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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.
| 1. |
If `xsina+ysin2a+zsin3a=sin4a``xsinb+ysin2b+zsin3b=sin4b``xsinc+ysin2c+zsin3c=sin4c`then the roots of the equation `t^3-(z/2)t^2-((y+2)/4)t+((z-x)/8)=0,a , b , c ,!=npi,`are`sina ,sinb ,sinc`(b) `cosa ,cosb ,cosc``sin2a ,sin2b ,sin2c`(d) `cos2a ,cos2bcos2c`A. `cos a, cos b, cos c`B. `sin a, sin b, sin c`C. `sin 2a, sin 2b, sin 2c`D. `cos 2a, cos 2b, cos 2c` |
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Answer» Correct Answer - A a, b, c are roots of equation. `x sin theta+ y sin 2 theta + z sin 3 theta = sin 4 theta` `rArr x sin theta+y (2 sin theta cos theta) + z (3 sin theta -4 sin^(3) theta)` `= 4 sin theta cos theta cos 2 theta` `rArr cos^(3) theta-z/2 cos^(2) theta-(y+2)/4 cos theta + (z-x)/8=0` |
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| 2. |
Consider the cubic equation `x^3-(1+cos theta+sin theta)x^2+(cos theta sin theta+cos theta+sin theta)x-sin theta. cos theta =0` Whose roots are `x_1, x_2 and x_3`A. 3B. 4C. 5D. 6 |
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Answer» Correct Answer - C Now if `1=sin theta`, we get `theta=pi//2` If `1= cos theta`, then `theta=0, 2pi` and if `sin theta=cos theta`, we get `tan theta=1`. therefore, `theta=pi/4, (5pi)/4` Therefore, the number of values of `theta` in `[0, 2pi]` is 5. |
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| 3. |
Cosider the cubic equation : `x^3-(1+costheta+sintheta)x^2+(costhetasintheta+costheta+sintheta)x-sinthetacostheta=0` whose roots are `x_1,x_2,x_3`. The value of `(x_1)^2+(x_2)^2+(x_3)^2` equalsA. 1B. 2C. `2 cos theta`D. `sin theta (sin theta+ cos theta)` |
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Answer» Correct Answer - B `x^(3)-(1+cos theta + sin theta) x^(2) +(cos theta sin theta + cos theta + sin theta)x-sin theta cos theta=0` Given cubic function is `f(x)=(x-1)(x-cos theta) (x- sin theta)` Therefore, roots are `1, sin theta`, and `cos theta`. Hence, `x_(1)^(2)+x_(2)^(2)+x_(3)^(2)=1+sin^(2) theta+cos^(2) theta=2` |
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| 4. |
`tan^(3)x-3tanx=0` |
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Answer» Correct Answer - `x=npiorx=(mpi+(pi)/(3))orx=(pi+(2pi)/(3)),"where m ", n , p inI` `tan^(3)x-3tanx=0rArrtanx(tan^(2)x-3)=0` `rArrtanx =0 or tanx=sqrt(3)ortanx=-sqrt(3)` `rArrx=npior tanx="tan "(pi)/(3)ortanx=-"tan "(pi)/(3)=tan(pi-(pi)/(3))` `rArrx=npiorx=mpi+(pi)/(3)orx=ppi+(2pi)/(3)`"where " m , n, p inI` |
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| 5. |
If `tan(A-B)=1a n dsec(A+B)=2/(sqrt(3))`, then the smallest positive values of A and B, respectively, are`(25pi)/(24),(19pi)/(24)`(b) `(19pi)/(24),(25pi)/(24)``(31pi)/(24),(31pi)/(24)`(d) `(13pi)/(24),(31pi)/(24)`A. `(25 pi)/(24), (19 pi)/24`B. `(19 pi)/24, (25 pi)/24`C. `(31 pi)/24, (13 pi)/24`D. `(13pi)/24, (31 pi)/24` |
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Answer» Correct Answer - A `tan (A-B)=1` `rArr A-B=n_(1) pi+pi/4 or A-B=pi/4, (3pi)/4, -(3pi)/4`, ... `sec(A+B) =2/sqrt(3)` `rArr A+B=2n_(2)pi pm pi/6 or A+B=pi/6, (11 pi)/6`, ... For the least positive values of A and B, we have `A+B=(11pi)/6, A-B=pi/4` `rArr B=(19 pi)/24, A=(25 pi)/24` |
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| 6. |
Solve the equation:`cos^2[pi/4(sinx+sqrt(2)cos^2x)]-tan^2[x+pi/4tan^2x]=1` |
