March 2011 - Career Point

A 1 B 1 = 2 2
AB = 2 2 – 2 = 2( 2 – 1)
OA = 2 – 1
taking origin as centre and OA as radius circle
will touches all four circle
∴ equation of circle is
x 2 + y 2 = ( 2 – 1) 2
x 2 + y 2 = 3 – 2 2
35. [B] y = f(x) =
1+
⇒
x
x
tan x
dy 1+
x tan x − x(tan
x + xsec
=
dx
2
(1 + x tan x)
=
1−
x
2
sec
2
(1 + x tan x)
y = x 2
x
2
y = cos 2 x
2
x)
31. [A]
(0, 2)
O
π/2 π
P(a, a 2 )
(2, 0)
Clearly a > 0
Also P lies on that side of line x + y = 2
Where origin lies
∴ a + a 2 – 2 < 0 ⇒ (a – 1) (a + 2) < 0
⇒ – 2 < a < 1 but a > 0
∴ 0 < a < 1
∴ a ∈ (0, 1)
32. [D]
Triangle is right angled at O(0, 0). Therefore
orthocentre is O(0, 0) and circumcentre is mid
⎛ a b ⎞
point of hypotense i.e. ⎜ , ⎟
⎝ 2 2 ⎠
∴ Distance between orthocentre and
1 2 2
circumcentre = a + b
2
33. [B]
Let a = 3K, b = 7 K and c = 8K
a + b + c
∴ s = = 9K
2
R abc s abcs
there = . = r 4∆
∆ 4s(
s − a)(
s − b)(
s − c)
3K.7K.8K 7 R 7
=
= ⇒ =
4.6K 2K K 2 r 2
34. [D] sinx (sin x + cos x) = K
⇒ sin 2 x + sin x cos x = K
⇒
1−
cos 2x
sin 2x + = K
2 2
⇒
1 (sin2x – cos2x + 1) = K
2
Q – 2 ≤ sin 2x – cos 2x ≤ 2
⇒
1− 2 sin 2x − cos 2x
+ 1 2 +1
≤
≤
2
2
2
⇒
1− 2 2 +1
≤ K ≤
2
2
dy = 0 ⇒ x 2 = cos 2 x
dx
There is only one point in (0, 2
π ) say x, at
dy
which = 0 dx
at x 1 – h ⇒ x 2 < cos 2 x
dy
∴ > 0 dx
& at x 1 + h ⇒ x 2 > cos 2 dy
x ∴ < 0 dx
∴ at x 1 slope change from + ve to – ve
∴ There is only one critical point
⎛ π ⎞
in ⎜0 , ⎟ at which f(x) has local maxima.
⎝ 2 ⎠
36. [B] 3cos 2 θ – 2 3 sinθ cosθ + 3sin 2 θ = 0
( 3 cosθ + sinθ) (cosθ – 3 sinθ) = 0
1
⇒ tan θ =
3
π
∴ θ = nπ + 6
or tanθ = – 3
or θ = nπ –
3
π
∴ | r – s| = | – 3 – 6| = 9
37. [B]
cot –1 n π π > ⇒ < cot
–1
π 6 6
{ Q cot –1 x ∈ 0 (a, π)}
⇒ – ∞ < π
n < 3
– ∞ < n < 3 π
– ∞ < n < 5.4
⇒ max. n = 5 {Q n ∈ N}
n < π, n∈ N
π
38. [A]
∴ tan π [x] = 0 ∀ x ∈ R since [x] ∈ Z
Period of {x} = 1
⇒ Period of sin 3π{x} = 1
Hence period of f(x) = 1
XtraEdge for IIT-JEE 118
MARCH 2011