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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.
| 301. |
Let R be a relation on the set N of naturalnumbers defined by `nRm n` is a factor of m(ie. nim) Then R isA. R is reflexive, symmetric but not transitiveB. R is transitive, symmetric but not reflexiveC. R is reflexive, transitive but not symmetricD. R is an equivalence relation |
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Answer» Correct Answer - C nRm `iff` n is a factor of m. implies m is divisible by n. Reflexivity We know that n is divisible by `nAAn in N` `(n, n) inRAAn inN` R is reflexive. Symmetric `n, m in N` Let n=2, m=6 m is divisible by n but n is not divisible by m. Hence R is not symmetric. Transtivity Let `(n,m)in R and (m,p) in R " then "(n, m) in R and (m, p)inRimplies(n, p)in R` or If m is divisible by n and p is divisible by m. Hence p is divisible by n. `(n, p)inR AAn, p in N` R is transitive relation on N. Hence R is reflexive, transitive but not symmetric. `therefore underset(0)overset(2a)intf(x)dx=2underset(0)overset(a)intf(x)dx if f(2a-x)=fx` |
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| 302. |
The relation S is defined on the set of integers Z as xSy if integer x devides integer y. ThenA. s is an equivalence relationB. s is only reflexive and symmetricC. s is only reflexive and transitiveD. s is only symmetric and transitive |
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Answer» Correct Answer - C The relation S is defined on the set of integers Z and xSy, if integer x divides integery. Reflexive : Since, every integer divides itself `therefore` integer x divides integer x implies xSx Hence, S is reflexive. Symmetric : Let `x, y in Z` such that xSy i.e., integer x divides integer y Now, this does not implies that integer y divides integer x. e.g. Take x = 2 and y = 4 Then, 2 divides 4 but 4 does not divides 2. Thus, S is not symmetric. Transitive : Let `x, y, z in Z` such that xSy and ySz. implies integer x divides integer y and integer y divides integer z implies integer x divides integer z implies xSz Hence, S is transitive. |
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| 303. |
If a non-empty set A contains n elements, then its power set contains how many elements ?A. `n^(2)`B. `2^(n)`C. 2nD. n+1 |
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Answer» Correct Answer - B No. of elements in power set of `A=2^(n)`. |
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| 304. |
If A={1, 2}, B={2, 3} and C={3, 4}, then what is the cardinality of `(AxxB)nn(AxxC)` ?A. 8B. 6C. 2D. 1 |
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Answer» Correct Answer - C `AxxB={(1,2),(1,3),(2,2),(2,3)}` `AxxC={(1,3),(1,4),(2,3),(2,4)}` Number of elements in `(AxxB)nn(AxxC)=2` |
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| 305. |
If `N_(a)={ax|x inN}`, then what is `N_(12)nnN_(8)` equal to?A. `N_(12)`B. `N_(20)`C. `N_(24)`D. `N_(48)` |
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Answer» Correct Answer - C Given, `N_(a)={ax|x inN}` `therefore N_(12)={12, 24, 36, 48,...}` `and N_(8)={8, 16, 24,...}` `thereforeN_(12)nnN_(8)={24, 48,....}` `=N_(24)` |
