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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
Solve : If `(1)/(cos theta)+(1)/(cos 3theta)=(1)/(cos 5theta)` |
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Answer» Correct Answer - `theta=(2n+1)(pi)/(14), n in Z` `(1)/(cos theta)+(1)/(cos 3theta)=(1)/(cos 5theta)` `rArr (cos 3theta+cos theta) cos 5theta = cos 3theta cos theta` `rArr 2cos 5thetas cos 3theta + 2cos 5 theta cos theta = 2 cos 3theta cos theta` `rArr cos 8 theta + cos 2 thetas + cos 6theta + cos theta + cos 4 theta = cos 4 theta + cos 2 theta` `rArr 2 cos 7 theta cos theta = 0` ` rArr cos 7 theta = 0` (as `cos theta = 0` is not possible) `therefore cos 7 theta = 0` `therefore 7theta = (2n + 1)(pi)/(2), n in Z` or `theta =(2n+1)(pi)/(14), n in Z` |
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| 52. |
Solve `(3 sin theta-sin 3 theta)/(sin theta)+(cos 3 theta)/(cos theta)=1`. |
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Answer» We have `(3 sin theta- sin 3 theta)/(sin theta)+(cos 3 theta)/(cos theta)=1` `:. (4 sin^(3) theta)/(sin theta)+((4 cos^(3) theta-3 cos theta))/(cos theta)=1` `rArr 4 sin^(2) theta+4 cos^(2) theta -3 =1` `rArr 1=1` This is always true except `theta= (npi)/2, n in Z` for which either `sin theta` or `cos theta` is zero. |
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| 53. |
Solve : `1+2cosec x=-("sec"^(2)(x)/(2))/(2)` |
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Answer» Correct Answer - `x=2n pi-(pi)/(2), n in Z` `1+2 cosec x =-("sec"^(2)(x)/(2))/(2)` `rArr 1+2((1+"tan"^(2)(x)/(2))/(2tan.(x)/(2)))=-((1+"tan"^(2)(x)/(2)))/(2)` `rArr 2tan.(x)/(2)+2+2"tan"^(2)(x)/(2)+tan.(x)/(2)+"tan"^(3)(x)/(2)=0` `rArr "tan"^(3)(x)/(2)+2"tan"^(2)(x)/(2)+3tan.(x)/(2)+2=0` `rArr (tan.(x)/(2)+1)("tan"^(2)(x)/(2)+tan.(x)/(2)+2)=0` `tan.((x)/(2))+1=0` `rArr tan.(x)/(2)=tan(-(pi)/(4))` `rArr x = 2n pi-(pi)/(2), n in Z` |
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| 54. |
Solve : `2sin(3x+(pi)/(4))=sqrt(1+8sin2x.cos^(2)2x),x in (0,2pi)` |
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Answer» Correct Answer - `x=(pi)/(12),(17pi)/(12)` `2sin(3x+(pi)/(4))=sqrt(1+8sin 2x.cos^(2)2x)` `rArr 2((sin3x + cos 3x)/(sqrt(2)))=sqrt(1+8 sin 2x cos 2x cos 2x)` `rArr sqrt(2)(sin 3x + cos 3x)^(2)=1+2(sin 6x + sin 2x)` `rArr 2(1+sin 6x)=1+2 sin 6x + 2 sin 2x` `rArr 2sin 2x=1` `rArr sin 2x = 1//2 = sin pi//6` or `2x=n//pi+(-1)^(n)pi//6, n in Z` or `x=(n pi)/(2)+(-1)^(n)(pi)/(12), n in Z` `therefore x=(6n+(-1)^(n))(pi)/(pi)/(12)` `therefore x=(pi)/(12),(5pi)/(12),(13pi)/(12),(17pi)/(12)` But for `x=(5pi)/(12)` and `(13pi)/(12), sin(3x+(pi)/(4))lt 0` `therefore x =(pi)/(12), (17pi)/(12)` |
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| 55. |
Which of the following are the solutons of equations `2sin 11x + cos 3x + sqrt(3) sin 3x = 0` ?A. `x=(n pi)/(7)-(pi)/(84), n in Z`B. `x=(n pi)/(4)+(7pi)/(48),n in Z`C. `x=(n pi)/(7)-(pi)/(63),n in Z`D. `x = (n pi)/(4)+(pi)/(24),n in Z` |
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Answer» Correct Answer - A::B `2 sin 11x + cos 3x + sqrt(3) sin 3x =0` `rArr (1)/(2)cos 3x + (sqrt(3))/(2)sin 3x + sin 11x = 0` `rArr sin (30^(@)+3x) + sin 11x =0` `rArr 2sin((pi)/(12)+7x)cos ((pi)/(12)-4x)=0` `rArr sin((pi)/(12)+7x)=0` or `cos((pi)/(12)-4x)=0` `rArr (pi)/(12)+7x = npi` or `4x-(pi)/(12)=n pi +(pi)/(2), n in Z` `rArr x = (n pi)/(7)-(pi)/(84)` or `x =(n pi)/(4)+(7pi)/(48), n in Z` |
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| 56. |
Find the number of solution of the equation `e^(sinx)-e^(-sinx)-4=0` |
