InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
Solve` (x^2 -x-1) (x^2-x-7) lt -5` |
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Answer» ` (x^2 -x-1) (x^2-x-7) +5 lt 0 ` Let `x^2-x=y` ` rArr (y-1)(y-7)+5 lt 0 ` `rArr y^2-8y + 12 lt 0 ` or `(y-6)(y-2) lt 0 ` `rArr 2 lt y lt 6 ` `rArr 2 lt x^2 -x lt 6 ` Now `x^2 - x - gt 2` or `(x-2) ( x+1) gt 0 ` or `(x-2)(x+1) gt 0 ` `rArr x in (- oo , -1 ) cup (2, oo) ` `x^2-x-6 lt 0 ` `rArr ( x-3)(x+2) lt 0 ` `rarr x in (-2 , 3)` Form (1) and (2) common values of `x are x in (-2, -1) cup (2,3)` |
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| 52. |
`||x| -3| gt 1.` |
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Answer» `||x| -3 gt 1.` `rArr |x|-3 gt -1 or |x| -3 gt 1` `rArr |x| lt 2 of |x| gt 4` `rArr -2 lt x lt 2 or x lt -4 or x gt 4 ` |
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| 53. |
Solve `|1+ (3)/(x)| gt 2` |
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Answer» `|1+ (3)/(x)| gt 2` or `-1le (x-3)/(x+1) le 1` `rArr (-4)/(x+1) le 0 " and " 0 le (x-3)/(x+1)+1` `rArr (-4)/(x+1) le 0 and 0 le (2x-2)/(x+1)` `rArr x gt -1 and { x lt -1 or x ge 1}` `rArr x ge 1` |
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| 54. |
Find the values of x of for which `sqrt(5-|2x-3|)` is defined |
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Answer» Correct Answer - `x in [-1,4]` We must have `5-|2x-3|ge 0 ` or `|2x-3|le 5` or `-5 le 2x -3 le 5 ` `rArr x in [-1,4]` |
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| 55. |
Solve `(2)/(x^2-x+1)-(1)/(x+1)-(2x-1)/(x^3+1) ge 0.` |
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Answer» `(2)/(x^2-x+1)-(1)/(x+1)-(2x-1)/(x^3+1) ge 0.` `(2(x+1)-(x^2-x+1)-(2x-))/((x+1)(x^2-x+1))ge 0 ` ` or -((x^2-x-2))/((x+1)(x^2-x+1)) ge 0 ` `(-(x-2)(x+1))/((x+1)(x^2-x+1))ge0` `or (2-x)/(x^2-x+1) ge 0 where x d ne -1` `or 2- x ge 0, x ne -1 " " (as x^2-x+1 gt 0 for AA x in R)` `rArr x le 2 , x ne -1 ` |
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| 56. |
Solve `x sqrt(x ) ge sqrt( x )-3` |
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Answer» `x+ sqrt(x) ge sqrt(x)-3` is meaningful only when `x gt 0 ` Now `x+sqrt(x ) gt sqrt(x)-3` `rArr x gt -3 ` From (1) and (2) ,we have `x ge 0` |
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| 57. |
Solve `|x^2-x-2|+|x+6|=|x^2-2x-8|` |
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Answer» Correct Answer - `x in (- oo ,-6 ] cup [-1,2]` `|x^2-x-2|+|x+6|=|x^2-2x-8|` or `|x^2-x-2|+|x+6|=(x^2-x-2)-(x+6)|` or `|x^2-x-2|+|-x-6|=|(x^2-x-2)+(-x-6)|` or `(x^2-x-2)(-x-6)ge 0 ` or `(x^2-x-2)(x+6) le 0 ` or `(x-2)(x+1)(x+6) le 0 ` or ` x in (-oo,-6] cup [-1,2]` |
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| 58. |
Solve `1 le |x-2| le 3` |
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Answer» `1 le |x-2| le 3` `rArr -3 le x -2 le -1 or 1 le x -2 le 3` `rArr -1 le x le 1 or 3 le x le 5` `rArr x in [-1 , 1 ] cup [ 3,5]` |
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| 59. |
Solve `|| x-1|-2 | lt 5 ` |
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Answer» `|| x-1|-2 | lt 5 ` `rArr -5 lt |x-1| lt 5` `rArr -3 lt |x-1 | lt 7 ` `0 le | x -1 | lt 7 ` `rArr -7 lt x -1 lt 7` `rArr -6 lt x lt 8 ` |
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| 60. |
Solve `|1-(|x|)/(1+|x|)| ge 1/2` |
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Answer» Correct Answer - `-1 le x le 1` |1-(|x|)/(1+|x|)|ge 1/2 or `|1/(1+x|x|)|ge1/2` or 1+|x|le2` or `|x|le 2 ` `-1 le x le 1` |
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| 61. |
Solve `|x-3| ge 2` |
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Answer» `|x-3| ge 2` `rArr x-3 le -2 or x -3 ge 2 ` `rArr x le 1 or x ge 5 ` |
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| 62. |
Solve `|(x-3)/(x+1)| le 1` |
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Answer» `|(x-3)/(x+1)| le 1` or `-1 le (x-3)/(x+1) le 1` `rArr (x-3)/(x+1)-1 le 0 " and " 0 le (x-3)/(x+1)+1` `rArr (-4)/(x+1) le 0 " and " 0 le (2x -2)/(x+1)` `rArr x lt -1 " and " { x lt - 1 or x ge 1}` `rArr x ge 1` |
