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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.
| 101. |
Solve `sin x-3 sin 2x + sin 3x = cos x -3 cos 2x + cos 3x`. |
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Answer» The given equation is `sin x-3 sin 2x + sin 3x=cos x-3 cos 2x+cos 3x` `rArr 2 sin 2x cos x-3 sin 2x=2 cos 2x cos x-3 cos 2x` `rArr sin 2x(2 cos x-3)=cos 2x(2 cos x-3)` `rArr sin 2x=cos 2x" "("As "cos x ne 3//2)` `rArr tan 2x=1` `rArr 2x=n pi+pi//4` `rArr x=(npi)/2+pi/8, n in Z` |
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| 102. |
`(sin3theta)/(2cos2theta+1)=(1)/(2)` if `(n in Z)`A. `theta = 2n pi+(pi)/(6)`B. `theta=2n pi-(pi)/(6)`C. `theta=n pi+(-1)^(n)(pi)/(6)`D. `theta-n pi-(pi)/(6)` |
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Answer» Correct Answer - C `(sin 3theta)/(2cos 2theta +1)=(1)/(2)` `rArr (3 sin theta - 4 sin^(3)theta)/(3-4 sin^(2)theta)=(1)/(2)` `rArr sin theta =(1)/(2)rArr theta = n pi+(-1)^(n)pi//6, n in Z` |
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| 103. |
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 number of values of `y in [0, 6pi]` isA. 5B. 3C. 4D. 6 |
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Answer» Correct Answer - D 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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| 104. |
Solve `4 cos^(2)x+6 sin^(2)x=5`. |
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Answer» Correct Answer - `x=n pi pm pi/4, n in Z` `4+2 sin^(2) x=5` or `sin^(2) x=1/2= sin^(2) pi/4` `rArr x=n pi pm pi/4, n in Z` |
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| 105. |
Solve for x and y `12sinx-2y^2=21-8y-5 cosx` |
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Answer» Correct Answer - `x=2npi + cos^(-1) (5/13), n in Z and y=2` `12 sin x +5 cos x=2y^(2) - 8y+21` `rArr sqrt(12^(2)+5^(2))(12/13 sin x + 5/13 cos x)=2(y^(2)-4y+4)+13` `rArr 13 cos(x-alpha)=2 (y-2)^(2)+13` where `cos alpha = 5//13`. Now `L.H.S. le 13 and R.H.S. ge 13` `rArr 13 cos(x-alpha)=2 (y-2)^(2)+13=13` `rArr cos (x-alpha)=1 and y=2` `rArr x-alpha=2n pi and y=2` `rArr x=2npi + cos^(-1) (5/13) and y=2`, where `n in Z` |
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| 106. |
Solve `(tan 3x - tan 2x)/(1+tan 3x tan 2x)=1`. |
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Answer» We have `(tan 3x-tan 2x)/(1+tan 3x tan 2x)=1` `rArr tan (3x-2x)=1` or `tan x=1` One of the principal solution of the equation is `x=pi/4`. Since period of `tan x` is `pi`, general solution is given by `x=npi+pi/4, n in Z`. But for these values of x, term `(tan 2x)` in original equation is not defined. So, `x=n pi+pi/4, n in Z` is not the solution of the equation. |
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| 107. |
Solve : `2+tan x. cot.(x)/(2)+cot x. tan.(x)/(2)=0`. |
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Answer» Correct Answer - `x = 2n pi pm(2pi)/(3), n in Z` `2+tan x.cot. (x)/(2)+cotx.tan.(x)/(2)=0` …(1) Let `an x.cot.(x)/(2)=y` Then equation (1) becomes `2+(y+(1)/(y))=0` `rArr y+(1)/(y)=-2` On solving, we get y = -1 or `tan x. cot.(x)/(2)=-1` or `(sin x)/(cos x)(cos.(x)/(2))/(sin.(x)/(2))=-1` or `(2sin.(x)/(2)cos.(x)/(2))/(cos x).(cos.(x)/(2))/(sin.(x)/(2))-1` or `(2"cos"^(2)(x)/(2))/(cos x)=-1` or `(1+cos x)/(cos x)=-1` or `cos x= -(1)/(2)` or `x=2n pi pm (2pi)/(3), n in Z` |
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| 108. |
Solve `7 cos^(2) theta+3 sin^(2) theta=4`. |
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Answer» We have `7 cos^(2) theta+3 sin^(2) theta=4` `rArr 7(1- sin^(2) theta)+3 sin^(2) theta=4` or `4 sin^(2) theta=3` or `sin^(2) theta =3/4=((sqrt(3))/2)^(2)` or `sin^(2) theta="sin"^(2) pi/3` or `theta= npi pm pi/3, n in Z` |
