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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. |
Solve:`(|x-3|)/(x-3)gt0,x in R` |
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Answer» Correct Answer - `(3,oo)` `(|x-3|)/(x-3)gt0 iff x-3 gt0iff xgt3 iff x in(3,oo)`. |
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| 2. |
Solve: `(2x-3)/(3x-7)gt0,x in R` |
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Answer» Correct Answer - `(-oo,(3)/(2))uu((7)/(3),oo)` `(2x-3lt0and3x-7lt0)or(2x-3gt0and3x-7gt0)` `rArr (xlt(3)/(2)andxlt(7)/(3))or(xgt(3)/(2)andxgt(7)/(3))` `rArr(xlt(3)/(2))or(xgt(7)/(3))rArr x in(-oo,(3)/(2))uu((7)/(3),oo)`. |
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| 3. |
Solve: `(x-3)/(x+4)gt0,x in R` |
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Answer» Correct Answer - `(-oo,-4)uu(3,oo)` `(x-3ltandx+4lt0)or(x-3gt0andx+4gt0)` `rArr(xlt3 andxlt-4)or(xgt3andxgt-4)` `rArr (xlt-4) or (xgt3) rArr x in(-oo,-4) uu(3,oo)`. |
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| 4. |
Find the solution set of the inequation `(|x-2|)/((x-2))gt0,ne2`. |
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Answer» Correct Answer - `(2,oo)` Clearly,`(x-2)gt0 rArr x gt 2 rArr x in (2,oo)` |
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| 5. |
Find the solution set of the inequation `(x+1)/(x+2)gt1`. |
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Answer» Correct Answer - `(-oo,-2)` `(x+1)/(x+2)-1gt0 rArr(x+1-x-2)/(x+2)gt0 rArr(-1)/(x+2)gt0rArrx+2lt0` `rArr x lt-2 rArr x in(-oo,-2)`. |
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| 6. |
IQ of a person is given by formula: IQ`=(M A)/(C A)xx100 , w h e r e M A`is mental age and CA is chronological age. If `80lt=I Qlt=140`for a group of 12 year children, find the range of their mental age. |
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Answer» When c = 12, we have `IQ=((m)/(12)xx100)=(25m)/(3)`. `therefore 80leIQle140rArr80le(25m)/(3)le140` `rArr 80le(25m)/(3)and (25m)/(3)le140` `rArr(3)/(25)xx80 lem and mle140xx(3)/(25)` `rArr(48)/(5)lem andm le(84)/(5)` `rArr 9.6 le m le 16.8`. Hence, the required mental age for a group of 12-year children is 9.6 years or more and 16.8 years or less. |
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| 7. |
Solve the sytem of inequation `x-2ge0,2x-5le3`. |
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Answer» Correct Answer - `x in [2,4]` `(x-2lge0and2x-5le3)` `rArr(xge2 andxle4)rArr2lexle4 rArrx in[2,4]` |
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| 8. |
Find the solution set of the inequation` (1)/(x-2)lt0`. |
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Answer» Correct Answer - `(-oo,2)` `(1)/((x-2))lt0rArrx-2lt0rArrxlt2rArrx in(-oo,2)` |
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| 9. |
Solve: `(1)/(x-1)le2,x in R` |