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Answer» Put `e^(sin x)=t` `rArr t^(2)-4t-1=0` `rArr t=e^(sin x)=2 pm sqrt(5)` Now `sin x in [-1, 1]`. Thus, `e^(sin x) in [e^(-1), e^(1)]` and `2 pm sqrt(5) notin [e^(-1), e^(1)]` Hence, there does not exist any solution. |
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| 57. |
The number of all possible values of `theta`, where `0 lt theta lt pi`, for which the system of equations `(y+z)cos 3 theta =(xyz) sin 3 theta ,x sin 3 theta =(2cos3theta)/y+(2sin3theta)/z and (x y z)sin3theta=(y+2z)cos3theta+ysin3theta` have a solution `(x_0,y_0,z_0)` wiith `y_0 z_0 !=0` is |
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Answer» Correct Answer - 3 We have `(y+z) cos 3 theta=(xyz) sin 3 theta` ...(1) `xyz sin 3 theta=(2 cos 3 theta)z+(2 sin 3 theta)y` ...(ii) `(xyz) sin 3 theta=(y+2z) cos 3 theta+y sin 3 theta` ...(iii) `:. (y+z) cos 3 theta=(2 cos 3 theta)z+(2 sin 3 theta)y` `=(y+2z) cos 3 theta+y sin 3 theta` `rArr y(cos 3 theta-2 sin 3 theta)=z cos 3 theta` and `y(sin 3 theta- cos 3 theta)=0` Since, `y, z ne 0, sin 3 theta-cos 3 theta=0` or `tan 3 theta=0` `rArr 3 theta=npi +pi/4, n in Z` `rArr theta=((4n+1)pi)/12, n in Z` `rArr theta=pi/12, (5pi)/12, (9pi)/12` |
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| 58. |
Find the general solution of the equation `sin2x+sin4x+sin6x=0`. |
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Answer» The given equation may be written as `(sin2x+sin4x+sin6x=0`. `rArr 2" sin "((6x+2x))/(2)" cos "((6x-2x))/(2)+sin4x=0` `[becausesinC+sinD=2"sin"((C+D))/(2)"cos"((C-D))/(2)]` `rArr2sin4xcos2x+sin4x=0` `rArrsin4x(2cos2x+1)=0` `rArrsin4x=0or cos2x=-(1)/(2)=-"cos"(pi)/(3)=cos(pi-(pi)/(3))="cos"(2pi)/(3)` `rArrsin4x=0orcos2x="cos"(2pi)/(3)` `rArr4x=npior2x=(2mpi+-(2pi)/(3))`, where m , `ninI` `rArrx=(npi)/(4)orx=(mpi+-(pi)/(3))` where m,`ninI`. Hence , the general solution is given by `x=(npi)/(4)orx=(mpi+-(pi)/(3))`, where m, `ninI`. |
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| 59. |
Solve the following equation: `tan^2x+(1-sqrt(3))tanx-sqrt(3)=0` |
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Answer» Correct Answer - `x=npi +pi/3, n in Z or x=mpi - pi/4; m in Z` `tan^(2)x + tan x - sqrt3 tan x - sqrt3 = 0` or `(tan x - sqrt3) (tan x + 1) = 0` or `tan x = sqrt3 or tan x = -1` `tan x = sqrt3 = tan (pi//3)` `rArr x = n pi + pi//3, n in Z` and `tan x = - 1 = tan (- pi//4)` `rArr x = m pi - pi//4, m in Z` |
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| 60. |
Find the general solution of each of the equations : (i) `4sin^(2)x=1` (ii) `2 cos^(2)x=1` (iii) `cot^(2)x=3` |
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Answer» (i) `4sin^(2)x=1rArrsin^(2)x=(1)/(4)=((1)/(2))^(2)="sin"^(2)(pi)/(6)` `rArrsin^(2)x="sin"^(2)(pi)/(6)` `rArrx={npi+-(pi)/(6)}` , where `ninI` Hence , the general solution is x = `(npi+-(pi)/(6))`, `n inI`. (ii) `2cos^(2)x=1rArrcos^(2)x=(1)/(2)=((1)/(sqrt(2)))^(2)="cos"^(2)(pi)/(4)` `rArr cos ^(2)x="cos"^(2)(pi)/(4)` `rArrx=(npi+-(pi)/(4))`, where `ninI`. Hence , the general solution is x = `(npi+-(pi)/(4)),ninI`. (iii) `cot^(2)x=3rArrtan^(2)x=(1)/(3)=((1)/(sqrt(3)))^(2)="tan"^(2)(pi)/(6)` `rArrtan^(2)x=" tan "^(2)(pi)/(6)` `rArrx=(npi+-(pi)/(6))`, where `ninI`. Hence , the general solution is `(npi+-(pi)/(6))`, where `ninI`. |
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| 61. |
If `log_10(sin x) + log_10(tany)+ log_10 2=0` and `coty= 2sqrt3 cos x,` then ordered pair `(x, y)` satisfying the equations simultaneously is(are) (A) `(pi/3 ,pi/3)` (B) `(pi/3 ,pi/6)` (C) `(pi/6 ,(2pi)/3)` (D) `(pi/3 , (7pi)/6)` |
