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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.
| 151. |
The number of distinct real roots of the equation `sin pi x=x^(2)-x+(5)/(4)` is |
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Answer» Correct Answer - B `sin pi x = x^(2)-x + (5)/(4)` `rArr sin pi x = (x-1//2)^(2)+1` `rArr sin pi x le 1` and `x^(2)-x + (5)/(4) ge 1` `therefore sin pi x = 1` and `x = (1)/(2)` |
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| 152. |
One of the general solutions of `sqrt(3) cos theta -3 sin theta =4 sin 2 theta cos 3 theta` isA. `(3n pm 1) pi//12, AA n in Z`B. `(4n pm 1) pi//9, AA n in Z`C. `(3n pm 1) pi//9, AA n in Z`D. `(3n pm 1) pi//3, AA n in Z` |
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Answer» Correct Answer - C We have `4 sin theta sin 2 theta sin 4 theta =3 sin theta -4 sin^(3) theta` or `sin theta [4 sin 2 theta sin 4 theta -3 +4 sin^(2) theta]=0` or `sin theta [2(cos 2 theta - cos 6 theta)-3+2 (1- cos 2 theta)]=0` or `sin theta (-2 cos 6 theta -1)=0` or `sin theta =0 or cos 6 theta =-1//2` `rArr theta = n pi or 6 theta =2 n pi pm 2 pi//3, AA n in Z` `= n pi or theta =(3n pm 1) pi//9, AA n in Z` |
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| 153. |
Solve `sin 3 theta-sin theta=4 cos^(2) theta-2`. |
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Answer» Correct Answer - `theta=(2n+1) pi/4, n in Z or theta =(4n+1) pi/2, n in Z` `sin 3 theta- sin theta=4 cos^(2) theta-2` `rArr 2 sin theta cos 2 theta=2 cos 2 theta` `rArr cos 2 theta=0` or ` sin theta=1` `rArr 2 theta=(2n+1) pi/2, n in Z` or `theta=(4n+1) pi/2, n in Z` or `theta=(2n+1) pi/4, n in Z`, or `theta=(4n+1) pi/2, n in Z` |
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| 154. |
Let `S_(1)` be the set of all those solution of the equation `(1+a) cos theta cos(2 theta-b)=(1+a cos 2 theta) cos (theta -b)` which are independent of a and b and `S_(2)` be the set of all such solutions which are dependent on a and b. Then The set `S_(1)` and `S_(2)` areA. `{n pi, n in Z}` and `1/2 {n pi +(-1)^(n) sin^(-1) (a sin b) +b, n in Z}`B. `{n pi/2, n in Z}` and `{n pi+(-1)^(n) sin^(-1) (a sin b), n in Z}`C. `{n pi/2, n in Z}` and `{n pi +(-1)^(n) sin^(-1) (a/2 sin b), n in Z}`D. none of these |
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Answer» Correct Answer - A `(1+a) cos theta cos (2 theta-b)=(1+a cos 2 theta) cos (theta-b)` `rArr cos theta cos (2 theta-b)+a cos theta cos (2 theta-b)` `= cos (theta-b) + a cos 2 theta cos (theta-b)`. `rArr cos (3 theta -b)+ cos (theta-b)+a { cos (3 theta-b)+ cos (theta-b)}` `=2 cos (theta-b)+a {cos (3 theta -b)+cos (theta+b)}` `rArr cos (3 theta-b) - cos (theta-b)=a cos (theta+b)-a cos (theta-b)` `rArr 2 sin (2 theta-b) sin theta =2 a sin theta sin b` `rArr sin theta=0 or sin (2 theta-b) =a sin b` if `sin theta=0` then `theta=n pi, n in Z rArr S_(1) = {n pi|n in Z|}` Also, `2 theta-b=n pi +(-1)^(n) sin^(-1) (a sin b), n in Z` `rArr S_(2)=1/2 {n pi +(-1)^(n) sin^(-1) (a sin b)+b, n in Z}` |
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| 155. |
Let `S_(1)` be the set of all those solution of the equation `(1+a) cos theta cos(2 theta-b)=(1+a cos 2 theta) cos (theta -b)` which are independent of a and b and `S_(2)` be the set of all such solutions which are dependent on a and b. Then Condition that should be imposed on a and b such that `S_(2)` in non empty isA. `|a/2 sin b| lt 1`B. `|a/2 sin b| le 1`C. `|a sin b| le 1`D. none of these |
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Answer» Correct Answer - C `sin (2 theta-b)=a sin b rArr |a sin b| le 1` |
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| 156. |