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Answer» Correct Answer - `(-oo,1)uu[(3)/(2),oo)` `(1)/(x-1)-2le0 rArr(1-2x+2)/(x-1)le0rArr(3-2x)/(x-1)le0` `rArr (3-2x le0 and x-1gt0)or(3-2xge0andx-1lt0)` `rArr(3le2x and xgt1) or(-2xge-3 and x lt1)` `rArr(xge(3)/(2) andxgt1) or (xle(3)/(2) andxlt1)` `rArr (xlt1) or(xge(3)/(2))=x in(-oo,1)uu[(3)/(2),oo)`. |
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| 10. |
Solve: `|x+1|+|x|gt3,x in R` |
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Answer» Correct Answer - `(-oo,-2)uu(1,oo)` Putting x +1 = 0 and x = 0, we get x = -1 and x = 0 as the critical points. These points divide the whole real line three parts, namely`(-oom-1),[-1,0)` and `[0,oo)`. Case I When`-oo lt x lt -1`. In this case,`x+1lt 0 and xlt0`. `therefore |x+1|=-(x+1) =-x-1 and |x|=-x`. `So,|x+1|+|x|gt3 rArr -x-1-xgt3 rArr -2xgt4rArr x lt-2`. `therefore " solution set in the case " =(-oo,-2)" "[therefore -oolt x lt-1]` Case II When `=1lexlt0`. In the case, `x+1ge0andxlt0`. `therefore |x+1|=x+1 and |x|=-x`. `So, |x+1|+|x|gt3 rArr x+1-xgt3 rArr 1 gt3`, which is abured. Case III When `0 le x lt oo`. In this case, `x+1 gt 0 and x ge 0`. `therefore |x+1| = x+1 and |x| = x`. So, `|x+1| +|x|gt 3 rArr x+1+xgt 3 rArr 2x gt2 rArr x gt 1`. ` therefore` solution set in the case `=(1,oo)`. Hence, solution set `=(-oo,-2)uu(1,00)`. |
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| 11. |
Solve: `(x-3)/(x+1)lt0,x in R` |
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Answer» Correct Answer - `(-1,3)` `(x-3lt0andx+1gt0) or(x-3gt0 and x+1lt0)` `rArr(xlt3and x gt-1) or (xgt3 andxlt-1)` `rArr -1ltxlt3rArrx in(-1,3)." "[xge3 andx lt-1 " is not ossible"]` |
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| 12. |
Solve each of the following inequations and represent the solution set on the number line. `3x+8gt2, " where " (i)x inZ," "(ii)x inR`. |
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Answer» Correct Answer - (i) `{-1, 0,1,2,3,4,…}` `(##RSA_MATH_XI_C06_E01_004_A01##)` (ii) `(-2, oo)` `(##RSA_MATH_XI_C06_E01_004_A02##)` `3xgt-6rArrxgt-2`. (i) All integers less than `-2 "are" -1,0,1,2,3,4,…` (ii) Set of all real numbers greater than `-2 " is "(-2,oo)` |
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| 13. |
Solve: `(3)/(x-2)gt2,x in R` |
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Answer» Correct Answer - `(2,(7)/(2))` `(3)/(x-2)-2gt0 rArr (3-2x+4)/(x-2)gt0rArr(7-2x)/(x-2)gt0` `rArr(7-2xlt0andx-2lt0) or(7-2x gt0and x-2gt0)` `rArr (7lt2x and xlt2)or(7gt2x andxgt2)` `rArr(xgt(7)/(2)andxlt2)or(xlt(7)/(2)andxgt2)` `rArr 2ltxlt(7)/(2) rArr x in(2,(7)/(2))" "[xlt(7)/(2)andxgt2" cannot hold"]` |
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| 14. |
Solve each of the following inequations and represent the solution set on the number line. `3-2xge4x-9, " where " x in R`. |
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Answer» Correct Answer - `(-oo,2]` `(##RSA_MATH_XI_C06_E01_007_A01##)` `3-2xge4x-9rArr-6x ge-12 rArrxle2`. `therefore" solution set " =(-oo,2]`. |
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| 15. |
Solve: `|(2x-1)/(x-1)|gt2,x in R` |
