Mathematics_Today_-_October_2016

39. In a triangle ABC, 2
    2 22

cosAC ac.

acac

− = +

+ −

Then

(a) B = π/3 (b) B = C

(c) A, B, C are in A.P.

(d) B + C = A

40. A line parallel to the line x – 3y = 2 touches the
    circle x^2 + y^2 – 4x + 2y – 5 = 0 at the point
    (a) (1, –4) (b) (1, 2)
    (c) (3, –4) (d) (3, 2)
    SOLUTIONS


41. (a) : The total number of ways, the person can
    select n coins from 2n + 1 coins is
    21
    1

21

2

21

3

nn n 21 n 255

CC C Cn

++ +++++=... +

⇒ 1 ++ + ++ =+^21 nn n++ +CC C 1 212 213 ...^21 n+Cn 255 1

⇒^1 + = ⇒ = ⇒ = ⇒ =

2

() 221 nn 256 2228 2 nn 8 4.

2. (b) : The word ARRANGE consist of 7 letters.
   Among these 7 letters there are two R’s and two A’s. No.
   of ways of arranging the letters such that two R’s are

always together =^6
2

!
!
∴ The numbers of words where two R’s do not come

together are^7
22

6

2

6

2

7

2

1 6

2

5

2

! 900

!!

!

!

!

!

!

!

− = ⎜⎛⎝ −⎞⎠⎟= ⋅ =.

3. (c) : After giving the largest fruit to the youngest boy,
   the remaining 5 fruits can be given to the remaining 5
   boys in 5! ways, i.e. in 120 ways.


4. (b) : We k n o w t h a t

n

r n r

n

r

r

n n

r

r

P n

r

C P

r

C

!!

= ⇒ =

==

∑∑
11
Now nr
r

n nnnn n

n

CC CCC C Cn

=

∑ =+ ++++−

1

0123 ( ... ) 0

=+++++( ... )− = −

nnnnCCCC 0123 nC Cn nn 021

5. (c) : We h a v e^474523
   1

5

CCj

j

+ −

=

∑

= +++++

(^47) CCCCCC 4 51 3 50 3 49 3 48 3 47 3
= + ++++()
47
4
47
3
48
3
49
3
50
3
51
CC CCCC 3
=++++()
48
4
48
3
49
3
50
3
51
CC CCC (^3)
=+++()
(^49) CC CC 4 49 3 50 3 51 3
=++=+=()()
(^50) CC C CC C 4 50 3 51 3 51 4 51 3 52 4

6. (a) : The required number of numbers are
   5! + (5! – 4!) = 120 + 96 = 216


7. (b) : To form a circle atleast 3 points are required.
   ∴ Number of circles which can be drawn by 9 points
   =^9 C 3 , but given that 4 points are concyclic.
   Therefore, instead of getting^4 C 3 number of circles we
   get only one circle.
   Hence total number of required circles is
   9 C 3 –^4 C 3 + 1 = 81


8. (b) : The arrangement of lectures of the 5 persons
   may be done as (AC)BDE, where C always deliver lecture
   after A. Thus total number of such arrangements are
   4! = 24


9. (b) : We h a v e 34 sinxx+=cos
   3
   2

1

2

⇒ sinxx+=cos 2

⇒ sin sinxxππ+=cos cos

33

2

⇒−cos⎝⎜⎛x π⎟⎞⎠=

3

2 , which is impossible.

As we know that −≤ ⎛ −

⎝⎜

⎞

1 cos x 3 ⎠⎟≤ 1

π.

10. (d) : We h a v e 522
    2

cosθ++=<<cos^2 θ 100 , θπ

⇒ 5cos2θ + 1 + cosθ + 1 = 0

⇒ 5(2cos^2 θ – 1) + cosθ + 2 = 0

⇒ 10cos^2 θ + cosθ – 3 = 0

⇒ (2cosθ – 1)(5cosθ + 3) = 0

∴ θ==π θπ−−⎜⎝⎛− ⎠⎟⎞= − ⎝⎛⎜ ⎠⎟⎞

3

3

5

3

5

and cos^11 cos

()∵ 0 <<θπ

11. (a) : We h a v e logcosθθtanθθ+=logsin cot 0
    ⇒−logcosθθtanθθlogsin tan = 0

⇒ log tan =

log cos

log tan

log sin

θ

θ

θ

θ

⇒ log sin =

log cos

θ

θ

1

⇒⇒logcosθsinθθθθ= 1 cos =sin ⇒tan =tanπ

4

∴ θπ=+nnZπ ∈

4

,.