Mathematics Times – July 2019

(Ben Green) #1

  1. 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.........


  1. If


/ 4
sin
/ 4

2
(1 x)(2 cos 2 )

dx
I
e x





   then find


27 I^2 equals ....

SECTION-


  1. b 2. b 3. c 4. c


SECTION-


  1. (b,c) 6. (a,b,c) 7. (a,c,d) 8. (a,b,c)

  2. (a,b,c) 10. (a,c) 11. (b,d) 12. (b,c,d)


SECTION-


  1. (0.75) 14. (0.50) 15. (3.00)

  2. (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., m2,
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,

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