IIT-JEE 2011 - Career Point

RQ || SP
Also PQ . RQ ≠ 0
∴ PQRS is not a square
⇒ PQRS is a parallelogram
35. The value of
Ans.[B]
Sol.
x lim →0
t ln(1
+ t)
dt
3 ∫
is
4
t + 4
x1 x 0
(A) 0 (B) 12
1
1
(C) 24
Use L'hospital rule in
1 t ln(1
x +
Lim
x → 0 3 ∫ 0 4
x t + 4
x ln(1
+ x)
= Lim
x → 0 4 2
+
= Lim
=
(x
4)3x
ln(1
+ x)
x
x → 0 .3(x
2 + 4 )
1 1 =
3.4 12
t)
dt
1
(D) 64
36. The number of 3 × 3 matrices A whose entries are
either 0 or 1 and for which the system
⎡x⎤
⎡1⎤
A
⎢ ⎥
=
⎢ ⎥
⎢
y
⎥ ⎢
0
⎥
has exactly two distinct solution, is -
⎢⎣
z⎥⎦
⎢⎣
0⎥⎦
(A) 0 (B) 2 9 – 1
(C) 168 (D) 2
Ans.[A]
SECTION – II
Multiple Correct Choice Type
This section contains 5 multiple choice questions. Each
question has four choices (A), (B), (C) and (D), out of
which ONE OR MORE may be correct.
37. Let ABC be a triangle such that ∠ ACB = 6
π and
let a, b and c denote the lengths of the sides
opposite to A, B and C respectively. The value(s)
of x for which a = x 2 + x +1, b = x 2 – 1 and c = 2x
+ 1 is (are)
(A) − ( 2 + 3)
(B) 1+ 3
(C) 2 + 3
(D) 4 3
Ans.[B]
Sol.
As sum of two sides is always greater than third
side, So x > 1
Now
2 +
π a b – c
Cos = 6 2ab
(a – b) 2ab – c
=
23 2 +
2ab
2 2
(a – b) – c
3 – 2 =
ab
3 – 2 =
(x
2
(x + 2)
2
2
2
+ x + 1)
2
– (2x + 1)
(x
2
2
–1)
–3
3 – 2 =
2
x + x + 1
x 2 + x + 1 =
3
2 – 3
x 2 + x + 1 = 3(2 + 3 )
x 2 + x – 5 – 3 3 = 0
(x – (1 + 3 ) (x + (2 + 3 )) = 0
x = 1 + 3 , – (2 + 3 )
So x = 3 + 1
38. Let A and B be two distinct points on the
parabola y 2 = 4x. If the axis of the parabola
touches a circle of radius r having AB as its
diameter, then the slope of the line joining A and
B can be -
1
1
(A) − (B)
r
r
2
2
(C) (D) −
r r
Ans.[C, D]
2
2
Sol. Let A ( t 1 ,2t1)
B ( t 2,2t
2 )
2(t
Slope =
2 − t1)
2
=
2 2
t 2 − t1
t 1 + t 2
Equation of circle will be
2 2
(x − t1
) (x − t 2)
+ ( y − 2t1
) ( y − 2t
2)
= 0
2 2 2 2
x + y – x(t1 + t 2)
– 2y(t1 + t 2)
+ 2 2
t 1 t 2 + 4t 1 t 2
= 0
As it touches x axis so
2 2 2
2 2 (t t )
t 1 t 2 + 4t 1 t 2 =
1 + 2
4
2 2
4 4 2 2
4t 1 t 2 + 16 t 1 t 2 = t 1 + t 2 + 2 t 1 t 2
2 2
( ) 2
t1 − t 2 = 16 t 1 t 2 . . . (1)
AB is diameter so
2 2 2
1 t 2)
( t − + 4 (t 1 – t 2 ) 2 = 4r 2 . . . (2)
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MAY 2010