Which of the following sets can be the subset of the general solutionof `1+c a s3x=2cos2x(n in Z)?``npi+pi/3`(b) `npi+pi/6``npi-pi/6`(d) `2npi`A. `n pi+pi/3`B. `n pi+pi/6`C. `n pi- pi/6`D. `2 n pi` |
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Answer» Correct Answer - B::C::D `1+cos 3x=2 cos 2x` or `1+4 cos^(2) x-3 cos x=2(2 cos^(2) x-1)` or `4 cos^(3) x-4 cos^(2) x-3 cos x+3=0` Let `cos x=y`, we have `4y^(3)-4y^(2)-3y+3=0` or `4y^(2) (y-1)-3(y-1)=0` or `(y-1) (4y^(2)-3)=0` `rArr y=1 or y^(2)=3/4` `rArr cos x=1 or cos^(2) x=3/4` `rArr cos x =1 or cos^(2) x= cos^(2) pi//6` `rArr x=2 n pi or x= n pi pm (pi//6), n in Z` |
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| 157. |
The general solution of `4 sin^4 x + cos^4x= 1` isA. `n pi pm alpha//2, alpha=cos^(-1) (1//5), AA n in Z`B. `n pi pm alpha//2, alpha=cos^(-1) (3//5), AA n in Z`C. `2 n pi pm alpha//2, alpha=cos^(-1) (1//3), AA n in Z`D. none of these |
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Answer» Correct Answer - A `4 sin^(4) x+cos^(4) x=1` `rArr (2 sin^(2) x)^(2) +1/4 (2 cos^(2) x)^(2)=1` or `(1-cos 2x)^(2) +1/4 (1+ cos 2x)^(2)=1` or `5 cos^(2) 2x-6 cos 2x+1=0` or `(cos 2x-1)(5 cos 2x-1)=0` or `cos 2x=1 or cos 2x=1/5` `rArr 2x=2npi or 2x=2npi pm alpha`, where `alpha=cos^(-1) (1/5), AA n in Z` |
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| 158. |
Find general value of `theta` which satisfies both `sin theta = -1//2` and `tan theta=1//sqrt(3)` simultaneously. |
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Answer» Here `sin theta lt 0` and `tan theta gt =0`, then `theta` lies in the third quadrant. Now `sin theta=-1/2` `rArr theta =pi+pi/6=(7pi)/6` Generalizing, we have `theta=2npi+(7pi)/6, n in Z`. |
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| 159. |
Find the general solution of `(1-2 cos theta)^(2)+(tan theta +sqrt(3))^(2)=0`. |
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Answer» Correct Answer - `theta=2n pi+(5 pi)/3, n in Z` `(1-2 cos theta)^(2)+(tan theta +sqrt(3))^(2)=0` `rArr (1-2 cos theta)^(2)=0` and `(tan theta +sqrt(3))^(2)=0` `rArr cos theta=1/2` and `tan theta=-sqrt(3)` Hence, `theta` is in the fourth quadrant, and `theta=2pi-pi/3=(5pi)/3` Hence, the general solution is `theta=2 npi+(5pi)/3, n in Z`. |
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| 160. |
Solve `log_(|sin x|) (1+cos x)=2`. |
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Answer» Correct Answer - No solution `log_(|sin x|) (1+cos x)=2` `rArr 1+cos x=sin^(2) x` `rArr 1+cos x=(1-cos x) (1+ cos x)` `rArr 1+cos x =0` or `1-cos x=1` `rArr cos x=-1` or `cos x=0` `rArr sin x=0` or `|sin x|=1` Both of these are not possible as base of logarithm cannot be either 0 or 1. |
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| 161. |
If both the distinct roots of the equation `|sinx|^2+|sinx|+b=0in[0,pi]`are real, then the values of `b`are`[-2,0]`(b) `(-2,0)``[-2,0]`(d) `non eoft h e s e`A. `[-2, 0]`B. `(-2, 0)`C. `[-2, 0)`D. none of these |
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Answer» Correct Answer - B Given that `|sin x|^(2)+|sin x|+b=0` `rArr |sin x|=(-1 pm sqrt(1-4b))/2` `rArr 0 le (-1 pm sqrt(1-4b))/2 lt 1" "rArr -2 lt b lt 0` |
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| 162. |
All the permissible value of b ,a=sin(2x-b)if `a=0` and `x=S_(2)` is a subset of `(0, pi)` are given byA. `b in (-n pi, 2npi), n in Z`B. `b in (-n pi, 2pi - npi), n in Z`C. `b in (-n pi, npi), n in Z`D. none of these |
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Answer» Correct Answer - B `a=0 rArr sin (2 theta-b)=0` `rArr 2 theta-b=n pi, n in Z` `rArr theta=(n pi+b)/2, n in Z, S_(2)=(n pi +b)/2, n in Z` `S_(2)` is subset of `(0, pi)` `:. 0 lt (n pi +b)/2 lt pi, n in Z` `:. 0 lt n pi +b lt 2pi rArr -n pi lt b lt -n pi +2pi` `rArr b in (-n pi, 2pi-npi), n in Z` |