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| 109. |
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 of `sin^(2) y+2 cos^(2) y` isA. `4//5`B. `9//5`C. 2D. none of these |
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Answer» Correct Answer - B 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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| 110. |
Solve `2^(cos 2x)+1=3.2^(-sin^(2) x)` |
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Answer» Correct Answer - `x=n pi, n in Z or x=n pi pm pi/2, n in Z` `2^(cos 2x)+1=3.2^(- sin^(2)x)` `rArr 2^(1-2 sin^(2) x)+1=3.2^(- sin^(2) x)` Let `2^(- sin^(2) x)=t` Now, given equation reduces to `2t^(2)+1=3t` `rArr t=1 and t=1//2` If `t=1` then `sin^(2) x=0` `rArr x=npi, n in I` If `t=1/2` then `sin^(2) x=1` `rArr x=npi pm pi/2, n in I`. |
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| 111. |
Solution of the equation `sin (sqrt(1+sin 2 theta))= sin theta + cos theta` is `(n in Z)`A. `n pi=pi/4`B. `n pi+pi/12`C. `n pi+pi/6`D. none of these |
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Answer» Correct Answer - A `sin (sqrt(1+sin 2 theta))= sin theta + cos theta` `rArr sin(sin theta + cos theta)= sin theta + cos theta` The equation of the form `sin x =x` `rArr x=0` `rArr 1+ sin 2 theta=0` `rArr sin 2 theta =-1` `rArr theta=npi -pi/4, n in Z` |
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| 112. |
Solev `(sin^(2) 2x+4 sin^(4) x-4 sin^(2) x cos^(2) x)/(4-sin^(2) 2x-4 sin^(2) x)=1/9`. |
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Answer» We have `(sin^(2) 2x+4 sin^(4) x-4 sin^(2) x cos^(2) x)/(4-4 sin^(2) x- sin^(2) 2x)=1/9` `rArr (4 sin^(4)x)/(4 cos^(2) x-4 sin^(2) x cos^(@) x)=1/9` `rArr sin^(4) x/cos^(4) x=1/9` `rArr tan^(2) x=(1/sqrt(3))^(2)=("tan"pi/6)^(2)` `rArr x=n pi pm pi/6` |
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| 113. |
Solve `sin^(4)x=1+tan^(8)x`. |
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Answer» Correct Answer - No solution `sin^(4) x=1 + tan^(8) x` Now, `1+tan^(8) x ge 1 and sin^(4) x le 1` `rArr L.H.S.=R.H.S.` Hence, for the given equation to be satisfied, `sin^(4) x=1 and 1+ tan^(8) x=1` `rArr sin^(2) x=1 and tan^(8) x=0` which never possible, since `sin x` and `tan x` vanish simultaneously. Therefore, the given equation has no solution. |
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| 114. |
If `alpha < beta < gamma` and `sin gamma cos alpha=1,` where `alpha,gamma in[pi,2 pi],` then the least integral value of `f(x) = | x - alpha| + | x - beta| + |x - gamma|` is |
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Answer» Correct Answer - C `sin gamma.cos alpha =1 alpha, gamma in [pi, 2pi]` `therefore sin gamma = cos alpha =1` `rArr gamma = pi//2, alpha = 2pi` (rejected) `(as alpha lt beta lt gamma)` Other possibility is `sin gamma = cos alpha =-1 rArr gamma = 3 pi//2, alpha = pi` `f(x)|_(min)=f(beta)=beta-alpha+0+gamma-beta` `=gamma -alpha` `=(3pi)/(2)-pi=(pi)/(2)` `f(x)ge pi//2 rArr` least integral value of f(x) is 2. |
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| 115. |
Find the number of solution of the equation `cot^(2) (sin x+3)=1` in `[0, 3pi]`. |
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Answer» Correct Answer - Six solutions We have `cot^(2) (sin x+3)=1="cot"^(2) pi/4` `rArr sin x+3=n pi pm pi/4` Now, `2 le sin x +3 le 4` `:. Sin x+3=pi-pi/4` or `sin x+3=pi+pi/4` `:. Sin x=(3pi)/4-3` ...(1) or `sin x=(5pi)/4-3` ...(2) For (1), we have two values of x in interval `(pi, 2pi)`. For (2), we have two values of x in each of the intervals `(0, pi), (2pi, 3pi)`. So, there are total six solutions. |