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Answer» `cos^(2) [pi/4(sin x+sqrt(2) cos^(2) x)]-tan^(2) [x+pi/4 tan^(2) x]=1` `rArr sin^(2) [pi/4 (sin x+ sqrt(2) cos^(2) x)]+tan^(2) [x+pi/4 tan^(2) x]=0` It is possible only when `sin^(2) [pi/4 (sin x+sqrt(2) cos^(2) x)]=0`...(i) and `tan^(2) [x+pi/4 tan^(2) x]=0` ...(ii) `:. pi/4 (sin x+ sqrt(2) cos^(2) x) = n pi, n in I` or `sin x +sqrt(2) cos^(2) x=4n` This equation has solution only for `n=0`. Thus, `sin x+sqrt(2) cos^(2) x=0` i.e., `sqrt(2) sin^(2) x- sin x-sqrt(2)=0` or `(sin x-sqrt(2)) (sqrt(2) sin x +1) =0` `:. sin x= - 1/sqrt(2)` `rArr x=2 kpi-pi//4, k in Z` Also these values of x satisfy Eq. (ii), therefore , the general solution of given equation is given by `x=2kpi - pi/4, k in Z` |
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| 7. |
Solve : `secx-tan x=sqrt(3)`. |
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Answer» We have `secx-tanx=sqrt(3)rArr(1)/(cosx)-(sinx)/(cosx)=sqrt(3)rArr(1-sinx)=sqrt(3)cosx` `rArrsqrt(3)cosx+sinx=1`. Thus , the given equation becomes `sqrt(3)cosx+sinx=1`. Dividing both sides of (i) by `sqrt((sqrt(3))^(2)+1^(2))` , i .e by 2, we get `(sqrt(3))/(2)cosx+(1)/(2)sinx=(1)/(2)` `rArrcosx"cos"(pi)/(6)+sinx" sin "(pi)/(6)=(1)/(2)` `rArrcos(x-(pi)/(6))="cos"(pi)/(3)` `rArr(x-(pi)/(6))=2npi+-(pi)/(3), "where "n in I[becausecostheta=cosalpharArrtheta=2npi+-alpha]` `rArr(x-(pi)/(6))=(2npi+(pi)/(3))or(x-(pi)/(6))=(2npi-(pi)/(3))` `rArrx=2npi+(+(pi)/(3)+(pi)/(6))orx=2npi+(-(pi)/(3)+(pi)/(6))"where "ninI` `rArrx=(2npi+(pi)/(2))orx=(2npi-(pi)/(6)), "where "ninI` `rArrx=(2npi-(pi)/(6)), "where"n inI` `[because"secx is not defined when x"=(2npi+(pi)/(2))]`. Hence , the general solution is `x=(2npi-(pi)/(6)), "where"ninI`. |
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| 8. |
`3^((1/2+log_3(cosx+sinx)))-2^(log_2(cosx-sinx))=sqrt(2)` |
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Answer» Correct Answer - `x=2n pi+(3pi)/(4),2n pi+(pi)/(12),n in Z` `3^(((1)/(2)+log_(3)(cos x + sin x)))-2^(log_(2)(cos x - sin x)=sqrt(2))` or `sqrt(3)(cos x + sin x) - (cos x - sin x) = sqrt(2)` or `sqrt(3) cos x + sqrt(3) sin x - cos x + sin x = sqrt(2)` or `(sqrt(3)+1)sin x + (sqrt(3)-1)cos x = (2sqrt(2))/(2)/(2)` or `((sqrt(3)+1)/(2sqrt(2)))sin x + ((sqrt(3)-1)/(2sqrt(2)))cos x = (1)/(2)` or `sin (x + (5pi)/(12))=(1)/(2)` or `x + (5pi)/(12)=2n pi pm (pi)/(3), n in Z` or `x = 2n pi pm(pi)/(3)-(5pi)/(12)` or `x = 2n pi +(3pi)/(4), 2n pi + (pi)/(12), n in Z` |
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| 9. |
(i) `sinx=(sqrt(3))/(2)` (ii) `cosx=1` (iii) `secx=sqrt(2)` |
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Answer» Correct Answer - (i) `x=npi+(-1)^(n)*(pi)/(3),n inI` (ii) `x=2npi, n in I` (iii) `x=(2npi+-(pi)/(4)), n in I` |
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| 10. |
(i) `sin2x=(1)/(2)` (ii) `cos3x=(1)/(sqrt(2))` (iii) `"tan"(2x)/(3)=sqrt(3)` |
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Answer» Correct Answer - (i) `x=(npi)/(2)+(-1)^(n)*(pi)/(12), n in I` (ii) `x=((2npi)/(3)+-(pi)/(12)), n inI` (iii) `x=(3n+1)(pi)/(2),n in I` |
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| 11. |
(i) `4cos^(2)x=1` (ii) `4sin^(2)x-3=0` (iii) `tan^(2)x=1` |
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Answer» Correct Answer - (i) `x=(npi+-(pi)/(3)), n in I` (ii) `x=(npi+-(pi)/(3)), n in I` (iii) x=(npi+-(pi)/(4)), n in I` |
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| 12. |
(i) `cos3x=cos2x` (ii) `cos5x=sin3x` (iii) `cosmx=sinnx` |
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Answer» Correct Answer - (i) `x=2npiorx=(2npi)/(5)"where"n in I` (ii) `x=((npi)/(4)+(pi)/(16))orx=(npi-(pi)/(4)),"where"n in I` (iii) `x=((4k+1)pi)/(2(m+n))orx=((4k-1)pi)/(2(m-n)),"where"k inI` |