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Answer» Correct Answer - `((3)/(4),1)uu(1,oo)` Using`|x|gta iff x lt -a x gt a`, we have `|(2x-1)/(x-1)|gt2 iff(2x-1)/(x-1)lt-2or(2x-1)/(x-1)gt2`. Case I When`(2x-1)/(x-1)lt-2` `(2x-1)/(x-1)+2lt0 rArr (2x-1+2x-2)/(x-1) lt 0 rArr (4x-3)/(x-1) lt 0`. `therefore (4x-3 lt0and x-1gt0)or(4x-3 gt0 and x-1lt0)` `rArr (xlt(3)/(4)andxgt1) or(xgt(3)/(4) and x lt1)` `rArr (3)/(4)ltxlt1 rArr x in((3)/(4),1)`. Case II When `(2x-1)/(x-1) gt 2` Then, `(2x-1)/(x-1) -2 gt 0 rArr (2x-1-2x+2)/(x-1) gt 0 rArr (1)/(x-1) gt0` `rArr x-1gt0rArr gt1 rArrx in(1,oo)`. `therefore "solution set " =((3)/(4),1)uu(1,oo)`. |
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| 16. |
Solve `-4xgt16, " when "x inZ`. |
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Answer» Correct Answer - `x in [-5,-6,-7,…]` `-4xgt16rArrx lt-4rArrx in[-5,-6,-7,-8,...}`. |
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| 17. |
Solve `(x-2)/(x+5)gt2.` |
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Answer» We have `(x-2)/(x+5)gt2` `rArr (x-2)/(x+5)-2gt0` `rArr(x-2-2x-10)/(x+5)gt0` `rArr (-(x+12))/(x+5)gt0` `rArr (x+12)/(x+5)lt0` [on mutipling both sides by -1]. `therefore " either "(x+12lt0 andx+5lgt0)or(x+12gt0andx+5lt0)`. Case I When `x+12lt0andx+5gt0`. In this case, `x+12lt0andx+5gt0`. `rArr xlt-12andxgt-5` `rArr -5ltxlt-12, " which is impossible " " "[therefore-12lt-5]` Case II when `x+12gt0andx+5lt0`. In this case, `x+12gt0andx+5lt0`. `rArrxgt-12 andxlt-5` `rArr -12ltxlt-5rArrx in(-12-5)`. `therefore` solution set`=(-12-5)`. |
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| 18. |
Solve: `(|x+2|-x)/(x)lt2,x in R` |
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Answer» Correct Answer - `(-oo,-2)uu(1,oo)` `(|x+2|-x)/(x)-2lt0 rArr(|x+2|-x-2x)/(x)lt0rArr (|x+2|-3x)/(x)lt0` Case I When ` x +2ge 0` `Then, `x ge -2 and |x+2|=x+2`. `therefore (|x+2|-3x)/(x)lt0 rArr (x+2-3x)/(x)lt0rArr(2-2x)/(x)lt0` `rArr (2)/(x)-2lt0 rArr (2)/(x)lt2 rArr 2x lgt2 rArr x gt1`. `therefore (xge-2 and xgt1)rArr x gt1rArrx in(1,oo)`. Case II When ` x+2 lt0`. Then ` x lt -2 and |x+2| = -(x+2)` `therefore (|x+2|-3x)/(x) lt0 rArr (-x-2-3x)/(x)lt0 rArr(-4x-2)/(x)lt0` `rArr (4x+2)/(x)gt0 rArr 4+(2)/(x)gt0rArr (2)/(x)gt-4`. `rArr -4xlt2 rArr x lt(-1)/(2)`. `therefore x lt-2 and x lt (-1)/(2) rArr x lt-2 rArr x in(-oo,-2)`. Hence, solution set `=(-oo,-2)uu(1,oo)`. |
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| 19. |
Find the solution set of the inequation `|x-1|lt2`. |
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Answer» Correct Answer - `(-1,3)` Using `|x|lta iff-altxlta,` we get `|x-1|lt2iff-2ltx-1lt2rArr-1ltxlt3 rArr x in(-1,3)`. |
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| 20. |
Solve `x+5gt4x-10, " when "x inR`. |
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Answer» Correct Answer - `x in (-oo,5)` `x+5gt4x-10rArr3xlt15rArrxlt5 rArr x in(-oo,5)`. |
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| 21. |
Solve `(3)/(x-2)lt1, " when "x inR`. |