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| 163. |
Number of roots of the equation`2^(tan(x-pi/4))-2(0. 25)^sin^(3((x-pi/4))/(cos2x))+1=0,i s_______` |
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Answer» `(sin^(2) (x-pi/4))/(cos 2x)=(1/2 (sin x-cos x)^(2))/(cos^(2) x-sin^(2) x)` `=(-1/2 (sin x - cos x))/(cos x + sin x)=-1/2 tan (x-pi/4)` Given equation reduces to `2^(tan(x-pi/4)-2(0.25)^(-1/2 tan(x-pi/4))+1=0` `rArr 2^(tan(x-pi/4)=1` `rArr x=pi//4`, which is not possible as `cos 2x=0` for this value for `x`, which is not defining the original equation. |
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| 164. |
Solve `2 sin^(2) x-5 sin x cos x -8 cos^(2) x=-2`. |
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Answer» If `cos x=0`, then `2 sin^(2) x=-2 or sin^(2) x=-1`, which is not possible. Clearly, `cos x ne 0` Now, `2 sin^(2) x-5 sin x cos x -8 cos^(2) x=-2` Dividing both sides by `cos^(2)x`, we get `2 tan^(2) x-5 tan x-8=-2 sec^(2) x` or `2 tan^(2) x-5 tan x-8+2 (1+ tan^(2) x)=0` or `4 tan^(2) x-5 tan x-6=0` or `(tan x-2) (4 tan x+3)=0` Now, `tan x-2 =0` Let `tan x=2=tan alpha` `rArr x=npi +alpha = npi +tan^(-1) 2, n in Z` or `4 tan x+3=0` `rArr tan x=-3/4 = tan beta` (let) `rArr x=mpi + tan^(-1) (-3/4)`, where `m in Z` |
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| 165. |
A general solution of `tan^(2) theta+ cos 2 theta=1` is `(n in Z)`A. `n pi-pi/4`B. `2npi+pi/4`C. `npi+pi/4`D. `n pi` |
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Answer» Correct Answer - A::C::D We have `tan^(2) theta=1-cos 2 theta=2 sin^(2) theta` or `cosec^(2) theta tan^(2) theta=2` or `(1+cot^(2) theta) tan^(2) theta=2` or `tan^(2) theta+1=2` or `tan^(2) theta=1` or `tan theta = pm 1` `rArr theta=n pi pm pi/4, n in Z` Moreover, `tan^(2) theta=2 sin^(2) theta` `rArr sin^(2) theta=0rArr theta = npi` |
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| 166. |
Let `(b cos x)/(2 cos 2x-1)=(b + sin x)/((cos^(2) x-3 sin^(2) x) tan x), b in R`. Equation has solutions ifA. `b in (-oo, 1/2)-{-1, 0, 1/3}`B. `b in (-oo, 1)-{-1, 0, 1/3}`C. `b in R-{-1, 0, 1/3}`D. none of these |
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Answer» Correct Answer - A `(b cos x)/(2 cos 2x-1)=(b+sin x)/((cos^(2) x-3 sin^(2) x) tan x)` `{:((1),2 cos 2x -1 ne 0,rArr,n ne npi pm pi/6", " n in Z),((2),tan x ne 0,rArr,x ne npi", " n in Z),((3),cos^(2) x-3 sin^(2) x ne 0,rArr,x ne npi pm pi/6", "n in Z):}` Also `2 cos 2x-1 =2 (cos^(2) x-sin^(2) x)-(cos^(2) x+sin^(2)x)` `= cos^(2) x-3 sin^(2) x` Now, the given equation reduces to `b sin x=b + sin x` `rArr sin x=b/(b-1)` Now, `-1 le sin x le 1` `rArr -1 le b/(b-1) le 1` `rArr b/(b-1) +1 ge 0` and `b/(b-1) -1 le 0` `rArr (2b-1)/(b-1) ge 0` and `1/(b-1) le 0` `rArr b in (-oo, 1/2] uu (1, oo)` and `b in (-oo, 1)` `sin x ne 0, pm 1/2 or b ne 0, -1, 1/3` When `b=1/2` then `sin x=-1`, which is not possible, as `tan x` is not defined for this value of `sin x`. So, `b in (-oo, 1/2)-{-1, 0, 1/3}` For any other value of b, `sin x` takes two values for `x in (0, pi)`. |
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| 167. |
Find common roots of the equations `2sin^2x+sin^2 2x=2a n dsin2x+cos2x=tanxdot` |
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Answer» We have `2 sin^(2) x+ sin^(2) 2x=2` ...(i) and `sin 2x+cos 2x=tan x` ...(ii) Solving Eq. (i), `sin^(2) 2x=2 cos^(2) x` `rArr 4 sin^(2) x cos^(2) x =2 cos^(2) x` `rArr cos^(2) x (2 sin^(2) x-1)=0` `rArr 2 cos^(2) x cos 2x=0` `rArr cos x=0 or cos 2x=0` `rArr x=(2n+1) pi/2 or x=(2n+1) pi/4, n in Z` ...(iii) Now solving Eq. (ii), `(2 tan x+1-tan^(2) x)/(1+tan^(2) x)=tan x` `rArr tan^(3) x+ tan^(2) x-tan x-1=0` `rArr (tan^(2) x-1) (tan x+1) =0` `rArr tan x= pm 1` `rArr x= n pi pm pi/4, n in Z` ...(iv) From Eqs. (iii) and (iv) , common roots are `(2n+1) pi/4`. |
