Inverse trigonometric functions & General Solution of Trigonometric Equations. ------------------ ----------------------------- ----------------- 1. Sin ( ܛ ܖૡ ૠ ) = a) √ૠ b) √ૠ c) ૠ d) ૡ ૠ Ans b. Solution : Method 1. Ans a: √17 > 1 a) is rejected. w.k.t Sin ( sin ଵ ଵ ) = ଵ d is rejected. If sin θ = 8/17 then cos θ = 15/17 GE ≠ 15/17 . c is rejected. b is the answer. OR Let sin ଵ ଵ = θ then sin θ = ଵ Cos θ =√1 െ sin ଶ θ = ට1 െ ሺ ଵ ሻ ଶ = ଵହ ଵ 15 ଶ +8 ଶ = 17 ଶ then G E =sin ଶ = ට ଵ ୡ୭ୱ ଶ = ට ଵ భఱ భళ ଶ = ට భళషభఱ భళ ଶ = ට మ భళ ଶ = ଵ √ଵ ------------------ ----------------------------- ----------------- 2. The value of Tan ( - ܖ܉ܜૡ ) = a) b) ଽ c) ଵ d) ଽ Ans: b. Solution: Let tan ଵ ଵ = θ then tan θ = ଵ Tan ( ସ - tan ଵ ଵ ) = Tan ( ସ - θ) = ଵ – ୲ୟ୬ ଵା ୲ୟ୬ = ଵ – భ ఴ ଵା భ ఴ = ଽ . ------------------ ----------------------------- ----- 3. The domain of the function f(x) = ܛ ܖቂ ቀ ܠቁቃ is a) 1 x 2 b) -1 x 2 c) 2 x 4 d) 1 x 5 Ans: a. Solution : we know that sin ଵ ݔis defined when - 1 x 1 -1 ଶ ቀ ୶ మ ଶ ቁ 1 2 ଵ ୶ మ ଶ 2 ଵ ଵ ଶ ୶ మ ଶ 2 1x ଶ 4 1x 2 ------------------ ----------------------------- ------
Transcript
Inverse trigonometric functions & General Solution of Trigonometric Equations. ------------------ ----------------------------- -----------------
1. Sin ( ) =
a) √ b) √
c) d)
Ans b. Solution : Method 1. Ans a: √17 > 1 a) is rejected. w.k.t Sin ( sin ) = d is rejected. If sin θ = 8/17 then cos θ = 15/17
a) 1 x 2 b) -1 x 2 c) 2 x 4 d) 1 x 5 Ans: a. Solution : we know that sin is defined when - 1 x 1 -1 1 2 2
2 1 x 4 1 x 2 ------------------ ----------------------------- ------
4. If cos ( 3 ) = tan ( ) then x =
a) 0 b) 1 c) -1 d) 0,1, -1 Ans: d . Solution : 1) inspection method GE is satisfied by all values . Hence ans is d. oR ii) cos = θ cos θ = x GE. cos3 θ = x 4 cos θ – 3 cos θ = x
4x - 3x = x 4( x - x )= 0 x( x - 1) = 0 x = 0,1, -1. ------------------ ----------------------------- ----------------- -
5. If = + β and - β ) = √ then + =
a) 1 + a b) a + b c) 1 + ab d) 1+ b Ans: c. Solution: α - β ) = √1 b α - β = cos √1 b ) = sin sin (α - β) = b Now sin α + cos β = sin α + 1 sin β 1 + sin α sin β = 1 + sin (α + β) sin (α - β) = 1 + a b. ------------------ ----------------------------- ---------------
6. If + then x satisfies a) 2x + 3x + 1 =0 b) 2x - 3x = 0 c) 2x + x - 1 =0 d) 2x + x + 1 =0
Ans: b Solution: ( x= 0 satisfies GE and b also. Hence ans b)
Or sin + cos 1 sin sin + cos 1 - sin
cos 1 - 2 sin = -2 θ (sin θ = x) 1 – x = cos (-2 θ ) = cos 2 θ = 1 – 2 sin2 θ 1 – x = 1 – 2x x = 2x 2x - x = 0
x = 0 or x = x = 0 is the solution
