1. Xtra Edge February 2012 - Career Point

OR
Let A → total of 8 in first throw
B → total of 8 in 2nd throw
Number of exhaustive cases when a pair of dice is
thrown = 6 × 6 = 36
Cases favourable to a total of 8 in each throw are
(2, 6), (6, 2), (3, 5), (5, 3), (4, 4)
Their number = 5
P(A) = P(B) = 5/36
5 5 25
P(A and B) = P(A). P(B) = × =
36 36 1296
12. dx
2x
dy + 2
1+
x
2
4x
y = 2
1+
x
2x
4x
Here P = , Q = 2
2
1+
x 1+
x
2x
∫ dx
+
2
I.F. = e 1 x = e
2
ln(
1+
x )
2
= 1 + x 2
Solution is : y (1 + x 2 2
4x
) = ∫ 2
+
1
y(1+x 2 4x ) =
3
3
+ c
x ⎡(
2 + x)
– 1⎤
13. ∫ e ⎢ ⎥ dx
2
⎣ ( 2 + x)
⎦
∫ ⎥ ⎥
⎡
⎤
x 1 ⎛ – 1 ⎞
e ⎢ + ⎜ ⎟
⎜ ⎟
dx
2
⎢⎣
2 + x ⎝ ( 2 + x)
⎠⎦
↓ ↓
f(x) f ′(x)
1
= e . c
2 x
x
+
+
2x
∫ 2 2
1−
x − ( x )
2
dx
OR
x
(1 + x 2 ) dx + c
Let x 2 = t. Then, d(x 2 dt
) = dt ⇒ 2x dx = dt ⇒ dx =
2 x
∴ I = ∫
dt
2
1− t − t
= ∫ − { + −1}
2
dt
t t
= ∫
1 1
− { + + − −1}
4 4
2
dt
t t
⇒ Ι = ∫
⎪⎧
⎛ 1 ⎞ 5⎪⎫
− ⎨⎜
+ ⎟ − ⎬
⎪⎩ ⎝ 2 ⎠ 4⎪⎭
2
dt
= ∫
t
dt
2
5 ⎛ 1 ⎞
− ⎜t
+ ⎟
4 ⎝ 2 ⎠
XtraEdge for IIT-JEE 97 FEBRUARY 2012
= ∫
⎛
⎜
⎝
2
5 ⎞
⎟
2 ⎟
⎠
⇒ I = sin –1
dt
⎛ 1 ⎞
− ⎜t
+ ⎟
⎝ 2 ⎠
2
⎛ t + 1/
2 ⎞
⎜
⎟ + C
⎝ 5 / 2 ⎠
= sin –1 ⎛ 2t + 1⎞
⎜
⎟ + C
⎝ 5 ⎠
= sin –1 ⎛ ⎞
⎜
2 + 1
⎟
⎜ ⎟
⎝ 5 ⎠
2
x
+ C
14. Divide by cos 2 x in N r & D r both
∫ + dx
x
(tan x – 2)(
2tan
x 1)
Let tan x = t
sec 2 x dx = dt
dt
∫ ( t – 2)(
2t
+ 1)
sec 2
1 dt
2 ∫ ( t – 2)(
t + 1/
2)
1 2 ⎡ 1 1 ⎤
. ∫ ⎢ – ⎥ dt
2 5 ⎣ t – 2 t + 1/
2⎦
1 ⎡
⎢log
5 ⎣
( t – 2)
⎤
⎥ + c
( t + 1/
2)
⎦
15. Position vector of A is
kˆ 3 î + jˆ + 2 ⇒ A (3, 1, 2)
Position vector of B is
kˆ î – 2 jˆ − 4 ⇒ B(1, –2, – 4)
d.r.s of the line AB are 3, –1, 1 – (– 2), 2 – (– 4)
i.e. 2, 3, 6
∴ Eq. of the required plane through B and
perpendicular line AB is
2 (x – 1) + 3 (y + 2) + 6 (z + 4) = 0
⇒ 2x + 3y + 6z + 28 = 0
16. →
a = kˆ î – 3jˆ
+ , →
b = kˆ î – jˆ + and →
c = kˆ 2 î – jˆ –
→
b × →
c =
î
1
2
jˆ
– 1
– 1
k ˆ
1
– 1
∴ L.H.S = →
a × ( →
b × →
c )
= kˆ 2 î + 3jˆ
+