
Solutions To AIEEE, 2008
PART A - MATHEMATICS
1. Let f : N ® Y be a function defined as f(x) = 4x + 3 where
Y = {y Î N :
y = 4x + 3 for some x Î N}. Show that f
is invertible and its inverse is
(1)
(2) ![]()
(3)
(4) ![]()
1. (3)
f(x) = 4x + 3
f¢(x) = 4 > 0
\ f(x) is strictly increasing function
Þ f(x) is one-one and onto
Þ f-1(x) exist
Putting f(x) = y
y = 4x + 3
Þ ![]()
2. Let R be the real line. Consider the following subsets of the plane R ´ R
S = {(x, y) : y = x + 1 and 0
< x < 2}
T = {(x, y) : x - y is an
integer}.
Which one of the following is true?
(1) Both S and T are equivalence relations on R
(2) S is an equivalence relation of R but T is not
(3) T is an equivalence relation on R but S is not
(4) Neither S nor T is an equivalence relation on R
2. (3)
3. The
conjugate of a complex number is
. Then that complex numbers is
(1)
(2) ![]()
(3)
(4) ![]()
3. (2)
. z
= ![]()
![]()
4. The quadratic equations
x2 = 6x + a = 0
and x2 - cx + 6 = 0
have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3.Then the common roots is
(1) 4 (2) 3
(3) 2 (4) 1
4. (3)
x2 - 6x + a = 0 (1)
x2 - cx + 6 = 0 (2)
Given second root equation(1)/second root of equation(2) = 4/3 = k
Let one root is a and other root is 4k of first equation
(x - a) (x - 4k) = 0,
Þ x2 - (a + 4k)x + 4k a = 0 (3)
(x - a) (x - 3k) = 0
Þ x2 - (a + 3k)x + 3ka = 0 (4)
By comparison (3) with (1) and (4) with (2)
a + 4k = 6, (5)
a + 3k = c, 3ka = 6 Þ a = 2/k
(5) Þ 2/k + 4k = 6 Þ k = 1, k = ½
k = ½ is not allowed as it gives
second root of second equation as 3 ´ ½ = 3/2 which is not an integer.
\ k = 1, a = 2.
5. Let A be a square matrix all of whose entries are integers. Then which one of the following is true?
(1) If det A ¹ ± 1, then A-1 exists and all its entries are non-integers
(2) If det A = ±1, then A-1 exists and all its entries are integers
(3) If det A = ±1, then A-1 need not exist
(4) If det A = ±1, then A-1 exists but all its entries are not necessarily integers
5. (2)
6. Let a, b, c be any real numbers. Suppose that there are real numbers x, y, z not all zero such that x = cy + bz, y = az + cx, and z = bx + ay. Then a2 + b2 + c2 + 2abc is equal to
(1) -1 (2) 0
(3) 1 (4) 2
6. (3)
(for non trivial
solution)
1 - a2 + c (-c - ab) + b (-ac - b) = 0
a2 + b2 + c2
+ 2abc = 1
7. How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which no two S are adjacent?
(1) 6 . 7 . 8C4 (2) 6 . 8 . 7C4
(3) 7 . 6C4 . 8C4 (4) 8 . 6C4 . 7C4
7. (3)
First we took M, I, I, I, P, P, I
and arranged then arrangement =
ways.
There are 8 spaces available for 4
S.
So, total number of arrangement
according to requirement = ![]()
= ![]()
8. 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) -12 (2) 12
(3) 4 (4) -4
8. (1)
Given a + ar = 12 (where a first
term and r common ratio)
ar2 + ar3
= 48
r2(a + ar) =
48
r2 = 4
r = ±2 Þ r = -2 (is
valid)
\ first
term is -12 and second
term is -12 ´ (-2) = 24
9. Let