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| 13. |
(i) `sec3x=-2` (ii) `cot4x=-1` (iii) `"cosec "3x=(-2)/(sqrt(3))` |
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Answer» Correct Answer - (i) `x=((2npi)/(3)+-(2pi)/(9)), n inI` (ii) `x=(4n+3)(pi)/(16), n in I` (iii) `x=(npi)/(3)+(-1)^(n)*(4pi)/(9), n in I` |
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| 14. |
Find the number of solution of the equation `1+e^cot^(2x)=sqrt(2|sinx|-1)+(1-cos2x)/(1+sin^4x)forx in (0,5pi)dot` |
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Answer» `L.H.S. =1+e^(cot^(2) x) ge 2` As `sqrt(2|sin x|-1) le 1` and `(1- cos 2x)/(1+sin^(4) x)=(2 sin^(2) x)/(1+sin^(4) x)=2/(1/(sin^(2) x)+sin^(2) x) le 1` `:. R.H.S.=sqrt(2|sin x|-1)+(1- cos 2x)/(1+sin^(4) x) le 2` Equation will be satisfied if `L.H.S.=R.H.S.=2`. This is possible when `cot^(2) x=0` and `|sin x|=1`. `rArr x=(2n+1) pi.2, n in Z` |
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| 15. |
If `cos 3x+sin (2x-(7pi)/6)=-2`, then x is equal to `(k in Z)`A. `pi/3 (6k+1)`B. `pi/3 (6k-1)`C. `pi/3 (2k+1)`D. none of these |
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Answer» Correct Answer - D `cos 3x+sin (2x-(7 pi)/6)=-2` `:. Cos 3x=-1` and `sin (2x -(7pi)/6)=-1` or `3x=(2n+1) pi and 2x-(7pi)/6=-pi/2+2 m pi , m, n in Z` or `x=(2n pi)/3+pi/3 and x=mpi+pi/3` or `x=2 kpi +pi/3, k in Z`. |
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| 16. |
The equation `x^3=3/4x=-(sqrt(3))/8`is satisfied by`x=cos((5pi)/(18))`(b) `x=cos((7pi)/(18))``x=cos((23pi)/(18))`(d) `x=cos((17pi)/(18))`A. `x=cos((5pi)/(18))`B. `x =cos((7pi)/(18))`C. `x=cos((23pi)/(18))`D. `x=-sin((7pi)/(9))` |
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Answer» Correct Answer - A::B Let `x = cos theta` `rArr 4 cos^(3) theta -3 cos. theta =-(sqrt(3))/(2)` `rArr cos 3 theta = cos.(5pi)/(6)` `rArr 3 theta = 2n pi pm (5pi)/(6), n in Z` `rArr theta =(2n pi)/(3)pm(5pi)/(18),, n in Z` Put `n=0, theta =(5pi)/(18)` `n=1, theta =(2pi)/(3)+(5pi)/(18)=(17pi)/(18)` `theta =(2pi)/(3)-(5pi)/(18)=(7pi)/(18)` |
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| 17. |
Number of roots of `cos^2x+(sqrt(3)+1)/2sinx-(sqrt(3))/4-1=0`which lie in the interval `[-pi,pi]`is2 (b) 4(c) 6 (d)8A. 2B. 4C. 6D. 8 |
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Answer» Correct Answer - B `1- sin^(2) x+ (sqrt(3)+1)/2 sin x -sqrt(3)/4-1=0` or `sin^(2) x-(sqrt(3)+1)/2 sin x + sqrt(3)/4=0` or `4 sin^(2) x-2 sqrt*3) sin x -2 sin x + sqrt(3)=0` On solving, we get `sin x=1//2, sqrt(3)//2` `rArr x=pi//6, 5 pi//6, pi//3, 2 pi//3` |
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| 18. |
Write the solution set of the equation `(2costheta+1)(4costheta+5)=0`in the interval `[0,2pi]` |
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Answer» `(2costheta+1)(4costheta+5) = 0` `=>(2costheta+1) = 0 or (4costheta+5) = 0` `=>costheta = -1/2 or cos theta = -5/4` But, minimum value of `cos theta` is `-1`, so `costheta != -5/4.` `:. costheta = -1/2` `=>theta = (2pi)/3,(4pi)/3.` |
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| 19. |
If `(1-tan theta)(1+tan theta)sec^2 theta+2^(tan^2 theta)=0` then in the interval `(-pi/2,pi/2),` the value of `theta` is |
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Answer» Given that `(1-tan theta) (1+tan theta) sec^(2) theta+2^(tan^(2) theta)=0` or `(1- tan^(2) theta) (1+tan^(2) theta)+2^(tan^(2) theta)=0` Let us put `tan^(2) theta=t`. Then `(1-t) (1+t)+2^(t)=0" "or" "1-t^(2)+2^(t)=0` It is clearly satisfied by `t=3`. thus, we get `tan^(2) theta=3` Therefore, `theta= pm pi//3` in the given interval. |
