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.
| 1. |
The solution set of the inequation `|(1)/(x)-2| lt 4`, isA. `(-oo, -1//2)`B. `(1//6, oo)`C. `(-1//2, 1//6)`D. `(-oo, -1//2) cup (1//6, oo)` |
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Answer» Correct Answer - D We have, `|(1)/(x)-2| lt 4, x ne 0` `rArr (|1-2x|)/(|x|) lt 4 rArr |2x-1| lt 4 |x|` Now, three cases arise: CASE I When `x ge 1//2`. In this case, we have `|2x-1| = 2x-1 " and " |x| =x` `therefore |2x-1| lt 4|x|` `rArr 2x-1 lt 4x rArr 2x gt -1 rArr x gt -(1)/(2)` But, `x ge (1)/(2), " Therefore," x ge (1)/(2)` is the solution in this case. CASE II When `0 lt x lt 1//2:` In this case, we have `|2x-1| = -(2x-1) " and "|x|=x` `therefore |2x-1| lt 4 |x|` `rArr -(2x-1) lt 4x rArr 6x gt 1 rArr x gt (1)/(6)` But, `0 lt x lt 1//2." Therefore, " x in (1//6, 1//2)` CASE III When `x lt 0`: In this case, we have `|x| = -x " and "|2x-1|=-(2x-1)` `therefore |2x-1|lt 4|x|` `rArr -(2x-1) lt -4x rArr 2x lt -1 rArr x lt -(1)/(2)` As `x lt 0. " Therefore, " x in (-oo, -1//2)` is the solution set in this case. Hence, `x in (-oo, -1//2) cup(1//6, oo)`. |
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| 2. |
Write the solution set of the inequation `|x-1|geq|x-3|dot`A. `(-oo, 2]`B. `[2, oo)`C. `[1, 3)`D. none of these |
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Answer» Correct Answer - B Here, x = 1 and x=3 are two critical points which divide the real line into these three parts, namely, `x lt 1, 1 le x lt 3 " and " x ge 3`. So, we discuss the following cases: CASE I When `x lt 1:` In this case, we have `|x-1| = -(x-1) " and " |x-3| = -(x-3)` `therefore |x-1| ge |x-3|` `rArr -(x-1) ge -(x-3) rArr 1 ge 3`, which is absurd. So, the inequation has no solution for `x lt 1` CASE II When `1 le x lt 3` In this case, we have `|x-1| = x-1 " and " |x-3| = -(x-3)` `therefore |x-1| ge |x-3|` `x-1 ge - (x-3) rArr 2x-4 ge 0 rArr x ge 2` But, ` 1 le x lt 3`. Therefore, `1 le x lt 3 " and " x ge 2 rArr x in [2, 3)` CASE III When ` x ge 3` In this case, we have `|x-1| = x-1 " and " |x-3|= x-3` `therefore |x-1| ge |x-3|` `rArr x-1 ge x-3 rArr -1 ge -3,` which is correct. So, the given inequation has all solutions satisfying `x ge3`. Hence, the solution set of the given inequation is `[2, oo)`. |
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| 3. |
The total number of integral points i.e. points having integral coordinates lying in the region represented by the inequations `|x-y| lt 3 " and " |x+y| lt 3` isA. 25B. 36C. 13D. 12 |
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Answer» Correct Answer - C It is evident from the Fig. 6 that there are 13 points in the region enclosed by `|x-y| lt 3 " and " |x+y| lt 3.` The coordinates of the points are `(-2, 0),(-1, 0), (0, 0),(1, 0),(2, 0),(0,1),(0, 2),(0, -1),(0, -2),(1,1), (-1, 1),(-1, 1) " and " (1, -1)` |