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Answer» Correct Answer - C `log_(10)(sin x)+log_(10)(tan y) + log_(10)2=0` `therefore 2 sin x tan y = 1` ….(1) where sin x gt 0, tan y gt 0 `cot y=2sqrt(3)cos x` ….(2) From (1) and (2), `2 sin x = 2sqrt(3)cos x` `therefore tan x = sqrt(3)` `rArr x = 2n pi + (pi)/(3), n in I` (as sin x gt 0) From (1), `tan y =(1)/(sqrt(3)rArr y=n pi + (pi)/(6), n in I` Hence, ordered pairs are `(pi//3, pi//6), (pi//3,7pi//6),(7pi//3,pi//6), (7pi//3,7pi//6)`. |
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| 62. |
Find the general values of x and y satisfying the equations `5sinx cosy=1`; `4tanx =tany` |
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Answer» We have `5 sin x cos y=1 and 4 tan x= tan y` or `5 sin x cos y=1 and 4 sin x cos y= sin y cos x` or `sin x cos y =1/5 and cos x sin y =4/5` `rArr sin x cos y + cos x sin y=1` and `sin x cos y - cos x sin y =- 3/5` `rArr sin (x+y)="sin"pi/2 and sin (x-y)= sin {sin^(-1)(-3/5)}` `rArr x+y=n pi +(-1)^(n) pi/2 n in Z` and `x-y=m pi + (-1)^(m) sin^(-1) (-3/5), m in Z` `rArr x=(m+n) pi/2+(-1)^(n) pi/4+(-1)^(m) 1/2 sin^(-1) (-3/5) , m, n in Z` and `y=(n-m) pi/2+(-1)^(n) pi/4 + (-1)^(m+1)1/2 sin^(-1) (-3/5), m, n in Z` |
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| 63. |
What is the general solution of the equation: `tan^2 theta + 2sqrt3 tan theta = 1?` |
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Answer» Correct Answer - `theta=n pi + pi/12, n in Z or theta = mpi - (5pi)/12; m in Z` `tan^(2) theta + 2 sqrt3 tan theta = 1` or `(tan theta + sqrt3)^(2) = 4` or `tan theta = 2 - sqrt3 or -2 - sqrt3` or `tan theta = tan 15^(@), - tan 75^(@)` `= "tan"(pi)/(12), tan (-(5pi)/(12))` From `tan theta = "tan" (pi)/(12)`, we get `theta = n pi + (pi)/(12), n in Z` From `tan theta = tan (-(5pi)/(12))`, we get `theta = m pi - (5pi)/(12), m in Z` |
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| 64. |
Solve `cos 4 theta+ sin 5 theta=2`. |
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Answer» The equation `cos 4theta+ sin 5 theta=2` is valid only when `cos 4 theta=1` and `sin 5 theta=1`. Thus, `4 theta=2npi and 5 theta=2m pi+pi//2, n , m in Z` `rArr theta=(2n pi)/4 and theta=(2m pi)/5+pi/10, n, m in Z` Putting n, `m=0, pm 1, pm 2`, ..., the common value in `[0, 2pi]` is `theta=pi//2`. Therefore, the solution is `theta=2k pi+pi//2, k in Z`. |
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| 65. |
Solve `1 + sin x "sin"^(2) x/2=0`. |
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Answer» `1+sin x "sin"^(2) x/2=0` or `2+2 sin x "sin"^(2) x/2=0` or `2+sin x(1- cos x)=0` or `4+2 sin x - sin 2x=0` or `sin 2x=2 sin x+4` The above result is not possible for any value of x as L.H.S. has maximum value 1 and R.H.S. has minimum value 2. Hence, there is no solution. |
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| 66. |
For what value of `k`the equation `sinx+cos(k+x)+cos(k-x)=2`has real solutions? |
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Answer» `sin x+cot (k+x)+cos (k-x)=2` `rArr 2 cos x. cos k+sin x=2` This equation is of the form `a cos x +b sin x=c` Here `a=2 cos k, b=1` and `c=2` Since for real solutions, `|c| le sqrt(a^(2)+b^(2))`, we have `2 le sqrt(1+4 cos^(2) k)" "or" "cos^(2) k ge 3/4` `rArr sin^(2) k le 1/4" "or" "sin^(2)k-1/4 le 0` or `(sin k+1/2) (sin k-1/2) le 0` or `-1/2 le sin k le 1/2` `rArr (-pi)/6 le k le pi/6` |
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| 67. |
Solve `cos^(50) x- sin^(50)x=1` |
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Answer» `cos^(50)x-sin^(50)x=1` or `cos^(50)x=1+sin^(50)x` `L.H.S. le 1` and `R.H.S. ge 1` Hence, we must have `cos^(50) x=1+sin^(50)x=1` or `sin x=0` or `x=n pi, n in Z` |
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| 68. |