( x = does not satisfies the equation ) ------------------ ----------------------------- ----------------- ------------
7. If sec-1x = cosec-1 y then sin-1( ) + sin-1( ) = a) Π b) π / 2 c) - π /2 d) - π
Ans: b. Solution: Given sec-1x = cosec-1 y cos ) = sin )
8. If tan ax – tan bx = 0 ( a ≠ b) then the values of x form a series in
a) AP b) G P c) H P d) AGP. Ans: a. Solution : tan ax = tan bx ax = n π + bx ax – bx = n π x ( a – b) = n π x = . put n = 1 ,2, x= , etc. which are in AP. ------------------ ----------------------------- ----------------- ------------
9. If ( ) + ( ) = ( ) then x =
a) 0 b) √ √ c) √ √ d) π / 2 . Ans: C
Solution: w.k.t sin x )+ sin ( y ) = sin ( x 1 + y √1 )
sin ( ) + sin ( ) = sin ( 1 )+ 1 ) )
= sin ( ( √5 + 4 √2 )/ 9 ) ) x =( √5 + 4 √2 )/ 9 Ans c ------------------ ----------------------------- ----------------- ------------
10. The value of 5 + - = a) b) c) d) . Ans: c Solution: GE = π + tan (
. ) - cot
= π + tan ( ) - cot = π - tan ( ) - cot
= π – [tan ( ) + cot ] = π - = Ans c ------------------ ----------------------------- ----------------- -----------
11. Two angles of a triangle are and , then the third angle is a) b) c) d) .
Ans: b. Solution: A + B + C = Π where A = cot 2 = tan
13. If - = then the values of x are a) 0 or ½ b) 1 or 1/3 c) -1 or -1/3 d) -1 ,1/2
Ans: a Solution type 1: Inspection method: GE is satisfied by a) type 2: GE = cos x – sin x = sin 1 x = – sin x - sin x = sin 1 x
sin 1 x = – 2 sin x ( 1 – x ) = sin – 2 sin x ) = cos 2 sin x )
= 1 – 2 sin ( sin x) = 1 – 2x 1 – x = 1 – 2x 2x x 2x x = 0
x ( 2x – 1) = 0 x= 0 , ½ ------------------ ----------------------------- ----------------- ------------ ----
14. If + = then x = a) ± √3 b) 1 c)
√ d) 0
Ans: b . Solution. Inspection method: ± √3 is not a solution since cos √3 is not defined. GE is satisfied by x = 1 ------------------ ----------------------------- ----------------- --------
15. The value of + + = a) 0 b) cot 26 c) cot 1 d) cot 1
a) 1 b) 3) 0 d) Ans: a) Solution: clearly when x = 1 GE = ans: a)
Or x = x
=
=
x (2 x x ( 2 + x ) x = 0 , 1. But x ≠ 0. - ----------- ----------------------------- ----------------- ------------ 17. The value of 2 = a) b) c) d)
Ans: b. searching method . put x = 1 then GE = 2tan √2 - 1] = 2 = . Put x = 1 in choices, Hence cot 1 = tan 1 = Hence answer must be either a or b Put x = -1 then GE = 2tan √2 - tan (π
) ] =
= 2tan √2 + 1 ] = - 2tan √2 - 1 ] = -2 (
) =
Put x= -1 in a and b Then cot 1 = π
=
b tan 1 ) = . Hence answer is b
------------------ ----------------------------- ----------------- ----------- 18. If 2 = √ , then x is in a) [-1, 1 ] b) [-
, 1] c) [-
√ ,
√ ] d) [ 0, 1] .