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| 168. |
Assume that `theta` is a rational multiple of `pi` such that `cos theta` is a distinct rational. The number of values of `cos theta` isA. 3B. 4C. 5D. 6 |
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Answer» Correct Answer - C `theta=k pi, k=p/q, p, q in I, q ne 0` `cos kpi` is a rational. Hence, `k=0, 1, 1//2, 1//3, 2//3` There are five values of `cos theta` for which `cos theta` is rational. |
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| 169. |
If `n_(1)` denotes the maximum number of roots of `sin theta = k_(1)` in `[0,2pi]` and `n_(2)` denotes the maximum number of roots of `cos theta=k_(2)` in `[0,2pi]`, thenA. `n_(1)+n_(2)=5`B. `n_(1)+n_(2)=4`C. `n_(1)+n_(2)=6`D. `n_(1)+n_(2)=3` |
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Answer» Correct Answer - A `sin theta = k_(1)` has maximum roots if `k_(1) = 0` , so `theta = pi, pi, 2pi` `cos theta = k_(2)` has maximum roots if `k_(2) ne pm 1`, so there are two values of `theta` `therefore n_(1)=3` and `n_(2)=2` |
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| 170. |
If `sin^(3) theta+sin theta cos theta+ cos^(3) theta=1`, then `theta` is equal to `(n in Z)`A. `2 n pi`B. `2n pi +pi/2`C. `2npi-pi/2`D. `npi` |
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Answer» Correct Answer - A::B `(sin^(3) theta+cos^(3)theta)-(1-sin theta cos theta)=0` or `(sin theta + cos theta)(1-sin theta cos theta)-(1-sin theta cos theta)=0` or `(1-sin theta cos theta) (sin theta + cos theta-1)=0` If `sin theta cos theta=1` `rArr 2 sin theta cos theta =2` `rArr sin 2 theta=2` (not possible) `rArr sin theta + cos theta=1` `rArr cos (theta-pi/4)=1/sqrt(2)` `rArr theta-pi/4=2npi pm pi/4, n in Z` `rArr theta=2npi` or `2npi+pi/2` |
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| 171. |
Solve `sin^(2) theta gt cos^(2) theta`. |
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Answer» Correct Answer - `n pi+pi/4 lt theta lt n pi + (3pi)/4, n in Z` `sin^(2) theta gt cos^(2) theta` `rArr cos 2 theta lt 0` `rArr (pi//2) lt 2 theta lt (3pi //2)` or `(pi//4) lt theta lt (3pi//4)` Taking general values, i.e., adding `2 n pi`, we get `2npi+pi//2 lt 2 theta lt 2 n pi +3 pi//2, n in Z` or `n pi + pi//4 lt theta lt n pi +3 pi//4` |
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| 172. |
The number of values of `theta` satisfying `4 cos theta+3 sin theta=5` as well as `3 cos theta + 4 sin theta = 5` isA. oneB. twoC. zeroD. none of these |
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Answer» Correct Answer - C On adding given equations, we get `7(cos theta+ sin theta)=10` or `cos theta+ sin theta=10/7` But maximum value of `cos theta+ sin theta=sqrt(1^(2)+1^(2))=sqrt(2) lt 10/7`. So, no `theta` is possible. |
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| 173. |
Solve `2 sin theta + 1=0`. |
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Answer» Correct Answer - `theta=n pi +(-1)^(n+1) pi/6, n in Z` `2 sin theta+1=0` `rArr sin theta =-1/2 = sin ((-pi)/6)` `rArr theta=n pi +(-1)^(n) ((-pi)/6), n in Z` `=n pi +(-1)^(n+1) pi/6, n in Z` |
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| 174. |
Let `(b cos x)/(2 cos 2x-1)=(b + sin x)/((cos^(2) x-3 sin^(2) x) tan x), b in R`. Equation has solutions ifA. InfiniteB. depends upon the value of bC. twoD. none of these |