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| 4. |
The solution of `6+x-x^(2) gt0`, isA. `-1 lt x lt 2`B. `-2 lt x lt 3`C. `-2 lt x lt -1`D. none of these |
| Answer» Correct Answer - B | |
| 5. |
The set of real values of `x` for which `(10x^2 +17x-34)/(x^2 + 2x - 3) < 8,` isA. `(-5//2, 2)`B. `(0, oo)`C. `(-1, oo)`D. none of these |
| Answer» Correct Answer - B | |
| 6. |
If `p,q,r,s,t` are numbers such that `p+q lt r+s` `q+r lt s+t` `r+s lt t+p` `s+t lt p+q` thn the largest and smallest numbers areA. p and q respectivelyB. r and t respectivelyC. r and q respectivelyD. q and p respectively |
| Answer» Correct Answer - A | |
| 7. |
The solution set of the inequation `|2x-3| lt x-1`, isA. `(4//3, 3//2) cup (3//2, 2)`B. `(4//3, 2)`C. `[3//2, 2)`D. none of these |
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Answer» Correct Answer - B CASE I when `2x-3 ge 0 i.e., x ge (3)/(2)` In this case, we have `|2x-3| = 2x-3` `therefore |2x-3| lt x -1` `rArr 2x-3 lt x-1` `rArr x-2 lt 0 rArr x lt 2 rArr x in [3//2, 2) " "[because x ge 3//2]` CASE II when `2x-3 lt 0 i.e., x lt 3//2` In this case, we have `|2x-3|= (2x-3)` `therefore |2x-3| lt x-1` `rArr -(2x-3) lt x-1` `rArr 3x-4 gt 0 rArr x gt (4)/(3) rArr x in (4//3, 3//2) " " [because x lt 3//2]` Thus, the set of the values of x satisfying the given inequation is `(4//3, 3//2)cup[3//2, 2_=(4//3, 2)` |
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| 8. |
The solution set of the inequation `(|x+3|+x)/(x+2) gt 1`, isA. `(-5, -2) cup (-1, oo)`B. `(-5, -2)`C. `(-1, oo)`D. none of these |
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Answer» Correct Answer - A We have `(|x+3|+x)/(x+2) gt 1` `rArr (|x+3|+x)/(x+2)-1 gt 0` `rArr (|x+3|+x-x-2)/(x+2) gt 0 rArr (|x+3|-2)/(x+2) gt 0` Now two cases arise. CASE I When `x + 3 ge 0 i.e. x ge -3` In this case, we have `|x+3| = x+3` `therefore (|x+3|-2)/(x+2) gt 0` `rArr (x+3-2)/(x+2) gt 0` `rArr (x+1)/(x+2) gt 0 rArr x in (-oo, -2) cup(-1, oo)` But `x ge -3.` ` therefore x in [-3, -2) cup (-1, oo)` CASE II When `x + 3 lt 0 i.e. x lt -3` In this case, we have `|x + 3| = -(x+3)` `therefore(|x+3|-2)/(x+2) gt 0` `rArr (-(x+3)-2)/(x+2) gt 0` `rArr (-x-5)/(x+2) gt 0 rArr (x+5)/(x+2) lt 0 rArr -5 lt x lt -2` But ` x lt -3. "So," x in (-5, -3)`. Hence, the solution set of the given inequation is `(-5, -2)cup (-1, oo)` |
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| 9. |
The solution set of the inequation `(x+4)/(x-3) le2`, isA. `(-oo, 3) cup (10, oo)`B. `(3, 10]`C. `(-00, 3) cup [10, oo)]`D. none of these |
| Answer» Correct Answer - C | |
| 10. |
The number of integral solutions of `(x+2)/(x^2+1)>1/2`is4 2.5 3. 3 4. 2 5. 6A. 4B. 5C. 3D. 2 |
| Answer» Correct Answer - C | |
