Mathematics_Today_-_October_2016

30. (a) : Let P(h, k) divides the chord AB in the ratio

2 : 1. OD⊥AB, where O is the centre of the circle
and D is the foot of the perpendicular drawn from the
centre O.

Now, OB== 10 ;.OD (^1)
∴ DB= 10 1− = 3 ∴AB== 2 ;.DB (^6)
∵ AP PB::= 21 ∴PB==^1 AB
3
2
∴ DP=DB−PB=321.− =
Now, from ΔODP, we have
OP^2 = OD^2 + DP^2 ⇒ h^2 + k^2 = 1 + 1 = 2
∴ locus of the point (h, k) is x^2 + y^2 = 2

31. (c) : The line y = x cuts the circle x^2 + y^2 – 2x = 0 at
    the points A(0, 0) and B(1, 1). Therefore the equation of
    the circle, whose diameter is AB, is
    (x – 0)(x – 1) + (y – 0)(y – 1) = 0
    ⇒ x^2 + y^2 – x – y = 0


32. (c) : Let the line 3x + y + 5 = 0 cut the circle x^2 + y^2 = 16
    at the points A and B, hence AB is a chord of the circle.

The mid-point of the chord AB is ⎜⎛⎝−−^3 ⎞⎠⎟
2

1

2

, (the foot

of the perpendicular from the centre to the chord) and

length of the chord AB=×^216 − =

25

10

2 27

2

.

∴ Centre of the circle, whose diameter is AB, is

⎛⎝⎜−−^3 ⎞⎠⎟ ==

2

1

2

1

2

27

2

,. and its radius AB

Thus the equation of the circle is

⎛⎜⎝xy+^3 ⎠⎞⎟ ++⎝⎛⎜ ⎞⎠⎟ =

2

1

2

27

2

22

⇒ x^2 + y^2 + 3x + y = 11

33. (b) : Given that, a
    cos^22 C ccos Ab
    22

3

2

+=

⇒ a(1 + cosC) + c(1 + cosA) = 3b
⇒ a + c + (acosC + ccosA) = 3b
⇒ a + c + b = 3b (∵ acosC + ccosA = b)
⇒ a + c = 2b, i.e. a, b, c are in A.P.

34. (b) : We have 3cosx + 4sinx = 2k + 1

⇒

+

+

+

= +

+

3

34

4

34

21

22 22 3422

cosxxsin k

⇒^3 +=+

5

4

5

21

5

cosxxsin k

⇒−cos(x ααα)=^21 k+ , cos ==, sin

5

3

5

4

5

where

∵ −≤ − ≤ ⇒−≤ 11121 + ≤

5

cos(x α) k 1

⇒ –5 ≤ 2 k + 1 ≤ 5 ⇒ –6 ≤ 2 k ≤ 4

Therefore integral values of k are –3, –2, –1, 0, 1, 2

∴ Required number is 6.

35. (d) : The word ‘COMBINE’ is consisting of 7 letters.
    There are 3 vowels namely- E, I and O.
    The two vowels for beginning and ending place can
    be arranged in^3 P 2 ways. For each of the^3 P 2 ways the
    remaining 5 letters can be arranged in 5! ways.
    Therefore required number is^3 P 2 × 5! = 720.


36. (c, d) : Given that nC 4 , nC 5 and nC 6 are in A.P.
    ∴⋅
    −

=

−

+

−

2

554466

n

n

n

n

n

n

!

!( )!

!

!( )!

!

!( )!

⇒

−

=

−−

+

×

2

55

1

45

1

()()()nnn 65

⇒

−

= −−+

−−

12

5

4530

() 45

()()

n ()()

nn

nn

⇒ n^2 – 9n + 50 = 12(n – 4) ⇒ n^2 – 21n + 98 = 0

⇒ n^2 – 14n – 7n + 98 = 0 ⇒ (n – 14)(n – 7) = 0

⇒ n = 14 or n = 7

37. (b, d) : We know that |cosx| ≥ 0
    ⇒ sinx ≥ 0. So, there is no x in (π, 2π).
    Now, if x = 2π, |cos2π| ≤ sin2π, which is not true.
    So, 0 ≤ x ≤ π
    If 0 ≤≤xxxxxπ then ≤ ⇒ ≤
    2

cos sin cos sin

⇒≥∴≤≤tanxx 1

42

ππ

If π π

2

≤≤xxx then cos ≤sin

⇒ –cosx ≤ sinx ⇒ tanx ≤ –1 (∵ cosx < 0)

∴ ππ< ≤

2

3

4

x

∴∈⎡

⎣⎢

⎤

⎦⎥

∪⎡

⎣⎢

⎤

⎦⎥

x ππ π π

42 2

3

4

,,.

Solution Sender of Maths Musing

SET-164

1. Akash Banerjee W.B.


2. Sk. Izajur Rahaman W.B.


3. N. Jayanthi Hyderabad


4. Khokon Kumar Nandi W.B.