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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 a,b,c are positive real numbers such that `a^(log_3(7));b^(log_7(11))=49 and c^(log_11(25))=sqrt(11)` find the vale of `a^((log_3(7))^2)+b^((log_7(11))^2)+c^((log_11(25))^2)` |
| Answer» Correct Answer - 469 | |
| 2. |
Find the values of `theta`, for which `cos3theta + sin3theta + (2sin2theta-3)(sintheta-costheta)` is always positive. |
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Answer» Given expression can be written as: `4cos^(3)theta-3costheta+3sintheta-4sin^(3)theta+(2sin2theta-3)(sintheta-costheta)` Applying given condtions, we get `rArr -4(sin^(3)theta-cos^(3)theta) + 3(sintheta-costheta) + (sintheta-costheta) (2sin2theta-3) gt 0` `rArr -4(sintheta-costheta)(sin^(2)theta + cos^(2)theta + sinthetacostheta)+3(sintheta-costheta) + (sintheta-costheta) (2sin2theta-3) gt 0` `rArr (sintheta-costheta) {-4 -4sintheta costheta+3+4sintheta-3} gt 0` `rArr -4(sintheta-costheta) gt 0` `rArr -4sqrt(2)sin(theta-pi/4) gt 0 rArr sin(theta-pi/4) lt 0 rArr 2npi - pi lt theta-pi/4 lt 2npi, n in I` `rArr 2npi - (3pi)/(4) lt theta lt 2npi + pi/4 rArr theta in (2npi -(3pi)/(4), 2npi + pi/4), n in I` Ans. |
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| 3. |
Express the following in logarithimic form: a) `81 = 3^(4)` b) `0.001 = 10^(-3)` c) `2=128^(1//7)` |
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Answer» a) `log_(3)81=4` b) `log_(10(0.001)=-3` c) `log_(128)2 = 1//7` |
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| 4. |
Solve the system of equations: `(log)_a x(log)_a(x y z)=48(log)_a y log_a(x y z)=12 , a >0, a!=1(log)_a z log_a(x y z)=84 ` |
| Answer» `(a^(4),a,a^(7))` or `(1/a^(4), 1/a, 1/a^(7))` | |
| 5. |
Findthe product of the positive roots of the equation `sqrt((2008))(x)^((log)_(2008)x)=x^2dot` |
| Answer» Correct Answer - `(2008)^(2)` | |
| 6. |
Find the value of `2log2/5 +3 log25/8 -log 625/128` |
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Answer» `2log2/3+3log25/8+log128/625` `log2^(2)/5^(2)+log(5^(2)/2^(3))^(3) + log 2^(7)/5^(4)` `=log 2^(2)/5^(2).5^(6)/2^(9).2^(7)/5^(4) = log 1=0` |
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| 7. |
Let `(x_0, y_0)`be the solution of the following equations:`(2x)^(1n2)=(3y)^(1n3)``3^(1nx)=2^(1ny)`The `x_0`is`1/6`(b) `1/3`(c) `1/2`(d) 6A. `1/6`B. `1/3`C. `1/2`D. 6 |
| Answer» Correct Answer - c | |
| 8. |
The value of `cos^2 10^0-cos10^0cos50^0+cos^2 50^0`is equal to`4/3`(b) `1/3`(c) `3/4`(d) `3`A. `3/2(1+cos20^(@))`B. `3/4`C. `3/4+cos20^(@)`D. `63/16` |
| Answer» Correct Answer - 2 | |
| 9. |
If ` sin2A= lambda sin 2B` prove that `(tan(A+B)/tan(A-B))=(lambda+1)/(lambda-1)` |