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| 306. |
If A is a subset of B, then which one of the following is correct ?A. `A^(c)subeB^(c)`B. `B^(c)subeA^(c)`C. `A^(c)=B^(c)`D. `AsubeAnnB` |
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Answer» Correct Answer - B Suppose U={a,b,c,d,e,f,g,h} A={a,b,c,d} `B={a,b,c,d,e}" Given "AsubeB` `A^(c)={e,f,g,h}, B^(C)={f,g,h}` Hence, `B^(C)subeA^(C)` |
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| 307. |
Let A={1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Then the number of subsets of A containing two or three elements isA. 45B. 120C. 165D. 330 |
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Answer» Correct Answer - C Given, A{1, 2, 3, 4, 5, 6, 7, 8, 9, 10} Set A has 10 elements. Number of sub sets Containing 2 and 3 elements is `10_(c_(2))+10_(c_(3))` `10_(c_(2))+10_(c_(3))=(10xx9)/(2)+(10xx9xx8)/(3xx2)` `=45+120=165` |
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| 308. |
Let P={1, 2, 3} and a relation on set P is given by the set R={(1,2),(1,3),(2,1),(1,1),(2,2),(3,3),(2,3)}. Then R is:A. reflexive, transitive but not symmetricB. Symmetric, transitive but not reflexiveC. Symmetric, reflexive but not transitiveD. None of the above |
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Answer» Correct Answer - A Given relation `R={(1,2),(1,3),(2,1),(1,1),(2,2),(3,3),(2,3)}` is reflexive and transitive only but not symmetric `because (3,1) and (3,2) cancelinR`. |
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| 309. |
Let `R={x|x inN,x" is a multiple of 3 and x"le100}` `S={x|x in N, x" is a multiple of 5 and x "le100}` What is the number of elements in `(RxxS)nn(SxxR)`?A. 36B. 33C. 20D. 6 |
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Answer» Correct Answer - A Let R = `R={x:x in N, x" is a multiple of 3 and "xle100}` and `S={x:x in N, x" is a multiple of 5 and "xle100}` `thereforeR={3, 6, 9, 12, 15, ....,99}` and `S = {5, 10, 15, …., 95, 100}` Now, `(RxxS)nn(SxxR)=(RnnS)xx(SnnR)` =(15, 30, 45, 60, 75, 90)`xx`(15, 30, 45, 60, 75, 90) `therefore` Number of elements in `(RxxS)nn(SxxR)` `=6xx6=36` |
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| 310. |
If `(log_(3)x)(log_(x)2x)(log_(2x)y)=log_(x)x^(2)`, then what is y equal to?A. 4.5B. 9C. 18D. 27 |
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Answer» Correct Answer - B `(log_(3)x)(log_(x)2x)(log_(2x)y)=log_(x)x^(2)`, `(logx)/(log3)xx(log2x)/(logx)xx(logy)/(log2y)=(logx^(2))/(logx)` `(logy)/(log3)=(2logx)/(logx)` logy=2log3 logy=log9 y=9 |
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| 311. |
Let A = { x | x |
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Answer» Correct Answer - D Given `A={x:xle9, x inN}={1,2,3,4,5,6,7,8,9}` Total possible multiple of 3 are ,brgt 3, 6, 9, 12, 15, 18, 21, 24, 27 But 3 and 27 are not possible because 3 and 27 can not be express as such that a+b+c is multiple of 3 `6rarr1+2+3` `9rarr2+3+4,5+3+1,6+2+1` `12rarr9+2+1,8+3+1,7+1+4,7+2+3`, 6+4+2,6+5+1,5+4+3` `15rarr9+4+2,9+5+1,8+6+1,8+5+2`, 8+4+3,7+6+2,7+5+3,6+5+4` `18rarr9+8+1,9+7+2,9+6+3`, `9+5+4,8+7+3,8+6+4,7+6+5` `21rarr9+8+4,9+7+5,8+7+6` `24rarr9+8+7` Hence, total largest possible subsets are 30. |