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Answer» Correct Answer - C `(b cos x)/(2 cos 2x-1)=(b+sin x)/((cos^(2) x-3 sin^(2) x) tan x)` `{:((1),2 cos 2x -1 ne 0,rArr,n ne npi pm pi/6", " n in Z),((2),tan x ne 0,rArr,x ne npi", " n in Z),((3),cos^(2) x-3 sin^(2) x ne 0,rArr,x ne npi pm pi/6", "n in Z):}` Also `2 cos 2x-1 =2 (cos^(2) x-sin^(2) x)-(cos^(2) x+sin^(2)x)` `= cos^(2) x-3 sin^(2) x` Now, the given equation reduces to `b sin x=b + sin x` `rArr sin x=b/(b-1)` Now, `-1 le sin x le 1` `rArr -1 le b/(b-1) le 1` `rArr b/(b-1) +1 ge 0` and `b/(b-1) -1 le 0` `rArr (2b-1)/(b-1) ge 0` and `1/(b-1) le 0` `rArr b in (-oo, 1/2] uu (1, oo)` and `b in (-oo, 1)` `sin x ne 0, pm 1/2 or b ne 0, -1, 1/3` When `b=1/2` then `sin x=-1`, which is not possible, as `tan x` is not defined for this value of `sin x`. So, `b in (-oo, 1/2)-{-1, 0, 1/3}` For any other value of b, `sin x` takes two values for `x in (0, pi)`. |
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| 175. |
The real roots of the equation `cos^7x+sin^4x=1`in the interval `(-pi,pi)`are __________, ________, and ________ |
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Answer» Correct Answer - `x= 0, pm pi/2` `cos^(7) x=1-sin^(4) x= (1-sin^(4) x)(1+sin^(2) x)` or `cos^(7) x = cos^(2) x(2-cos^(2) x)` or `cos^(2) x(cos^(5)x+cos^(2) x-2)=0` or `cos x =0` `rArr x=pm pi//2` in the given interval or `cos^(5) x+cos^(2)x-2=0`, which holds when `cos x=1`, hence `x=0`. Thus, there are total three solutions : `0, pm pi//2`. |
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| 176. |
Number of solutions of the equation `sin x + cos x-2sqrt(2) sin x cos x=0` for `x in [0, pi]` isA. 3B. 0C. 1D. 2 |
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Answer» Correct Answer - D `sin x + cos x - 2 sqrt(2) sin x cos x=0` `rArr sin x+cos x=2sqrt(2) sin x cos x` `rArr 1+sin 2x=2 sin^(2) 2x` `rArr 2 sin^(@) 2x- sin 2x-1=0` `rArr (2 sin 2x+1) (sin 2x-1)=0` `rArr sin 2x=-1//2 or sin 2x=1` `rArr 2x=11 pi//6, pi//2` `rArr x=11pi//12, pi//4` |
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| 177. |
Let `0A. `6pi`B. `7pi`C. `8pi`D. `4pi` |
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Answer» Correct Answer - A `3+3 cos theta =2 -2 cos^(2) theta` `rArr 2 cos^(2) theta+3 cos theta+1=0` `rArr (2 cos theta+1) (cos theta+1) =0` `:. Cos theta=-1 or cos theta=-1/2` If `cos theta=-1` then `theta=pi, 3 pi, 5 pi, 7 pi, 9pi`, ... If `cos theta=-1//2` then `theta=(2pi)/3, (4pi)/3, (8pi)/3, (10 pi)/3, (14 pi)/3, ...` Solution in increasing order are `(2pi)/3 lt pi lt (4pi)/3 lt (8 pi)/3 lt 3pi lt (10 pi)/3 lt (14 pi)/3 lt 5pi`, ... `:. theta_(3)+theta_(7)=(4pi)/3+(14 pi)/3=6pi` |
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| 178. |
`sin^2 n theta- sin^2 (n-1)theta= sin^2 theta` where `n` is constant and `n != 0,1` |
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Answer» Correct Answer - `theta=k pi, (p pi)/(n-1), ((2q+1)pi)/(2n); k, p, q, n in Z` `sin^(2)n theta-sin^(2) (n-1) theta=sin^(2) theta` or `sin(n theta-(n-1) theta) sin (n theta+(n-1) theta)= sin^(2) theta` or `sin theta sin ((2n-1) theta)=sin^(2) theta` or `sin theta=0` or `sin((2n-1) theta)= sin theta` `rArr theta=k pi`, or `(2n-1) theta=m pi +(-1)^(m) theta, k m in Z` `rArr theta= kpi`, or `(2n-1) theta=mpi+theta`, when `m` is even and `(2n-1) theta=mpi - theta`, when `m` is odd `rArr theta=kpi`, or `theta=mpi//(2n-2)`, when `m` is even and `theta=mpi//2n`, when `m` is odd `rArr theta=kpi` or `theta=p pi//(n-1)`, and `theta=(2q+1)pi//2n`, where `k, p, n q in Z` |
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| 179. |
Number of values of `p`for which equation `sin^3x+1+p^3-3psinx=0(p >)`has a root is ________ |
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Answer» Correct Answer - 1 `sin^(3) x+p^(3)+1=3p sin x` `rArr (sin x p+1) (sin^(2) x+1+p^(2)-sin x-p-p sin x)=0` therefore, either `sin x+p+1=0rArr p rArr p=-(1+sin x)` or `sin x =1=p` Hence, only value of `p(p gt 0)` is possible which is given by `p=1`. |