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Answer» Given `sin2A = lambda sin2B` `rArr (sin2A)/(sin2B) = lambda/1` Applying componendo and dividendo, `(sin2A+sin2B)/(sin2B-sin2A) = (lambda+1)/(1-lambda)` `rArr ((2sin(2A+2B)/(2))cos(2A-2B)/(2))/(2cos(2B+2A)/(2)sin(2B-2A)/(2)) = (lambda+1)/(1-lambda)` `rArr (sin(A+B)cos(A-B))/(cos(A+B)sin(-(A-B)))=(lambda+1)/(1-lambda) rArr (sin(A+B)cos(A-B))/(cos(A+B)x-sin(A-B)) = (lambda+1)/(-(lambda-1))` `rArr (sin(A+B)cos(A-B))(cos(A+B)sin(A-B))=(lambda+1)/(lambda-1) rArr tan(A+B) cot(A-B)=(lambda+1)/(lambda-1)` `rArr (tan(A+B))/(tan(A-B)) = (lambda+1)/(lambda-1)` |
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| 10. |
The value of sin 10.sin 30. sin 50.sin 70 isA. `1/36`B. `1/32`C. `1/18`D. `1/16` |
| Answer» Correct Answer - 4 | |
| 11. |
solve the equation for `x , 5^(1/2)+5^(1/2 + log_5 sinx) = 15^(1/2 + log_15 cosx)` |
| Answer» `x=2npi+pi/6, n in I` | |
| 12. |
If `Sigma_(0)^(n=1)(r+2)/(r+1)= prod-(r=10)^(99)log_(r)(31)`, lies two successive integers whose sum is equal toA. 5B. 7C. 9D. 100 |
| Answer» Correct Answer - B | |
| 13. |
If `sin^ 4 x/2+cos^4 x/3 =1/5` thenA. `tan^(2)x = 2/3`B. `sin^(8)8+(cos^(8)x)/27 = 1/125`C. `tan^(2)x=1/3`D. `(sin^(8)x)/(8)+(cos^(8)x)/(27)+2/125` |
| Answer» Correct Answer - A,B | |
| 14. |
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. `13/9`B. `8/9`C. `20/9`D. `2/3` |
| Answer» Correct Answer - A | |
| 15. |
If `x^(2)-4x+5-siny=0, y in [0,2pi]`, then- A) `x=1, y=0`, B) `x=1, y=pi/2` , C) `x=2, y=0`, `x=2, y=pi/2` |
| Answer» Correct Answer - D | |
| 16. |
Solve the equation `sin^(4)x + cos^(4)x =7/2 sinx.cosx.` |
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Answer» `sin^(4)x+cos^(4)x=7/2sinx.cosx rArr (sin^(2)x +cos^(2)x)^(2) =7/2sinx.cosx` `rArr 1-1/2(sin2x)^(2)=7/4(sin2x) rArr 2sin^(2)2x+7sin2x-4=0` `rArr (2sin2x-1)(sin2x+4) =0 rArr sin2x =1/2` or `sin2x = -4` (Which is not possible) `rArr 2x = npi+(-1)^(n)pi/6 , n in I` i.e, `x=(npi)/(2)+(-1)^(n)pi/12, n in I` Ans. |
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| 17. |
For non-negative integer n, let f(n) `(sum_(k=0)^(n)sin((k+1)/(k+2)pi)sin((k+2)/(n+2)pi))/(sum_(k=0)^(n)sin^(2)((k+1)/(n+2)pi))` Assuming `cos^(-1)x` takes values in `[0,pi]`, which of the following options is/are correct?A. `sin(7cos^(-1)f(5))=0`B. `f(4)=sqrt(3)/2`C. `underset(ntoinfty)"lim"f(n)=1/2`D. If `alpha=tan(cos^(-1)f(6))`, then `alpha^(2)+2alpha-1=0` |
| Answer» Correct Answer - 1,2,4 | |
| 18. |
Find all value of `theta`, between `0 & pi`, which satisfy the equation; `cos theta * cos 2 theta * cos 3theta = 1/4`. |
| Answer» `pi.8, pi/3, (3pi)/8, (5pi)/8, (2pi)/8, (7pi)/8` | |
| 19. |
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` |
| Answer» Correct Answer - C,D | |
| 20. |
Find antilog of `5/6` of the base 64. |
| Answer» Correct Answer - 32 | |
| 21. |
If `x ne (npi)/2, n in I` and `(cosx)^(sin^(2)x-3sinx+2)=1`, then find the general solutions of x. |