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| 312. |
If `log_(8)m+log_(8).(1)/(6)=(2)/(3)`, then m is equal toA. 24B. 18C. 12D. 4 |
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Answer» Correct Answer - A `log_(8)m+log_(8).(1)/(6)=(2)/(3)` `implieslog_(8)(m.(1)/(6))=(2)/(3)` `implies(8)(2)/(3)=m.(1)/(6)` `impliesm=24` |
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| 313. |
What is the decimal number representation of the binary number `(11101.001)_(2)`?A. `30.125`B. `29.025`C. `29.125`D. `28.025` |
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Answer» Correct Answer - C `11101.001` `=1xx2^(4)+1xx2^(3)+1xx2^(2)+0xx2^(1)+1xx2^(0).0xx2^(-1)+0xx2^(-2)+1xx2^(-3)` `=(16+8+4+0+1).(0+0+(1)/(8))=29.125` |
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| 314. |
The decimal number `(57.375)_(10)` when converted to binary number takes the form:A. `(111001.011)_(2)`B. `(100111.110)_(2)`C. `(110011.101)_(2)`D. `(111011.011)_(2)` |
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Answer» Correct Answer - A `(111001.011)_(2)=[2^(5)xx1+2^(4)xx1+2^(3)xx1+2^(2)xx0+2^(1)xx0+2^(0)xx1].[2^(-1)xx0+2^(-2)xx1+2^(-3)xx1]=(57.375)_(10)` |
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| 315. |
The decimal representation of the number `(1011)_(2)` in binary system is:A. 5B. 7C. 9D. 11 |
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Answer» Correct Answer - D `(1011)_(2)=2^(3)xx1+2^(2)xx0+2^(1)xx1+2^(0)xx1` `=8+0+2+1=(11)_(10)` |
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| 316. |
If A={x:x is a multiple of 3} and B={x:x is a multiple of 4} and C={x:x is a multiple of 12}, then which one of the following is a null set?A. `(A//B)uuC`B. `(A//B)//C`C. `(AnnB)nnC`D. `(AnnB)//C` |
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Answer» Correct Answer - D A={x:x is a multiple of 3} `therefore` A={3, 6, 9, 12, 15, 18, 24,….} `therefore (AnnB)/(C)` `"B={x : x is a multiple of 4}"` `thereforeB={4,8,12,16,20,24,28,32,....}` C={x : x is a multiple of 12} `therefore C={12, 24, 36, 48, 60, 72, 84, 96,....}` `AnnB={12,24,....}= (AnnB)/(C)`. |
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| 317. |
What is the binary equivalent of decimal number `(0.8125)_(10)`?A. `(0.1101)_(2)`B. `(0.1001)_(2)`C. `(0.1111)_(2)`D. `(0.1011)_(2)` |
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Answer» Correct Answer - A Work with option `(0.1101)_(2)=1xx2^(-1)+1xx2^(-2)+1xx2^(-4)` `=(1)/(2)+(1)/(4)+(1)/(16)=(13)/(16)=(0.8125)_(10)`. Hence option (a) gives the binary equivalent of given decimal number. |
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| 318. |
A, B, C and D are four sets such that `AnnB=CnnD=phi`. Consider the following : 1. `AuuC` and `BuuD` are always disjoint. 2. `AnnC and BnnD` are always disjoint Which of the above statements is/are correct ?A. 1 onlyB. 2 onlyC. Both 1 and 2D. Neither 1 nor 2 |
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Answer» Correct Answer - B Let A={1,2} B={3,4,0} C={5,6,0} D={7,8} Such that `(AnnB)=(CnnD)=phi` `implies(AuuC)={1,2,5,6,0}` `implies(BuuD)={3,4,7,8,0}` `implies(AuuC)nn(BuuD)={0}` `"So "(AuuC) and (BuuD)` are not always dispoint `implies (AnnC) and phi and (BnnD)=phi` `"So "(AnnC) and (BnnD)` are always disjoint. |