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| 116. |
Solve `4cot2theta=cot^2theta-tan^2theta` |
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Answer» `4/(tan 2 theta)=1/(tan^(2) theta)-tan^(2) theta` or `(4(1-tan^(2) theta))/(2 tan theta)=(1- tan^(4) theta)/(tan^(2) theta)" "["put "tan 2 theta=(2 tan theta)/(1-tan^(2) theta)]` or `(1-tan^(2) theta)[2 tan theta-(1+tan^(2) theta)]=0` or `(1-tan^(2) theta)(tan^(2) theta-2 tan theta+1)=0` or `(1-tan^(2) theta) (tan theta-1)^(2) =0` or `tan theta= pm 1` `rArr theta= n pi pm pi/4, n in Z` |
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| 117. |
Find number of solution of the equation `2 sin x+5 sin^(2) x+8sin^(3)x+... oo=1` for `x in [0, 2pi]`. |
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Answer» Correct Answer - Two solutions We have `2 sin x+5 sin^(2)x+8sin^(3)x+...=1` ...(1) `:. 2sin^(2)x+5 sin^(3) x+...=sin x` ...(2) Substracting (2) from (1), we get `2 sin x+3(sin^(2)x+sin^(3)x+...)=1-sin x` `:. 2sin x+ (3 sin^(2) x)/(1- sin x)=1-sin x` `rArr 2 sin x-2 sin^(2)x+3 sin^(2)x=1-2 sin x+sin^(2)x` `rArr sin x=1/4` So there will be two solutions. |
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| 118. |
The solutions of the equation `1+(sin x - cos x)"sin" pi/4=2 "cos"^(2) (5 x)/2` is/areA. `x=(n pi)/3+pi/8, n in Z`B. `x=(npi)/2+(5pi)/16, n in Z`C. `x=(npi)/3+pi/4, n in Z`D. `x=(npi)/2+(7pi)/8, n in Z` |
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Answer» Correct Answer - A::B `1+(sin x - cos x) "sin" pi/4=2 "cos"^(2) (5pi)/2` ...(i) `rArr 1+(sin x- cos x) 1/sqrt(2) =1 + cos 5x` `rArr cos 5x+ cos (x+pi/4)=0` `rArr 2 cos(3x+pi/8) cos (2x-pi/8)=0` `rArr cos (3x+pi/8) =0, or cos (2x-pi/8)=0` ...(ii) `3x+pi/8=n pi +pi/2 or 2x- pi/8 = n pi +pi/2` ...(iii) or `x=(n pi)/3+pi/8 or x=(n pi)/2+(5 pi)/16, n in Z` |
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| 119. |
For `x in (0,pi)` the equation `sinx+2sin2x-sin3x=3` hasA. infinitely many solutionsB. three solutionsC. one solutionD. no solution |
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Answer» Correct Answer - D `sin x +2 sin 2x- sin 3x=3` `= sin x+4 sin x cos x -3 sin x+4 sin^(3) x=3` `rArr sin x [-2+4 cos x+4(1-cos^(2) x)]=3` `rArr sin x [2-(4 cos^(2) x-4 cos x+1)+1]=3` `rArr 3-(2 cos x-1)^(2) = 3 cosec x` Now `R.H.S. ge 3` But `L.H.S. lt 3` Hence, no solution. |
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| 120. |
Show that `x=0` is the only solution satisfying the equation `1+sin^(2) ax=cos x`, where a is irrational. |
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Answer» The given equation is `1+sin^(2) ax = cos x` or `(1- cos x) + sin^(2) ax=0` or `2 sin^(2) (x//2)+ sin^(2) ax=0` Since both the terms on the L.H.S. are non-negative, their sum will be zero only if each one is zero. Thus, `sin (x//2)=0 and sin ax =0` `rArr x//2=n pi and ax =m pi, AA n, m in Z` `rArr x=2n pi and x=m pi//a AA n, m in Z` `rArr 2 n pi = m pi//a rArr 2 an =m` ...(ii) Since m and n are integer and a is irrational, Eq. (ii) is possible only if `m=0` and n`=0`. Hence, `x=0` is the only solution. |
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| 121. |
The number of solution of `sin x+sin 2x+sin 3x` `=cos x +cos 2x+cos 3x, 0 le x le 2pi`, isA. 7B. 5C. 4D. 6 |
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Answer» Correct Answer - D We have `(sin x+ sin 3x)+sin 2x=(cos x+cos 3x)+cos 2x` or `2 sin 2x cos x+ sin 2x=2 cos 2x cos x + cos 2x` or `sin 2x(2 cos x +1)=cos2x(2 cso x +1)` or `(2 cos x+1) (sin 2x- cos 2x)=0` or `cos x=-1//2 or sin 2x-cos 2x=0` `rArr x=2 n pi pm (2 pi//3) or tan 2x=1=tan (pi//4)` `rArr x=2n pi pm (2pi//3) or x=(4n+1) pi//8, n in Z` But here `0 le x le 2pi`. Hence, `x=pi//8, 5pi//8, 2pi//3, 9 pi//8, 4pi//3, 13 pi//8`. |