Then which one of the following is true?
(1) f is differentiable at x = 0 and at x = 1
(2) f is differentiable at x = 0 but not at x = 1
(3) f is differentiable at x = 1 but not at x = 0
(4) f is neither differentiable at x = 0 nor at x = 1
9. (2)
LHD =
=
= a finite value
which lies between -1 and 1.
RHD = sin(¥) = a finite
value which lies between -1 and 1.
RHD ¹ LHD.
\ f(x)
is differentiable at x = 0 but not at x = 1.
10. How many real solutions does the equation x7 + 14x5 + 16x3 + 30x - 560 = 0 have?
(1) 1 (2) 3
(3) 5 (4) 7
10. (1)
f(x) = x7 + 14x5 + 16x3 + 30x - 560
f¢(x) = 7x6 + 70x4 + 48x2 + 30 > 0
Þ f(x) is strictly increasing function
Þ hence f(x) cuts x axis only at one point
11. Suppose the cubic x3 - px + q has three distinct real roots where p > 0 and q > 0. Then which one of the following holds?
(1) The cubic has minima at
and maxima at ![]()
(2) The cubic has minima at both
and -![]()
(3) The cubic has maxima at both
and -![]()
(4) The cubic has minima at
and maxima at -![]()
11. (4)
|
f¢(x) = 0 4x2 - p = 0 x = ± |
|
12. The
value of
is
(1)
(2) ![]()
(3)
(4) ![]()
12. (2)
I = 
=
= ![]()
= x + ln(sinx - cosx) + c
=
x + logsin(x - p/4) + k where k = log
+ c.
13. The area of the plane region bounded by the curves x + 2y2 = 0 and x + 3y2 = 1 is equal to
(1)
(2) ![]()
(3)
(4) ![]()
13. (3)
x = -2y2 x = 1 - 3y2
x1 = -2y2 x2 = 1 - 3y2
x1 = x2 (for cutting points)
Þ -2y2 =
1 - 3y2
y2 = 1
y = ±1
Area = ![]()
Area =
=
=
=
.
14. Let
and
.
Then which one of the following is true?
(1)
and J < 2 (2)
and J > 2
(3)
and J < 2 (4)
and J > 2
14. (1)

\ I
< 2/3

J < 2.
15. The differential equation of the family of circles with fixed radius 5 units and centre on the line y = 2 is
(1) (y - 2)y¢2 = 25 - (y - 2)2 (2) (y - 2)2y¢2 = 25 - (y - 2)2
(3) (x - 2)2y¢2 = 25 - (y - 2)2 (4) (x - 2)y¢2 = 25 - (y - 2)2
15. (2)
Equation of circle
(x - h)2 + (y - 2)2
= 25
Differentiating with respect to x
2(x - h) + 2(y - 2)y¢ = 0
(x - h) = -(y - 2) y¢
Þ (y
- 2)2
y¢2 + (y - 2)2 = 25.
16. The solution of the differential equation
![]()
satisfying the condition y(1) = 1 is
(1) y = xlnx + x2 (2) y = xe(x - 1)
(3) y = xlnx + x (4) y = ln x + x
16. (3)
![]()
I.F. = ![]()
y.
y = xlogx + x as y(1) = 1 Þ c = 1.
17. The perpendicular bisector of the line segment joining P(1, 4) and Q(k , 3) has y-intercept -4. Then a possible value of k is
(1) 2 (2) -2
(3) -4 (4) 1
17. (3)
|
Slope of PQ = Since
the perpendicular bisector passing through R and S Slope
of RS = We
know slope of RS ´ slope of PQ = -1 Þ k = ± 4 Þ k = -4. |
|
18. The point diametrically opposite to the point P(1, 0) on the circle x2 + y2 + 2x + 4y - 3 = 0 is
(1) (-3, 4) (2) (-3, -4)
(3) (3, 4) (4) (3, -4)
18. (2)
|
Equation of circle x2 + y2 + 2x + 4y - 3 = 0 Centre (-1, -2) Since P and Q are diametrically
opposite. \ O
is the mid point of P and Q. \ (x,
y) = (-3, -4) |
|
19. A parabola has the origin as its focus and the line x = 2 as the directrix. Then the vertex of the parabola is at