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| 319. |
What is `(1000000001)_(2)-(0.0101)_(2)` equal to ?A. `(512.6775)_(10)`B. `(512.6875)_(10)`C. `(512.6975)_(10)`D. `(512.0909)_(10)` |
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Answer» Correct Answer - B `(1000000001)_(2)` `=1xx2^(0)+0xx2^(1)+0xx2^(2)+......+1xx2^(9)` `=1+0+0+......+512` `=(513)_(10)` `(0.0101)_(2)=0xx2^(-1)+1xx2^(-2)+0xx2^(-3)+1xx2^(-4)` `=(1)/(4)+(1)/(16)=(5)/(16)=(0.3125)_(10)` `(1000000001)_(2)-(0.0101)_(2)=513-0.312` `=(512.6875)_(10)` |
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| 320. |
If set A = {2, 3, 4, 5, 6}, state which of the following statements arc true and which are false :(i) 2 ∈ A(ii) 5, 6 ∈ A(iii) 3, 4, 7 ∈ A(iv) 2, 8 ∈ A |
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Answer» (i) True (ii) True (iii) False (iv) False |
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| 321. |
The relation R defined on the set of natural numbers as {(a,b) : a differs from b by 3} is given byA. `{(1,4),(2,5),(3,6),...}`B. `{(4,1),(5,2),(6,3),...}`C. `{(1,3),(2,6),(3,9),...}`D. None of these |
| Answer» Correct Answer - B | |
| 322. |
Let `R = {(3, 3), (6, 6), (9, 9), (6, 12), (3, 9), (3, 12), (3, 6)}` is a relation on set `A = {3, 6, 9, 12}` then R isA. an equivalence relationB. reflexive and symmetric onlyC. reflexive and transitive onlyD. reflexive only |
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Answer» Correct Answer - C Here, (3, 3), (6, 6), (9, 9), (12, 12) So, it is Reflexive and (3, 6), (6, 12), (3, 12) So, it is Transitive Here, reflexive and transitive only. |
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| 323. |
The relation `R={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)}` on a set A={1, 2, 3} isA. reflexive but not symmetricB. reflexive but not transitiveC. symmetric and transitiveD. Neither symmetric nor transitive |
| Answer» Correct Answer - A | |
| 324. |
The number of equivalence relations that can be defined on set {a, b, c}, isA. 5B. `3!`C. `2^(3)`D. `3^(3)` |
| Answer» Correct Answer - A | |
| 325. |
Prove that a relation R defined on `NxxN` where `(a, b)R(c, d) ad = bc` is an equivalence relation. |
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Answer» R defined on `N xx N` such that (a, b) R (c, d) `iff` ad = bc Reflexivity Let (a, b) `in N xx N` `implies a, b in N implies ab = ba` implies (a, b) R (a, b) `therefore` R is reflexive on, `N xx N`. Symmetry Let (a, b), (c, d) `in N xx N`, then (a, b) R (c, d) implies ad = bc implies cb = da implies (c, d) R (a, b) `therefore` R is symmetric on `N xx N` Transitivity Let `(a, b), (c, d), (e, f), in N xx N`. Then, (a, b) R (c, d) implies ad = bc ... (i) (c, d) R (e, f) implies cf = de ... (ii) From Eqs. (i) and (ii), (ad) (cf) = (bc) (de) implies af = be implies (a, b) R (e, f) `therefore` R is transitive relation on `N xx N`. `therefore R` is equivalence relations on `N xx N`. |
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| 326. |