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| 63. |
Solve `0 lt | x-3| le 5` |
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Answer» `0 lt | x-3| le 5` `rArr -5 le x -3 lt 0 or 0 lt x -3 le 5` `rArr -2 le x lt 3 or 3 lt x le 8` `rArr x in [ -2 ,3 ) cup ( 3,8]` |
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| 64. |
Solve `||x-2|-3|lt 5 ` |
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Answer» Correct Answer - `-6 lt x lt 10 ` `||x-2|-3|lt 5` or `-5 lt |x-2|-3 lt 5 ` `-2 lt |x-2| lt 8` or `-8 lt x -2 lt 8` `or -6 lt x lt 10 ` |
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| 65. |
Solve `|x^2+x-6 |lt 6 ` |
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Answer» Correct Answer - `x in (-4,-1) cup (0,3)` `|x^2+x-6|lt 6` `rArr -6 lt x^2 +x -6 lt 6` Now `-6 lt x^2 +x -6 ` or `x(x+1) gt 0 ` `rArr x in (-oo, -1)cup(0,oo)` `x^2+x-6 lt 6` or `x^2+x-12 lt 0 ` or ` (x+4)(x-3) lt 0 ` `rArr x in (-4,3)` From (1) and (2) common values are `x in(-4 ,-1 ) cup (0,3)` |
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| 66. |
Sove `((x+2)(x^2-2x+1))/(-4 +3x-x^2) ge 0 ` |
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Answer» Correct Answer - `x in (-oo ,-2] cup {1}` `((x+2)(x^2-2x+1))/(-x^2+3x-4) ge 0 ` or `((x+2)(x-1)^2)/(x^2-3x+4) ,e 0 ` Now `x^2- 3x+4 gt 0 AA x in R` `therefore (x+2)(x-1)^2 le 0 ` `rArr x in (-oo , -2 ] cup {1}` |
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| 67. |
Find the value of 1/x for the following values of x : (a) [-5 , -1] (b) (3,6) (c ) (-2 , 3) (d ) (-oo , -2] ( e) [-3,4] |
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Answer» Correct Answer - (a) ` (1/5, 1/2) ( b) (-5,-1/5) ( c) (0,1/3 ) (d) [ -1/2 , 0 ) ( e) (-oo - 1/3 ] cup [ 1/4 , oo) ` (a) `2 lt x lt 5` ` rArr 1/2 gt 1/xgt1/5` `rArr 1/(x)in(1/5,1/2)` (b) `-5 le x lt -1` `rArr -1/5 ge 1/x gt-1` `rArr 1/(x)in(-1,-1/5]` (c) `x gt3 or 3 lt x lt oo` `x le -2 or - oo lt x le-2` `rArr 0 gt1/xge-1/2` `rArr 1/x""in[-1/2,0)` (e ) `x in [-3 4]or -3 le x le 4` Now for `1/x `to get defined ,we must have `-3 le x lt 0 ` or `0 lt x le 4` `rArr -1/3 ge 1/xgt -ooor 1/4 le 1/x""lt oo` `rArr 1/x""in(-oo,1/3]cup[1/4,oo)` |
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| 68. |
Solve `(x)/(x+2) le (1)/(|x|)` |
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Answer» `(x)/(x+2) le (1)/(|x|)` `or (x|x|)/(x+2) le 1` or ` (x|x|)/(x+2) le 1` `or (x|x| - x- 2 )/(x+2) le 0` Case (i) `x ge 0 ` or `((x^2)-x-2)/(x+2) le 0` or `0 le x le 2 (as x ge 0)` Case(ii) `x lt 0 ` `(-x^2-x-2)/(x+2) ge 0 ` `rArr -2 lt x lt 0 (as x^2 + x+2 gt 0 AA x in R, and x lt 0 )` From (1) and (2) `x in (-2,2]` |
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| 69. |
Find the values of 1/x for the following values of x (2,5) (b ) [-5 ,-1] ( c) `( 3, oo) ` (d) `(-3,oo) ` ( e) `(-oo , 4) ` |
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Answer» Correct Answer - `(a) [1,5] (b) [ 9 , 36 ] ( c) [0,9] ( d) [0,oo) (e) [0,oo,) ` `-5 le x le -1` `rArr -2 lt x lt 0 or0 le x le 3` `For -2 lt x lt 0, x^2 in (0,4)` and for `0le x le 3, x^2 in [0,9] " "(1)` From (1) and (2) `x^2 in[0,9] " " (2)` Alternatively ,`x in (-2 ,3]` now least value of `x^2` is 0 which Greatest value of `x^2` is 9 for x =3 `rArr x^2 in [0,9]` (d) `(-3,oo)` Here least value of `x^2` is 0 for x=0 and when x goes up to `rArr x^2 in [0,oo)` ( e) `(-oo , 4)` Here least value of `x^2` is 0 for x =0 for x=0 and`x^2 rarr oo` when `xrarr -oo` Hence `x^2in [0,oo)` |
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| 70. |
Find the number of integal values of x satisfying `sqrt(-x^2+10x-16) lt x -2` |
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Answer» Correct Answer - 3 `sqrt(-x^2+10x-16) lt x-2 ` We must have `-x^2+10 x -16 gt 0 ` or `x^2- 10 x+16 le 0 ` `rArr 2 le x le 8 ` Also `-x^2+10x- 16 lt x^2 -4x +4` or `2x^2-14x+20 gt 0` or `x^2- 7 x + 10 gt 0 ` `rArr x gt 5 or x lt 2` From (1) and (2),`5 lt x le 8 rArr x = 6,7,8` |