The general solution of the equation `8 cos x cos 2x cos 4x = sin 6x//sin x` isA. `x=(n pi//7)+(pi//21), AA n in Z`B. `x=(2pi//7)+(pi//14), AA n in Z`C. `x=(n pi//7)+(pi//14), AA n in Z`D. `x=(n pi)+(pi//14), AA n in Z` |
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Answer» Correct Answer - C The given equation is `8 sin x cos x cos 2x cos 4 x= sin 6x (sin x ne 0)` `rArr sin 8x= sin 6x` `rArr 2 cos 7x sin x =0` As `sin x ne 0, cos 7x=0` or `7x=n pi +pi//2, n in Z` i.e. `x= n pi//7+pi//14, n in Z` |
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| 69. |
Find the number of real solution of the equation `(cos x)^(5)+(sin x)^(3)=1` in the interval `[0, 2pi]` |
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Answer» Correct Answer - Three solutions `cos^(5) x le cos^(2) x and sin^(3) x le sin^(2) x` So, `cos^(5)x+sin^(3) x=1` is possible only when `cos^(5) x= cos^(2) x` and `sin^(3) x= sin^(2) x`, which is possible only at `x=0, pi//2, and 2pi`. |
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| 70. |
If the equation `tan (P cot x)=cot (P tan x)` has a solution in `x in (0, pi)-{pi/2}`, then prove that `P le pi/4`. |
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Answer» `tan (P cot x)=cot (P tan x)` or `tan (P cot x)= tan (pi/2 - P tan x)` or `P cot x=pi/2-P tan x` or `P(tan x+ cot x) = pi/2` where `P gt 0` Now. `tan x+ cot x ge 2` or `2P le P (tan x+ cot x) = pi/2 or P le pi/4` |
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| 71. |
The -number of solutions of the equation `cos(pisqrt(x-4)cos(pi sqrtx)=1` is |
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Answer» We have `cos(pi sqrt(x)) cos (pi sqrt(x-4))=1`, where `x ge 4` `rArr cos (pi sqrt(x))=1` and `cos (pisqrt(x-4))=1` `rArr pisqrt(x)=2m pi` and `(pi sqrt(x-4))=2n pi" "(m, n in Z)` `rArr x=4 m^(2) and x-4=4n^(2)` Now, `m^(2)-n^(2)=1` `rArr m=1, n=0` `:. x=4` So, `x=4` is the only solution of the equation. |
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| 72. |
Find the number of solutions of the equation `sin^4x+cos^4x-2sin^2x+3/4sin^2 2x=0` in the interval `[0,2pi]` |
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Answer» Correct Answer - Four solutions `sin^(4)x+cos^(4)x-2 sin^(2)x+3/4 sin^(2) 2 x=0` `rArr (sin^(2)+cos^(2) x)^(2)-2 sin^(2)x cos^(2)x-2 sin^(2) x +3/4 . 4 sin^(2) x. cos^(2) x=0` `rArr 1-2 sin^(2) x+sin^(2) x cos^(2) x=0` `rArr sin^(4)x + sin^(2) x-1=0` `rArr sin^(2) x=(sqrt(5)-1)/2` `rArr sin x= pm sqrt((sqrt(5)-1)/2)` Thereare two values of x for `sin x= sqrt((sqrt(5)-1)/2)` and two values of `x` for `sin x=-sqrt((sqrt(5)-1)/2)`. so, there are four solutions. |
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| 73. |
If `3sinx+4cosa x=7`has at least one solution, then find the possible values of `adot` |
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Answer» We have `3 sin x+4 cos ax=7` which is possible only when `sin x=1` and `cos ax =1` or `x=(4n+1) pi/2` and `ax=2mpi, m, n in Z` or `(4n+1) pi/2=(2m pi)/a` or `a=(4m)/(4n+1), m, n in Z` |
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| 74. |
The number of solutions of the equation `sin^3xcosx+sin^2xcos^2x+sinxcos^3x=1`in the interval `[0,2pi]`is/are0 (b) 2(c) 3 (d)infinite |
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Answer» Correct Answer - A `sin^(2) x cos x+sin^(2) x cos^(2) x+ sin x cos^(3)x=1` or `sin x cos x (sin^(2) x+sin x cos x + cos^(2)x)=1` or `(sin 2x)/2 (1+(sin 2x)/2)=1` or `sin 2x(2+sin 2x)=4` or `sin^(2) 2x+2 sin 2x-4 =0` or `sin 2x=(-2 pm sqrt(4+16))/2=-1 pm sqrt(5)` This is not possible since `-1 le sin 2x le 1`. Hence, the given equation has no solution. |
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| 75. |
Solve `3(sec^(2) theta+tan^(2) theta)=5`. |