Statement-1 If Sets A and B have three and six elements respectively, then the minimum number of elements in `A uu B` is 6. Statement-2 `AnnB=3`.A. Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for Statement-1B. Statement-1 is true, Statement-2 is true, Statement-2 is not a correct explanation for Statement-1C. Statement-1 is true, Statement-2 is falseD. Statement-1 is false, Statement-2 is true |
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Answer» Correct Answer - A `n(AuuB)=n(A)+n(B)-n(AnnB)` `=3+6-n(AnnB)=9-n(AnnB)` As maximum number of element in `(AnnB) = 3` `therefore` Minimum number of elements in `(AnnB)=9-3=6` Both statements are true, Statement-2 is a correct explanation for Statement-1. |
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| 327. |
If A = {2, 3, 5}, B = {2, 5, 6}, then `(A - B) xx (AnnB)` isA. `{(3,2),(3,3),(3,5)}`B. `{(3,2),(3,5),(3,6)}`C. `{(3,2),(3,5)}`D. None of these |
| Answer» Correct Answer - C | |
| 328. |
In order that a relation R defined on a non-empty set A is an equivalence relation, it is sufficient, if RA. is reflexiveB. is symmetricC. is transitiveD. possesses all the above three properties |
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Answer» Correct Answer - D By definition for equivalent relation. R should be reflexive, symmetric, transitive. |
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| 329. |
If `A ={ theta : 2cos^2 theta + sintheta |
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Answer» `because 2cos^(2)theta+sinthetale2` `therefore 2(1-sin^(2)theta)+sinthetale2` `implies 2sin^(2)theta-sinthetage0` `implies sintheta(2sintheta-1)ge0` `implies sintheta(sintheta-(1)/(2))ge0` `therefore sinthetale0andsinthetage(1)/(2)` Now, the values of `theta` which lie in teh interval `(pi)/(2)lethetale(3pi)/(2)[because B={theta:(pi)/(2)lethetale(3pi)/(2)}]` So, `theta` satisfy `sin theta le 0` in the interval `(pi)/(2)lethetale(5pi)/(6)`. `therefore AnnB={theta:pilethetale(3pi)/(2)}` and `AnnB={theta:(pi)/(2)lethetale(5pi)/(6)}` Hence, `AnnB={theta:(pi)/(2)lethetale(5pi)/(6)orpilethetale(3pi)/(2)}` `={theta:thetain[(pi)/(2)(5pi)/(6)]uu[pi,(3pi)/(2)]}` |
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| 330. |
Let `N`denote the set of all natural numbers and R be the relation on `NxN`defined by `(a , b)R(c , d) a d(b+c)=b c(a+d)dot`Check whether R is an equivalence relation on `NxNdot` |
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Answer» Reflexive Since, `(a,b)R(a,b)iffab(b+a)=ba(a+b),AAa,binN` is true. Hence, R is reflexive. Symmetric (a, b) R (c, d) `iff ad(b+c)=bc(a+d)` `iff bc(a+d)=ad(b+c)` `iff cb(d+a)=da(c+b)` `iff (c,d)R(a,b)` Hence, R is symmetric. Transitive Since, `(a,b)R(c,d)iffad(b+c)=bc(a+d)` `iff (b+c)/(bc)=(a+d)/(ad)` `if (1)/(c)+(1)/(b)=(1)/(d)+(1)/(a)` `iff (1)/(a)-(1)/(b)=(1)/(c)-(1)/(d)` `therefore (a,b)R(c,d)iff(1)/(a)-(1)/(b)=(1)/(c)-(1)/(d)" ... (i)"` and similarly (c,d) R (e,f) `iff (1)/(c)-(1)/(d)=(1)/(e)-(1)/(f)" ... (ii)"` From Eqs. (i) and (ii), `(a,b)R(c,d)and(c,d)R(e,f)iff(1)/(a)-(1)/(b)=(1)/(e)-(1)/(f)iff(a,b)R(e,f)` So, R is transitive. Hence, R is an equivalence relation. |
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| 331. |
Simplify: `(sin 8theta cos theta - sin 6theta cos 3theta)/(cos 2theta cos theta - sin 3theta sin4theta)` |