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| 180. |
Number of solutions (s) of the equation `(sin x)/(cos 3x) +(sin 3x)/(cos 9x)+(sin 9x)/(cos 27 x)=0` in the interval `(0, pi/4)` is __________. |
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Answer» Correct Answer - 6 `(sin x)/(cos 3x)+(sin 3x)/(cos 9x)+(sin 9x)/(cos 27 x)=0` or `(2 sin x cos x)/(2 cos 3x cos x)+(2 sin 3x cos 3x)/(2 cos 9x cos 3x)+(2 sin 9x cos 9x)/(2 cos 27 x cos 9x)=0` or `(sin (3x-x))/(2 cos 3x cos x)+(sin (9x-3x))/(2 cos 9x cos 3x)+(sin (27x -9x))/(2 cos 27 x cos 9x)=0` or `(tan 3x-tan x)+(tan 9x-tan 3x)+(tan 27x-tan 9x)=0` or `tan 27 x-tan x=0` or `tan x=tan 27 x` `rArr 27x=n pi +x, n in I` or `x=(npi)/26, n in I`. or `x=pi/26, (2pi)/26, (3pi)/26, (4pi)/26, (5pi)/26, (6pi)/26` Hence, there are six solutions. |
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| 181. |
The general solution of `cos x cos 6x =-1` isA. `x=(2n+1) pi, n in Z`B. `x=2n pi, n in Z`C. `x= n pi, n in Z`D. none of these |
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Answer» Correct Answer - A `cos x cos 6x=-1` or `2 cos x cos 6x=-2` or `cos 7x+ cos 5x=-2` or `cos 7x =-1 and cos 5x=-1` The value of x satisfying these two equations simultaneously and lying between 0 and `2pi` is `pi`. Therefore, the general solution is `x=2npi+pi, n in Z`. Thus, `x=(2n+1) pi, n in Z` |
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| 182. |
Number of solutions of the equation `4(cos^(2) 2x+ cos 2 x +1)+tan x (tan x-2sqrt(3))=0` in `[0, 2pi]` is |
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Answer» Correct Answer - C `(4 cos^(2) 2x+4 cos 2x+1) + (tan^(2) x-2 sqrt(3) tan x+3)=0` `rArr (2 cos 2x+1)^(2) + (tan x-sqrt(3))^(2)=0` `rArr cos 2x = -1/2 and tan x = sqrt(3)` `rArr 2 cos^(2) x-1=-1/2 and tan x=sqrt(3)` `rArr cos x = pm 1/2 and tan x = sqrt(3)` `rArr x=pi/3, (4pi)/3` `:.` Number of solution of equation `=2` |
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| 183. |
Solve `cos theta=1//3`. |
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Answer» Correct Answer - `theta=2npi pm cos^(-1) 1/3, n in Z` `cos theta=1//3` or `cos theta=cos^(-1) (cos 1//3)` `rArr theta=2n pi pm "cos"^(-1) 1/3, n in Z` |
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| 184. |
Solve `tan x+tan 2x+tan 3x = tan x tan 2x tan 3x, x in [0, pi]`. |
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Answer» We have `tan x + tan 2x +tan 3x=tan x tan 2x tan 3x` We know that `tan A+tan B+tan C=tan A tan B tan C`. iff `x=n pi, n in Z`. So form given equation, we have `x+2x+3x=n pi, n in Z` `:. x=npi//6, n in Z` `:. x=pi//6, 2 pi//6, 3pi//6, 4pi//6, 5pi//6, 6pi//6` or `x=pi//6, pi//3, pi//2, 2pi//3, 5pi//6, pi` But for `x=pi//6, 5pi//6, tan 3x` is not defined. And for `x=pi//2, tan x` is not defined. `:. x=pi//3, 2 pi//3, pi` |
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| 185. |
The sum of all the solutions in `[0, 4pi]` of the equation `tan x+ cot x+1= cos (x+pi/4)` isA. `3 pi`B. `pi//2`C. `7pi//2`D. `4pi` |
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Answer» Correct Answer - C `tan x + cot x +1 = cos (x+pi/4)` or `tan x+cot x=cos(x+pi/4)-1` Now `tan x +cot x le-2` and `cos(x+pi/4)-1 ge -2` It implies that equality holds when both are `-2`. Thus, `cos(x+pi/4)=-1` `rArr x+pi/4=(2m+1)pi, m in Z` `rArr x=(3pi)/4` or `(11pi)/4` Therefore, the sum of the solutions is `(3pi)/4+(11pi)/4=(7pi)/2` |
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| 186. |
If `log_(0.5) sin x=1-log_(0.5) cos x,` then the number of solutions of `x in[-2pi,2pi]` is |
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Answer» Correct Answer - 2 `log_(0.5) sin x=1-log_(0.5) cos x, x in [-2pi, 2pi]` `sin x gt 0` and `cos x gt 0` `sin x cos x=1//2` `:. sin 2x=1` therefore, there are 2 value of x. |