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| 122. |
If `x+y=pi//4` and `tan x+tan y=1`, then `(n in Z)`A. `sin x=0` alwaysB. when `x=npi+pi//4` then `y=-npi`C. when `x=npi` then `y=npi +(pi//4)`D. when `x=npi+pi//4` then `y=npi -(pi//4)` |
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Answer» Correct Answer - B::C From `tanx+tan y=1`, we have `(sin x)/(cos x)+(sin y)/(cos y)=1` or `sin x cos y+ sin y cos x= cos x cos y` or `2 sin (x+y) =2 cos x cos y` or `2 sin (x+y) = cos (x+y) + cos (x-y)` or `2 sin (pi/4) = cos (pi/4) + cos (c-y)` or `cos (x-y)=1/sqrt(2)= cos (pi/4)` `rArr x-y=2n pi pm (pi//4), AA n in Z` ...(i) Also we have `x+y= pi//4` ...(ii) From Eqs. (i) and (ii), we have `x= n pi + (pi//4) and y= -n pi, AA n in Z` or `x = npi and y=- n pi + pi//4, AA n in Z` |
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| 123. |
Solve: `cottheta``+cos e ctheta=sqrt(3)` |
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Answer» Correct Answer - `theta=2n pi+pi/3, n in Z` `cot theta + cosec theta = sqrt(3)` or `(cos theta)/(sin theta)+1/(sin theta)=sqrt(3)` or `cos theta +1 = sqrt(3) sin theta` or `sqrt(3) sin theta - cos theta=1` ...(i) or `2 cos (theta +pi/3)=-1` or `cos (theta + pi/3)=-1/2` or `cos (theta + pi/3)= "cos" (2 pi)/3` `rArr theta +pi/3=2 n pi pm (2pi)/3, n in Z` or `theta = 2n pi pm (2pi)/3 - pi/3, n in Z` `:. theta =2npi +pi/3, n in Z` or `theta =2 n pi - pi =(2n-1) pi, n in Z` But `theta` cannot be equal to `(2n-1)pi` as it makes `sin theta=0`. Hence, `theta=2n pi +pi/3, in Z`. |
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| 124. |
Find the most general solution of `2^1|cosx|+cos^2x+|cosx|^(3+oo)=4` |
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Answer» We have `2^(1+|cos x|+cos^(2)x+|cos x|^(3)+...oo)=4` or `2^(1+|cos x|+|cos x|^(2)+|cos x|^(3)+...oo)=4` or `2^(1/(1-|cos x|))=2^(2)` (sum of infinite G.P.) or `1/(1-|cos x|)=2` or `|cos x|=1/2 or cos x = pm 1/2` `rArr x=n pi pm pi/3, n in Z` |
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| 125. |
find the value of theA. `pm sqrt(n pi), n in {0, 1, 2, ...}`B. `pm sqrt(n pi), n in {1, 2, ...}`C. `pi/2+2npi, n in {..., -2, -1, 0, 1, 2 ...}`D. `2npi, n in {..., -2, -1, 0, 1, 2, ...}` |
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Answer» Correct Answer - A `(fogogof) (x)= sin^(2) (sin x^(2))` `(gogof) (x) = sin (sin x^(2))` `:. Sin^(2) (sin x^(2))= sin (sin x^(2))` `rArr sin (sin x^(2)) [sin (sin x^(2))-1]=0` `rArr sin (sin x^(2))=0` or `1` `rArr sin x^(2) = npi` or `2 mpi+pi//2`, where `m, n in I` `rArr sin x^(2)=0` `rArr x^(2) =n pi rArr x= pm sqrt(npi), n in {0, 1, 2, ...}`. |
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| 126. |
Solve `cos x + cos 2x+...+ cos (nx) =n, n in N`. |
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Answer» Correct Answer - `x=0` We have `cos x + cos 2x +...+ cos (nx) =n` Now, `cosx+cos 2x+...+ cos (nx) le n` So, `cos x = cos 2x = ... = cos nx=1` `:. x=0` |
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| 127. |
Solve `sin theta+cos theta=sqrt(2) cos A`. |
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Answer» Correct Answer - `theta=2n pi +pi/4 pm A, n in Z` `sin theta + cos theta=sqrt(2) cos A` `rArr (1//sqrt(2)) sin theta +(1//sqrt(2)) cos theta=cos A` or `cos (theta-pi//4)=cos A` `rArr theta-pi//4=2npi pm A` or `theta=2npi+pi//4 pm A, AA n in Z` |
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| 128. |
Find the principal solutions of each of the following equations : (i) `sinx=(1)/(2)` (ii) `cosx=(1)/(sqrt(2))` |