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Answer» Correct Answer - `theta= npi pm pi/6, n in Z` `sec^(2) theta+tan^(2) theta=5/3` or `tan^(2) theta=1/3 = tan^(2) (pi/6)` `rArr theta=n pi pm pi/6, n in Z` |
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| 76. |
Solve `sec theta-1=(sqrt(2)-1) tan theta`. |
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Answer» Correct Answer - `theta=2npi or theta=2n pi +pi/4, n in Z` `sec theta-1=(sqrt(2)-1) tan theta` `rArr ((1-cos theta)/(cos theta))=((sqrt(2)-1) sin theta)/(cos theta)` or `2 sin^(2) (theta//2)=(sqrt(2)-1)2 sin (theta//2) cos (theta//2)` or `sin (theta//2)=0, tan (theta//2)=(sqrt(2)-1)=tan (pi//8)` `rArr theta//2=n pi or theta/2=npi +pi/8, n in Z` `rArr theta=2n pi or theta=2 n pi +pi/4` |
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| 77. |
Solve `7 cos^(2)x+sin x cos x-3=0`. |
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Answer» Correct Answer - `x=n pi +(3pi)/4 and x=k pi +"tan"^(-1) 4/3 (k, n in Z)` `7 cos^(2) x+sin x cos x-3=0` Dividing by `cos^(2)x`, we get `7+tan x-3 sec^(2) x=0` or `7+tan x-3 (1+tan^(2) x)=0` or `3 tan^(2) x-tan x-4 =0` or `(tan x+1) (3 tan x-4)=0` or `tan x=-1` and `tan x=4/3` `rArr x=npi +(3pi)/4, n in Z` and `x=kpi+tan^(-1) (4/3), k in Z` |
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| 78. |
Solve the equation `2 (cos x+cos 2x)+sin 2x (1+2 cos x)=2 sin x` for `x in [-pi, pi]`. |
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Answer» The given equation is `2(cos x+2 cos^(2) x-1)+2 sin x cos x (1+2 cos x) -2 sin x=0` `rArr 2(2 cos^(2) x+ cos x-1)+2 sin x(2 cos^(2) x+ cos x -1)=0` `rArr (sin x+1) (2 cos^(2) x+ cos x-1) =0` `rArr sin x=-1 or 2 cos^(2) x+2 cos x- cos x-1 =0` `rArr x=2npi -pi/2, n in Z or (2 cos x-1) (cos x+1)=0` `rArr x=2n pi - pi/2, n in Z or x=2npi pm pi/3 or x=2n pi + pi, n in Z` For `-pi le x le pi, x= -pi/2, pm pi/3, pm pi` |
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| 79. |
Solve `4 cos theta-3` sec `theta=tan theta`. |
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Answer» We have `4 cos theta-3 sec theta = tan theta` or `4 cos theta-3/(cos theta)= (sin theta)/(cos theta)` or `4 cos^(2) theta-3= sin theta` or `4(1- sin^(2) theta)-3= sin theta` or `4 sin^(2) theta + sin theta-1=0` or `sin theta=(-1 pm sqrt(1+16))/8=(-1 pm sqrt(17))/8` Now, `sin theta=(-+sqrt(17))/8`. Thus, `sin theta=sin alpha`, where `sin alpha=(-1+sqrt(17))/8` `rArr theta=n pi +(-1)^(n) alpha`, where `sin alpha =(-1 + sqrt(17))/8 and n in Z` Also, `sin theta=(-1-sqrt(17))/8` `rArr sin theta= sin beta`, where `sin beta=(-1-sqrt(17))/8` `rArr theta=n pi +(-1)^(n) beta, n in Z` where `sin beta =(-1-sqrt(17))/8` |
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| 80. |
General solution of `sin^(2) x-5 sin x cos x -6 cos^(2)x=0` isA. `x=n pi-pi//4, n in Z` onlyB. `npi+tan^(-1) 6, n in Z` onlyC. both (a) and (2)D. none of these |
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Answer» Correct Answer - C Dividing the given equation by `cos^(2) x`, as `cos x =0` does not satisfy the equation, we have `tan^(2) x-5 tan x -6 =0` or `(tan x+1) (tan x-6)=0` or `tan x=-1 or tan x=6` If `tan x=-1=tan (-pi//4)`, then `x=npi-pi//4, AA n in Z` and, if `tan x=6=tan alpha` (say) `rArr alpha = tan^(-1) 6`, then, `x=n pi +alpha =n pi+tan^(-1) 6, AA n in Z` Hence, `x=npi-(pi//4), n pi + tan^(-1) 6, n in Z`. |
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| 81. |
If `secxcos5x+1=0,w h e r e0 |
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Answer» `secx*cosx+1=0` `(cos5x)/cosx=-1` `cos5x=-cosx` `cos5x+cosx=0` `2cos((5x+x)/2)*cos((5x-x)/2)=0` `2 cos3x*cos2x=0` `cos3x=0` `3x=pi/2,3/2pi` `x=pi/6,pi/2` `cos2x=0` `2x=pi/2` `x=pi/4` |