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| 187. |
Solve `cos theta+cos 7 theta+cos 3theta+cos 5 theta=0`, |
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Answer» Correct Answer - `theta=(n pi)/8, n in Z` `(cos theta+cos 7 theta)+(cos 3 theta+ cos 5 theta)=0` or `2 cos 4 theta (cos theta+ cos theta)=0` or `4 cos 4 theta cos 2 theta cos theta=0` or `4xx 1/(2^(3) sin theta) (sin 2^(3) theta)=0` `rArr sin 8 theta=0` or `theta=npi//8, n in Z` |
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| 188. |
Solve `3tan^(2) theta-2 sin theta=0`. |
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Answer» Correct Answer - `theta=n pi, n pi +(-1)^(n) pi/6, n in Z` `tan^(2) theta-2 sin theta=0` or `3(sin^(2) theta)/(cos^(2) theta)-2 sin theta=0, cos theta ne 0` or `3 sin^(2) theta-2 sin theta (1- sin^(2) theta)=0` or `3 sin^(2) theta-2 sin theta (1- sin^(2) theta)=0` or `sin theta (2 sin^(2) theta+3 sin theta -2)=0` or `sin theta (2 sin theta-1) (sin theta+2)=0` or `sin theta=0, 1-2` (rejected) `rArr theta=npi, n pi +(-1)^(n) pi/6, n in Z` |
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| 189. |
Solve `(sin 10^(@))^(tan x+tan 3x)=tan 15^(@)+tan 30^(@)+ tan 15^(@). Tan 30^(@), x in (0, pi]`. |
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Answer» Correct Answer - `x=pi/4, (3 pi)/4, pi` `R.H.S.=1` `rArr tan x+tan 3x=0` `rArr sin 4x=0` `rArr x=(n pi)/4, n in Z` `rArr x=pi//4, 3pi//4, pi` |
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| 190. |
Number of solution(s) satisfying the equation `1/(sinx)-1/(sin2x)=2/(sin4x)`in `[0,4pi]`equals0 (b)2 (c) 4(d) 6 |
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Answer» Correct Answer - C `1/(sin x)-1/(sin 2x)=2/(sin 4x)` `rArr (sin 2x- sin x)/(sin x sin 2x)=2/(2 sin 2x cos 2x)` `rArr 2 sin 2x cos 2x-2 sin x cos 2x=2 sin x` `rArr sn 4x-sin 3x+sin x=2 sin x` `rArr sin 4x=sin 3x+sin x` `rArr 2 sin 2x cos 2x=2 sin 2x cos x` `rArr 2x=2npi pm x" "("as "sin 2x ne 0)` `rArr x=(2n pi)/3, n in I" "("as "x=2k pi" is not in domain")` `n=1, 2, 4, 5, ...` Thus, four solutions are `(2pi)/3, (4pi)/3, (8pi)/3, (10 pi)/3`. |
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| 191. |
Number of solutions of equation `2"sin" x/2 cos^(2) x-2 "sin" x/2 sin^(2) x=cos^(2) x-sin^(2) x` for `x in [0, 4pi]` isA. 6B. 8C. 10D. 12 |
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Answer» Correct Answer - C `(cos^(2) x-sin^(2) x) (2 "sin" x/2-1)=0` `cos 2x=0 or "sin" x/2=1/2` `:.` Number of solution `=8+2=10` |
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| 192. |
The number of solution the equation `cos(theta)cos(pitheta)=1`has0 (b)2 (c) 4(d) 2 |
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Answer» Correct Answer - C `cos (theta) cos (pi theta)=1` `rArr cos (theta)=1 and cos (pi theta)=1` ...(i) or `cos (theta) =-1 and cos (pi theta)=-1` ...(ii) If `cos (theta)=1 rArr theta=2 mpi` and `cos (pi theta) =1 rArr theta =2k, k in Z` which is possible only when `theta=0`. Equation (ii) is not possible for any `theta` as for `cos (theta)=-1, theta` should be odd multiple of `pi`, i.e., irrational and for `cos(pi theta)=-1, theta` should be odd integer, i.e., rational. Both the conditions cannot be satisfied. Therefore, `theta=0` is the only solution. |
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| 193. |
Find the general solutions of the following equations:`tan2theta=0`(ii) `tan(theta/2)=0`(iii) `tan(3theta)/(4)=0` |
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Answer» (i) `tan 2theta = 0` `=>tan 2theta = tan 0` `=>2theta = npi+0` `=> theta = (npi)/2` (ii)`tan (theta/2) = 0` `=>tan (theta/2) = tan 0` `=>(theta/2) = npi+0` `=> theta = 2npi` (iii)`tan ((3theta)/4) = 0` `=>tan ((3theta)/4) = tan 0` `=>((3theta)/4) = npi+0` `=> theta = (4npi)/3` |