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Answer» Correct Answer - `x in (-oo,2)uu(5,oo)` `(3)/(x-2)-1lt0 rArr(3-x+2)/(x-2)lt0rArr(5-x)/(x-2)lt0`. `therefore (5-xlt0andx-2gt0)or(5-xgt0 andx-2lt0)` `rArr(xgt5 and xgt2)or(xlt5 and xlt2)` `rArr(xgt5 and xlt2)rArrx in(-oo,2)uu(5,oo)`. |
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| 22. |
Solve `|4-x|lt2`. |
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Answer» We have, `|x|ltaiff-altxlta`. `therefore|4-x|lt2` `iff-2lt4-xlt2` `iff-2lt4-xand4-xlt2` `iff -2-4-xand-xlt2-4` `iff-6lt-xand-xlt-2` `iff 6gt x andxgt2`. `iff 2ltxlt6`. `therefore` solution set `={x inR:2ltxlt6}=(2,6)`. |
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| 23. |
Solve: `(1)/(|x|-3)le(1)/(2),x in R` |
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Answer» Correct Answer - `*(-oo,-5]uu(-3,3)uu[5,oo)` Case I When `x ge 0` Then, `|x|=x` `therefore (1)/(|x|-3)le(1)/(2)rArr(1)/(x-3)-(1)/(2)le0rArr(2-x+3)/(2(x-3))le0rArr(5-x)/(x-3)le0`. `therefore (5-xle0 and x-3gt0) or(5-x ge0 and x-3lt0)` `rArr (xge5 and x gt3) or(x le5 and x lt3)` `rArr(xge5) or (xlt3)` `rArr (0lex lt3) or(xge5)` `rArr x in [0,3) uu[5,oo)" "[therefore x ge 0]` Case II When `x lt 0`. Then, `|x| = -x` `therefore(1)/(|x|-3)le(1)/(2) rArr(1)/(x-3)le(1)/(2)rArr(-1)/(x+3)le(1)/(2)rArr (1)/(2)+(1)/(x+3)le0` `rArr (x+3+2)/(x+3)ge0 rArr (x+5)/(X+3)ge0` `therefore (x+5 ge0and x+3gt0) or(x+5le0 and x+3 lt0)` `rArr(xgt-5 and xgt-3) or (xle-5 and x lt-3)` `rArr (xgt-3) or (xle-5)`. Bu `xlt0`. `therefore(-3ltxlt0)or(xle-5)rArr x in(-3,0)uu(-oo,-5]`. Hence, ` x in(-oo,-5]uu(-3,0)uu[0,3)uu[5,oo)` `x in(-oo,-5] uu(-3,3)uu[5,oo)`. |
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| 24. |
Solve: `|4x-5|le(1)/(3),x in R` |
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Answer» Correct Answer - `[(7)/(6),(4)/(3)]` Using `|x|leaiff-alexlea`, we get `|4x-5|le(1)/(3)rArr-(1)/(3)le4x-5le(1)/(3)rArr(14)/(3)le4x and4x le(16)/(3)` `rArr(7)/(6)lex and x le(4)/(3) rArr(7)/(6)le x le (4)/(3) rArr x in [(7)/(6),(4)/(3)]`. |
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| 25. |
Solve `|3-4x|ge9, x in R`. |
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Answer» We have, `|x|gea iff x le-aorxgea`. `therefore |3-4x|ge9 iff3-4x le-9or3-4x ge9` `iff -4xle-9-3 or -4xge 9-3` `iff -4xle-12or -4x ge6` `iffx ge3 or le(-3)/(2)` `iff x le(-3)/(2)orxge3` `iff x in(-oo, (-3)/(2)]orx in[3,oo)`. `therefore` solution set `=(-oo,(-3)/(2)]uu[3,oo)`. |
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| 26. |
Solve `|3x-2|le(1)/(2),x inR.` |
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Answer» We have, `|x|lea iff-alexlea`. `therefore|3x-2|le(1)/(2)iff-(1)/(2)le3x-2le(1)/(2)` `iff(-1)/(2)le3x-2and 3x-2le(1)/(2)` `iff(-1)/(2)+2le3xand3xle(1)/(2)+2` `iff(3)/(2)le3x and 3xle(5)/(2)` `iff(1)/(2)lex and xle(5)/(6)` `(1)/(2)lexle(5)/(6)`. `therefore` solution set `={x inR:(1)/(2)lexle(5)/(6)}=[(1)/(2),(5)/(6)]`. |
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| 27. |
Solve `(x)/(x-5)gt(1)/(2), " when "x inR`. |