Ans: c. Solution w.k. t. the range of RHS is [ , ]
19. The solution set of the equation = 2 is a) { 1, 2} b) { -1, 2} c) {-1, 1, 0} d) {1, ½ , 0} Ans: c) Solution : clearly is not defined. reject a) and b) .
consider d) = but 2 = .
=
≠ hence reject d) Only possibility is ans c. or
= 2 =
x =
x( ) = 2x x - = 0 x( 1 – x ) = 0 x = 0, ±1
------------------ ----------------------------- ----------------- ------------ 20. If in a triangle ABC , C is 900, then + =
a) b) c) d) Ans: c) Solution: ABC is a right angled triangle . Put a= 3 , b= 4 and c= 5 . Then C= 90 Then tan + tan = tan + tan = tan
+ tan =
= tan + tan = where m = 1. Ans: c. ------------------ ----------------------------- ----------------- ------------
21. The value of tan[ ) + )] = a) a b) c) b d)
Ans: b) . Solution Inspection method. Step1. Put a = 1 , b= 0 Then GE = tan [tan 1)] = 1. When we put a=1, b= 0 in choices answer is either a or b Step2. a=0, b= 1 Then GE = tan [tan 1)+ tan 1) ] = tan [ ] =∞ When we put a=0, b= 1 in choices Ans b) ( 1/a = 1/0 = ∞ ).
---- -------------- ----------------------------- ----------------- ------------ 22. If , , ....an is an AP with common difference d , then
tan { + +.......+ } =
a)
b)
c)
d)
Ans : b ) Solution: Given , , ....an are in AP - a1 = a3 – a2 = ...... an - a n-1 = d
Consider tan + tan +.......+ tan
= + tan +.......+ tan
= tan a - tan a + tan a - tan a + ........... + tan a - tan a = tan a - tan a = tan
=
= tan
=tan
GE: tan tan
) =
Hence answer is b. ------------------ ----------------------------- ----------------- ------------
23. If + = then x = a) 0 b) 1 c) √2 -1 d) 1 Ans: d). Solution: Inspection method: Clearly x = o is not a solution since tan 0 =0 And cot 0 = LHS = ≠ .
When x= 1 LHS of GE = + ≠ X = I is not a solution When x= √2 -1 , LHS of GE = + = √2 -1 is not a solution. x = -1 is only a solution ------------------ ----------------------------- ----------------- -
24. If + + = , then ∑ X √1 x = a x y z b) 2xyz c) 2( x+ y + z) d) x y z Ans b Solution: It is a standard result . Or Inspection method. Step 1. put x= y = 1 and z = 0. Then ∑ sin x = π and choice b) and d) matches with this value. Step 2. Put x =
√ and z = 1 . Then
∑ sin x = π and choice b) matches with this value b is the answer.
------------------ ----------------------------- --------- 25. If tan( x + y ) = 33 and x = then y =
a) b) tan 30 c) tan d) tan Ans: d) Solution: x + y = tan 3 3 y =tan 3 3 - tan 3 = tan
a) ( b) , cos c) cos (d) , π - cos Ans: (d ). Solution: GE = 5cos 2 θ + 2 cos + 1 = 0
5( 2 cos θ -1) + ( 1 + cos θ ) + 1 =0 10 x + x - 3 =0 where x = cos θ 10 x + 6x - 5x - 3 =0 2x ( 5x + 3 ) -1 ( 5x + 3 ) =0 x = ,
cos θ = ½ = cos , and cos θ = = cos( π - cos-1 (3/5))
Solution is , π - cos-1 (3/5)) .
