
∫cosxdx<∫ 1 dx
x x
0 0
J < 2
10. The area of the plane region bounded by the curves x + 2y2 = 0 and x + 3y2 = 1 is equal to
4 5
(1) (2)
3 3
1 2
(3) (4)
3 3
1
Sol (1) y2 =− x
2
1
y2 = (1−x)
3
Page 6 AIEEE 2008 SOLUTIONS MATHS
y
1
y2 = (1- x)
3
x
1
y2 = – x
2
On solving
1 1
− x = (1−x)
2 3
1 1
− x = (1−x)
2 3
1 1 1
⇒ x− x=
3 2 3
−1
x = ×6
3
=−2
1
∴ y2 =− x(−2)
2
y2 =1
y=±1
Required Area
=∫1(x −x )dy
1 2
0
=2∫1[(1−3y2)−(−2y2)]dy
0
=2∫1(1−y2)dy
0
 y31
=2y− 
 3 
0
AIEEE 2008 SOLUTIONS MATHS Page 7
 1
=21− 
 3
4
=
3
∫ sinxdx
11. The value of 2 is
 π
sinx− 
 4
π π
(1) x−log|cos(x− )|+c (2) x+log|cos(x− )|+c
4 4
π π
(3) x−log|sin(x− )|+c (4) x+log|sin(x− )|+c
4 4
Sol (4) 2 ∫ sinx dx
 π
sinx− 
 4
 π π
sinx− + 
 4 4
= 2 ∫ dx
 π
sinx− 
 4
 π π π  π
sinx− cos +sin cosx− 
= 2 ∫  4 4 4  4dx
 π
sinx− 
 4
π π  π
= 2 cos ∫dx+ 2sin ∫cotx− dx
4 4  4
 π
= x+logsinx−  +c
 4
Page 8 AIEEE 2008 SOLUTIONS MATHS
12. AB is a vertical pole with B at the ground level and A at the top. A man finds that the angle of
elevation of the point A from a certain point C on the ground is 60°. He moves away from the pole
along the line BC to a point D such that CD = 7 m. From D the angle of elevation of the point A is
45°. Then the height of the pole is
7 3 1 7 3 1
(1) m (2) m
2 3+1 2 3−1
7 3 7 3
(3) ( 3+1)m (4) ( 3−1)m
2 2
Sol (3) Let height of pole is y m
A
y
60° 45°
x 7m
B C D
y
∴ tan60°=
x
y
⇒ x=
3
y
& tan 45° =
x+7
⇒ x + 7 = y
y
⇒ +7=y
3
 1 
⇒ y1− =7
 3
7 3
⇒ y=
3−1
7 3
y= ( 3+1)m
2
AIEEE 2008 SOLUTIONS MATHS Page 9
13. How many real solutions does the equation x7 + 15x5 + 16x3 + 30x – 560 = 0 have ?
(1) 5 (2) 7
(3) 1 (4) 3
Sol (3) f(x) = x7 + 14 x5 + 16 x3 + 30x – 560
∴ f′(x)>0
⇒ f(x) is increasing
⇒ f(x) = 0 has only one solution
 1
(x−1)sin if ≠1
14. Let f(x)= x−1 Then which one of the following is true ?
 0 if x=1
(1) f is differentiable at x = 1 but not at x = 0
(2) f is neither differentiable at x = nor at x = 1
(3) f is differentiable at x = 0 and at x = 1
(4) f is differentiable at x = 0 but not at x = 1
1 1 1
Sol (4) f′(x)=sin − cos
x−1 x−1 x−1
f(x) is differentiable at x = 0
but not differentiable at x = 1
15. The first two terms of a geometric progression add up to 12. The sum of the third and the fourth
terms is 48. If the terms of the geometric progression are alternately positive and negative, then
the first term is
(1) 4 (2) –4
(3) –12 (4) 12
Sol (4) a + ar = 12
a(1 + r) = 12
ar2 + ar3 = 48
ar2 (1 + r) = 48
r2 = 4
r = ± 2
⇒ r = –2
∴ a(1 – 2) = 12
a=−12
Page 10 AIEEE 2008 SOLUTIONS MATHS