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| 20. |
The number of solutions of the equation `cos^(2)((pi)/(3)cos x - (8pi)/(3))=1` in the interval `[0,10pi]` isA. 1B. 3C. 5D. 7 |
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Answer» Correct Answer - C `cos^(2)((pi)/(3)cos. X-(8pi)/(3))=1` `rArr (pi)/(3)cos. X-(8pi)/(3)=k pi, (k in Z)` `rArr cos x = 3k + 8, k in Z` `rArr cos x = -1`, when `k = -3` `rArr` there are 5 solutions for `x in [0, 10pi]` |
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| 21. |
The number of solutions of the equation `|2 sin x-sqrt(3)|^(2 cos^(2) x-3 cos x+1)=1` in `[0, pi]` isA. 2B. 3C. 4D. 5 |
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Answer» Correct Answer - B `|2 sin x -sqrt(3)|^()=1` Case I : `2 cos^(2) x-3 cos x+1 =0` `cos x=1/2, 1` `x=0, pi/3` But at `x=pi/3, L.H.S. =0^(@)` `:. x=pi/3` (rejected) `:. x=0` is a solution Case II : `2 sin x-sqrt(3) =1, -1` `:. sin x=(sqrt(3)+1)/2, (sqrt(3)-1)/2` `sin x=(sqrt(3)-1)/2` `:.` x has 2 values in `[0, pi]` `:.` Total number of solutons `=2+1=3` |
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| 22. |
If roots of the equation `2x^2-4x+2sintheta-1=0` are of opposite sign, then `theta` belongsA. `(pi/6, (5pi)/6)`B. `(0, pi/6) uu ((5pi)/6, 2pi)`C. `((13 pi)/6, (17 pi)/6)`D. `(0, pi)` |
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Answer» Correct Answer - B Roots will be of opposite sign if product of roots is negative `rArr sin theta lt 1/2` |
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| 23. |
The value of x in `(0,pi/2)` satisfying the equation, `(sqrt3-1)/sin x+ (sqrt3+1)/cosx=4sqrt2` is -A. `pi/12`B. `(5pi)/12`C. `(7pi)/24`D. `(11 pi)/36` |
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Answer» Correct Answer - A::D `(sqrt(3)-1)/(2sqrt(2) sin x)+(sqrt(3)+1)/(2sqrt(2) cos x)=2` `"sin" pi/12 cos x + "cos" pi/12 sin x= sin 2x` `sin 2x= sin (x+pi/12)` `:. 2x=x+pi/12 or 2x=pi-x- pi/12` `x=pi/12 ot 3x=pi/12` `:. X=pi/12 or (11 pi)/36` |
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| 24. |
One root of the equation `cos x-x+1/2=0` lies in the interval (A) `[0,pi/2]` (B) `[-pi/2,0]` (C) `[pi/2,0]` (D) noneA. `(0, pi/2)`B. `(- pi/2, 0)`C. `(pi/2, pi)`D. `(pi, (3pi)/2)` |
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Answer» Correct Answer - A Let `f(x)=cos x-x+1/2` `f(0)=1+1/2 gt 0` `f(pi/2)=0- pi/2 + 1/2=(1-pi)/2 lt 0` Therefore, one root lies in the interval `(0, pi/2)`. |
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| 25. |
The equation ` cos^8 x + b cos^4 x + 1 = 0` will have a solution if b belongs to :A. `(-oo, 2]`B. `[2, oo)`C. `(-oo, -2]`D. none of these |
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Answer» Correct Answer - C `cos^(8)x+b cos^(4) x+1=0` `rArr b=-(cos^(4) x+1/(cos^(4) x)) le -2 AA x in R` `rArr b in (-oo, -2]` |
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| 26. |
The number of solutions of the equation `1+cosx+cos2x+sinx+sin2x+sin3x=0,`which satisfy the condition `pi/2 |
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Answer» Correct Answer - 2 Given `pi/2 lt |3x - pi/2| lt pi` `rArr pi/2 lt (3x - pi/2) le pi` or `-pi le (3x - pi/2) lt (-pi)/2` `rArr x in [(-pi)/6, 0) uu (pi/3, pi/2]` Now, `1+cos x + cos 2x + sin x + sin 2x + sin 3x =0` `rArr 2 cos^(2) x+ cos x + sin 2x +2 sin 2x cos x=0` `rArr cos x(2 cos x +1) + sin 2x (2 cos x + 1) =0` `rArr (cos x + sin 2x) (2 cos x+1)=0` `rArr cos x(1+2 sin x) (2 cos x+1) =0` `rArr cos x=0` or `sin x = (-1)/2` (as for given interval, `cos x gt 0`) `rArr x=pi/2` or `x= (-pi)/6` Hence, there are 2 solutions. |
