IIT-JEE 2010 - Career Point

9.
y ⎛ x
If for the differential equation y′ = + φ x ⎟ ⎞
⎜
⎝ y ⎠
the
x
general solution is y =
log | Cx |
then f (x / y) is
given by -
(A) – x 2 / y 2 (B) y 2 / x 2
(C) x 2 / y 2 (D) – y 2 / x 2
This section contains 2 questions (Questions 10 to 11).
Each question contains statements given in two
columns which have to be matched. Statements (A, B,
C, D) in Column I have to be matched with statements
(P, Q, R, S, T) in Column II. The answers to these
questions have to be appropriately bubbled as
illustrated in the following example. If the correct
matches are A-P, A-S, A-T, B-Q, B-R, C-P, C-Q and
D-S, D-T then the correctly bubbled 4 × 5 matrix
should be as follows :
P Q R S T
A
B
C
D
P
P
P
P
Q
Q
Q
Q
R
R
R
R
S
S
S
S
T
T
T
T
10. The domain of the functions
Column-I
Column-II
(A) sin –1 (x/2 – 1) (P) (3 – 2π, 3 – π) ∪ (3,4]
+ log (x – [x])
2
(B) e x + 5sin π / 16 − x (Q)(0, 4) – {1, 2, 3}
(C) log 10 sin (x – 3) (R)[– π/4, π/4]
2
+ 16 − x
(D) cos –1 1− 2x
4
2
(S) [– 3/2, 5/2]
(T) None
11. Column-I Column-II
I denotes an integral
(A)
∫ π x log sin x dx (P) I = (π/8) log 2
0
(B)
∫ ∞ 2
log (x+x –1 dx − π
) (Q) I = log 2
0
2
1+
x
2
(C)
∫ π / 4
log (1+ tan x)dx (R) I = π log 2
0
(D)
∫ π log (1 – cos x)dx (S) I = π log 2
0
(T) None
This section contains 8 questions (Questions 12 to 19).
The answer to each of the questions is a SINGLE-
DIGIT INTEGER, ranging from 0 to 9. The
appropriate bubbles below the respective question
numbers in the OMR have to be darkened. For
example, if the correct answers to question numbers X,
Y, Z and W (say) are 6, 0, 9 and 2, respectively, then
the correct darkening of bubbles will look like the
following :
X Y Z W
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
12. Suppose X follows a binomial distribution with
parameters n = 6 and p. If 9P(X = 4) = P(X = 2), find
4p.
13. If Q is the foot of the perpendicular from the point
x − 5 y + z − 6
P(4, –5,3) on the line = = then
3 − 42
5
100 (PQ) 2 is equal to
457
14. If a = (0, 1, –1) and c = (1, 1, 1) are given vectors,
then |b| 2 where b satisfies a × b + c = 0 and a . b = 3
is equal to
15. ABC is an isosceles triangle inscribed in a circle of
radius r. If AB = AC and h is the altitude from A to
BC. If the triangle ABC has perimeter P and area ∆
then lim 512r ∆
h→0
3 is equal to
p
16. If f(x) = sin x, x ≠ n π, n = 0, ± 1, ± 2, .....
= 0 otherwise
and g(x) = x 2 + 1, x ≠ 0, 2
= 4 x = 0
= 5 x = 2
then lim g(f(x)) is .....
x→0
17. If y = (1 + 1/x) x 2 y 2(2)
+ 1/8
then
is equal to
(log 3/ 2 −1/3)
18. If the greatest value of y = x/log x on [e, e 3 ] is u then
e 3 /u is equal to
19. If z ≠ 0 and 2 + cos θ + i sin θ = 3/z, then find the
value of 2(z + z ) – |z| 2 .
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
XtraEdge for IIT-JEE 62 DECEMBER 2009