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Answer» `(1/2[sin9theta+sin7theta-sin9theta-sin3theta])/(1/2[cos3theta+costheta-costheta+cos7theta])` `(sin7theta-sin3theta)/(cos3theta+cos7theta)` `(2cos5thetasin2theta)/(2cos5thetacos2theta)` `tantheta`. |
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| 332. |
The set (A ∪ B ∪ C) ∩ (A ∩ B′ ∩ C′)′ ∩ C′ is equal to(A) B ∩ C′ (B) A ∩ C (C) B ∪ C′ (D) A ∩ C′ |
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Answer» The correct choice is (A). Since (A ∪ B ∪ C) ∩ (A ∩ B′ ∩ C′)′ ∩ C′ = (A ∪ (B ∪ C)) ∩ (A′ ∪ (B ∪ C)) ∩ C′ = (A ∩ A′) ∪ (B ∪ C) ∩ C′ = φ ∪ (B ∪ C) ∩ C′ = B ∩ C′ ∪ φ = B ∩ C′ |
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| 333. |
Two set A and B are such that n(A ∪ B) = 21, n(A′ ∩ B′) = 9, n(A ∩ B) = 7, find n(A ∩ B)’. |
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Answer» ∴ n(A′ ∩ B′) = n(A ∪ B)’ = 9 we know that, u = n(A ∪ B)’ + n(A ∪ B) = 9 + 21 = 31 ∴ n(A ∩ B)’ =u - n(A ∪ B) = 30 - 7 = 23 |
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| 334. |
If A and B are disjoint, then A ∩ B A) μ B) A C) B D) Φ |
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Answer» Correct option is D) Φ |
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| 335. |
The value of parameter `alpha`, for which the function `f(x) = 1+alpha x, alpha!=0` is the inverse of itselfA. `-2`B. `-1`C. 1D. 2 |
| Answer» Correct Answer - B | |
| 336. |
Let the function `f ; R-{-b}->R-{1}` be defined by `f(x) =(x+a)/(x+b),a!=b,` thenA. f is one-one but not ontoB. f is onto but not one-oneC. f is both one-one and ontoD. `f^(-1)(2) = a - 2b` |
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Answer» Correct Answer - C::D `f:R-{-b}rarrR-{1}` `f(x)=(x+a)/(x+b)" "[aneb]` Let `x_(1), x_(2) in D_(f)` `f(x_(1)) = f(x_(2))` implies `(x_(1)+a)/(x_(1)+b)=(x_(2)+a)/(x_(2)+b)` `impliesx_(1)x_(2)+bx_(1)+ax_(2)+ab=x_(1)x_(2)+ax_(1)+bx_(2)+ab` `impliesb(x_(1)-x_(2))=a(x_(1)-x_(2))` `implies (x_(1)-x_(2))(b-a)=0` `implies x_(1)=x_(2)" "[because aneb]` `therefore` f is one-one function. Now, let `y=(x+a)/(x+b)` `xy+by=x+a` `x(y-1)=a-by` `x=(a-by)/(y-1)andf^(-1)(y)=(a-by)/(y-1)` `because y inR-{1}` `therefore` x is defined, `AAy inR-{1}` `f^(-1)(2)=(a-2b)/(2-1)=a-2b` |
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| 337. |
Let f and g be real valued functions defined as `f(x)={{:(7x^(2)+x-8",",xle 1),(4x+5",",1lt x le 7),(8x+3",",x gt7):} " " g(x)={{:(|x|",",xlt -3),(0",",-3le x lt 2),(x^(2)+4",",xge 2):}` The value of (gof) (0) + (fog) (-3) isA. -8B. 0C. 8D. 16 |
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Answer» Correct Answer - B `(gof)(0)=g(f(0))=g(7(0)^(2)+0-8)` `=g(-8)=|-8|=8` and `(fog)(-3)=f(g(-3))=f(0)=7(0)^(2)+0-8=-8` `therefore (gof)(0)+(fog)(-3)=-8+8=0` |
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| 338. |
Let `P={theta:sintheta-costheta=sqrt2cos theta}and Q={theta:sintheta+costheta=sqrt2sintheta}` be two ses. Then,A. `PsubQ and A - P ne phi`B. `QcancelsubP`C. `PcancelsubQ`D. P = Q |