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Answer» As `x ne (npi)/2 rArr cosx ne 0,1,-1` So, `(cosx)^(sin^(2)x-3sinx+2) = 1 rArr sin^(2)x-3sinx+2=0` `therefore (sinx-2)(sinx-1)=0 rArr sinx=1,2` Where `sinx=2` is not possible and `sinx=2` is not possible and `sinx=1` which is also not possible as `x ne (npi)/(2)` `therefore` no general solutions is possible. Ans. |
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| 22. |
If `cosx+cosy+cosalpha=0` and `sinx + siny + sinalpha=0`, then `cot((x+y)/2)=`A. `sinalpha`B. `cosalpha`C. `cotalpha`D. `2sinalpha` |
| Answer» Correct Answer - C | |
| 23. |
Solve: `costhetacos2thetacos3theta=1/4`, where `0 lethetalex`. |
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Answer» `1/2 (2costhetacos3theta) cos2theta=1/4 rArr (cos2theta+ cos4theta) cos2theta=1/2` `rArr 1.2[2cos^(2)2theta+2cos4thetacos2theta]=1/2 rArr 1+cos4theta+2cos4thetacos2theta=1` `therefore cos4theta =0` or `(1+2cos2theta)=0` Now from the first equation: `2cos4theta=0=cos(pi//2)` `therefore 4theta=(n+1/2)pi rArr theta=(2n+1)pi/8, n in I` for `n=0, theta=pi/8, n=1, theta=(3pi)/8, n=2, theta=(5pi)/8, n=3, theta=(7pi)/8` `(therefore 0 le thetalepi)` and from the second equation. `cos2theta=-1/2=-cos(pi//3) = cos(pi-pi//3) = cos(2pi//3)` `therefore 2theta=2kpi+-2pi//3 therefore theta=kpi+ pi//3, k in I` Again for `k=0, theta=pi/3, k=1 , theta=(2pi)/(3) (therefore 0 le theta le pi)` `therefore theta=pi/8, pi/3, (3pi)/8, (5pi)/8, (2pi)/8, (7pi)/8` Ans. |
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| 24. |
Find the general solution of the equation, `2+tanx.cotx/2.tanx/2=0` |
| Answer» `x=2npi+-(2pi)/3, n in I` | |
| 25. |
The maximum value of the expression `1/(sin^2 theta + 3 sin theta cos theta + 5cos^2 theta)` is |
| Answer» Correct Answer - 2 | |
| 26. |
Q. The value of is equal `sum_(k=1)^13(1/(sin(pi/4+(k-1)pi/6)sin(pi/4+k pi/6))` is equalA. `3-sqrt(3)`B. `2(3-sqrt(3))`C. `2(sqrt(3)-1)`D. `2(2+sqrt(3))` |
| Answer» Correct Answer - C | |
| 27. |
Find the set of values of x for which `(tan3x-tan2x)/(1+tan3x.tan2x)=1`. |
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Answer» We have, `(tan3x-tan2x)/(1+tan3x.tan2x)=1 rArr tan(3x-2x)=1 rArr tanx=1` `rArr tanx=tanx/4 rArr x=npi+pi/4,n in I` (using `tantheta=tanalpha ltimplies theta= npi+alpha)` But for this value of x, `tan 2x` is not defined. Hence the solution set for `x` is `phi`. |
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| 28. |
Solve for `x:(a) log_(0.3)(x^(2)+8) gt log_(0.3)(9x)`, b) `log_(7)( (2x-6)/(2x-1)) gt 0` |
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Answer» a) `x in (1,8)` b) `x in (-infty, 1//2)` |
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| 29. |
Find the solution set of equation `5(1+log_(5)cosx) = 5//2` |