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Answer» (i) The given equation is sin `x=(1)/(2)`. We know that `"sin"(pi)/(6)=(1)/(2)andsin(pi-(pi)/(6))=(1)/(2)`. `therefore "sin"(pi)/(6)=(1)/(2)and"sin"(5pi)/(6)=(1)/(2)`. Hence , the principal solutions are `x=(pi)/(6)andx=(5pi)/(6)`. (ii) The given equation is cos x `=(1)/(sqrt(2))`. We know that `"cos"(pi)/(4)=(1)/(sqrt(2))andcos(2pi-(pi)/(4))=(1)/(sqrt(2))`. `therefore"cos"(pi)/(4)=(1)/(sqrt(2))and "cos"(7pi)/(4)=(1)/(sqrt(2))`. Hence , the principal solutions are `x=(pi)/(4)and x = (7pi)/(4)`. (iii) The given equation is tan `x=(1)/(sqrt(3))`. We know that tan `(pi)/(6)=(1)/(sqrt(3))and tan (pi+(pi)/(6))=(1)/(sqrt(3))`. `therefore "tan"(pi)/(6)=(1)/(sqrt(3))and" tan"(7pi)/(6)=(1)/(sqrt(3))`. Hence , the principal solutions are `x=(pi)/(6) andx=(7pi)/(6)`. |
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| 129. |
Solve the equation `sqrt3cos x + sin x = sqrt2 `. |
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Answer» We have, `sqrt(3) cos x + sin x=sqrt(2)` ...(1) Dividing both sides by `sqrt((sqrt(3))^(2)+1^(2))=2`, we get `sqrt(3)/2 cos x+1/2 sin x=1/sqrt(2)` `rArr cos(x-pi/6)=1/sqrt(2)` `rArr cos(x-pi/6)="cos" pi/4` `rArr x-pi/6=2n pi pm pi/4, n in Z` `rArr x= 2n pi pm pi/4+pi/6` `rArr x=2npi +pi/4+pi/6 or x=2npi - pi/4+pi/6` `rArr x=2n pi +(5 pi)/12 or x=2npi-pi/12`, where `n in Z` |
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| 130. |
The solutions of the system of equations `sin x sin y=sqrt(3)/4, cos x cos y= sqrt(3)/4` areA. `x=pi/3+pi/2 (2n+k), n, k in I`B. `y=pi/6+pi/2 (k-2n), n, k in I`C. `x=pi/6+pi/2 (2n+k), n, k in I`D. `y=pi/3+pi/2 (k-2n), n, k in I` |
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Answer» Correct Answer - A::B::C::D `sin x sin y=sqrt(3)/4` and `cos x cos y = sqrt(3)/4` Then, `cos x cos y + sin x sin y=sqrt(3)/2` `rArr cos (x-y)=sqrt(3)/2` `rArr x-y=2n pi pm pi/6, n in I` ...(i) and `cos x cos y-sin x sin y=0` `rArr cos (x+y)=0` `rArr x+y= k pi +pi/2, k in I` ...(ii) From Eqs. (i) and (ii), we get `2x=2npi+k pi pm pi/6+pi/2` `rArr x=pi/2(2n +k) pm pi/12+pi/4` `:. x=pi/2 (2n+k)+pi/3 or x=pi/2 (2n+k)+pi/6` `:. y=pi/6+pi/2 (k-2n) and y=pi/3+pi/2 (k-2n)` |
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| 131. |
Find the number of integral values of k for which the equation `7 cos x+5 sin x=2k+1` has at least one solution. |
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Answer» Correct Answer - Eight values We know that `a cos theta+b sin theta=c` has solution only when `|c| le sqrt(a^(2)+b^(2))`. Then for the given equation, we must have `|2k+1| le sqrt(74)` `rArr -sqrt(74) lt 2k +1 lt sqrt(74)` `rArr -8 lt 2k +1 lt 8` (for integral solutions) `rArr k=-4, -3, -2, -1, 0, 1, 2, 3` Thus, eight values of k will satisfy the given inequality. |
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| 132. |
If the sum of all the solutions of the equation `8 cosx.(cos(pi/6+x)cos(pi/6-x)-1/2)=1` in `[0,pi]` is `k pi` then k is equal toA. `20//9`B. `2//3`C. `13//9`D. `8//9` |
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Answer» Correct Answer - C `8 cos x (cos (pi/6+x) cos (pi/6-x)-1/2)=1` `rArr 8 cos x [cos^(2) pi/6- sin^(2) x-1/2]=1` `rArr 8 cos x [1/4-(1-cos^(2) x)]=1` `rArr 8cos^(3)x-6 cos x =1` `rArr 4 cos^(3) x-3 cos x=1/2` `rArr cos 3 x=1/2="cos" pi/3` `3x=2npi pm pi/3, n in Z` Since `x in [0, pi]`, possible values are `3x=pi/3, 2pi pm pi/3` `:. x=pi/9, (5pi)/9, (7pi)/9` sum of solution `=(13 pi)/9` `:. k=13/9` |
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| 133. |
The values of a for which the equation `sqrtasinx-2cosx=sqrt2+sqrt(2-a)` has solutions areA. `p gt 0`B. `p le 3`C. `1 le p le 2`D. `sqrt(5)-1 le p le 2` |