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| 27. |
For `0 lt theta lt pi/2`, the solutions of `sigma_(m-1)^(6)"cosec"(theta+((m-1)pi)/4)" cosec "(theta+(mpi)/4)=4sqrt(2)` is (are):A. `pi//4`B. `pi//6`C. `pi//12`D. `5pi//12` |
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Answer» Correct Answer - C::D `sum_(m=1)^(6) cosec (theta +((m-1))/4 pi) cosex (theta +(mpi)/4)=4sqrt(2)` `rArr 1/(sin (pi//4))[(sin (theta+pi//4-theta))/(sin theta. sin (theta+pi//4))+(sin ((theta+pi//2)-(theta+pi//4)))/(sin (theta+pi//4).sin (theta+pi//2))+..+(sin ((theta+3pi//2)-(theta+5pi//4)))/(sin (theta+3pi//2). sin (theta+5pi//4))]=4sqrt(2)` `rArr 1/(sin (pi//4))[(sin (theta+pi//4) cos theta-cos (theta+pi//4) sin theta)/(sin theta. sin (theta+pi//4))+(sin (theta+pi//2) cos (theta+pi//4)-cos (theta+pi//2) sin (theta+pi//4))/(sin (theta+pi//4). sin (theta+pi//2))+... (sin (theta+3pi//2)cos (theta+5pi//4)-cos (theta+3pi//2) sin (theta+5pi//4))/(sin (theta+3pi//2). sin (theta+5pi//4))]=4sqrt(2)` `rArr sqrt(2) [cot theta-cot (theta+pi//4)+cot(theta+pi//4)-cot (theta+pi//2)+...+cot (theta+5pi//4)-cot(theta+3pi//2)]=4sqrt(2)` `rArr tan theta+cot theta=4` `rArr tan theta=2 pm sqrt(3)` `rArr theta=pi/12 or (5pi)/12` |
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| 28. |
The smallest positive `x`satisfying the equation `(log)_(cosx)sinx+(log)_(sinx)cosx=2`is`pi/2`(b) `pi/3`(c) `pi/4`(d) `pi/6`A. `pi//2`B. `pi//3`C. `pi//4`D. `pi//6` |
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Answer» Correct Answer - C Let `log_(cos x) sin x=t`, then the given equation is `t+1/t=2` or `(t-1)^(2)=0` or `t=1` or `log_(cos x) sin x=1 or sin x = cos x` `rArr tan x=1` or `x=pi//4` |
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| 29. |
If `2 sin^(2)(x-(pi)/(3))-5sin(x-(pi)/(3))+2lt0`, then belongs tpA. `(((12n-5)pi)/(6),((4n+1)pi)/(2)),n in Z`B. `(((6n-7)pi)/(6),((2n+1)pi)/(2)),n in Z`C. `(((4n+1)pi)/(6),n pi),n in Z`D. `((4n+1)(pi)/(2),(12n+7)(pi)/(6)),n in Z` |
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Answer» Correct Answer - D `2sin^(2)(x-(pi)/(3))-5 sin(x-(pi)/(3))+2 lt 0` `rArr (2 sin(x-(pi)/(3))-1)(sin(x-(pi)/(3))-2)lt0` `rArr (1)/(2)lt sin (x-(pi)/(3))lt2` `rArr (1)/(2)lt sin (x-(pi)/(3))le 1` `therefore x-(pi)/(3)in (2n pi + (pi)/(6), 2n pi+(5pi)/(6))` `x in (2n pi+(pi)/(2), 2n pi+(7pi)/(6)), n in I` |
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| 30. |
If the equation `x^(2)+12+3sin(a+bx)+6x=0` has atleast one real solution, where `a, b in [0,2pi]`, then the value of a - 3b is `(n in Z)`A. `2n pi`B. `(2n+1)pi`C. `(4n-1)(pi)/(2)`D. `(4n+1)(pi)/(2)` |
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Answer» Correct Answer - C `x^(2)+12+3 sin (a+bx)+6x =0` `rArr (x+3)^(2)+3+3sin (a+bx)=0` `rArr (x+3)^(2)+3=-3 sin (a+ bx)` `L.H.S. ge 3` but `R.H.S. le 3` `L.H.S. = R.H.S. = 3` `therefore x=-3` and `sin (a + bx) =-1` `rArr sin (a-3b)=-1` or `a-3b=(4n-1)(pi)/(2), n in Z` |
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| 31. |
In each of the following , find the general value of x satisfying the equation : (i) `sinx =(-sqrt(3))/(2)` (ii) `cosx=(-1)/(2)` (iii) `cotx=-sqrt(3)` |
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Answer» (i) `sinx=(-sqrt(3))/(2)=-"sin"(pi)/(3)= sin(pi+(pi)/(3))="sin"(4pi)/(3)`. `thereforesinx " sin"(4pi)/(3)rArrx={npi+(-1)^(n)*(4pi)/(3))`, where `n in I`. Hence , the general solution is x = `{npi+(-1)^(n)*(4pi)/(3))`, where `ninI`. (ii) `cosx=(-1)/(2)=-"cos "(pi)/(3)=cos(pi-(pi)/(3))="cos "(2pi)/(3)`. `therefore cos x="cos"(2pi)/(3)rArrx=(2npi+-(2pi)/(3))` , where `ninI`. Hence , the general solution is x `=(2npi+-(2pi)/(3))` , where `nin I`. (iii) `cotx=-sqrt(3)` `rArrtanx =(-1)/(sqrt(3))=-"tan"(pi)/(6)="tan"(pi-(pi)/(6))="tan"(5pi)/(6)` `rArr"tan " x = "tan " (5pi)/(6)` `rArrx=(npi+(5pi)/(6))`, where `ninI`. |