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Answer» Correct Answer - `x in (-oo,-5)uu(5,oo)` `(x)/(x-5)-(1)/(2)gt0rArr(2x-x+5)/(2(x-5))gt0rArr(x+5)/(x-5)gt0`. `therefore (x+5lt0 and x-5lt0)or(x+5gt0and x-5gt0)` `rArr(xlt-5 and xlt5)or(xgt-5 and xgt5)` `rArr(xlt-5)or(xlt5)rArrx in(-oo,-5)uu(5,oo)`. |
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| 28. |
Solve: `|3x-7|gt4,x in R` |
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Answer» Correct Answer - `(-oo,1)uu((11)/(3),oo)` `"Using" |x|gta rArr x lt -aorxgta`, we get `|3x-7|gt4 rArr 3x -7lt-4or 3x-7gt4` `rArr3xlt3 or 3xgt11 rArrxlt1orxgt(11)/(3)` `rArrx in(-oo,1)uu((11)/(3),oo)`. |
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| 29. |
Solve each of the following system of equations in`Rdot``1lt=|x-2|lt=3` |
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Answer» 1)`1<=|x-2|` `|x-2|>=1` `x-2>=1` `x>=3,x<=1` `x in(-oo,1]uu[3,oo)` 2)`|x-2|<=3` `|x-2|<=3` `-3<=x-2<=3` `-1<=x<=5` `x in[-1,5]` `x in [-1,1]uu[3,5]`. |
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| 30. |
Solve `(2)/(|x-3|)gt5,x inR.` |
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Answer» Clearly, `x-3ne0 "and therefore",xne3`. We have, `(2)/(|x-3|)gt5" "...(i)` Since `|x-3|` is positive, we may multiply both sides of (i) by `|x-3|`. This gives `2gt5|x-3|` `iff (2)/(5)gt|x-3|` `iff|x-3|gt(2)/(5)` `iff (-2)/(5)ltx-3lt(2)/(5)" " [therefore|x|lta iff-altxlta]` `iff (-2)/(5)ltx-3and x-3lt(2)/(5)` `iff (-2)/(5)+3ltx and xlt(2)/(5)+3` `iff (13)/(5)ltx and xlt(17)/(5)`. `iff (13)/(5)ltxlt(17)/(5)`. Also, as shown above, `x ne 3`. `therefore` solution set `={x inR :(13)/(5)ltxlt(17)/(5)}-{3}` `=(2.6,3.4)-{3}=(2.6,3)uu(3,3.4)`. |
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| 31. |
Solve: `|5-2x|le3,x in R` |
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Answer» Correct Answer - `[1,4]` `"Using" |x|leaiff-alexlea`, we get `|5-2x|le3 rArr-3le5-2x le 3` `rArr -3 le5-2x and 5-2xle3` `rArr 2xle8 and2xge2 rArrxle4 xge1` `rArr 1 le xle4 rArrx in[1,4]`. |
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| 32. |
Solve the following systems of linear inequations: `(4)/(x+1)le3le(6)/(x+1),xgt0` |
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Answer» Correct Answer - `(1)/(3)le xle1` `(4)/(x+1)le3rArr (4)/(x+1)-3le0rArr(1-3x)/(x+1)le0`. `therefore (1-3x le0 and x+1gt0) or(1-3x ge0 andx+1lt0)` `rArr (xge(1)/(3) and xgt-1)or(xle(1)/(3)and xlt-1)` `rArr xge(1)/(3)orxlt-1rArrx in[(1)/(3),oo)uu(-oo,-1)` Again `(6)/(x+1)ge3 rArr (6)/(x+1)-3 ge0 rArr(3-3x)/(x+1)ge0 rArr (1-x)/(1+x)ge0`. `therefore (1-xge0 and 1+xgt0) or(1-xle0and1+xlt0)` `rArr (xle1 andxgt-1) or(xge1 andxlt-1)` `-1 ltxle1 rArr x in(-1,1]`. `therefore "solution set"={(-oo,-1)uu[(1)/(3),oo)}nn(-1,1]=[(1)/(3),1]`. |
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| 33. |
Solve each of the following inequations and represent the solution set on the number line. `(2x-1)/(12)-(x-1)/(3)le(3x+1)/(4)," where " x in R`. |