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Answer» Correct Answer - D We must have `0 le p le 2` Also `sqrt(2)+sqrt(2-p)le sqrt(p + 4)` `rArr 2sqrt(4-2p)le 2p` `rArr 4-2p le p^(2)` `rArr p^(2)+2p-4le 0` `rArr p in [sqrt(5)-1,2]` |
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| 134. |
The total number of solutions of `cos x= sqrt(1- sin 2x)` in `[0, 2pi]` is equal toA. 2B. 3C. 5D. none of these |
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Answer» Correct Answer - A `cos x=sqrt(1- sin 2x)=|sin x-cos x|` (a) `sin x lt cos x` `rArr x in [0, pi/4) uu ((5pi)/4, 2pi]` ...(i) Then the given equation is `cos x = cos x - sin x` or `sin x=0` `rArr x=2pi` [ from Eq. (i)] (b) `sin x ge cos x` `rArr x in (pi/4, (5pi)/4)` `rArr cos x=sin x-cos x` or `tan x=2` `rArr x=tan^(1-) 2` ...(ii) Hence, there are two solutions. |
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| 135. |
If `xa n dy`are positive acute angles such that `(x+y)`and `(x-y)`satisfy the equation `tan^2theta-4tantheta+1=0,`then`x=pi/6`(b) `y=pi/4`(c) `y=pi/6`(d) `y=pi/4`A. `x=pi/6`B. `y=pi/4`C. `y=pi/6`D. `x=pi/4` |
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Answer» Correct Answer - C::D `(x+y)` and `(x-y)` satisfy the equation `tan^(2) theta-4 tan theta+1=0`. Thus, `tan (x+y)+tan (x-y)=4` and `tan (x+y)tan (x-y)=1` `:. Tan2x=tan ((x+y)+(x-y))` or `tan 2x= (tan (x+y)+tan (x-y))/(1-tan (x+y) tan (x-y))` `:. Tan2x=oo` or `2x=90^(@)` or `x=45^(@)=pi/4` `:. y=pi/6` |
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| 136. |
If `cos(x+pi/3)+cos x=a`has real solutions, thennumber of integral values of `a`are 3sum of number of integral values of `ai s0`when `a=1`, number of solutions for `x in [0,2pi]`are 3when `a=1,`number of solutions for `x in [0,2pi]`are 2A. number of integral values of a are 3B. sum of number of integral values of a is 0C. when `a=1`, number of solution for `x in [0, 2pi]` are 3D. when `a=1`, number of solutions for `x in [0, 2pi]` are 2 |
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Answer» Correct Answer - A::B::D `cos (x+pi//3)+cos x=a` or `1/2 cos x-(sqrt(3)/2) sin x + cos x =a` or `(3/2) cos x- (sqrt(3)/2) sin x=a` `rArr -sqrt((9/4+3/4)) le a le sqrt((9/4+3/4))` `rArr -sqrt(3) le a le sqrt(3)` ...(i) Hence, there are three integral values of `a=-1, 0, 1` whose sum is 0. For `a=1`, the given equation is `(sqrt(3)//2) cos x - (1//2) sin x =1//sqrt(3)`. Thus, `cos (x+pi/6)=1/sqrt(3)` `rArr x+pi/6=2n pi pm alpha`, where `alpha=cos^(-1) (1/sqrt(3))` `rArr x=2npi - pi/6 pm alpha` Hence, the solution for `a=1` in `[0, 2pi]` are `cos^(-1) (1//sqrt(3))-11pi//6, 11pi//6-cos^(-1) (1//sqrt(3))`. |
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| 137. |
Solve : `2cos^(2)x+3sinx=0`. |
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Answer» We have `2cos^(2)x+3sinx=0` `rArr2(1-sin^(2)x)+3sinx=0` `rArr2sin^(2)x-3sinx-2=0` `rArr2sin^(2)x-4sinx+sinx-2=0` `rArr2sinx(sinx-2)+(sinx-2)=0` `rArr(sinx-2)(2sinx+1)=0` `rArr(sinx-2)=0or(2sinx+1)=0` `rArrsinx=2or(2sinx+1)=0` `2sinx+1=0[becausesinx=2` is not possible] `rArrsinx=-(1)/(2)=-"sin"(pi)/(6)=sin(pi+(pi)/(6))="sin"(7pi)/(6)` `rArrsinx="sin"(7pi)/(6)` `rArrx={npi+(-1)^(n)*(7pi)/(6)}`, where `ninI`. Hence , the solution is given by `x={npi+(-1)^(n)*(7pi)/(6)}`, where `ninI`. |
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| 138. |
If `x+y=2pi//3` and `sin x//sin y=2`, then theA. number of values of `x in [0, 4pi]` are 4B. number of values of `x in [0, 4pi]` are 2C. number of values of `y in [0, 4pi]` are 4D. number of values of `y in [0, 4pi]` are 8 |
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Answer» Correct Answer - A::C `x+y=2pi//3` or `y=(2pi//3)-x` `:. sin x=2 sin ((2pi)/3-x)` `=2[(sqrt(3)/2)cos x+(1/2) sin x]` `=sqrt(3) cos x + sin x` `rArr cos x =0` `rArr x=npi+pi/2, n in Z` `rArr y=(2pi)/3- n pi-pi/2=pi/6-n pi` Hence, for `x in [0, 4pi], x=pi//2, 3pi//2, 5pi//2, 7pi//2` and for `y in [0, 4pi], y=pi//6, 7pi//6, 13pi//6, 19pi//6` |