29. If The general solution of
= 3 is given by θ =
a) 2n π ± (b) n π ±
(c). 2n π ± (b) n π ± Ans: (d)
Solution: GE =
= 3 tan θ = 3
tan θ =±√3 = tan( ± ) θ = n π ± ------------------ -----------------------------
30. If Sec x cos5x + 1 =0 where 0<x < 2 π , then x is equal to : a) , b)
c) d) none of these Ans: c Solution: GE : +1 = 0 cos5x + cosx = 0
2 cos 3x cos 2x =0 Cos 3x = 0 or cos2x =0 3x = or 2x = or
X= or or or x = or inspection method. ------------------ ----------------------------- ----------------- ------------
31. The solution of the equation 2x = 1+ x is a) x = ( 2n + 1 ) , n Є Z (b) x = n π , n Є Z
.(C) x= ( 2n + 1 ) , n Є Z (d) no solution. Ans: d Solution:GE: sin 2x = 1+ cos x Here cos x 0 The max value of LHS is 1 and minimum Vaue of RHS is 1 . This is possible only when
cos x = 0 and sin 2x = 1 x = and 2x =
X = .This is not possible . Hence solution does not exists. ------------------ ----------------------------- ----------------- ------------
32. The most general value of θ satisfying the equation + √ tan θ - 1) 2 = 0 are given by
a) n π ± b) n π + ( -1) n
c). 2n π + (d) 2n π + Ans: c.
Solution: a b = 0 a = 0 and b =0 1 2 sin θ = 0 and √3 tan θ - 1 = 0
sin θ = - ½ and tan θ = √
G.S. is θ = m π + ( -1) m and θ = mπ + . Both holds good only when m is odd m = 2n + 1 G.S. is θ = ( 2n + 1)π = 2n π + ------------------ ----------------------------- ----------------- ------------
33. The most general solution of θ satisfying Tan θ + tan( + θ ) = 2 is / are
a) n π ± , n Є Z . b) 2n π + , n Є Z
.c) 2n π ± , n Є Z d) ) 2n π +( -1) n , n Є Z Ans: a Solution: GE: Tan θ + tan( + θ ) = 2
Tan θ + tan 135 θ = 2 Tan θ +
= 2
Tan θ + tan θ - 1 + tan θ = 2 + 2 tan θ tan θ = 3 tan θ = ±√3 θ = n π ± , n Є I
------------------ ----------------------------- 34. The general solution of x for which cos2x , , and sin 2x are
in Ap , are given by a) n π , n π + b) n π , n π + c) n π + , d) none
of these. Ans: b . Solution: a,b,c are in AP a + c = 2b Cos2x + sin 2x = 1 Sin2x = 1 – cos2x = 2 sin2x
2 sinx cos x - 2 sin2x = 0 2 sinx( cos x – sinx) =0 sinx = o or cosx = sinx
sinx = 0 or tanx = 1 X = n π or x = n π + , n Є Z. ------------------ ----------------------------- -----------------
35. If √ sec θ + tan θ =1 then the general solution of θ= a) n π + b) 2n π + c) 2n π - d) 2n π ± Ans : c.
GE : √ + = 1 cos θ – sin θ = √2
√
cos θ - √
sin θ = 1
cos( θ + ) = cos( 0 )
G.S. is θ + = 2 n π θ = 2 n π - . ------------------ ----------------------------- ----------------- 36. The equation sin cos + 1 =0 has a) one solution b) two solutions c) infinite solutions d) no solution. Ans: d GE: sin3x = -2 < - 1 . Hence solution does not exists. ------------------ -----------------------------
37. The general solution of √ ( + ) = 2 is a) x = n π b) x = ( 4n + 1)
c) x = ( 4n + 1) d) n π + . Ans: b. Solution clearly x = satisfies the GE. ------------------ ----------------------------- 38. If – π x Π ,– π y Π and cos x + cosy = 2 then general solution is x = a) 2 n π + y b) 2 n π - y
c) n π + y d) n π + ( -1) n y. Ans a. Solution : the Max. value of cosx is 1 . Hence cos x + cosy = 2 cosx = 1 and cosy =1
x = 0 and y = 0 hence x – y = 0 Hence cos( x – y) = cos 0 X – y = 2 n π x = 2 n π + y ------------------ ----------------------------- -----------------
