March 2011 - Career Point

(0,3/2)
(0,1)
(–1,0) (1,0)
22. [9]
Centre of given circle be C(–4,5) and its radius r =1
Let the distance between two point A and B be
d(AB) so 2a = max{d(PQ)} 2 where P ≡ (x, y) and
Q ≡ (–2, 3)
= [d(CQ) + r] 2 2
2 2
= ( (4 − 2) + (5 − 3) + 1)
= ( 2 2 + 1) 2
and 2b = min{d(PQ)} 2 = [d(CQ) – r] 2
2
2 2
= [ (4 − 2) + (5 − 3) −1]
= ( 2 2 − 1)
so 2(a + b) = ( 2 2 + 1) + (2 2 −1)
= 2(8 + 1) = 18 ⇒ a + b = 9
1
23. [9] Volume of cone v 1 = π r 2 h
3
2
2
2
25. [1]
Lt
Lt
x →
=
1–
g( x)
= x→1+
g(
x)
g(1)
h(1)
+ 1 f (1) f (1) + (1) + 1
= =
h
3 + 3 2 2 + 6
h( 1) 1 (1) (1) 1
So
+ f + +
= f (1) =
h …..(1)
3
4
Lim
Now as g(1) = ( 1 ln x)
Lt 2
x→1
(ln x)
ln x
x→
1 +
2
2
ln x
= e = e
…….(2)
h( 1) + 1
So from (1) & (2) = e 2 ⇒ h(1) = 6e 2 –1
6
& f(1) = 2e 2
So 2g(1) + 2f(1) – h(1) = 2e 2 + 4e 2 – 6e 2 + 1 = 1
26. [6] for continuous at x = 0
Lim α +
+
x→0
sin[x]
x
=
⎡
2
1 ⎤
Lim β + ⎢–
+ x .... ⎥ = 2
⎣ 3! 5!
⎦
–
x→0
α = β – 1 = 2 ⇒ α = 2, β = 3
y
x
h
3
r
From figure
x
h − y
r
= h
h
x
i.e. h – y = x or y = h(1
− )
r
r
If v be the volume of the cylinder, then
3
v = πx 2 y = πx 2 x
2 x
h(1
− ) = π h(
x − )
r
r
dv ⎛ 3x
⎞
= πh x ⎜2
− ⎟⎠
dx ⎝ r
– + –
0 2r
3
2r
Hence x = gives a maximum of v
3
2
⎛ 2r
⎞ ⎞
∴ v 2 = ⎜
2 4
πh ⎜ ⎟ 1 − ⎟ = .πr 2 h
⎝ 3 ⎠ ⎝ 3 ⎠ 27
v1
1/ 3
∴ =
v2
4 / 27
9
= 4
Thus 4v 1 : v 2 = 9 : 1
24. [1]
f –1 (5) = α so f(α) = 5
2α 3 + 3α 2 + α – 1 = 5 ⇒ 2α 3 + 3α 2 + α – 6 = 0
(α – 1) (2α 2 + 5α + 6) = 0 ⇒ α = 1
Hence f –1 (5) = 1
O 2
Hence required Area = 2 × 3 = 6 sq. units
27. [8]
Let the equation of line through P(λ, 3) be
x − λ y − 3
= = r ⇒ x = λ + r cos θ
cosθ
sin θ
and y = 3 + r sin θ
Line meets the ellipse x 2 2
+ y = 1
16 9
Such that 9x 2 + 16y 2 = 144 at A and D
⇒ 9(λ + r cos θ) 2 + 16(3 + r sin θ) 2 = 144
⇒ 9(λ 2 + r 2 cos 2 θ + 2λr cosθ)
+ 16 (9 + r 2 sin 2 θ + 6r sin θ) = 144
⇒ (9cos 2 θ + 16 sin 2 θ) r 2 +
(18λ cos θ + 96 sin θ)r + 9λ 2 = 0
9λ
∴ PA.PD =
2
9cos θ + 16sin
Since line meet the axes at B and C
3λ
So, PB.PC =
sin θcosθ
from (i) & (ii) ⇒ λ ≥ 8
2
2
θ
… (i)
... (ii)
XtraEdge for IIT-JEE 94
MARCH 2011