| 11. |
If `4 le x le 9`, thenA. `(x-4)(x-9) le 0`B. `(x-4)(x-9) ge 0`C. `(x-4)(x-9) lt 0`D. `(x-4)(x-9) gt 0` |
| Answer» Correct Answer - A | |
| 12. |
`(1)/(2) ((3x)/(5) + 4 )ge (1)/(3) (x - 6)`A. `[120, oo]`B. `(-oo, 120]`C. `[0, 120]`D. `[-120, 0]` |
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Answer» Correct Answer - B We have, `(1)/(2) ((3x)/(5) + 4) ge (1)/(3)(x-6)` `rArr (1)/(2) ((3x+20)/(5) ) ge (1)/(3)(x-6)` `rArr (3x + 20)/(10) ge (x-6)/(3)` `rArr 3(3x + 20) ge 10 (x-6) " " ["Multiplying both sides by 30 i.e the LCM of 10 and 3"]` `rArr 9x+60 ge 10x-60` `rArr 9x-10x ge -60-60` `rArr -x ge -120 ` `rArr x le 120` `rArr x in (-oo, 120]` Hence, the solution set of the given inequation is `(-oo, 120]`, REMARK The solution set of simultaneous inequations is the intersection of their solution sets. |
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| 13. |
If `(3(x-2))/(5)gt=(5(2-x))/(3)`, then x belongs to the intervalA. `(2, oo)`B. `[2, oo)`C. `(-oo, 2]`D. none of these |
| Answer» Correct Answer - B | |
| 14. |
If `|3x + 2| lt 1`, then x belongs to the intervalA. `(-1, -1//3)`B. `[-1, -1//3]`C. `(-oo, -1)`D. `(-1//3, oo)` |
| Answer» Correct Answer - A | |
| 15. |
if `|(2)/(x-4)| gt 1`, then x belongs to the internalA. `(2, 6)`B. `(2, 4) cup (4, 6)`C. `(-oo, 2)`D. `(6, oo)` |
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Answer» Correct Answer - B We have, `|(2)/(x-4)| gt 1` `rArr (2)/(|x-4|) gt 1` `rArr 2 gt |x-4|` ` rArr |x-4| lt 2` `rArr 4-2 lt x lt 4+2 " " [because |x-a| lt r hArr a-r lt x lt a+r]` `rArr 2 lt x lt 6` `"But", (2)/(x-4)` is not meaningful for x = 4. `therefore 2 lt x lt 6 " and " x ne 4 rArr x in (2, 4)cup(4,6)` |
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| 16. |
`-5 le (2-3x)/4 le 9`A. `(-oo, 22//3)`B. `[-34//3, 22//3]`C. `[22//3, oo)`D. `(-oo, -34//3]` |
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Answer» Correct Answer - B We have, `-5 le (2-3x)/(4)le 9` `rArr -20 le 2-3x le 36` `rArr -22 le -3x le 34` `rArr -34 le 3x le 22` `rArr -34 le 3x le 22 rArr -(34)/(3) le x le (22)/(3) rArr x in [-34//3, 22//3]` |
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| 17. |
The number of integral solutions of `x^2+9 |
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Answer» Correct Answer - D We have, `x^(2) + p lt (x+3)^(2) lt 8x + 25` `rArr x^(2) + 9 lt x^(2) + 6x + 9 lt 8x + 25` `rArr x^(2) + 9 lt x^(2) + 6x + 9 " and "x^(2) + 6x + 9 lt 8x + 25` `rArr 6x gt 0 " and " x^(2) -2x-16 lt 0` `rArr x gt 0 " and " 1-sqrt(17) lt x lt 1 + sqrt(17)` `rArr 0 lt x lt 1 + sqrt(17) rArr x = 1, 2, 3, 4, 5` |
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| 18. |
Solution set of `(5x-1)lt (x+1)^(2)lt (7x -3)` isA. (1, 4)B. [2, 4]C. (2, 4)D. `(-oo, 1)cup(2, oo)` |