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| 139. |
If `0 le x le 2pi`, then the number of real values of x, which satisfy the equation `cos x + cos 2x + cos 3x + cos 4x=0`, isA. 5B. 7C. 9D. 3 |
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Answer» Correct Answer - B `cos x + cos 2x + cos 3x + cos 4x=0, 0 le x lt 2pi` `rArr (cos x + cos 4x) + (cos 2x + cos 3x)=0` `rArr 2 "cos"(5x)/2 "cos" (3x)/2+2 cos (5 x)/2 "cos" x/2 =0` `rArr 2 "cos" (5x)/2 [2 cos x cos x/2]=0` `rArr "cos" (5x)/2 =0` or `cos x=0` or `"cos" x/2 =0` `rArr x=((2n+1)pi)/5` or `x=(2n+1) pi/2` or `x=(2n+1) pi, n in Z` `rArr x={pi/5, (3pi)/5, pi, (7pi)/5, (9pi)/5, pi/2, (3pi)/2}` `:.` Number of solution is `7` |
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| 140. |
The expression `cos 3 theta + sin 3 theta + (2 sin 2 theta-3) (sin theta- cos theta)` is positive for all `theta` inA. `(2 n pi-(3pi)/4, 2npi+pi/4), n in Z`B. `(2npi-pi/4, 2npi +pi/6), n in Z`C. `(2npi-pi/3, 2npi +pi/3), n in Z`D. `(2npi-pi/4, 2npi+(3pi)/4), n in Z` |
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Answer» Correct Answer - A::B `cos 3 theta+sin 3 theta+(2 sin 2 theta-3) (sin theta - cos theta) gt 0` or `4(cos theta - sin theta) (cos^(2) theta+ sin theta cos theta+ sin^(2) theta-sin theta cos theta) gt 0` or `-4sqrt(2) sin (theta - pi/4) gt 0` or `sin (theta - pi/4) lt 0` `rArr (2n-1) pi lt theta - pi/4 lt 2npi, n in I` `rArr 2n pi - (3pi)/4 lt theta lt 2n pi + pi/4, n in I` |
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| 141. |
If `sin A = sin B and cos A = cos B,` find all the values of A in terms of B. |
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Answer» `sin A- sin B=0` and `cos A-cos B=0` `rArr 2"sin" (A-B)/2 "cos" (A+B)/2=0` and `2 "sin" (A+B)/2 "sin" (B-A)/2 =0` We observe that the common factor gives `"sin" (A-B)/2=0`. Thus, `(A-B)/2 = npi, n in Z` or `A-B=2n pi, n in Z` or `A=2n pi +B, n in Z` |
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| 142. |
The number of ordered 5-tuple `(u, v, w, x, y)` where `(u, v, w, x, y in [1, 11])` which satisfy the inequality `2^(sin^2u+3cos^2v).3^(sin^2w+cos^2x).5^(cos^2y)>=720` isA. 216B. 246C. 432D. 432 |
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Answer» Correct Answer - C Given `2^(sin^(2)u+3cos^(2)v).3^(sin^(2)w + cos^(2)x).5^(cos^(2)y)ge 2^(4).3^(2).5^(1)` The equality holds when `sin^(2)u=sin^(2)w=cos^(2)v=cos^(2)x` `=cos^(2)y=1` `therefore u,w in {(pi)/(2),(3pi)/(2),(5pi)/(2),(7pi)/(2)}` and `x, y, v in {pi, 2pi, 3pi}` Number of ordered 5-tuples `=4^(2)xx3^(3)=432` |
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| 143. |
One of the general solutions of `sqrt(3) cos theta -3 sin theta =4 sin 2 theta cos 3 theta` isA. `m pi+pi//18, m in Z`B. `mpi//2 +pi//6, AA m in Z`C. `m pi//3 + pi//18, m in Z`D. none of these |
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Answer» Correct Answer - C 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` or `cos (theta+pi//6)= sin 5 theta=cos (pi//2-5 theta)` `rArr theta+pi//6=2n pi pm (pi//2-5 theta), n in Z` `rArr theta=(n pi//3)+(pi//18)` or `theta=(-n pi//2)+(pi//6), AA n in Z` |
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| 144. |
The general solution of the equation, `2cot.(theta)/(2)=(1+cot theta)^(2)` is `(n in Z)`A. `n pi+(-1)^(n)(pi)/(4)`B. `n pi+(-1)^(n)(pi)/(6)`C. `n pi+(-1)^(n)(pi)/(3)`D. none of these |