39. The equation 3 x + 10cosx – 6 =0 is satisfied if a) x = n π ± cos b) x = 2n π ± cos
c) x = n π ± cos b) x = 2n π ± cos Ans: b. Solution GE.: 3( 1 - cos x ) + 10cosx – 6 =0 3 – 3t + 10 t - 6 =0 where t = cosx
3t - 10 t +3 =0 3t - 9t - t +3 =0
3t( t – 3 ) -1 ( t – 3) =0 ( 3t -1) (t-3)=0 cosx = t = 3 > 1 No solution , Cosx = 1/3 = cos α where α = cos
X = 2n π ± cos i.e. b). ------------------ ----------------------------- ----------------- 40. If tan2x = tan , then the value of x =
a) √ b)
c) √ d) None of these Ans: a) Solution: GE: tan2x = tan
2x = n π + 2x - nπ x– 2 = 0
X = √ ------------------ ----------------------------- ----------------- 41. If the equation cos3x x + sin3x x= 0., then x = a) ( 2n+1) b) ( 2n - 1) c) d) n π , . Ans: a:) Solution: put n = 0 in answers a,b,c and d , Substitute x=0, x= , x= . GE is not satisfied when x =0 Hence c and d are not
correct answers. GE is not satisfied when x = . Hence b is not the correct answer. Hence a is the answer. Or
( cos 3x - sin 3x ) +3 ( cos3x cosx + sin3x sinx )} =0
{cos6x + 3 cos2x} =0 2x = A
{cos3A + 3 cosA} =0 cos A 0 cos 2x 0
cos2x = 0 2x = ( 2n + 1) x = ( 2n + 1) ------------------ ----------------------------- ----------------- ------------ 42. If x ≠ and – = 1 , then all solutions of x are given by a) 2n π + b) ( 2n + 1) π -
c) 2n π + 1 d) None of these. Ans : d. Solution: Since x ≠ , cos x ≠ 0,1,-1. Hence only possibility is
sin x – 3 sinx 2 = 0 sin x – 2 sinx – sinx 2 0 ( sinx - 1 ) ( sinx – 2) =0
Sin x = 1 or 2 which is not possible as x ≠ and 2 > 1 Hence d is the answer. ------------------ ----------------------------- ----------------- 43. If 1 + sin x + x + x +..........∞ = 4 + 2√ with 0<x <
π and x ≠ , then x =
a) b) c) or d) or
Ans d.
Solution : put sinx = r then
GE: 1 + r + r + ........ ∞ = 4 + 2√3 ( in GP.)
With | | < 1 since 0<x < π and x ≠ .
S = = 4 + 2√3
= 4 + 2√3 1 – sinx = √
x √ √
= √
1 – sinx = √
sinx = 1 - √ = √ = √ x = 60 or 120 Ans: d. ------------------ ----------------------------- ----------------- ------- 44. The general solution of of = cos2x – 1 is a) 2n π b) c) n π d) (2n + 1) π Ans: c Solution: when n=1 a) π b) c) 2 π d is 3 π
Among these GE is not satisfied by b) b is not the correct answer. Since GE is satisfied for π , 2π , 3 π also. General solution is θ = n π . Or
tan x = cos2x – 1 = - ( 1 – cos2x)
= - 2sin x sin x = - 2sin x cos x sin x ( 1 + 2cos x ) = 0 sin x = 0 sinx = 0 x= n π where n Є Z ------------------ ------------------------------------------------- 45. Cot θ =sin2 θ ( θ ≠ n π ),if θ = a) , b) , c) only d) only Ans b) Solution : when θ = , in GE
LHS = RHS = 1 . GE is satisfied by θ = also . and is not a solution
θ = , . Or GE
= 2 sin θ cos θ
cos θ = 2 sin θ cos θ cos θ ( 1 - 2 sin θ ) = 0
Cos θ =0 θ =
sin θ = ½ sin θ = ±√
θ = ± ------------------ ------------
46. If sin( cotθ ) =cos( tanθ) then the value of θ =
a)n π + b ) 2n π ± c) n π - d) 2n π ± Ans: a) Solution: we know that sin = cos π cotθ = π tanθ cotθ = tanθ tan θ = 1 θ = n π + . ------------------ ----------------------------- ----------------- ------------ 47. If tan θ + 3 cot θ = 5 sec θ then θ = a) n π + ( -1)n , nЄ Z . b) n π + ( -1)n , nЄ Z
c) n π + ( -1)n+ 1 , nЄ Z or n π + ( -1)n , nЄ Z
d) n π + ( -1)n , nЄ Z or n π + ( -1)n , nЄ Z Ans: b). Solution. GE:
+3
=
sin θ + 3 cos θ = 5 sin θ 1 + 2 cos θ = 5sin θ which is satisfied for x = only.