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| 32. |
If the equation `k sin x + sqrt(k-2)cos x+(tan alpha+cot alpha)=0, 0 lt alpha lt(pi)/(2)`, possesses real solution, then k belongs toA. `(-oo,-3]uu[2,oo)`B. `[-3,2]`C. `[0,2)`D. R |
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Answer» Correct Answer - A For real solution `sqrt(k^(2)+k-2)ge|tan alpha + cot alpha| ge 2` `rArr k^(2)+k-6 ge 0` `rArr k in (-oo, -3] uu[2,oo)` |
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| 33. |
In each of the following , find the general value of x satisfying the equation : (i) `sinx=(1)/(sqrt(2))` (ii) `cosx =(1)/(2)` (iii) `tanx=(1)/(sqrt(3))` |
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Answer» (i) Given : sin `x=(1)/(sqrt(2))`. The least value of x in `[0, 2pi[" for which sin"x=(1)/(sqrt(2))is x = (pi)/(4)`. `therefore"sin x = sin"(pi)/(4)rArrx=npi+(-1)^(n)*(pi)/(4)`, where ` n in I`. Hence , the general solution is x = `npi+(-1)^(n)*(pi)/(4)`, where ` n in I`. (ii) Given `cos x =(1)/(2)`. The least value of x in `[0, 2pi["for which cos "x=(1)/(2) is x=(pi)/(3)`. `therefore" cos x cos "(pi)/(3)rArrx=(2nx+-(pi)/(3))`, where ` n in I`. Hence , the general solution is `x=(2nx+-(pi)/(3))`, where ` n in I` (iii) Given : `tanx=(1)/(sqrt(3))`. The least value of x in `[0 , 2pi[" for which tan "x=(1)/(sqrt(3))is (pi)/(6)`. `therefore tanx ="tan "(pi)/(6)rArrx=(npi+(pi)/(6))`, where `n in I` . Hence , the general solution is `x=(npi+(pi)/(6))`, where `ninI`. |
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| 34. |
The number of solutions of equation `sin.(5x)/(2)-sin.(x)/(2)=2` in `[0,2pi]` is |
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Answer» Correct Answer - A Given equation can hold only if `sin.(5x)/(2)-1` and `sin.(x)/(2)=-1` i.e. `(5x)/(2)=2n pi +(pi)/(2)` and `(x)/(2)=2p pi -(pi)/(2)` (n `p in I`) For some possible p and n if there exists a solution, we must have `10 p pi-(5pi)/(2)=2n pi + (pi)/(2)` oe `10 p - 2n = 3` L.H.S. is even, R.H.S. is odd Hence, not possible for any p and n. |
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| 35. |
Find the general solution of the trignometric equation `3^(1/2+log_(3)(cosx+sinx))-2^(log_(2)(cosx-sinx))=sqrt(2)`A. `2n pi+(5pi)/(4)`B. `n pi-(pi)/(4)`C. `n pi+(-1)^(n)(pi)/(4)`D. `2n pi+(pi)/(4)` |
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Answer» Correct Answer - A `because A.M. ge G.M. therefore (2^(sin x)+2^(cos x))/(2)ge sqrt(2^(sin x).2^(cos x))` `therefore 2^(sin x)+2^(cos x)ge 2. sqrt(2^(sin x + cos x))` But minimum value of cos x + sin x is `- sqrt(2)` `thereofre 2^(sin x)+2^(cos x)ge 2. sqrt(2^(-sqrt(2)))=2^(1-(1)/(sqrt(2)))` But the given equation is `2^(sin x)+2^(cos x)=2^(1-(1)/(sqrt(2)))`, which can hold only if `2^(sin x)=2^(cos x)=2^(-(1)/(sqrt(2)))` `rArr x = 2n pi + (5pi)/(4), n in Z` |
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| 36. |
`cosx+sinx=1` |
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Answer» Correct Answer - `x=(2npi+(pi)/(2))orx=2npi , "where "ninI` `cosx+sinx=1rArr(1)/(sqrt(2))cosx+(1)/(sqrt(2))sinx=(1)/(sqrt(2))` `rArrcosx " cos"(pi)/(4)+sinx " sin"(pi)/(4)=(1)/(sqrt(2))` `rArrcos(x-(pi)/(4))="cos "(pi)/(4)` `rArrx-(pi)/(4)=2npi+(pi)/(4)orx-(pi)/(4)=2npi-(pi)/(4)` `rArrx=2npi+(pi)/(2)orx=2npi " where " n inI`. |
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| 37. |
`cosx-sinx=-1` |