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Answer» Taking log to base 5 on both sides in given equations: `(1+log_(5)cosx). Log_(5)5 = log_(5)(5//2) rArr log_(5)5 + log_(5)cosx=log_(5)5 - log_(5)2` `rArr log_(5)cos-log_(5)2 rArr cosx=1//2 rArr x=2npi+-pi/3, n in I` Ans. |
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| 30. |
If the value of the expression `sin 25^@* sin 35^@* sin 85^@` can be expressed as `(sqrta+sqrt b)/c`where a,b,c N and are in their lowest form, find the value of `(a+b+c)` |
| Answer» Correct Answer - 24 | |
| 31. |
Find the general solutions of equation `sin^(4)x + cos^(4)x=sinx cosx` |
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Answer» Using half-angle formulae, we can represent given equation in the form: `rArr ((1-cos2x)/2)^(2)+((1+cos2x)/2)^(2) = sinx cosx` `rArr (1-cos2x)^(2) + (1+cos2x)^(2)=4sinx cosx` `rArr sin^(2)2x +sin2x = 2` `rArr sin2x = 1` or `sin2x =-2` (which is not possible) `2x=2npi+pi/2, n in I` `rArr x=npi + pi/4, n in I` |
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| 32. |
Determine the smallest positive value of `x` which satisfy the equation `sqrt(1+sin2x)-sqrt(2)cos3x=0` |
| Answer» Correct Answer - `x=pi/16` | |
| 33. |
If `1/6 sintheta, costheta` and `tantheta` are in G.P. then the general solution for `theta` is-A. `2npi+-pi/3`B. `2npi+-pi/6`C. `npi+-pi/3`D. none of these |
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Answer» Correct Answer - a Since, `1/6sintheta,costheta, tantheta` are in G.P. `rArr cos^(2)theta=1/6(sintheta.tantheta) rArr 6cos^(3)theta+cos^(2)theta-1=0` `therefore (2costheta-1)(3 cos^(2)theta+2costheta+1)=0` `rArr costheta=1/2` (other values of `costheta` are imaginary) `rArr costheta=cospi/3 rArr theta=2npi+-pi/3, n in I`. |
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| 34. |
Let `P={theta:sintheta-costheta=sqrt2cos theta}and Q={theta:sintheta+costheta=sqrt2sintheta}` be two ses. Then,A. `P sub Q` and `Q-P ne phi`B. `Q cancel(sub) P`C. `P cancel(sub)Q`D. `P =Q` |
| Answer» Correct Answer - D | |
| 35. |
Solve for x: `(2/3)^((|x|-1)/(|x|+1)) gt 1` |
| Answer» Correct Answer - `10pi` cm | |
| 36. |
If `cos(alpha+beta)=4/5; sin(alpha-beta)=5/13` and `alpha,beta` lie between `0&pi/4` then find the value of `tan2alpha` |
| Answer» Correct Answer - `56/73` | |
| 37. |
Find the general solution of the trignometric equation `3^(1/2+log_(3)(cosx+sinx))-2^(log_(2)(cosx-sinx))=sqrt(2)` |
| Answer» `x=2npi+pi/12` | |
| 38. |
Find the general solution values of `theta` for which the quadratic `(sintheta)x^(2)+(2costheta)x+(costheta+sintheta)/2` is the square of a linear function. |
| Answer» `2npi+pi/4` or `(2n+1)pi-tan^(-1)2, n in I` | |
| 39. |
If `sintheta=-1/2` and `tantheta=1/sqrt(3)` then `theta` is equal to,A. `30^(@)`B. `150^(@)`C. `210^(@)`D. none of these |
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Answer» Let us first find out `theta` lying between 0 and `360^(@)`. Since `sintheta=-1/2 rArr theta=210^(@)` or `330^(@)` and `tantheta=1/sqrt(3) rArr theta=30^(@)` or `210^(@)` Hence, `theta=210^(@)` or `(7pi)/6` is the value satisfying both. |