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| 82. |
Find number of solutions of equation `sin^(2) theta- 4/(sin^(3) theta-1)=1- 4/(sin^(3) theta-1), theta in [0, 6pi]`. |
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Answer» Correct Answer - Three solutions We have `sin^(2) theta- 4/(sin^(3) theta-1)=1- 4/(sin^(3) theta-1)`, where `sin theta ne 1` `:. sin^(2) theta=1, sin theta ne 1` `:. sin theta=-1` Thus, there are three values of `theta` one each in intervals `[0, 2pi], [2pi, 4pi]` and `[4pi, 6pi]` |
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| 83. |
General solution of `tan theta+tan 4 theta+tan 7 theta=tan theta tan 4 theta tan 7 theta` isA. `theta=n pi//12`, where `n in Z`B. `theta=n pi//9`, where `n in Z`C. `theta = npi+pi//12`, where `n in Z`D. none of these |
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Answer» Correct Answer - D From the given equation, we have `(tan theta+tan 4 theta)/(1-tan theta tan 4 theta)=-tan 7 theta` or `tan (theta+4theta)=-tan 7 theta` or `tan 5 theta = tan (-7theta)` `rArr 5 theta= n pi -7 theta` or `theta = npi//12`, where `n in Z`, but `n ne 6, 18, 30`, ... |
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| 84. |
The number of values of `x`in the interval `0,5pi`satisfying the equation `3sin^2x-7sinx+2=0`is`0`(b) `5`(c) `6`(d) `10` |
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Answer» `3sin^2x-7sinx+2 = 0` `=>3sin^2x-6sinx-sinx+2 = 0` `=>(sinx-2)(3sinx-1) = 0` `=>sinx = 2 or sinx = 1/3` But, `sin x` can not be more than `1`. `:. sinx = 1/3` It means, value of `sinx` is positive. It means, `x` lies in first and second quadrant. So, there are `2` values from `0` to `2pi`. So, from `0` to `5pi`, there will be `5` values for `x` that will satisfy the given equation. |
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| 85. |
The number of values of `theta`in the interval `(-pi/2,pi/2)`satisfying the equation `(sqrt(3))^(sec^2theta)=tan^4theta+2tan^2theta`is2 (b)4 (c) 0(d) 1A. 2B. 4C. 0D. 1 |
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Answer» Correct Answer - A `tan^(4) theta+2 tan^(2) theta=(tan^(2) theta+1)^(2)-1=(sec^(2) theta)^(2) -1=sec^(4) theta-1` Puutting `sec^(2) theta=t` we get `(sqrt(3))^(t)=t^(2)-1` `rArr t=2` is the only solution as `t gt 1` Hence, there will be 2 values of `theta` in given interval. |
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| 86. |
If `tan^2x-5secx=1`is satisfied by exactly seven distinct values of `x in [0,((2n+1)pi)/2],n in N ,`then the greatest value of `n`is________. |
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Answer» Correct Answer - 7 Here `(2 sec x +1) (sec x - 3)=0` which gives two values of `theta` in each of `[0, 2pi], (2pi, 4pi], (4pi, 6pi]` and one value in `(6pi, 6pi +(3pi)/2)`. |
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| 87. |
If `cos4x=a_0+a_1cos^2x+a^2cos^4x`is true for all values of `x in R ,`then the value of `5a_0+a_1+a_2`is_______ |
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Answer» Correct Answer - 5 `cos 4x=2 cos^(2) 2x-1` `=2(2 cos^(2) x-1)^(2)-1` `=2(4 cos^(4) x+1-4 cos^(2) x)-1` `=8 cos^(4) x-8 cos^(2) x+1` `:. a_(0)=1, a_(1)=-8, a_(2)=8` `:. 5a_(0)+a_(1)+a_(2)=5` |
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| 88. |
The equation `sin^4x-2cos^2x+a^2=0`can be solved if`-sqrt(3)lt=alt=sqrt(3)`(b) `sqrt(2)lt=alt=""sqrt(2)``-""1lt=alt=a`(d) none of theseA. `-sqrt(3) le a le sqrt(3)`B. `-sqrt(2) le a le sqrt(2)`C. `-1 le a le 1`D. none of these |
