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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. |
Which of the following is a singleton set ?A. {0}B. `{phi}`C. { xx is an even prime number }D. All of these |
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Answer» Correct Answer - D {0} contains one elements is ( d) `{phi}`contains one elements i.e `phi` {x: x is an even prime number }={2} contains one element , I,e 2. `therefore` All of the above are singleton sets . Hence , the correct option is ( d) |
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| 2. |
`( A cap B ) cup ( A cap C)`=__________`A. AB. `A cup B`C. `A cup C`D. All of these |
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Answer» Correct Answer - D `(A cap B ) cup ( A cap C) = B cup C ` `A= A cup B =A cup C = B cup C` Hence , the correct option is (d) |
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| 3. |
All fat boys in your colony. |
| Answer» Correct Answer - Not a set | |
| 4. |
From the above Venn diagram ,find `n(P-Q)+n(Q-P)`=__________.A. 10B. 4C. 6D. 8 |
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Answer» Correct Answer - D P-Q={21,7,28,35} `rArr n(Q-P)= 4` `Q-P=[24,30,36,42]` `rArr n(Q-P)=4` `therefore n(P-Q)+n(Q-P )=4+4 = 8 ` Hence , the correct option is (d) |
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| 5. |
Write dovvn all the possible subsets of {x, y, z} |
| Answer» Correct Answer - {x}, {y}, {z}, {x, y}, {x, z}, {y, z}, {x, y, z}, `phi` | |
| 6. |
`If ={ x: x in W, x le 10 } and B = { x: x in N x le 10 } " then find " n( A cup B)` The following steps are involved in solving the above problem.Arrange them in sequential order (A) `rArr n(A cup B)=11` ( B) `A cup B=`{0,1,2,3,4,5,6,7,8,9, 10 } cup {1,2,3,4,5,6,7,8,9,10} Form the given data A={1,2,34,5,6,7,8,9,10} and B={1,2,3,4,5,6,7,8,9,10} `therefore A cup B ={0,1,2,3,4,5,6,7,8,9,10}`A. CABDB. CBDAC. CBDAD. DCBA |
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Answer» Correct Answer - B ( c) ,( D),( B) and ( A) is the required sequential order Hence , the correct option is ( b) |
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| 7. |
If =O={0} then the number of all possible subsets of O is __.A. 2B. 3C. 1D. 4 |
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Answer» Correct Answer - A If a set has n elements, then number of all possible subsets`=2^n` Since O contains only one element `rArr ` Number of subsets `=2^1=2` Hence , the correct options is (d) |
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| 8. |
In the question above how many do not like ice creams ?A. 40B. 44C. 148D. 36 |
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Answer» Correct Answer - A Number of those who like ice creams = 3x+2x= 5x = 40 `therefore `Required number = 80 - 5 x = 40 Hence the correct option is ( a) |
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| 9. |
Write `A cup B` in the set builder form for the above question |
| Answer» Correct Answer - {x : x is a natural number and `x lt 100`.} | |
| 10. |
In the question above ,find the percentage of those who play only caromsA. 0.45B. 0.4C. 0.5D. 0.55 |
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Answer» Correct Answer - A The required percentage `=(9/20 xx 100 )%= 45%` Hence , the correct option is ( a) |
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| 11. |
If Y={a,e,{I,o} u} then which of the following is a correct statement ?A. `{I,o} sub Y`B. `{I ,0} in Y `C. `{i.o} in Y`D. `{{I,o} } in Y ` |
