lsEcnu,..- ..J Two balls are drawn at rand 0
II
d 5 red ba s. 01 1
2· An urn contains 6 white, 4 black an ·
pobability that :
(i) both the balls are red
(ii) one hall is red, the other is black
(iii) one hall is white.
Solution. Total number of balls == 6 + 4 +
5 == IS.
sc2
(i) P (Both balls are red) = !5C2
5X4
2Xl - -2
=
15 X14 21
(ii) P (One ball is red, the other is black) 2Xl
5c x 4c
I I
= 1sc
2
5X4 20 4
= 15 X14
- 15 X 7
-.
- 21
2Xl
6c x 9c
( iii) P (one ball is white) = I I
1sc
2
6X9 18 6 X9
= 15Xl4 ==15x7== 35·
2Xl
4. Using properties of determinants, prove that :
b+c c+a a+b a b c
q+r r+p p+q =2 p q r
y+z z+x x+y X y Z
L .H.S. = Sum of 23 i.e. 8 deterrni·r1ants .
Solution.
b c a h a a b C b 'h a h
q r p + q p !J + Cf ,.
y Z X _\' X X
q + Cf p q
y y X \'
•
,,\ I
,/ ( ,, ,,
+ r., r.. /
1
·I r J) /J t'I' I q+r f} q
. ' \' ..
., \ X ;'. .x y
=
h
CJ
c
r
"
I' +
('
r ,,
(I I>
l/ IOther rJ,,ts. vunish ]
V z. X : ..\ y
, a h c a b <.' ,, 1~
(- 1)~ lA, ("
= /J lJ r + (- I )2 /> ll r = 2 /J q r
.\ )' "·.,
X y Z
X y z
f sin4 2x dx.
r
1.t:1•11IUIIIC :
sin 4 2.x = (sin2 2x)2 = ( I - c~s 4x
1
= 0- 2cos 4x + cos 2 4x)
4
= ~ (I - 2 cos 4x +
1
+ c~s Bx)
3 l
= - cos 4x + cos 8x
1.
8
8 2
= f si n 4
2x dx
= g
3 I I· dx -2l Jcos 4x dx + gl Jcos 8x dx
3 l sin 4x l sin 8x
= -x--· - - + - - - + c
8 2 4 8 8
3 l . 1
= - x - - sin 4x + - si n 8x + c.
8 64 8 2
13. Prove that the curves x = y2 and xy = k cut at right angles if 8k = t.
... (2)
Solution. The given curves are :
x = y2 ... (1 ) and xy = k
113
From(2) . using ( 1), wey2y
get =: k => y3 = k => y= k .
Putting in (l ), x = k 213. 213 113
Thus the given curves intersect at P (k , k ).
D1ff. (l ) w.r.t. x, -1
dy
dy = 2_v ·
2v- dx
1 == • J.x
dy] - ~
111 1 == dt p - 2k 1.\ · 1
l)i!f· (2) W.r.t. X,
J,· + \' ( 1) == 0
t - ·
, / V
II
d 5 red ba s. 01 1
2· An urn contains 6 white, 4 black an ·
pobability that :
(i) both the balls are red
(ii) one hall is red, the other is black
(iii) one hall is white.
Solution. Total number of balls == 6 + 4 +
5 == IS.
sc2
(i) P (Both balls are red) = !5C2
5X4
2Xl - -2
=
15 X14 21
(ii) P (One ball is red, the other is black) 2Xl
5c x 4c
I I
= 1sc
2
5X4 20 4
= 15 X14
- 15 X 7
-.
- 21
2Xl
6c x 9c
( iii) P (one ball is white) = I I
1sc
2
6X9 18 6 X9
= 15Xl4 ==15x7== 35·
2Xl
4. Using properties of determinants, prove that :
b+c c+a a+b a b c
q+r r+p p+q =2 p q r
y+z z+x x+y X y Z
L .H.S. = Sum of 23 i.e. 8 deterrni·r1ants .
Solution.
b c a h a a b C b 'h a h
q r p + q p !J + Cf ,.
y Z X _\' X X
q + Cf p q
y y X \'
•
,,\ I
,/ ( ,, ,,
+ r., r.. /
1
·I r J) /J t'I' I q+r f} q
. ' \' ..
., \ X ;'. .x y
=
h
CJ
c
r
"
I' +
('
r ,,
(I I>
l/ IOther rJ,,ts. vunish ]
V z. X : ..\ y
, a h c a b <.' ,, 1~
(- 1)~ lA, ("
= /J lJ r + (- I )2 /> ll r = 2 /J q r
.\ )' "·.,
X y Z
X y z
f sin4 2x dx.
r
1.t:1•11IUIIIC :
sin 4 2.x = (sin2 2x)2 = ( I - c~s 4x
1
= 0- 2cos 4x + cos 2 4x)
4
= ~ (I - 2 cos 4x +
1
+ c~s Bx)
3 l
= - cos 4x + cos 8x
1.
8
8 2
= f si n 4
2x dx
= g
3 I I· dx -2l Jcos 4x dx + gl Jcos 8x dx
3 l sin 4x l sin 8x
= -x--· - - + - - - + c
8 2 4 8 8
3 l . 1
= - x - - sin 4x + - si n 8x + c.
8 64 8 2
13. Prove that the curves x = y2 and xy = k cut at right angles if 8k = t.
... (2)
Solution. The given curves are :
x = y2 ... (1 ) and xy = k
113
From(2) . using ( 1), wey2y
get =: k => y3 = k => y= k .
Putting in (l ), x = k 213. 213 113
Thus the given curves intersect at P (k , k ).
D1ff. (l ) w.r.t. x, -1
dy
dy = 2_v ·
2v- dx
1 == • J.x
dy] - ~
111 1 == dt p - 2k 1.\ · 1
l)i!f· (2) W.r.t. X,
J,· + \' ( 1) == 0
t - ·
, / V