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Answer» Correct Answer - C We have, `(5x-1) lt (x+1)^(2) lt (7x-3)` `rArr 5x-1 lt (x+1)^(2) " and " (x+1)^(2) lt (7x-3)` `rArr (x+1)^(2)-(5x-1) gt 0 " and " (x+1)^(2) - (7x -3) lt 0` `rArr x^(2) - 3x + 2 gt 0 " and " x^(2) - 5x + 4 lt 0` `rArr (x-1)(x-2) gt o " and " (x-1) (x-4) lt 0` `rArr x in (-oo, 1)cup(2, oo) " and " x in (1, 4)` `rArr x in (2, 4)` |
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| 19. |
`if (5x)/(4) + (3x)/(8) gt (39)/(8) " and " (2x-1)/(12)-(x-1)/(3)lt(3x+1)/(4)`, then x belongs to the internalA. `(3, oo)`B. `(0, oo)`C. `(-oo, 3)`D. `(-oo, 0)` |
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Answer» Correct Answer - A We have, `(5x)/(4) + (3x)/(8) gt (39)/(8) " and " (2x-1)/(12)-(x-1)/(3)lt(3x+1)/(4)` `rArr (10x+3x)/(8) gt (39)/(8) " and " (2x-1-4x+4)/(12) lt (3x + 1)/(4)` `rArr (13x)/(8) gt (39)/(8) " and " (-2x+3)/(12) lt (3x+1)/(4)` `rArr 13x gt 39 " and " -2x + 3 lt 9x + 3` `rArr x gt 3 " and " -11x lt 0` `rArr x gt 3 " and " x gt 0` `rArr x in (3, oo) " and " x in (0, oo) rArr x in (3, oo)` |
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| 20. |
`Sol v e(8x^2+16 x-51)/(2x^2+5x-12)>3`A. `(3//2, 5//2)`B. `(-4, -3)`C. `(-4, -3)cup (3//2, 5//2)`D. none of these |
| Answer» Correct Answer - C | |
| 21. |
Solution set of `(5x-1)lt (x+1)^(2)lt (7x -3)` isA. `phi`B. `{1}`C. `{2}`D. `{3}` |
| Answer» Correct Answer - D | |
| 22. |
The solution set of `x^(2) +x + |x| +1 lt 0`, isA. `(0, oo)`B. `(-oo, 0)`C. RD. `phi` |
| Answer» Correct Answer - D | |
| 23. |
The solution set of the inequation `(x^2-3x+4)/(x+1) > 1, x in RR`, isA. `(3, oo)`B. `(-1, 1) cup (3, oo)`C. `[-1, 1]cup [3, oo)`D. none of these |
| Answer» Correct Answer - B | |
| 24. |
The solution set of the inequation `(|x-2|)/(x-2) lt 0`, isA. `(2, oo)`B. `(-oo, 2)`C. RD. `(-2, 2)` |
| Answer» Correct Answer - B | |
| 25. |
The solution set of the inequation `|x+(1)/(x)| lt 4`, isA. `(2-sqrt(3), 2 +sqrt(3)) cup (-2-sqrt(3), -2 + sqrt(3))`B. `R-(2-sqrt(3), 2+sqrt(3))`C. `R-(-2-sqrt(3), -2 + sqrt(3))`D. none of these |
| Answer» Correct Answer - A | |
| 26. |
The solution set of the inequation `(1)/(|x|-3) lt (1)/(2)` isA. `(-oo, -5)cup(5, oo)`B. `(-3, 3)`C. `(-oo, -5) cup (-3, 3) cup (5, oo)`D. none of these |
| Answer» Correct Answer - C | |
| 27. |
The solutiong set of `|x^(2)-10| le 6`, isA. `(2, 4)`B. `(-4, -2)`C. `(-4, -2) cup (2, 4)`D. `[-4, -2] cup [2, 4]` |
| Answer» Correct Answer - D | |
| 28. |
The solution set of the inequation `|x-1|+|x-2|+|x-3|>= 6` isA. `[0, 4]`B. `(-oo, -2) cup [4, oo)`C. `(-oo, 0]cup[4, oo)`D. none of these |