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Answer» Correct Answer - B Since { i, o} is an element of Y. `therefore {I,o} in Y` Hence , the correct option is (b) |
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| 12. |
In the question above ,find `(A-B) cap ((B-A)`A. {2,4}B. AC. BD. `phi` |
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Answer» Correct Answer - D `(A-B) cap (B-A)=phi ` Hence , the correct option is (d) |
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| 13. |
Write the set builder from for the given set in the above question . |
| Answer» Correct Answer - W = {x : x is in woman prime minister of India.} | |
| 14. |
If X={1,3,5,7,9,} then the cardinal number of X is `"______"`. |
| Answer» Correct Answer - 5 | |
| 15. |
Write the difference of the sets containing the letters of the words MATHEMATICS and SOCIALAMANA. {M,A,T}B. {L,M,N}C. {T,H,E}D. {I,C,S} |
| Answer» Correct Answer - C | |
| 16. |
If two sets are disjoint , then _______.A. they have one element in commonB. they have 0 as one element in commonC. they have no element in commonD. they have two element in common |
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Answer» Correct Answer - C Disjoint sets have no elyments in common Hence, the correct option is (c). |
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| 17. |
If A and B are two disjoint sets `n(A)+n(B)=24` then find `n(A cup B)`A. 16B. 18C. 24D. Cannot say |
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Answer» Correct Answer - C n(A)+n(B)=24 A and B are disjoint `therefore n(A cap B) =0` `n(A cup B)+n(A cap B)=n(A)+n(B)` `rArr n(A cup B)=24` Hence , the correct option is ( c) |
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| 18. |
X and y are disjoint sets. If n(X) = 40 and n(Y)= 28, then find n(X-y)+n(Y-X) |
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Answer» `n(X-Y)=n(X)-n( X cap Y)` `n(Y-X)=n(y)-n(X cap Y)` x and Y are disjoint sets ,ie. `n(X cap Y )=0` `therefore n(X-Y)=n(X) and n(Y-X)=n (Y)` `therefore n(X-Y)+n(Y-X)=n(X)+n(Y)` = 40 + 28 = 68 |
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| 19. |
If `n(A)=8, n(B)=6` and the sets A and B are disjoint, then find `n(A uu B)`. |
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Answer» Given n(A)=8 and n(B)=6. A and B are disjoint `rArr A cap B = phi rArr n(A cap B)=0` `therefore n(A cup B)=n(A) +n(B)-n(A cap B)=8 +6 -0 = 14` |
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| 20. |
The cardinal number of the set containing letters of the word MOONROCKA. 8B. 6C. 4D. 2 |
| Answer» Correct Answer - B | |
| 21. |
Find the cardinal number of a set containg woman prime ministers of India. |
| Answer» Correct Answer - 1 | |
| 22. |
If A and B are disjoint sets, then `n(A cap B) = ?`A. 4B. 2C. 1D. 0 |
| Answer» Correct Answer - D | |
| 23. |
If A={ 2,3,4} and B ={3,5} and , then `A cap B` has only one element |
| Answer» Correct Answer - 1 | |
| 24. |
A bar graph is drawn to the scale 1 cm = 4x units .The length of the bar represeting a quantity 1000 units is 1.25 cm . Find xA. 200B. 175C. 250D. 275 |
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Answer» Correct Answer - A Given 1 cm = 4x units 1.25 cm = 1000 units 1 cm `=1000/1.25 untis ` 1 cm = 8000 units `rArr 4x = 8000 rArr x= 200` Hecnce , the correct option is ( a) |
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| 25. |
X= { The units digit of the sum of 10 consective natural numbers } Find X.A. {5}B. {2}C. {3}D. {0} |
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Answer» Correct Answer - A 10 consecutive natural numbers will have their units digits as 0, 1, 2, 3, 4, 6, 7, 8, 9. ` therefore ` Units digits of their sum = 5 ltbr. `therefore ` X={5}` Hence , the correct option is (a) |