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Answer» Correct Answer - `(0,oo)` `(##RSA_MATH_XI_C06_E01_011_A01##)` `(2x-1)-4(x-1)lt3(3x+1)rArr-2x+3lt9x+3rArr-11x lt0rArrx gt0`. `therefore " solution set " =(0,oo)` |
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| 34. |
Solve the following systems of linear inequations: `1le|x-2|le3` |
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Answer» Correct Answer - `[-1,1]uu[3,5]` (i) `|x-2|ge1 rArr (x-2) le-1 or(x-2)ge1` `rArr x le1 orxge3 rArr x in(-oo,1]uu[3,oo)`. `(ii) |x-2|le3 rArr -3lex-2le3rArr -1lexle5 rArr x in[-1,5]`. `therefore " required solution set ={(-oo,1]uu[3,oo)}nn[-1,5]=[-1,1]uu[3,5]`. |
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| 35. |
Solve each of the following inequations and represent the solution set on the number line. `(1)/(4)((2)/(3)x+1)ge(1)/(3)(x-2)," where " x in R`. |
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Answer» Correct Answer - `(-oo,5.5]` `(##RSA_MATH_XI_C06_E01_010_A01##)` `(1)/(4)*((2x+4))/(3)ge(x-2)/(3)rArr2x+3ge4x-8rArr-2xge-11rArrx le(11)/(2)` `therefore " solution set " =(-oo,5.5]` |
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| 36. |
Solve each of the following inequations and represent the solution set on the number line. `(x)/(4)lt((5x-2))/(3)-((7x-3))/(5)," where " x in R`. |
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Answer» Correct Answer - `(4,oo)` `(##RSA_MATH_XI_C06_E01_012_A01##)` `(2x-1)-4(x-1)lt3(3x+1)rArr-2x+3lt9x+3rArr-11x lt0rArrx gt0`. `rArr 15x-16xlt-4rArr-xlt -4 rArrxgt4` `therefore " solution set " =(4, oo)` |
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| 37. |
Solve the system of inequations `-5le(2-3x)/(4)le9`. |
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Answer» We have ` -5le(2-3x)/(4)le9rArr -5le(2-3x)/(4)le9`. Now, `-5le(2-3x)/(4)rArr -20le2-3x` [mutliplying both sides by 4] `rArr -22le-3x` [adding -2 on both sides] `rArr 22le3x` [multiplying both sides by -1] `rArr 3xle22` `rArr x le(22)/(3)` [dividing both sides by 3] `rArr x in (-oo,(22)/(3)]`. And, `(2-3x)/(4)le9 rArr 2-3x le 36` [multiplying both sides by 4] `rArr -3x le34` `rArr x ge 9-(34)/(3)` [dividing both sides by -3] `rArrx in[(-34)/(3),oo)`. `therefore " solution set " =(-oo,(22)/(30)]nn[(-34)/(3),oo)=[(-34)/(3),(22)/(3)]`. |
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| 38. |
Solve each of the following system of equations in`Rdot``(|x-1|-1)/(|x-2|-2)lt=0.` |
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Answer» `(|x-1|-1)/(|x-2|-2)<=0` `x<1` `(1-x-1)/(2-x-2)<=0` `1<=0` `x in phi` `x in[1,2]` `(x-1-1)/(2-x-2)<=0` `(x-2)/(-x)<=0` `(x-2)/x>=0` `x in(-oo,0)uu[2,oo)` `x in phi` `x>=2` `(x-1-1)/(x-2-2)<=0` `(x-2)/(x-4)<=0` `x in[2,4)`. |
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| 39. |
`(|x-2|-1)/(|x-2|-2)le0` |