answer is b). ------------------ ----------------------------- ----------------- -- 48. If tanx + tan4x + tan7x = tanx tan4x tan7x then x = a) n π / 3 b) n π / 4 c) n π / 6 d) n π / 12
Ans d). Solution : w. k.t if tan x + tany + tanz = tanx tany tanz then ] tan( x + y + z ) = 0.
tan12x = 0 12x = n π x = n π /12. ------------------ ----------------------------- ----------------- 49. The general solution of sinx + sin7x = sin4x in ( 0, ) are
a) , b) , c) , d) , Ans d) Solution: GE : sin7x + sin x – sin4x =0 2 sin4x cos 3x – sin 4x =0
sin4x( 2 cos3x - 1) =0 sin4x =0 4x = n π or x = n π / 4 = π / 4 in ( 0, )
cos 3x = = cos ( ) in ( 0, )
3x = x = x = , Ans d ------------------ ----------------------------- -----------------
50. The number of solutions of cosx = | | , 0 x 3 π is a) 3 b) 2 c) 4 d) 5
Ans: a. Solution: clearly 1 0
cosx = 1 + sinx ( | | = x if x 0 ) cos x – sin x = 1
By inspection x = 0, 2π , 3 π /2 are 3 solutions in [0,3π] Ans a ------------------ ----------------------------- ----------------- -----
51. The set of values of x for which –
= 1 is
a) φ b) { n π + , n Є Z } c) d) { 2n π + , n Є Z } Ans a. Solution: Clearly tan x = 1 = tan ( )
X = n π + , n Є Z . But for no values of x , sin2x is not defined , so that the equation has no solution. ------------------ ----------------------------- ----------------- 52. The general solution of cos7 θ cos 5 θ = cos3θ cosθ is a) , b) , c) , d ) , Ans c. Solution : GE: cos7 θ cos 5 θ = cos3θ cosθ
( cos12 θ + cos2 θ ) = ( cos 4 θ + cos2 θ ) cos12 θ – cos 4 θ = 0 -2 sin 8θ sin 4θ =0 8θ = n or 4θ = n θ = , Ans c
-------------- ----------------------------- ----------------- ------------ 53. If sec2x + + cosec2x = 4 , then the value of x is
a) b) c) d) Ans d. Clearly by inspection method d) is the answer. ------------------ ----------------------------- ----------------- ------
54. The equation 2 sin θ cos θ = x 2 + has a) one real solution b) no solution c) two real solutions d) θ = n π . ans b) Solution : GE : sin 2 θ = x 2 + . clearly sin 2 θ > 0. If x 1 solution does not exists. If 0<x< 1 then x < 1 but >1 hence x 2 + > 1 Solution does not exists. no solution ------------------ ----------------------------- ----------------- ------------ 55. If A and B are acute angles such that sinA = sin2B and 2 cos2A = 3 cos2B then A is
a) b) c) d)
Solution : Clearly by inspection method A= .