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Answer» Correct Answer - CConsider 4 cases: Case 1: x<1 (1-x)+(2-x)+(3-x)>=6; 6-3x>=6; 3x<=0; x<=0 (since last is always <1, all x<=0 all solutions) Case 2: 1<=x<2 (x-1)+(2-x)+(3-x)>=6; 4-x>=6; x<=-2 (no solutions here) Case 3: 2<=x<3 (x-1)+(x-2)+(3-x)>=6; x>=6 (again, no solutions here) Case 4: x>=3 (x-1)+(x-2)+(x-3)>=6; 3x-6>=6; 3x>=12; x>=4 (again, all x>=4 are solutions) Summary: The solutions are: ( − ∞ , 0 ] ∪ [ 4 , ∞ ) |
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| 29. |
The solution set of the inequation `(4x + 3)/(2x-5) lt 6`, isA. `(5//2, 33//8)`B. `(-oo, 5//2)cup (33//8, oo)`C. `(5//2, oo)`D. `(33//8, oo)` |
| Answer» Correct Answer - B | |
| 30. |
The solution set of the inequation `0 lt |3x+ 1|lt (1)/(3)`, isA. `(-4//9, -2//9)`B. `[-4//9, -2//9]`C. `(-4//9, -2//9)-[-1//3]`D. `[-4//9-2//9]-[-1//3]` |
| Answer» Correct Answer - C | |
| 31. |
The solution set of `x^(2) + 2 le 3x le 2x^(2)-5`, isA. `phi`B. `[1, 2]`C. `(-oo, -1] cup [5//2, oo)`D. none of these |
| Answer» Correct Answer - A | |
| 32. |
The solution set of the inequation `x^(2) + (a +b) x +ab lt 0, " where" a lt b,` isA. `(a, b)`B. `(-oo, a) cup (b, oo)`C. `(-b, -a)`D. `(-oo, -b) cup (-a, oo)` |
| Answer» Correct Answer - C | |
| 33. |
Writhe the set of values of `x`satisfying `|x-1|lt=3 a n d |x-1|lt=1.`A. `[2, 4]`B. `(-oo, 2] cup [4, oo)`C. `[-2, 0]cup [2, 4]`D. none of these |
| Answer» Correct Answer - C | |
| 34. |
The set of real values of x satisfying `||x-1|-1|le 1`, isA. `[-1, 3]`B. `[0, 2]`C. `[-1, 1]`D. none of these |
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Answer» Correct Answer - A Since x-1 changes its sign as x passes through 1. So, following cases arise. CASE I When `x lt 1` In this case, we have `|x-1|= -(x-1)` `therefore ||x-1|-1| le1` `rArr |-(x-1)-1| le 1` `rArr |-x| le1` `rArr |x| le 1` `rArr -1 le x le 1 " "[because |x| le a hArr -a le x le a]` `rArr -1 le x lt 1 " " [because x lt 1]` `therefore x in [-1, 1])` CASE II When `x ge 1` In this case, we have |x-1| = x -1 `therefore ||x-1|-1| le 1` `rArr |x-1-1| le 1` `rArr |x-2| le 1 rArr 2-1 le x le 2 + 1 rArr x in [1, 3]` Hence, `x in [-1, 3]` |
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| 35. |
the greatest negative integer satisfying `x^2+4x-774` isA. `-4`B. `-6`C. `-7`D. none of these |
| Answer» Correct Answer - D | |
| 36. |
If `9^(x+1) + (a^(2)-4a-2) 3^(x) + 1 lt 0 "for all" x in R,` thenA. `a in R`B. `a in R^(+)`C. `a in [1, oo)`D. `a in R -(2)` |
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Answer» Correct Answer - D We have, `9^(x+1) + (a^(2)-4a-2) 3^(x) + 1 lt 0 "for all" x in R,` `rArr 9y^(2) + (a^(2)-4a-2) y+1 gt 0 "for all " y gt 0, "where" y = 3^(x)` `rArr y(9y + (1)/(y) + a^(2) -4a-2) gt 0 "for all" y gt 0` `rArr (3sqrt(y)-(1)/(sqrt(y)))^(2) + (a-2)^(2) gt 0 "for all" y gt 0` `rArr a ne 2 rArr a in R -(2)` |
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