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Answer» We have either `x-2ge0orx-2lt0`. Case I When `x-2 ge0`. In this case, `x ge 2 and |x-2|=x-2` So, the given inequation becomes `(x-2-1)/(x-2-2)le0rArr(x-3)/(x-4)le0`. `therefore (x-3ge0andx-4lt0)or(x-3le0andx-4gt0)` `rArr (xge3 andxlt4)or(xlt3 andx gt4)` `rArr 3lexlt4" "[thereforexlt3 andx gt4 " isnot possible"]` `rArr x in[3,4)" "["this include " xge2]`. Case II When `x-2lt0` In this case, `xlt 2 and |x-2|=-(x-2)=-x+2`. So, the given inequation becomes `(-x+2-1)/(-x+2-2)le0rArr(-x+1)/(-x)le0` `rArr(x-1)/(x)le0` [mulitplying num. and denom.by -1] `therefore(x-1le0andxgt0)or(x-ge0 and xlt0)` `rArr(xle1 andxgt0)or(xge1andx lt0)` `rArr 0ltxle1" "[thereforex ge1andx lt0 " is not possible"]` `rArrx in(0,1]" "["this include " xlt2]`. Hence, from the above two cases, we get Solution set `=(0,1]uu[3,4)`. |
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| 40. |
The longest side of a trangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side. |
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Answer» Let the minimum length of the shortest side be x cm. Then, the longest side `=3x cm` and the third side `=(3x -2)` cm. `therefore x +3x +(3x-2)ge 61 rArr 7x ge 63 rArr x ge 9`. Hence, the minimum length of the shortest side is 9 cm. |
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| 41. |
The cost and revenue functions of a product are given by `C(x) = 20x +4000 and R(x) = 60x + 2000` respectively, where x is the number of items produced and sold. How many items must be sold to realise some profit? |
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Answer» Let the required number of items sold be x and let P(x) be the profit function, Then, profit function = (revenue function) - (cost function) `rArr P(x)=(60x+2000)-(20x+4000)` `rArr P(x) = 40x -2000`. For some profit, we have `P(x)gt 0`. `therefore 40x-2000 gt 0` `rArr 40x gt 2000` `rArr x gt 50` Hence, the manufacturer must sell more than 50 items ot realise some profit. |
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| 42. |
Solve the system of inequations: `(6x)/(4x-1)lt(1)/(2),(x)/(2x+1)ge(1)/(4)`. |
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Answer» The first inequation of the given system is `(6x)/(4x-1)lt(1)/(2)` `(6x)/(4x-1)-(1)/(2)lt0` `rArr (12x-4x+1)/(2(4x-1))lt0` `rArr (8x+1)/(2(4x-1))lt0rArr(8x+1)/(4x-1)lt0." "...(i)` `therefore (8x+1gt0and 4x-1lt0) or (8x+1lt0and 4x-1gt0)` `rArr(8xgt-1and 4xlt1) or(8xlt-1 and 4xgt1)` `rArr (xgt(-1)/(8)and xlt(1)/(4))or(xlt(-1)/(8)and xgt(1)/(4))` `rArr (-1)/(8)ltxlt(1)/(4)" "[therefore xlt(-1)/(8)and x gt(1)/(4)" is not possible"]` `rArr x in((-1)/(8),(1)/(4))`. The equation inequation of the given system is `(x)/(2x+1)ge(1)/(4)` `rArr(x)/(2x+1)-(1)/(4)ge0` `rArr (4x-2x-1)/(4(2x+1))ge0rArr(2x+1)/(2x+1)ge0" "...(ii)` `therefore (2x-1le0and 2x+1lt0) or(2x-1ge0and2x+1gt0)` `rArr (xle(1)/(2)and xlt(-1)/(2))or(xge(1)/(2)and x gt(-1)/(2))` `rArr (xlt(-1)/(2))or(xge(1)/(2))` `rArr x in(-oo,(-1)/(2))orx in[(1)/(2),oo)` `rArr x in(-oo,(-1)/(2))uu[(1)/(2),oo)` `therefore " solution set " =((-1)/(8),(1)/(4))nn{(-oo,(-1)/(2))uu[(1)/(2),oo)}=phi`. Hence, the given system of inequation has no solution. |
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| 43. |