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Answer» Correct Answer - B `2cot.(theta)/(2)=(1+cot thetas)^(2)` `rArr (2.2 cos theta//2.cos theta//2)/(2 sin theta//2.cos theta//2)=(1+cos theta)^(2)` `rArr (2(1+cos theta))/(sin theta)=cosec^(2)theta + 2 cos theta` `rArr sin theta =1//2` `rArr theta = n pi +(-1)^(n)(pi)/(6), n in Z` |
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| 145. |
Solve `cos 2x=|sin x|, x in (-pi/2, pi)`. |
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Answer» Correct Answer - `x=-pi/6, pi/6, (5pi)/6` If `sin x gt 0`, then we have `2 sin^(2)x+sin x -1 =0` `rArr sin x=1/2`, `:. x=pi/6, (5pi)/6` If `sin x lt 0` then we have `2sin^(2) x- sin x-1=0` `rArr sinx=-1/2` `:. x=-pi/6` |
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| 146. |
Let `tanx-tan^2x >0`and `|2s inx| npi,n in Z`(b) `x > npi-pi/6,n in Z``xA. `x gt npi, n in Z`B. `x gt npi-pi//6, n in Z`C. `x lt n pi-pi//4, n in Z`D. `x lt npi+pi//6, n in Z` |
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Answer» Correct Answer - A::D `tan x-tan^(2) x gt 0` or `tan x(tan x-1) lt 0` or `0 lt tan x lt 1` or `0 lt x lt pi//4` `rArr n pi lt x lt n pi+pi//4, n in Z` (generalizing) `|sin x| lt 1/2 rArr -1/2 lt sin x lt 1/2` `rArr -pi//6 lt x lt pi//6` `rArr -pi//6+n pi lt x lt pi/66+n pi, n in Z` (generalizing) Then the common values are `npi lt x lt npi+pi//6`. |
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| 147. |
Which of the following set of values of x satisfies the equation `2^((2 sin^(2) x-3sinx +1))+2^((2-2 sin^(2)x+3 sin x))=9` ?A. `x=n pi pm pi/6, n in I`B. `x= n pi pm pi/3, n in I`C. `x=n pi, n in I`D. `x=2n pi +pi/2, n in I` |
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Answer» Correct Answer - A::D `2^((2sin^(2)x-3 sin x+1))+2^(3-(2 sin^(2)x-3 sin x+1))=9` Let `2^((2 sin^(2)x-3 sin x+1))=t` `rArr t+8/t=9` or `t^(2) -9t+8=0` or `t=1, 8` `rArr 2 sin^(2) x-3 sin x+1=3` or `2 sin^(2) x-3 sin x+1=0` `rArr sin x = -1/2, sin x = 1/2, sin x =1` |
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| 148. |
Find the values of `theta` which satisfy `r sin theta=3` and `r=4 (1 + sin theta), 0 |
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Answer» We have `r sin theta =3` and `r=4 (1+ sin theta)` Eliminating r, we get `4 sin^(2) theta +4 sin theta -3 =0` `:. (2 sin theta-1) (2 sin theta +3)=0` `rArr sin theta =1/2, -3/2` (not possible) `rArr theta=pi/6, (5 pi)/6` Thus, we have two solutions. |
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| 149. |
`sin 3 alpha = 4 sin alpha sin(x + alpha) sin(x-alpha)`A. `n pi pm pi//4, AA n in Z`B. `n pi pm pi//3, AA n in Z`C. `n pi pm pi//9, AA n in Z`D. `n pi pm pi//12, AA n in Z` |
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Answer» Correct Answer - B We have `sin 3 alpha =4 sin alpha (sin^(2) x-sin^(2) alpha)` or `3 sin alpha -4 sin^(3) alpha =4 sin alpha sin^(2) x-4 sin^(3) alpha` or `3 sin alpha =4 sin alpha sin^(2)x` If `sin alpha ne0, sin^(2) x=3//4=(sqrt(3)//2)^(2)= sin^(2) (pi//3)`, Therefore `x= n pi pm pi//3, AA n in Z` If `sin alpha =0`, i.e., `alpha =n pi`, equation becomes an identity. |
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| 150. |
If `cos 2theta=(sqrt(2)+1)(cos theta-(1)/(sqrt(2)))`, then the general value of `theta(n in Z)`A. `2n pi pm (pi)/(6)`B. `n pi+(pi)/(12)`C. `n pi+(7pi)/(36)`D. none of these |
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Answer» Correct Answer - D `cos 2theta = (sqrt(2)+1)(cos. theta-(1)/(sqrt(2)))` `rArr (2 cos^(2)theta-1)=(sqrt(2)+1)/(sqrt(2))(sqrt(2)cos theta -1)` `rArr (sqrt(2)cos theta -1)(sqrt(2)cos theta +1)=(sqrt(2)+1)/(sqrt(2))(sqrt(2)cos theta -1)` `rArr cos. theta =(1)/(sqrt(2))` or `cos. theta =(1)/(2)` `rArr theta = 2n pi pm (pi)/(4)` or `theta = 2n pi pm(pi)/(3), n in Z` |
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