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| 194. |
Number of roots of the equation `(3+cos"x")^2=4-2sin^8x ,x in [0,5pi]a r e`_________ |
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Answer» Correct Answer - 3 `4 le L.H.S. le 16` `2 le R.H.S. le 4` Hence, equality can occur only when both sides are 4, which is possible if `x=pi, 3pi, 5 pi`. That is, there are three solutions. |
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| 195. |
Solve `cos theta+cos 3 theta-2 cos 2 theta=0`. |
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Answer» We have `cos theta+ cos 3 theta- 2 cos 2 theta=0` or `2 cos 2 theta cos theta-2 cos 2 theta=0` or `2 cos 2 theta( cos theta-1) =0` or `cos 2 theta=0 or cos theta-1 =0` `rArr 2 theta=(2n+1) pi/2, n in Z or theta=2 mpi, m in Z` or `theta=(2n+1) pi/4, n in Z or theta =2mpi, m in Z` |
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| 196. |
If `sintheta,1,cos2theta` are in G.P., then find the general values of `theta` |
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Answer» Correct Answer - `theta=npi+(-1)^(n-1) pi/2` `1^(2)= sin theta cos 2 theta` or `1- sin theta (1-2 sin^(2) theta)=0` or `2 sin^(3) theta- sin theta+1=0` or `(sin theta+1) (2 sin^(2) theta-2 sin theta+1)=0` or `sin theta=-1` the other factor gives imaginary roots `rArr theta=n pi +(-1)^(n) (-pi/2)` `=n pi-(-1)^(n) pi/2= n pi+(-1)^(n-1) pi/2` |
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| 197. |
Let `theta in [0,4pi]`satisfy the equation `(sintheta+2)(sintheta+3)(sintheta+4)=6.`If the sum of all the values of `theta`is of the form `kpi`, then the value of `k`is6 (b)5 (c) 4(d) 2A. 6B. 5C. 4D. 2 |
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Answer» Correct Answer - B `(sin theta+2) (sin theta+3) (sin theta+4) =6` `L.H.S. ge 6` and `R.H.S. =6` Therefore, equality only holds if `sin theta=-1 or theta =3pi//2, 7 pi//2` `:.` Sum `=5pi =lt k=5` |
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| 198. |
Solve the inequality `sin2x gtsqrt(2)sin^2x+(2-sqrt(2))cos^2x` |
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Answer» Correct Answer - `x in (n pi +pi/8, n pi+pi/4), n in Z` `sin 2x gt sqrt(2) sin^(2) x+(2 -sqrt(2)) cos^(2)x` `rArr 2 sin x cos x gt sqrt(2) sin^(2) x+(2-sqrt(2))cos^(2) x` `rArr tan^(2)x-sqrt(2) tan x+(sqrt(2)-1) lt 0` `rArr (tan x-1) (tan x-(sqrt(2)-1)) lt 0` `rArr (sqrt(2)-1) lt tan x lt 1` `rArr pi//8 lt x lt pi//4` `:. x in (n pi + pi/8, n pi +pi/4)` where `n in Z`. |
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| 199. |
If `4sin^4x+cos^4x=1,t h e nxi se q u a lto(n in Z)``npi`(b) `npi+-sin^(-1)sqrt(2/5)``(2npi)/3`(d) `2npi+-pi/4`A. `n pi`B. `npi pm sin^(-1) sqrt(2/5)`C. `(2 n pi)/3`D. `2n pi pm pi/4` |
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Answer» Correct Answer - A::B We have `4 sin^(4)x+cos^(4)x=1` or `4 sin^(4)x=1-cos^(4)x` `=(1-cos^(2) x) (1+ cos^(2) x)` `=sin^(2) x (2-sin^(2) x)` or `sin^(2)x[5 sin^(2) x-2]=0` or `sin x=0` or `sin x= pm sqrt(2//5)` `rArr x=n pi` or `x=n pi pm sin^(-1) sqrt(2/5), n in Z` |
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| 200. |
Prove that the general solution of `sin"theta"=sin"alpha"`is given by : `theta=npi+(-1)^nalpha,n in Zdot` |
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Answer» `sin theta = sin alpha` `=>sin theta - sinalpha= 0` `=>2sin((theta-alpha)/2)cos((theta+alpha)/2) = 0` `=>sin((theta-alpha)/2) = 0 and cos((theta+alpha)/2) = 0` `=> (theta-alpha)/2 = mpi and (theta+alpha)/2 = (2m+1)pi/2` `=>theta = 2mpi+alpha and theta = 2mpi+pi-alpha` So, the general solution will be `theta = npi +(-1)alpha.` |
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