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Answer» Correct Answer - `x=(2npi+(pi)/(2))orx=(2n-1)pi, "where "n in I` `cosx-sinx=-1` `rArr(1)/(sqrt(2))cosx-(1)/(sqrt(2))sinx=(-1)/(sqrt(2))` `"cos"(pi)/(4)cosx-"sin"(pi)/(4)sinx=-"cos"(pi)/(4)=cos(pi-(pi)/(4))` `rArrcos(x+(pi)/(4))="cos"(3pi)/(4)` `rArrx+(pi)/(4)=2npi+-(3pi)/(4)` `rArrx+(pi)/(4)=2npi+(3pi)/(4)orx+(pi)/(4)=2npi-(3pi)/(4)` `rArrx=2npi+(pi)/(2)orx=(2n-1)pi" where"n in I`. |
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| 38. |
`cotx+tanx=2"cosec "x` |
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Answer» Correct Answer - `x=2npi+-(pi)/(3), "where " n in I` The given equation is `(cosx)/(sinx)+(sinx)/(cosx)=(2)/(sinx)rArrcos^(2)x+sin^(2)x=2cosx` `rArrcosx=(1)/(2)="cos "(pi)/(3)` `rArrx=2npi+-(pi)/(3), "where "n inI`. |
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| 39. |
`2tanx-cotx+1=0` |
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Answer» Correct Answer - `x=npi+(3pi)/(4)orx=mpi+(-1)^(n)(pi)/(2),"where m "ninI` The given equations is `2tan^(2)x+tanx-1=0` `rArr(tanx+1)(2tanx-1)=0` `rArrtanx=-1ortanx=(1)/(2)` `rArrtanx="tan"(3pi)/(4)ortanx=tan("tan"^(-1)(1)/(2))` `rArrx=npi+(3pi)/(4)orx=mpi+"tan"^(-1)(1)/(2), "where m "n in I`. |
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| 40. |
Find the number of solution of `[cosx]+|sinx=1inpilt=xlt=3pi`(where `[]`denotes the greatest integer function). |
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Answer» We have `[cos x]+|sin x|=1` Now, `-1 le cos x le 1` `:.` Possible values of `[cos x]` are `0, 1, -1` Case I : `[cos x]=-1` or `cos x in [-1, 0)` `:. -1 +|sin x|=1` `rArr |sin x|=2` (not possible) Case II : `[cos x]=0` or `cos x in [0, 1)` `:. |sin x|=1` `rArr x=(3pi)/2, (5pi)/2` Case III : `[cos x]=1` or `cos x=1` `:. |sin x|=0` `:. x=2pi` Hence, there are three solutions. |
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| 41. |
If `tan a theta-tan b theta=0`, then prove that the values of `theta` forms an A.P. |
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Answer» We have `tan a theta = tan b theta` `rArr a theta = n pi + b theta, n in Z` `rArr (a -b) theta = n pi` `rArr n pi//(a -b) = theta` Thus, the values of `theta` form an A.P. with common difference `pi//(a -b)` |
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| 42. |
If `cos p theta+cos q theta=0`, then prove that the different values of `theta` are in A.P. with common difference `2pi// (p pm q)`. |
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Answer» `cos p theta = - cos q theta = cos (pi - q theta)` `rArr p theta = 2 n pi -+ (pi - q theta)` `rArr (p -+ q) theta = (2n -+ 1) pi` `rArr theta = ((2n -+1)pi)/((p pm q)), n in Z` `rArr theta = (r pi)/(p -+ q), " where " r = -3, -1, 1, 3`.... `rArr theta = ...., (-3pi)/(p-+q) , (-pi)/(p-+q), (pi)/(p-+q), (3pi)/(p-+q)`,.... The above series is an A.P., of common difference `= (2pi)/(p -+q)` |
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| 43. |
Let `k` be sum of all x in the interval `[0, 2pi]` such that `3 cot^(2) x+8 cot x+3=0`, then the value of `k//pi` is ___________. |
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Answer» Correct Answer - 5 `3cot^(2) x +8 cot x+3=0` `:. cot x = (-4 pm sqrt(7))/3` Both roots are negative. `:. Pi/2 lt x_(1), x_(2) lt pi` `:. Pi lt x_(1) + x_(2) lt 2pi` Product of roots of roots `=cot x_(1) cot x_(2)=1` `rArr cot x_(1) cot ((3pi)/2- x_(1))=1` `rArr x_(1)+x_(2)=(3pi)/2` Similarly, `x_(1)+x_(2)=(7pi)/2` |
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| 44. |