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Answer» Correct Answer - B We have `sin^(4) x-2 cos^(2) x+a^(2)=0` `:. a^(2)=2 cos^(2) x- sin^(4) x` `=2-2 sin^(2) x-sin^(4) x` `=3-(sin^(2) x+1)^(2)` Now `0 le sin^(2) x le 1` `:. 1 le sin^(2) x + 1 le 2` `:. 1 le (sin^(2) x+1)^(2) le 4` `:. -4 le -(sin^(2) x+1)^(2) le -1` `:. -1 le 3 -(sin^(2) x+1)^(2) le 2` `:. a^(2) le 2` `:. -sqrt(2) le a le sqrt(2)` |
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| 89. |
The equation `2sin^3theta+(2lambda-3)sin^2theta-(3lambda+2)sintheta-2lambda=0`has exactly three roots in `(0,2pi)`, then `lambda`can be equal to0 (b)2 (c) 1(d) `-1` |
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Answer» Correct Answer - A::C::D The equation becomes `(sin theta -2) (sin theta + lambda) (2 sin theta+1)=0` `rArr lambda= pm 1, 0`. |
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| 90. |
The number of roots of the equation `sin(2x+pi/18) cos(2x-pi/9)=-1/4` in `[0, 2pi]` isA. 2B. 4C. 6D. 8 |
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Answer» Correct Answer - B `2sin(2x+(pi)/(18)).cos(2x-(2pi)/(18))=(-1)/(2)` `therefore 2 sin (2x + 10^(@))cos (2x-20^(@))=(-1)/(2)` `therefore sin (4x-10^(@))+sin 30^(@)=(-1)/(2)` `therefore sin (4x-10^(@))=-1=sin 270^(@)` Now period of `sin (4x - 10^(@))` is `pi//2` For `x in (0, pi//2)` `therefore 4x = 280^(@)` `therefore x = 70^(@)` `therefore` for x in `[0, 2pi]`, there are four solutions |
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| 91. |
find all the possible triplets `(a_(1), a_(2), a_(3))` such that `a_(1)+a_(2) cos (2x)+a_(3) sin^(2) (x)=0` for all real x. |
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Answer» We have `a_(1)+a_(2) cos (2x) +a_(3) sin^(2) x=0` for all real x `:. a_(1)+a_(2) (1-2 sin^(2) x)+a_(3) sin^(2) x=0` `rArr (a_(1)+a_(2))+(-2a_(2)+a_(3)) sin^(2) x=0` Since this is true for all real x, we must have `a_(!)+a_(2)=0` and `-2a_(2)+a_(3)=0` `:. a_(2)=-a_(1)` and `a_(3)= -2a_(1)` Thus, there exists infinite triplets `(a_(1), -a_(1), -2a_(1))` |
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| 92. |
Find the number of pairs of integer `(x , y)`that satisfy the following two equations:`{cos(x y)=xtan(x y)=y`1 (b) 2(c) 4 (d)6A. 1B. 2C. 4D. 6 |
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Answer» Correct Answer - A `(sin (xy))/(cos (xy))=y` or `sin (xy)=xy` or `xy=0` `:. x=0 or y=0` But `x=0` is not possible `:. y=0` and `x=1`, i.e. `(1, 0)`. |
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| 93. |
Solve `sin^3thetacostheta-cos^3thetasintheta=1/4dot` |
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Answer» `sin^(3) theta cos theta-cos^(3) theta sin theta =1/4` or `4 sin theta cos theta(sin^(2) theta - cos^(2) theta)=1` or `2 sin 2 theta (-cos 2 theta)=1` or `- sin 4 theta=1` or `sin 4 theta=-1` `rArr 4 theta=2 n pi - pi/2` `rArr theta= (n pi)/2-pi/8, n in Z` |
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| 94. |
Solve `sqrt(3) cos theta-3 sin theta =4 sin 2 theta cos 3 theta`. |
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Answer» We have `sqrt(3) cos theta -3 sin theta=2 (sin 5 theta- sin theta)` or `(sqrt(3)/2) cos theta-(1/2) sin theta= sin 5 theta` `rArr cos (theta+pi/6)= sin 5 theta = cos (pi/2- 5 theta)` `rArr theta + pi/6 = 2n pi pm (pi/2 - 5theta)` or `theta=(n pi)/3+pi/18 or theta=- (n pi)/2 +pi/6, AA n in Z` |
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| 95. |
Consider the system of equations `x cos^(3) y+3x cos y sin^(2) y=14` `x sin^(3) y+3x cos^(2) y sin y=13` The value/values of x is/areA. `pm 5sqrt(5)`B. `pm sqrt(5)`C. `pm 1//sqrt(5)`D. none of these |