Since when A= , B = from sinA = sin2B . These values
satisfies 2 cos2A = 3 cos2B also. a = is the ans. ------------------ ----------------------------- ----------------- -------
56. Let n be a positive integer such that sin + cos = √ ,then a) n = 4 b) n= 1,2,3,4,....... c) n = 2 d) n = 6 Ans d. Solution: clearly by inspection n = 6 . Or
sin + cos = √ squaring 1 + 2 sin cos =
sin = - 1 = which is true when n = 6.
sin = n =6 only. ------------------ ----------------------------- ----------------- ------------ 57. If sin 400 = k and cosx = 1 – 2 then the value of x in ( 00, 3600 ) are a) 40o and 140o b) 80o and 280o c) 40o and 220o d) 80o and 260o Ans: b. Solution : Now cosx = 1 – 2 sin 2400 = cos800 .
x = 800 and 3600 - 800 = 2800 . x = 800 , 2800 in ( 00, 3600 )
------------------ ----------------------------- ----------------- 58. If 3 cos2x - 2√ cos x sinx - 3 sin2x = 0 then x = a) + b) + c) d) n π + Ans: a) Solution : GE: 3 cos2x - √3 sin2x =0 3 cos2x = √3 sin2x = √3 tan2x = √3 .
2x = n π + x = + . ------------------ ----------------------------- ----------------- ------- 59. The most general solution of + =0 a) 2n π + , n ЄZ , b) n π + , n ЄZ
c) 2n π - , n ЄZ d) n π - , n ЄZ Ans: b. Solution: GE: log tanx = - log cotx log tanx = - log = log tanx
log tanx = log tanx sinx = cosx Tanx = 1 x = n π + , n ЄZ ans :b . ------------------ ----------------------------- ----------------- ------- 60. The general solution of 3 tan( θ – 15 o)= tan ( θ + 15 o ) is a) θ = ± b) n ± c) + ( -1) n d) 2n -
Ans: c Solution: 3 tan( θ – 15 o)= tan ( θ + 15 o )
3 – –
=
3 sin θ – 15 cos θ 15 = sin θ 15 cos θ 15 2 sin θ – 15 cos θ 15 + sin ( -300) =0
sin2 θ + sin ( -300) + sin ( -300) =0 Sin 2 θ = 1 2 θ = n π + ( -1) n θ = + ( -1) n ------------------ ----------------------------- ----------------- -------
61. The most general solution of √ tan θ +1 = 0 & √ secθ -2 =0 is a) 2n π - b) 2n π +
c) n π + ( -1)n d) 2n π + Ans b)
Solution: Given tan θ = √
and cos θ = √ θ in IV
quadrant take θ = 2 π – =
general solution is θ = 2n π + Ans b. ------------------ ----------------------------- ----------------- ------------ 62. If cosx + cosy = 1 and cosxcosy = ¼ then the general solutions are a) x = 2n π ± , y = 2k π ± b) 2n π ± , y = 2k π ±
c) x = 2n π ± , y = 2k π ± b) n π ± , y = k π ± Ans: b. Solution: clearly by inspection method x = , y= is a solution . Hence ans is b). Or
Cosx – cosy = cosx cosy 4 cosxcosy = 1 4 =0
Now cosx + cosy = 1 and cosx - cosy = 0 cosx = ½ and cosy = ½
x = 2n π ± , y = 2k π ± ans : b. ------------------ ----------------------------- ----------------- ------- 63. =
a b) - c) d) Ans : c Solution: put x= 0 then GE :
In choices only a and c results in . Hence answer is either a or c.