Solve that following equations :`"tantheta"+"tan"2"theta"+sqrt(3)"tanthetatan"2"theta"=""sqrt(3)` |
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Answer» `sintheta/costheta+(cos2theta)/cos2theta+sqrt3sintheta/costheta*(sin2theta)/cos2theta=sqrt3` `(sinthetacos2theta+sin2thetacostheta+sqrt3sinthetasin2theta)/(costhetacos2theta)=sqrt3` `sin(theta+2theta)+sqrt3(sinthetasin2theta-costhetacos2theta)=0` `sin3theta-sqrt3(costhetacostheta-sinthetasin2theta)=0` `sin3theta-sqrt3cos3theta=0` `tan3theta=sqrt3` `theta=pi/9` General solution `theta=pi/9pmnpi`. |
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| 45. |
Solve that following equations :`sec x cos 5x+1=0,` `0 lt x le pi/2` find the value of x |
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Answer» `secx*cos5x+1=0` `(cos5x)/cosx=-1` `cos5x=-cosx` `cos5x+cosx=0` `2cos((5x+x)/2)*cos((5x-x)/2)=0` `2ocs3x*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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| 46. |
Find the number of solutions for the equation `sin 5x+sin 3x+sin x=0` for `0 le x le pi`. |
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Answer» Correct Answer - Three solutions `sin 3x + (sin 5x + sin x) = 0` `rArr sin 3x + (2 sin 3x cos 2x) = 0` `rArr sin 3x = 0` or `cos 2x = -(1)/(2) = "cos"(2pi)/(3), n in Z` `rArr x = npi//3` or `x = 2n pi -+ (2pi)/(3)` Then `x = 0, pi//3, 2pi//3`, Hence, there are three solutions |
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| 47. |
Solve `sin 6 theta=sin 4 theta-sin 2 theta`. |
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Answer» Correct Answer - `theta = (n pi)/4 or theta = m pi pm pi/6; m, n in Z` `sin 6 theta = sin 4 theta - sin 2 theta` or `sin 6 theta + sin 2 theta = sin 4 theta` or `2 sin 4 theta cos 2 theta = sin 4 theta` or `sin 4 theta = 0 or cos 2 theta = 1//2` `rArr 4 theta = n pi or 2 theta = 2m pi -+ pi//3, m, x in Z` or `theta = n pi//4 or theta = m pi -+ pi//6, m, n in Z` |
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| 48. |
The number of solution of `sin^(4)x-cos^(2) x sin x+2 sin^(2)x+sin x=0` in `0 le x le 3 pi` isA. 3B. 4C. 5D. 6 |
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Answer» Correct Answer - B `sin^(4) x-cos^(2) x sin x+ 2 sin^(2) x+ sin x=0` or `sin x [sin^(3) x-cos^(2) x+2 sin x+1]=0` or `sin x [sin^(3) x-1+sin^(2) x+2 sin x+1]=0` or `sin x [sin^(3)x+sin^(2) x+2 sin x]=0` or `sin^(2) x=0 or sin^(2) x+sin x+2=0` (not possible for real x) or `sin x=0` Hence, the solutions are `x=0, pi, 2pi, 3pi`. |
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| 49. |
The number of solution of `sin^(4)x-cos^(2) x sin x+2 sin^(2)x+sin x=0` in `0 le x le 3 pi` is |
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Answer» Correct Answer - 4 `sin^(4)x- cos^(2) x sin x+2 sin^(2) x+ sin x=0` or `sin x[sin^(3) x- cos^(2)x+2 sin x+1]=0` or `sin x[ sin^(3)x+sin^(2) x+2 sin x]=0` or `sin^(2) x [sin^(2) x+sin x+2]=0` or `sin x=0`, where `x=0, pi, 2 pi, 3pi` Hence, there are four solutions. |
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| 50. |
Solve `cos theta+cos 2theta+cos 3theta=0`. |
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Answer» Correct Answer - `theta=(2n+1) pi/4 or theta = 2npi pm (2pi)/3, n in Z` `cos theta + cos 2 theta + cos 3 theta = 0` or `(cos theta + cos 3 theta) + cos 2 theta = 0` or `2 cos theta cos 2 theta + cos 2 theta = 0` or `cos 2 theta (2 cos theta + 1) = 0` or `cos 2 theta = 0 or 2 cos theta + 1 = 0` `rArr 2 theta = (2n + 1) pi//2 or cos theta = - 1/2, n in Z` or `theta = (2n + 1) pi//4 or theta = 2n pi -+ 2pi//3, AA n in Z` |
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