March 2011 - Career Point

18. [8]
∴ for 4 moles of A 4+ no. of moles of N 2 H 4
required = (4 – n) moles
∴ for 10 –2 moles of A 4+ no. of moles of N 2 H 4
( 4 − n)
required = × 10 –2 moles
4
( 4 − n)
∴ × 10 –2 = 2.5 × 10 –3
4
or (4 – n) = 1
or n = +3
A 3+ x =(p –3)
A p+
I 2
I –
x = 2
∴ for 2 moles of A 3+ no. of moles of I 2 required
= (p – 3) moles
∴ for 10 –2 moles of A 3+ no. of moles of I 2
( p − 3)
required = × 10 –2 moles
2
( p − 3)
∴ × 10 –2 = 5 × 10 –3
2
or p – 3 = 1
or p = +4
k = t
1 ln
a
0
a
0
– x
= 10
1 ln 5/2
Let expiry time of the drug, t e = k
1 ln 2
ln 2
or t e = 10
ln5/ 2
10×
0.693
or t e =
=
(ln5 − ln 2)
10×
0.693
or t e =
=
(2.303−
2ln 2 )
10×
0.693
(ln10 − 2ln 2 )
10×
0.693
or t e =
=
(2.303 −1.386)
or t e ≈ 8 months
10×
0.693
(2.303−
2×
0.693)
6.93
0.917
= 7.56
19. [6]
24 .5×
0.1
20 × N = 24.5 × 0.1 ⇒ N = = 0.1225
20
mass of K 2 Cr 2 O 7 dissolved
294
per litre of its solution = 0.1225 × 6
= 6.0025 gram
~ 6 gram
MATHEMATICS
1. [A]
required = total – All different digits
= 6 6 – 6! = 6(6 5 – 5!)
2. [B]
Let lnx = t 2
x =
2
t
e ⇒ dx =
∫ 2
t
2
1
2
2
t
e
. 2t dt
2
. e
t
. dt =
∫ 2
t(2te
) 1
1
1
2
t
) dt
2
2
= ( t e
t –
∫ 2
e
t 2 dt = 2e 4 – e – x
3. [B]
Sum = 6, Product = – c
α = 2,
3 (2) 2 – d 2 = –24
d 2 = 36 ⇒ d = 6, – 6
I st term = – 4 or 8
S = 2
n [–8 + (n –1) 6] = n (3n –7)
or
S = 2
n [16 + (n –1) (–6)] = n (11– 3n)
4. [D] AA T = I ∴ x + y = 0
5. [D] Given equation (1 + x) n – 1 = 0
⇒ (1 + x) n = 1
'n' distinct roots of equation (1 + x) n = 1 lie on a
circle of radius 1 unit with centre (–1, 0)
Now, (–1, 2 ) lies on director circle of circle
(x + 1) 2 + y 2 = 1
⎛
arg ⎜
z r − ( −1
+
⎝ ( −1)
− ( −1
+
⎛
arg
⎜
⎝
(–1, 0)
2i)
⎞
⎟
π
=
2i)
⎠ 4
(0, 0)
− ( −1)
⎟ ⎞ π 2π = ⇒ = 8 ⇒ n = 8
− ( −1)
⎠ 4 π / 4
z r
6. [C]
Normal to y 2 = 4cx at 't '
y + xt = 2ct + ct 3 .....(1)
Normal at (at 2 1 + b, 2at 1 ) to y 2 = 4a(x – b)
y + xt 1 = 2at 1 + at 1 3 + bt 1 .....(2)
XtraEdge for IIT-JEE 99
MARCH 2011