(as 1 n^1 n & n^1 n) n( ) ( ) n n n
9.Sol: (a,b,d)
Points on L 1 and L 2 are respectively
A(1 , 2 , 2 ) and B(2 , , 2 )
So,
(^) AB (2 1)iˆ ( 2 )ˆj (2 2 )kˆ
and vector along their shortest distance is
2 i j kˆ 2 ˆ ˆ.
Hence,
2 1 2 2 2
2 2 1
^
1 2
and
9 9
Hence,
8 2 2
, ,
9 9 9
A
and
4 2 2
, ,
9 9 9
B
Mid point of
2 1
, 0,
3 3
AB
10.Sol:
Bag 1 Bag 2 Bag 3
Red balls 5 3 5
Green Balls 5 5 3
Total 10 8 8
(a) P(Ball is Green) = P B P G B P B( ) ( / ) 1 1 ( 2 )
P G B P B P G B( / ) 2 ( ) ( / ) 3 3
3 5 3 5 4 3 39
10 10 10 8 10 8 80
(b) P(Ball chosen is Green/Ball is from 3rd
Bag)
3
8
;
(c,d) P(Ball is from 3rd Bag/Ball chosen is
reen)
3 3
1 1 2 2 3 3
( ) ( / )
( ) ( / ) ( ) ( / ) ( ) ( / )
P B P G B
P B P G B P B P G B P B P G B
(^1)
3
( )
10
P B
(^2)
3
( )
10
P B ;
9.Sol:
10.Sol:
11.Sol:
3
4 3
(^4) 10 8 4
( )
10 3 5 3 5 4 3 13
10 10 10 8 10 8
P B
.
11.Sol: (A,B). There are three tasks here, first,
to write down a differential equation from the
given geometric property of the curve ,
second, to find its general solution and the third,
to identify a particular solution using the given
intial condition, viz., the point (1,0) lies on it,
i.e.y (^1) 0.
The equation of the tangent to T at a point
P x y 0 , 0 on it is
y y y x x 0 ' 0 (1)
So the point YPis0,y y x 0 ' 0 . The condition
YYP translates as
2 2 2
x y x 0 ' 0 1 (2)
for all x y 0 , 0 . Hence the differential equation
satisfied by is
xy' 1 x^2 (3)
Because of continuity, the same sign must hold
throughout. To find out which sign it is, note
first that has no points for which x>1. The
tangent at (1,0) is given to cut the y-axis at a
point which is at a distance 1 from (1,0).
So, this point must be the origin (0,0). Hence
the tangent to at (1,0) is the x-axis. But this
cannot be the curve , as otherwise, the
x- axis will also be the tangent to at any
P x 1 , 0 with 0 x 1 1 and then YP0, 0
will be at a distance less than 1 from P.
We conclude that there is some x 1 0,1 at
which y x 1 0 and hence y x 1 0 as
the curve is in the positive quadrant. We now