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| 71. |
Solve `(1)/(|x|-3) lt 1/2` |
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Answer» Correct Answer - ` x in (-oo ,5 ) cup (-3 , 3) cup (5,oo)` `(1)/(|x|-3)-1/2 lt 0` `or (2-|x|+3)/(|x|-3) lt 0 ` `or (|x|-5)/(|x|-3) gt 0 ` `rArr |x| gt 5 or |x| lt 3` ` rArr x lt 5 or x lt -5 or -3 lt x lt -3` `rArr x in (-oo , -5)cup(-3,3) cup (5,oo)` |
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| 72. |
Number of intergal values of x satisfying the inequality `(x^2+6x-7)/(|x+2||x+3|) lt 0 `isA. 5B. 6C. 7D. 8 |
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Answer» Correct Answer - A We have `(x^2+6x-7)/(|x+2||x+3|) lt 0 ` Clearly ,` ne -2,-3` Now `x^2+6x-7 lt 0 ` `rArr (x+7)(x-1) lt 0 ` `rArr in (-7,1)-{-3,-2}` So ,intergers are {-6 ,-5 -4,-1,0} |
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| 73. |
If `-4lex lt 2 ` then ||x+2|-3 lies in the inervalA. (1,3]B. [1,3]C. [0,3]D. `[0,oo)` |
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Answer» Correct Answer - C We have `-4 le x lt 2` `rArr -2 le x +2 lt 4 ` `rArr 0 le x+2 lt 4 ` `rArr -3 le |x+2|lt 4 ` `rArr 0 le ||x+2|-3| le 3 ` |
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| 74. |
In a group of 500 people, 350 speak Hindi and 300 speak English. It is given that each person speaks at least one language. (i) How many people can speak both Hindi and English? (ii) How many people can speak Hindi only? (iii) How many people can speak English only? |
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Answer» Let H be the set of people who speak Hind, and E be the set of people who speak English. `therefore n(H cup E)= 500 ,n(H)= 350,n(E)=300` We have to find `n(H cap E)`. Now `n (H cup E)=n(H)+n(E)-n(H cap )` `therefore 500=350 + 300 -n (H cap E)` `rArr 500=650 -n (H cap E)` `therefore n(H cap E )=150` Thus, 150 people can speak both Hindi and English. Number of people who speak Hindi only `= n(H) - n(H cap E) ` = 350 150 = 200 Number of people who speak English only `= n(E) n(H cap E)` = 300 - 150 =150 |
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| 75. |
If A={a,e,I,o,u} and B={a,I,e,c,d } then `n(A-B)` is ______. |
| Answer» Correct Answer - C | |
| 76. |
If A={a,e,I,o,u,a,e,i} then n (A) ______A. 4B. 8C. 16D. 5 |
| Answer» Correct Answer - D | |
| 77. |
If Y={{ a,e},{I,o},u} then which of the following is a corrcet statements ?A. `{a,e} sub y `B. `{a,e } ne Y`C. `{{a,e}} ne Y`D. `{a,e} ne Y` |
| Answer» Correct Answer - B | |
| 78. |
In a office ,the ratio of the percentages of employees who like only tea, percentage of employees who like only coffee ,perscentage of employees who like neither of the drinks is 8 : 7: 6: 4 Find the percentage of employees who like neither of the deinks.A. 0.12B. 0.08C. 0.2D. 0.16 |
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Answer» Correct Answer - D Let the percentage of employes who like only tea be 8 x % The percentage of employes who like only coffee = 7 x % The percentage of employees who like both the drinks = 6 x % The pecentage of employes who like neither of the drinks = 6 x % The percentage of employes who like neither of the drinks = 4 x % `n( mu)=-n( A cap B)` = 8 x %+7 x%+6 x%+4 x % = 100 % 25 x = 100 x= 4 `therefore ` 4 x = 16 Hence , the correct option is ( d) |
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| 79. |
In a group of 36 persons ,20 take coffee but not tea. 16 take tea coffee .Find the number of persons who take niether tea nor coffee.A. 2B. 1C. 0D. 3 |
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Answer» Correct Answer - C 20 take only coffee. 16 take only tea , `therefore` Number of persons who take either coffer or tea = 36 = Number of persons in the group `therefore `Required number =0 Hence , the correct option is (c ) |
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