Put x = 300 . then GE: tan √ = tan √3 =
In choices a reduces to 150 and c reduces to 600. Hence c is the answer.
a) 2x b) c) 2x d) 2 √1 x Ans: b) Solution : put x = 1 in GE: = 2 reject d. Put x = in GE: tan( 450 + 300 ) + tan ( 450 – 300 )
= tan ( 750 ) + tan 150 = 2 √3 + 2 - √3 = 4 In choices b) gives 4 . hence b is the correct answer. ------------------ ----------------------------- ----------------- ------ 65. If sin-1( ) + ) = , then x = a) 4 b) 5 c ) 1 d) 3 Ans: d. Solution: cosec ) = sin-1( )
GE : sin-1( ) + sin-1( ) = sin-1( ) = cos-1( ) clearly x =
3 . sin-1( ) = cos-1( ) ------------------ ----------------------------- ----------------- ------- 66. The value of cos( 2 + ) at x = 1/5 is
a) √ b) √ c) d) 2√6 Ans: b Solution: cos( 2cos x + sin x ) = cos ( + cos x ) = - sin (cos x )
= - sin (sin √1 x ) = - √1 x
At x = 1/5 ans is - 1 = - = - √ .
------------------ ----------------------------- ----------------- --- 67. The general solution of cosecx + cotx = √ is a) x = 2n π ± b) x = + -
c) x = - d) x = 2n π ± - Ans: d. Solution: GE: cosecx + cotx = √3
1 + cosx = √3 sinx
cosx + √3 sinx = - 1 cosx + √ sinx = -
cos( x + ) = cos G.S. is x = 2n π ± - ans :d.
------------------ ----------------------------- ----------------- 68. If α, β and γ are the roots of the equation x3 + m x2 + 3x + m = 0 , then the general solution of + + = a) ( 2n + 1) b) n π c) d) depend upon the value of x . Ans: b Solution : consider x3 + m x2 + 3x + m = 0 ∑ α = -b/a = -m , ∑ α = c/a = 3 ; α β γ = -d/a = -m
Now tan α + tan β + tan γ = tan – –
=tan –
= tan-1(0) =0.
G.S. is tan α +tan β +tan γ = n .where n Є Z . Ans b ------------------ ----------------------------- ----------------- ------------ 69. The equation 3 cosx + 4 sinx = 6 has a) finite solution b) infinite solution c) one solution d) no solution . Ans: d Here r =
√ = > 1 where a= 3, b= 4 c = 6.
Hence no solution. ------------------ ----------------------------- ----------------- ------------ 70. The value of x satisfying the equation sinx +
=
√
is given by a) 100 b) 300 c) 450 d) 600 Ans: d Solution : �Clearly by inspection x = 600 satisfies the equation .
because sin 600 +
= √ + √
= √
------------------ ----------------------------- ----------------- ------------ 71. If log 5(1 + sinx) + log 5(1 - sinx) = 0 then x in ( 0, ) is a) b) 0 c) d No value Ans d Solution: Clearly If log 5 [ (1 + sinx)( 1 – sinx)] =0
1 – sin2x = 0 cos2x =1 cosx = ± 1 x = 0, 0, ) no value.
72. If y = cosx and y = sin 5x then x = a) b) c) d) Ans: c Solution: clearly sin5x = cosx sin5x = sin ( - x)
5x = - x 6x = x = ans C ------------------ ----------------------------- ----------------- ------------ 73. The value of ) ) + ) ) =
a) b) c) d) 0 Ans: d : = 2 ‐ .
cos cos ) ) = cos cos 2π ) ) =
sin sin ) = sin sin 2π ) = - GE: = 0
Ans: d ------------------ ----------------------------- ----------------- ------------ 74. If θ = )) , then one of the possible value of θ is a) b) c) d) Ans: a) Solution: 600 ≡ 600 + 7200 = 1200
θ = sin sin 600 = sin sin 120 = sin sin 180 60 ) = sin sin 60 ) = 60 = ------------------ ----------------------------- ----------------- ------------ 75. The value of sin( 2 . = a) 0.96 b) 0.86 c) 0.94 d) 0. 84. Ans a) Solution: 0.8 = sin( 2 sin 0.8 = sin ( 2 θ ) where θ = sin
= 2 sin θ cos θ = 2 . . = = 0.096 . Ans a) ------------------ ----------------------------- ----------------- ------------