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| 40. |
If `cottheta=4/3`, then find the value of `sintheta, costheta` and `"cosec "theta` in first quadrant. |
| Answer» `3/5, 4/5, 5/3` | |
| 41. |
6. Find the positive integers p, q, r, s satisfying `tan (pi/24)=(sqrt(p)-sqrt(q))(sqrt(r)-(s))` |
| Answer» `p=3, q=2, r=2, s=1` | |
| 42. |
If `sintheta+sin^(2)theta=1`, then prove that `cos^(12)theta+3cos^(10)theta+3cos^(8)theta+cos^(6)theta-1=0` |
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Answer» Given that `sintheta=1-sin^(2)theta=cos^(2)theta` LHS `=cos^(6)theta(cos^(2)theta+1)^(3)-1=sin^(3)theta(1+sintheta)^(3) -1=(sintheta+sin^(2)theta)^(3)-1=1-1=0` |
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| 43. |
Find the value of `theta`, which satisfy `3-2 cos theta-4 sintheta-cos 2theta + sin 2theta=0`. |
| Answer» `theta=2npi`or `2npi+pi/2, n in I` | |
| 44. |
Find the maximum value of `1+sin(pi/4+theta) + 2cos(pi/4-theta)`.A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - d We have , `1+sin(pi/4+theta)+2cos(pi/4-theta)` `=1+1/sqrt(2)(costheta+sintheta) + sqrt(2)(costheta+sintheta) = 1+1/sqrt(2(+sqrt(2))(costheta+sintheta)` `=1+(1/sqrt(2)+sqrt(2)).sqrtcos(theta-pi/4)` `therefore` maximum value`=1+(1/sqrt(2)+sqrt(2)).sqrt(2)=4` |
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| 45. |
Prove that: `-4le5costheta+3cos(theta+pi/3)+3 ge10`, for all values of `theta`. |
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Answer» We have, `5costheta+3costheta(theta+pi/3)=5costheta+3costhetacospi/3-3sinthetasinpi/3=13/2costheta-(3sqrt(3))/(2)sintheta` Since `-sqrt((13/2)^(2)+(-3sqrt(3)/2)^(2)) le13/2 costheta-(3sqrt(3))/(2)sintheta le sqrt((13/2)^(2)+(-3sqrt(3))/(2)^(2))` `rArr -7 le 13/2 costheta-(3sqrt(3))/(2)sintheta le7` `rArr -7 le5costheta+3cos(theta+pi/3)le7` for all `theta`. `rArr -7+3 le5costheta+3cos(theta+pi/3)+3 ge7 +3` for all `theta` `rArr -4 le5 costheta+3cos(theta+pi/3)+3 le10` for all `theta` |
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| 46. |
If `sintheta+"cosec "theta=2`, then find the values of `sin^(8)theta + "cosec"^(8)theta` |
| Answer» Correct Answer - 2 | |
| 47. |
Minimum value of the expression `cos^2theta-(6 sintheta cos theta) + 3 sin^2theta + 2`, isA. `4+sqrt(10)`B. `4-sqrt(10)`C. 0D. 4 |
| Answer» Correct Answer - B | |
| 48. |
If `costheta=1/2 a n d pi |
| Answer» Correct Answer - 8 | |
| 49. |
Solve the equation : `2sin^(2)theta+sin^(2)theta=2` for `theta in (-pi,pi)`. |
| Answer» `theta= {-pi/4, -(3pi)/4, -pi/2, pi/4, (3pi)/4, pi/2}` | |
| 50. |
For `theta in(0,pi/2) `, the maximum value of sin(`theta+pi/6)+cos(theta+pi/6)` is attained at `theta`=A. `pi/12`B. `pi/6`C. `pi/3`D. `pi/4` |
| Answer» Correct Answer - A | |