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| 26. |
A bar graph is drawn to the scale of 1 cm = 2 m units. The length of the bar representing a quanti ty of 875 units is 1.75 cm. Find m .A. 125B. 225C. 250D. 375 |
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Answer» Correct Answer - C Given 1 cm = 2m unitis `rarr` (1) And 1/75 cm = 875 units `rArr ` 1 cm = 875/1.75 units `rArr 1 cm = 500 units rarr (2)` From Eqs ( 1) and ( 2) we get 2m= 500 `therefore m = 250 ` Hence , the correct option is ( c) |
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| 27. |
Solve `|2x-3|+|x-1|=|x-2|dot` |
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Answer» |2x-3|+|x-1|=|x-2|=|(2x-3)-(x-1)| `rArr (2x -3)(x-1) le 0 ` `rArr 1 le x le 3//2` |
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| 28. |
Solve `x^2+x-4|=|x^2-4|+|x|dot` |
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Answer» `|(x^2-4)+x|=|x^2-4|+ |x|` `rArr x(x^2-4) ge 0` `rArr x in [ -2 , 0 ] cup [ 2 , oo)` |
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| 29. |
If `|s in x+cos x|=|s in x|+|cos x|(s in x ,cos x!=0)`, then in which quadrant foes `x`lie? |
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Answer» Here we have |sin x + cos x | = | sin x | + | cos x | . It implies that sin x and cos x must have the same sign Therefore , x lies in the first or third quadrant . |
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| 30. |
Solve `|(x+1)/x|+|x+1|=((x+1)^2)/(|x|)dot` |
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Answer» `|(x+1)/(x)|+|x+1|=((x+1)^2)/(|x|)` or `|(x+1)/(x)|+|x+1|=|(x+1)/x+(x+1)|` `or ((x+1))/(x)xx(x+1) ge 0 ` `or (x+1)^2/(x)ge 0 ` `rArr x in (0,oo) cup {-1}` |
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| 31. |
Is `|t a n x+cos x| |
| Answer» Since tan x and cot x have always the same sign `| tan x + cot x | lt | tan x| + | cot x | ` does not hold true for any value of x. | |
| 32. |
Solve `0 lt |x| lt 2` |
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Answer» But given `|X| gt 0 rArr x ne 0 ` Now ` 0 lt |x| lt 2 ` `rArr x in (-2,2),x ne 0 ` `rArr x in (-2 , 2) - {0}` |
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| 33. |
Solve `x^2-4|x|+3 |
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Answer» `x^2-4 |x|+3lt 0 ` `rArr ( |x |-1 )(|x|-3)lt 0 ` `rArr 1 lt |x| lt 3` `rArr -3 lt x lt -1 or 1 lt x lt 3` `rArr x in (-3,-1 ) cup (1,3)` |
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| 34. |
Solve `|3x-2|=xdot` |
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Answer» |3x-2|=x Case (i) When `3x -2 ge 0 or x ge 2//3` For which we have 2-3x= x or x=1 (ii) Case , When `3x-2 lt 0 or x lt 2//3` For which we have `2-3x = x or x = 1//2` Hence , solution set is { 1/2 , 1} |
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| 35. |
Find all possible values of `(x^2+1)/(x^2-2)`. |
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Answer» `(x^2+1)/(x^2-2)` or `yx^2-2y= x^2+1` or `x^2=(2y+1)/(y-1)` Now `x^2 ge 0 rArr (2y+1)/(y-1) ge 0` Now `x^2 ge 0 rArr (2y +1)(y-1) le 0` `rArr y le -1//2 or y gt 1` |
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| 36. |
Find all possible values of (i) `sqrt(|x|-2)` (ii) `sqrt(3-|x-1|)` (iii) `Sqrt(4-sqrt^2))` |
