Mathematics_Today_-_October_2016

12. (b) : We h a v e

22

2

21

2

−cosx= tanxx⇒−⎝⎜⎛ tan ⎟⎞⎠=cosx

⇒−⎛⎝⎜ ⎞⎠⎟=

−

+

21

2

1

2

1

2

2

2

tan

tan

tan

x

x

x

⇒−⎛⎝⎜ ⎞⎠⎟ −

+

+

⎛

⎝

⎜

⎜

⎜

⎞

⎠

⎟

⎟

⎟

1 =

2

2

1

2

1

2

0

2

tan

tan

tan

x

x

x

Either 1

2

−tanx= (^0)
∴ tanxx==tan ⇒ =+nnZ, ∈
2
1
42 4
π π π
⇒ x n=+ 2 nz∈⇒x n=+ nZ∈
2
41
2
π ππ,(),
or 2
1
2
1
2
02
22
10
2
− +^2

• = ⇒−+=
  tan
  tan
  tan tan
  x
  x
  xx
  ⇒ = ± −⋅⋅
  ⋅
  tanxi= ±
  2
  11421
  22
  17
  4
  2
  ⇒ no solution

13. (d) : Here a^2 – 4a + 6 = (a – 2)^2 + 2 ≥ 2

∴−+=

∈

min

aR

{,aa^2 }1461

Now, sinsinxx+=cos 1 ⇒^1 x+=cosx

2

1

2

1

2

⇒ sin⎝⎜⎛xxn+ππ π⎞⎠⎟= ⇒ += +π − nπ

44 4

1

4

sin ( )

⇒ xn=+π ()−−∈ 1 nππ,nZ

44

14. (b) : We have 1 – 2sin^2 θcos^2 θ = –λ

⇒− 1 1 =−

2

sin^22 θλ

⇒− − 1 1 =−⇒+=−

4

14 3

4

1

4

( cos θλ) cos 4 θλ

∵ –1 ≤ cos4θ ≤ 1

∴− ≤^1 ≤ ⇒ − ≤+ ≤ +

4

1

4

4 1

4

3

4

1

4

3

4

1

4

4 3

4

1

4

cos θθcos

⇒≤−≤⇒−≤≤−^1

2

111

2

λλ

15. (c) : From ΔABD, we have

BD

BAD

AD

sin∠ sinB

=

From ΔACD, we have CD

CAD

AD

sin∠ sinC

=



  

∵ BD : CD = 1 : 3

∴ AD ∠∠BAD =

B

AD CAD

C

sin

sin

: sin

sin

13 :

⇒ ∠∠= ⇒ ∠

∠

sin =

sin

:sin

sin

: sin

sin

BAD CAD BAD

ππ CAD

34

13 2

3

1

3

⇒ ∠

∠

sin =

sin

BAD

CAD

1

6

16. (c) : Let the sides of the triangle ABC be 4x, 5x and 7x

∴ = + −

⋅⋅

cosA () () ()xxx=−

xx

547

25 4

1

5

222

⇒ the angle A is an obtuse angle

Thus, ΔABC is obtuse-angled

17. (b) : We have (a + b + c)(b + c – a) = λbc
    ⇒ (b + c)^2 – a^2 = λbc ⇒ b^2 + c^2 – a^2 = (λ – 2)bc

⇒ bca+ − = − ⇒ = −

bc

A

22 2

2

2

2

2

2

λλcos

∵ –1 ≤ cosA ≤ 1

∴−≤ 1 −^2 ≤⇒−≤−≤ ⇒ ≤ ≤

2

λ 122204 λλ

18. (b) : Let bc=+=23 2, 23 2− and A=° 60

∴ ⎛⎝⎜ − ⎞⎠⎟= −

+

tan BC bccot =°==°cot tan

bc

A

22

4

43

30 1 45

⇒ B – C = 90°. Again, B + C = 120° (∵ A = 60°)

Therefore the other two angles are B = 105° and

C = 15°.

19. (b) : Here A=π−⎛⎝⎜ππ π+ ⎞⎠⎟==°
    43

5

12

75.

Now a

A

b

B

b

sin sin sin sin

= ⇒ +

°

=

°

31

75 45

∴ = °

°

b sin +=

sin

(^45) ()
75
31 2