Mathematics Times – July 2019

12.Sol: Suppose that the persons A to E are
seated clockwise around the circle as shown
below

i.e., a b c    0

 

Minimum value of

( (c a b c  ))

 

= Minimum value of

2

c

also given

( )

3 2

c a b

a b

 





i.e., 9(  ) 18

^   ^2

so

(^22)
c 6(  2 4)

Therefore, minimum value of c^2  18
We treat the persons as vertices of a pentagon.
The problem amounts to colouring these five
vertices with three colours, r,b,g so that no
two adjacent vertices get the same colour.
Clearly, no colour can appear more than
two times. So 2 colours appear twice each
and the remaining colour appears only once.
This lone colour can be chosen in 3 ways and
the vertex to be coloured with in 5 ways. So
these are 15 such pairings where a vertex get
the lone colour. Take any one such, e.g. where
A gets red as shown. Then either B and D get
green and C and E the remaining colour blue
or vice versa. So for each one out of these 15
pairings, there are two possible distributions.
Hence the total number colourings is
15 2 30 .
13.Sol: Given that
 
/2
5
0
3 cos
cos sin
I d
 

 



 (1)

using the property

( ) ( )

b b

a a

f x dx f a b x dx  

we have

 

/2

5

0

3 sin

cos sin

I d

 



 





 (2)

by adding (1), (2) we get

 

/2

4

0

3

2

cos sec

d

I

 

 







 

/2 2

4

0

sec

3

(1 tan

  d









Let 1 tan t

^ sec^2  d 2( 1)t dt

The given integral is written in terms of ‘t’,

we get

(^4)
1
2( 1)
2 3
t dt
I
t
 

 

3 4

1

6 (t t dt)



  



i.e.,

2 3

1

2 6 6 0 0 1 1

2 3 2 3

I t t

  

          

     

    

^ I0.50

Given that A and B are independent events.

i.e., P A B P A P B(  ) ( ) ( )



12.Sol:

13.Sol:

14.Sol:

6

6 6 6

a b c

  ab c

We have A a B b A B c ,  ,  

^ ( , , ) (3,2,1)a b c  ,^6 C C C 1 ^52 ^31 ^180 ;

(4,3,2),^6 C C C 2 ^42 ^21  180 ;

(6,1,1),^6 C 1  6 ;

(6,2,2),^6 C 2  15 ;

= (6,3,3) ,^6 C 3  20 ;

=(6,4,4) ,^6 C 4  15 ;

= (6,5,5) ,^6 C 5  6.

Total 360 62 422  .