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Answer» `sqrt(|x|-2)` we know that square roots are defined for non- negative values only . It implies that we must have `|x|-2 le 0 ` Thus `sqrt(|x|-2) ge 0 ` (ii) `sqrt(3-|x-1|)` is defined when `3-|x-1| le 0 ` But the maximum value of 3-|x-1| is 3 , when |x-1| is 0 Hence for `sqrt(3-|x-1|)` to get defined , `0 le 3- |x-1| le 3 ` Thus , `sqrt(3-|x-1|)in [0,sqrt(3)]` Alternatively , `|x-1| ge 0` `rArr -|x-1| le 0 ` `rArr 3-|x-1|le3` But for `sqrt(3-|x-1|)` to get defined ,we must have `0 le 3 -|x-1| le 3 ` `rArr 0 le sqrt(3-|x-1| le sqrt(3)` (iii) `sqrt(4-sqrt(x^2))=sqrt(4-|x|)` `|x| ge 0 ` `rArr - |x| le 0 ` `rArr 4-|x| le 4 ` But for `sqrt(4-|x| )` to get defined `0 le 4 - |x| le 4 ` `therefore 0 le sqrt(4-|x|) le 2 ` |
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| 37. |
Solve`sqrt(x+3-4sqrt(x-1))+sqrt(x+8-6sqrt(x-1))=1` |
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Answer» `sqrt(x+3-4sqrt(x-1))+sqrt(x+8-6sqrt(x-1))=1` `or sqrt(x-1-4sqrt(x-1)+4)+sqrt(x-1-6sqrt(x-1))=1` `or sqrt(|sqrt(x-1)-2|^2)+ sqrt(|sqrt(x-1)-3|^2)=1` or `|sqrt(x-1)-2|+|sqrt(x-1)-3|=1` or `|sqrt(x-1)-2|+|sqrt(x-1)-3|=(sqrt(x-1)-2)-(sqrt(x-1)-3)` `rArr sqrt(x-1)-2 ge 0 and sqrt(x-1)-3 le 0 ` `rArr 4 le x-1 le9` `rArr 5 lex le 10` |
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| 38. |
Find the mode of the data 2, 4, 6, 4, 6, 7, 6, 7, aud 8A. 4B. 6C. 7D. 8 |
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Answer» Correct Answer - B `{:("Observation","Frequency"),(2,1),(4,2),(6,3),(7,2),(8,1):}` 6 has the highest frequency. `therefore` Mode = 6 Hence , the correct option is ( b) |
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| 39. |
The mode of the unimodal data 7 , 8, 9, 8, 9, 10, 9 ,10, 11 , 10, 11, 12, and xis 10. Find the value of x.A. 10B. 9C. 8D. 11 |
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Answer» Correct Answer - A The give obervations are 7,8,9,8,9,10,9,10,11,10,11,12 and x . `{:("Observation","Frequency"),(7,1),(8,2),(9,3),(10,2),(11,1),(12,1),(x,1):}` And also given that mode is 10 and the given date is unimodal. `therefore ` x= 10 Hence , the correct option is (a ) . |
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| 40. |
In a class, there are 100 students. Of them, 60 students attend music classes, 40 students attend dance classes, and 20 students attend both the classes. Find the number of students who attend neither of the classes. The following steps are involved in solving the above problem. Arrange them in sequential order (A) `n(M cup D)=n(M)+n(D)-n(M cup D) rarr n(M cap D)rarr n(M cup D)=60 + 40 -20 =100 -20 = 80 ` ( B) `therefore`Number of students who attend neither of the classes = 100 - 80 = 20 ( C) `n(M)=60 ,n(D)=40 and n(M cap D)= 20 ("given")` (D) Let n(M) be the number of students who attend music classes and n(D) be the number of students who attend dance classes.A. DACBB. DCBAC. ACDBD. DCAB |
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Answer» Correct Answer - D DCAB is the required sequenial order Hence , the correct option is (d) |
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| 41. |
The shaded region in the given figure is A. `A cap (B cup C)`B. `A cup (B cap C)`C. `A cap (B-C)`D. `A-(B cup C)` |
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Answer» Correct Answer - D The Shaded region in the figure is clearly `A-(B cup C)` |
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| 42. |
If a N = `{ ax : x in N}` then the set `4N cap 6N` isA. 8NB. 10 NC. 12 ND. Non of these |
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Answer» Correct Answer - C `4N={4,8,12,16,20,24….} and 6N ={6,12,18,24…} Hence ,`4N cap 6N`={ 12,24,36 …} = 12 N` |
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| 43. |