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Answer» Correct Answer - D `P:sintheta-costheta=sqrt(2)costhetaimpliestantheta=sqrt(2)+1` `Q:sintheta+costheta=sqrt(2)sinthetaimpliestantheta=(1)/(sqrt(2)-1)=sqrt(2)+1` `therefore P = Q` |
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| 339. |
If `tan[theta/2]=sqrt[[1-e]/[1+e]]tan[phi/2]` then prove that `cosphi=[costheta-e]/[1-ecostheta]` |
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Answer» Here, `tan[theta/2] = sqrt((1-e)/(1+e))tan(phi/2)` we know, `tan (theta/2) = +-sqrt((1-costheta)/(1+costheta))` So, our equation becomes, `=>+-sqrt((1-costheta)/(1+costheta)) = sqrt((1-e)/(1+e))(+-sqrt((1-costheta)/(1+costheta)))` `=>((1-costheta)/(1+costheta)) = ((1-e)/(1+e))((1-costheta)/(1+costheta))` `=>(1+e)(1-costheta)(1+cosphi) = (1-e)(1+costheta)(1-cosphi)` `=>(1+e)(1+cosphi-costheta-costhetacosphi) = (1-e)(1-cosphi+costheta-costhetacosphi)` `=>1+cosphi-costheta-cosphicostheta+e+ecosphi-ecostheta-ecosthetacosphi = 1-cosphi+costheta-cosphicostheta-e+ecosphi-ecostheta+ecosthetacosphi` `=>2cosphi+2e = 2costheta+2ecosthetacosphi` `=>cosphi+e = costheta+ecosthetacosphi` `=>cosphi(1-ecostheta)) = costheta - e` `=>cosphi = ( costheta - e)/(1-ecostheta)` |
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| 340. |
If P and Q any two sets, then Q – P = (a) Q ∪ P′ (b) Q ∩ P′ (c) Q′ ∩ P′ (d) Q′ ∩ P′ |
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Answer» For all x ∈ Q – P ⇒ x ∈ Q and x ∈ P ⇒ x ∈ Q and x ∈ P′ ⇒ x ∈ Q ∩ P′ ∴ Q – P = Q ∩ P′ |
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| 341. |
If n (A) = 120, n (B) = 250 and n (A – B) = 52 , then find n (A ∪ Β). |
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Answer» n (A – B) = n (A) – n ( A ∩ Β) ⇒ 52 = 120 – n (A ∩ Β ) ⇒ n (A ∩ B) = 120 – 52 = 68 Now, n (A ∪ B) = n (A) + n (B) – n ( A ∩ Β) = 120 + 250 – 68 = 302 |
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| 342. |
If X and Y are two sets and X′ denotes the complement of X, then X ∩ (X ∪ Y)′ is equal to A. XB. Y C. ϕ D. X ∩ Y |
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Answer» C. Φ Given: X and Y are two sets To find: X ∩ (X ∪ Y)′ X ∩ (X ∪ Y)′ {∵ A’ ∪ B’ = (A ∩ B)’} = X ∩ (X’ ∩ Y′) = (X ∩ X’) ∩ (X ∩ Y’) = Φ ∩ (X ∩ Y’) {∵ A’ ∩ A = Φ} = Φ {∵ Φ ∩ A = Φ} Hence, X ∩ (X ∪ Y)′= Φ |
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| 343. |
Let P and Q be two sets, then what is ( P ∩ Q′) ∪ (P ∪ Q)′ equal to ? |
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Answer» ( P ∩ Q′) ∪ (P ∪ Q)′ = ( P ∩ Q′) ∩ (P′ ∩ Q′) = ( P ∪ P′) ∩ (P ∪ Q′) ∩ (Q′ ∪ P′) ∩ (Q′ ∪ Q′) = ξ ∩ { Q′ ∪ (P ∩ P′)} ∩ Q′ = ξ ∩ { Q′ ∪ ξ) ∩ Q′ = ξ ∩ Q′ ∩ Q′ = ξ ∩ Q′= ξ |
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| 344. |
If A, B and C are three finite sets, then what is [(A ∪ B) ∩ C]′ equal to ?(a) (A′ ∪ B′) ∩ C′ (b) A′∩ (B′ ∩ C′) (c) (A′ ∩ B′) ∪ C′ (d) (A ∩ B) ∩ C |
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Answer» (c) (A′ ∪ B′) ∪ C′ Given, A, B and C are three finite sets, then [ (A ∪ B) ∩ C ]′ = (A ∪ B)′ ∪ C′ = (A′ ∪ B′) ∪ C′ |
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| 345. |
If the functions f and g defined from the set of real number R to R such that `f(x) = e^(x)` and g(x) = 3x - 2, then find functions fog and gof. Also, find the domain of the functions `(fog)^(-1)` and `(gof)^(-1)`. |
