- Let AP ( ; )a d denote the set of all the term
of an infinite arithmetic progression with first
term a and common difference d > 0. If
AP(1;3)AP(2;5)AP(3, 7)AP a d( ; )
then a d equals.........
- If
/ 4
sin
/ 4
2
(1 x)(2 cos 2 )
dx
I
e x
then find
27 I^2 equals ....
SECTION-
- b 2. b 3. c 4. c
SECTION-
- (b,c) 6. (a,b,c) 7. (a,c,d) 8. (a,b,c)
- (a,b,c) 10. (a,c) 11. (b,d) 12. (b,c,d)
SECTION-
- (0.75) 14. (0.50) 15. (3.00)
- (10.00) 17. (157.00) 18. (4.00)
SECTION-
SECTION-
SECTION-
SECTION-
1.Sol: Given that y = mx + 1 is a chord which
intersect the given circle. We know
perpendicular bisector of a chord in a circle
passes through the center of circle. That is
2 1
3
y
x m
x my m 2 3 0 (1)
Substitute , the equation of a chord in eq(1),
we get x- coordinate of mid point of chord.
i.e.
x m mx ( 1) 2m 3 0
x m m(1 ^2 ) 3 3 0
Since
3
5
x , the above equation is rewritten
as
(^3) (1 (^2) ) 3 3 0
5
m m
(^) m m^2 5 6 0
i.e., m2,
Therefore 2 m 4
2.Sol: The problem asks for the minimum values
of two functions
SECTION-
1.Sol:
2.Sol:
( ) and ( ) over the
semi - open interval 0, 2. But neither
^ nor is given explicitly as a function of
. So we first have to determine them from
the given matrix equation,as follows.
M I M ^1 (1)
We have a matrix of order 2. For such matrix,
there is an easy explicit formula for the inverse,
as follows
1
a b 1 d b
c d c a
(2)
where ad bc is the determinant. As a
result if we are given that
0
0
a b d b
c d c a
(3)
then equating the entries in the first row and
the second column of the matrices on the two
sides gives
b
b
and hence
(4)
and further , equating the entries in the first
row and the first column we get
d
a
,
which because of the last equation, becomes
d a d
(5)
Thus we have very simple way of determining
the functions ( ) and ( ) by merely
looking at the entries of M and doing a little
calculation to find its detreminant. Specifically,