For sets A, B,C , show that `(A-B) cup (A-C) = A -(B cup C)` |
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Answer» Let `x in (A-B) cup (A-C) ` ` rArr x in ( A-B) and x in (A-c)` `rArr ( x in A and x in B ) and ( x in A and x in C)` `rArr x in A and x in B) and ( x in A and notin C)` `rArr x in A and ( x notin B and x notin C) ` `rArr x in A and x notin ( B cup C) ` `rArr x in A - ( B cup C)` ` rArr ( A-B) cap ( A-C) sub A - ( B cup C)` Now let `y in A-(B cup C)` `rArr y in A and y notin ( B cup C)` `rArr y in A and ( y notin B and y notin C)` `rArr y in ( A- B) and y in ( A-C)` `rArr y in ( A - B ) cap ( A-c)` `rArr A-( B cup C) sub ( A-B) cap ( A-C)` From (1) and (2) we get `A-(B cup C) = ( A - B ) cap (A-C) |
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| 44. |
Solve the following (i) | x-2 | = (x-2) (ii) |x+3 |=-x-3 (iii) `|x^2-x|=x^2-x ` (iv) `|x^2-x-2|=2 +x-x^2` |
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Answer» (i) `|x-2|= (x-2), if x-2 le 0 of or x ge 2 ` (ii) `| x+3 | =-x-3, if x+3 le 0 or x le -3` (iii) `|x^2-x|=x^2-x, if x^2- x ge 0 ` or `x(x-1) ge 0 ` `rArr x in (-oo ,0] cup [1, oo)` (iv) `|x^2 - x -2 |=2 + x - x^2` or `x^2-x -2 le 0 ` or `(x-2)(x+1) le 0 ` `or -1 le x le 2 ` |
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| 45. |
Find the values of `a`for which eht equation `||x-2|+a|=4`can have four distinct real solutions. |
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Answer» `|x-2|+ a= pm 4` `rArr |x-2|= pm 4-a ` For four real roots `4-a lt 0 and -4-a lt 0` `rArr a in (-oo,-4)` |
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| 46. |
If n (A) =3 ,n(3)=6 and `A sube B` .Then the number of elements in `A cup B` is equal toA. 3B. 9C. 6D. Non of these |
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Answer» Correct Answer - C Since `A sube B, A sub B=B ` So, n `(A cup B) = n(B) = 6 ` |
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| 47. |
Two finite sets have m and `n ( m gt n)` elements .The number of subes of the first set is 112 more than that of the second set. The value of mn isA. 18B. 28C. 32D. 36 |
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Answer» Correct Answer - B According to the question `2^m-2^m= 112 ` `rArr 2^n (2^(m-n)-1)= 2^4 xx 7` Comparing exponets on both sides ,we get `2^n-2^4 and 2^(m-n)-1= 7 ` Now ,`2^n= 2^4` `rArr n=4` And `2^(m-n)=8` `rArr 2^(m-n)=2^3` `rArr m-n =3` `therefore m=7` |
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| 48. |
Solve `|x|= x^2-1` |
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Answer» `x^2-1 = |x|` ` rArr x^2-1 = x when x ge 0 ` or `x^2 =-x when x lt 0` `rArr x^2 -x-1 =0 rArr x= (1+ sqrt(5))/(2) ("as "x ge 0 )` `x^+x-1=0 rArr x =(-1-sqrt( 5))/(2) ("as " x lt 0 )` |
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| 49. |
If set A and B are defined as `A = {(x,y)|y = 1/x, 0 ne x in R}, B = {(x,y)|y = -x , x in R,}`. ThenA. `A cap B=A`B. `A cup B=B`C. `A cup B= phi`D. `A cup B-A` |
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Answer» Correct Answer - C For `A cap B`, We have `-x=1/x` `rArr -x^2 = 1 ` `rArr x^ 2 =-1 ` which is not possible `therefore A cap B = phi ` |
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| 50. |
Solve `|x^2-1|+x^2-4| gt 3` |
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Answer» Correct Answer - `-1 lt x lt 1 or x lt -2 or x gt 2 ` We have `|x^2-1|+x^2-4| gt 3 ` `rArr |x^2-1| +|4-x^2| gt |x^2 -1+4-x^2|` `rArr x^2 -1 )(x^2-4) lt 0 ` `rArr x^2 lt 1 or x^2 gt 4` `rArr -1 lt x lt 1 or x lt -2 or x gt 2` |
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