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Answer» Correct Answer - A::B::C::D `(fog)x=f(3x-2)=e^(3x-2)` and `(gof)x=g(e^(x))=3e^(x)-2` Let `(fog)x=yimpliese^(3x-2)=y` `implies 3x-2=log_(e)yimpliesx=(2+log_(e)y)/(3)` `implies(fog)^(-1)(y)=(2+log_(e)y)/(3)` implies `y gt 0` So, domain of `(fog)^(-1)` is `(0, oo)`. Now, again let `(gof) x = 3e^(x) - 2` `implies y = 3e^(x)-2impliese^(x)=(y+2)/(3)` `therefore x = log_(e)((y+2)/(3))` `implies (gof)^(-1)(y)=log_(e)((y+2)/(3))` Clearly, `y+2gt0impliesygt-2` `therefore` Domain of `(gof)^(-1)` is `(-2, oo)`. |
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| 346. |
Given that E = {2, 4, 6, 8, 10}. If n represents any member of E, then, write the following sets containing all numbers represented by(i) n + 1 (ii) n2 |
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Answer» Given E = {2, 4, 6, 8, 10} (i) Let A = {x | x = n + 1, n ∈ E} Thus, for 2 ∈ E, x = 3 4 ∈ E, x = 5, and so on. Therefore, A = {3, 5, 7, 9, 11}. (ii) Let B = {x | x = n2, n ∈ E} So, for 2 ∈ E, x = (2)2 = 4, 4 ∈ E, x = (4)2 = 16, 6 ∈ E, x = (6)2 = 36, and so on. Hence, B = {4, 16, 36, 64, 100} |
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| 347. |
70 trees were planted by Parth and 90 trees were planted by Pradnya on the occasion of Tree Plantation Week. Out of these 25 trees were planted by both of them together. How many trees were planted by Parth or Pradnya? |
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Answer» i. Let P be the trees planted by Parth and Q be the trees planted by Pradnya ∴ n(P) = 70 and n(Q) = 90 Total number of trees planted by Parth and Pradnya = n(P ∩ Q) = 25 ii. Number of trees planted by Parth or Pradnya = n(P ∪ Q) = n(P) + n(Q) – n(P ∩ Q) = 70 + 90 – 25 = 135 ∴ A total of 135 trees were planted by Parth or Pradnya |
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| 348. |
If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10};find(i) A ∪ B(ii) A UC(iii) B ∪ C(iv) B ∪ D(v) A ∪ B ∪ C(vi) A ∪ B ∪ D(vii) B ∪ C ∪ D |
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Answer» A = {1, 2, 3, 4], B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10} |
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| 349. |
If a set A and n elements then find the number of elements in its power set P(A). |
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Answer» The power set of set A is a collection of all subsets of A. For example: if the set A is {1, 2} then all possible subsets of A would be {} (empty set), {1}, {2}, {1, 2} Hence power set of A that is P(A) will be {ϕ, {1}, {2}, {1,2}} Now if the number of elements in set A is n then the number of elements in power set of A P(A) is 2n |
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| 350. |
Show that `A nnB = A nnB`implies `A = B` |
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Answer» let a be an element `a in A` then `a in (A uu B)` `=> a in (A nn B)` (given) so, `a in B` if `b in B` then `b in (A uu B)` `=> b in (A nn B) => b in A` that means for all `a in A, a in B & b in B, b in A` so, `A=B` hence proved |
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