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Answer» Correct Answer - A The given equations are `x cos^(3)y+3x cos y sin^(2) y=14` ...(i) and `x sin^(2) y+3x cos^(2) y sin y=13` ...(ii) Adding Eqs. (i) and (ii), we have `x(cos^(3) y+3 cos y sin^(2) y+3 cos^(2) y sin y+ sin^(3) y)=27` or `x(cos y+ sin y)^(3)=27` or `x^(1//3) (cos y + sin y) =3` ...(iii) Subtracting Eq. (ii) from Eq. (i), we have `x(cos^(3)y+3 cos y sin^(2) y-3 cos^(2) y sin y- sin^(3) y)=1` or `x(cos y- sin y)^(3)=1` or `x^(1//3) (cos y- sin y)=1` ...(iv) Dividing Eq. (iii) by (iv), we get `cos y+sin y=3 cos y-3 sin y` or `tan y=1//2` Case I : `sin y=1//sqrt(5) and cos y =2//sqrt(5)` `y=2n pi +alpha`, where `0 lt alpha lt pi//2` and `sin alpha =1//sqrt(5)` i.e., y lies in the first quadrant From Eqs. (iii) `x^(1//3) (3//sqrt(5))=3 or x=5 sqrt(5)` Case II : `sin y=-1//sqrt(5) and cos y=-2//sqrt(5)` `y=2npi+(pi+alpha)`, where `0 lt alpha lt pi//2` and `sin alpha = -1 //sqrt(5)` i.e., y lies in the third quadrant. Therefore, from Eq. (iii), `x^(1//3) (-3//sqrt(5))=3 or x=-5sqrt(5)`. Thus, `sin^(2) y+2 cos^(@) y=1//5+8//5=9//5`. Also there are exactly six values of `y in [0, 6pi]`, there in 1st quadrant and three in 3rd quadrant. |
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| 96. |
Find the number of solution of the equation `sqrt(cos 2x+2)=(sin x + cos x)` in `[0, pi]`. |
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Answer» We have `sqrt(cos 2x+2)=(sin x + cos x)` Squaring both sides, we get `2+cos 2x=1 + sin 2x` `:. Sin 2x-cos 2x =1` Again, squaring both sides, we get `1-sin 4x =1` `:. Sin 4x =0` `rArr 4x=n pi, n in Z` or `x=(npi)/4`, where `x in [0, pi]` `:. x=0, pi/4, pi/2, (3pi)/4, pi` But only `x=pi/4` and `pi/2` satisfy the original equation. 2. Never cancel the terms containing unknown terms which are in product on the two sides. It may cause the loss of a genuine solution. |
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| 97. |
The sum of all roots of `sin(pi(log)_3(1/x))=0`in `(0,2pi)`is`3/2`(b) 4(c) `9/2`(d) `(13)/3`A. `3//2`B. 4C. `9//2`D. `13//3` |
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Answer» Correct Answer - C `pi log_(3) (1/x)=k pi, k in I` `log_(3) (1/x)=k or x=3^(-k)` The possible values of k are `-1, 0, 1, 2, 3,...` `S=3+1+1/3+1/3^(2)+1/3^(3)+...oo=3/(1-1/3)=9/2` |
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| 98. |
Solve `(sin^3x/2-cos^3x/2)/(2+sinx)=(cosx)/3` |
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Answer» `("sin"^(3) x/2-"cos"^(3)x/2)/(2 + sin x)=(cos x)/3` or `(("sin"x/2-"cos"x/2)(1+"sin"x/2"cos"x/2))/(2(1+"sin" x/2"cos"x/2))=(cos x)/3` or `"sin"x/2-"cos"x/2=2/3 cos x` or `1-sin x=4/9 cos^(2) x` (On squaring) or `4/9 sin^(2) x-sin x+1-4/9=0` or `4 sin^(2) x-9 sin x+5=0` or `(4 sin x-5) (sin x -1)=0` or `sin x=1` `rArr x=2 npi+pi/2, n in Z` |
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| 99. |
Solve `sin 2x+cos 4x=2`. |
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Answer» Correct Answer - No solution `sin 2x+cos 4x=2` It is possible only when `sin 2x=1` and `cos 4x=1` `2x=2npi+pi/2 and 4x=2mpi` `:. X= npi +pi/4 and x= (mpi)/2`, where `m, n in Z` Then, solution is given by `(n pi +pi/4, n in I) nn ((mpi)/2, m in Z)= phi` |
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| 100. |
Solve `sin 2x=4 cos x`. |
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Answer» We have `sin 2x=4 cos x` `:. 2 sin x cos x=4 cos x` Here, we should not cancel `cos x`. Now, `2 sin x cos x -4 cos x=0` `:. cos x(2 sin x-4)=0` We have `cos x =0 or sin x=2`, which is not possible. `:. x=(2n+1) pi/2, n in Z` 3. The solution of the equation should not contain such values of angles which make any of the terms undefined or infinite in the original equation. Domain should not change while simplifying the equation. If it changes, necessary corrections must be made. |
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