Solve each of the following inequations and represent the solution set on the number line. `(5x-8)/(3)ge(4x-7)/(2) " where " x in R`. |
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Answer» Correct Answer - `(-oo,(5)/(2)]` `(##RSA_MATH_XI_C06_E01_008_A01##)` `10x-16ge12x-21rArr-2xge-5rArrxle(5)/(2)` `therefore " solution set "=(-oo,(5)/(2)]` |
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| 44. |
Solve each of the following inequations and represent the solution set on the number line. `((2x-1))/(3)ge((3x-2))/(4)-(2-x)/(5)," where " x in R`. |
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Answer» Correct Answer - `(-oo,2]` `(##RSA_MATH_XI_C06_E01_013_A01##)` `20(2x-1)ge15(3x-2)-12(2-x)rArr40x-20ge45x -30-24+12x` `rArr -17x ge -34 rArr x le 2`. `therefore " solution set " = (-oo, 2]` |
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| 45. |
Find all pairs of consecutive even positive integers both of which are larger than 5 such that their sum is less than 23. |
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Answer» Let the required consecutive even integers be x and x + 2. Then, `xgt5 and x +(x+2)lt23` `rArrxgt5 and 2xlt21` `rArr 5ltx and x lt 10.5` `rArr 5 lt x lt 10.5` `therefore` x can take the even integral value 6, 8 and 10. Hence, the required pairs of even integers are (6, 8), (8, 10) and (10, 12). |
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| 46. |
Solve `|x+1|gt4,x inR`. |
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Answer» We have,`|x|gtaiffxlt-aorxgta`. `therefore |x+1|gt4 iffx+1lt-4orx+1gt4` `iff xlt-4-1orxgt4-1` `iff x lt-5orxgt3` `iff x in(-oo,-5)orx in(3,oo)`. `therefore` solution set `=(-oo,-5)uu(3,oo)`. |
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| 47. |
Solve `|x|gt4, " when "x inR`. |
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Answer» Correct Answer - `x in (-oo,-4)uu(4,oo)` Using `|x|gta iff x. lt-aorxgta`, we get `|x|gt4 iffxlt-4orxgt4` `iff x in(-oo,-4)uu(4,00)`. |
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| 48. |
To receive grade A in a cource, one must obtain an average of 90 marksor more in five papers each of 100 marks. If Shikha scored 87, 95, 92 and 94marks in first four papers find the minimum marks that she must score in thelast paper to get grade A in the course. |
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Answer» Correct Answer - 82 marks Let the required marks be x. Then, `(89+93+95+91+x)ge(90xx5)rArrxge82`. |
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| 49. |
In the first four papers each of 100 marks, Rishigot `95 ,72 ,73 ,83`marks. If he wants an average of greater than or equalto `75`marks and less than `80`marks, find the range of marks he should score in thefifth paper. |
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Answer» Let marks in `5^(th)` paper be x Average=`(95+72+73+53+x)/5` `=(323+x)/5` `75<=(323+x)/5<80` `375<=323+x<400` `52<=x<77` `[52,77)`. |
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