JEE Mathematics Relations and Functions MCQs Set 05

Answer: (b) \( A\times(B\cup C) \)

Explanation:
1. The first elements are a and b, so the set is \( A\times X \) for some set X.
2. The second elements are c, d, e, which is \( B\cup C=\{c,d,e\} \).
3. Hence the set is \( A\times(B\cup C) \).

Teacher's Note:
1. Read off the set of first elements and the set of second elements.
2. A Cartesian product is a set of ordered pairs; the options with cap or cup alone contain single elements.

Answer: (b) 18

Explanation:
1. Number of elements = \( n(A)\cdot n(B)\cdot n(C) \).
2. This is 3 x 2 x 3 = 18.

Teacher's Note:
1. Multiply the sizes of the three sets.
2. The product has ordered triples.

Answer: (b) \( R_4=\{(1,3),(2,5),(2,4),(7,9)\} \)

Explanation:
1. A relation from X to Y must be a subset of \( X\times Y \).
2. In \( R_4 \), the pair (2,4) has 4 which is not in Y, and (7,9) has 7 which is not in X.
3. The other three sets have all first entries in X and second entries in Y (R1 = {(1,3),(3,5),(5,7)}).

Teacher's Note:
1. Check each pair: first element in X, second in Y.
2. R1 gives only the pairs (1,3), (3,5), (5,7).

Answer: (b) \( R=\{(1,2),(2,2),(3,3),(2,1),(2,3),(3,2)\} \)

Explanation:
1. The differences of squares are 1-4=-3, 1-9=-8, 4-9=-5 and their negatives. Only (1,3) and (3,1) violate the condition.
2. So R contains every pair except (1,3) and (3,1); in particular it contains (1,1), (2,2), (3,3).
3. The set in option (a) lacks (1,1), so it is not R; the domain, range and co-domain are all {1, 2, 3}.

Teacher's Note:
1. List all nine pairs and test each.
2. Only the pairs (1,3) and (3,1) fail.

Answer: (b) Reflexive but not symmetric

Explanation:
1. Reflexive: a divides a.
2. Not symmetric: 2 divides 4 but 4 does not divide 2.
3. It is transitive (a | b and b | c give a | c), so it is reflexive and transitive but not symmetric.

Teacher's Note:
1. Test with the pair (2, 4).
2. Divisibility is transitive.

Answer: (b) Transitive only

Explanation:
1. Not reflexive: x < x is false.
2. Not symmetric: 1 < 2 but not 2 < 1.
3. Transitive: x < y and y < z give x < z.

Teacher's Note:
1. Check x R x first.
2. Strict inequality is transitive.

Answer: (b) \( x\,R_1\,y\iff x^2=y^2 \)

Explanation:
1. x^2 = y^2 is reflexive, symmetric (equality is symmetric) and transitive.
2. The other three fail symmetry: for example 2 >= 1 but not 1 >= 2; 2 divides 4 but not conversely; 1 < 2 but not 2 < 1.

Teacher's Note:
1. An equivalence relation is reflexive, symmetric and transitive.
2. Equality of a function value is always an equivalence.

Answer: (b) Reflexive and transitive but not symmetric

Explanation:
1. a >= a, so it is reflexive.
2. a >= b and b >= c give a >= c, so it is transitive.
3. (2,1) is in the relation but (1,2) is not, so it is not symmetric.

Teacher's Note:
1. Test with a = 2, b = 1.
2. Check all three properties separately.

Answer: (b) Symmetric

Explanation:
1. If alpha is perpendicular to beta, then beta is perpendicular to alpha: symmetric.
2. A line is not perpendicular to itself: not reflexive.
3. If alpha is perpendicular to beta and beta to gamma, then alpha is parallel to gamma, not perpendicular: not transitive.

Teacher's Note:
1. Test each property with the lines x = 0 and y = 0.
2. Two lines perpendicular to a third are parallel.

Answer: (b) Reflexive and symmetric but not transitive

Explanation:
1. Reflexive: 1 + a^2 > 0 for every a.
2. Symmetric: 1 + ab = 1 + ba.
3. Not transitive: a = 1, b = -1/2, c = -4 gives 1 + ab = 1/2 > 0 and 1 + bc = 3 > 0 but 1 + ac = -3 < 0.

Teacher's Note:
1. Look for a counterexample to transitivity using one positive and two negative numbers.
2. a = 1, b = -1/2, c = -4 works.

Answer: (b) 2 only

Explanation:
1. Statement 1 is false: divisibility is reflexive and transitive but not symmetric (2 divides 4, 4 does not divide 2).
2. Statement 2: a set is a subset of itself, so S is not related to S (not reflexive). With S1 = {1,2} and S2 = {1}: S1 is not a subset of S2, but S2 is a subset of S1, so the pair is not symmetric. With S1 = {1}, S2 = {2}, S3 = {1}: S1 R S2 and S2 R S3 but not S1 R S3 (not transitive). (All three checks were confirmed by brute force over all subsets of A.)
3. Hence only statement 2 is correct.

Teacher's Note:
1. Test statement 1 with a pair such as (2, 4).
2. Use small subsets as counterexamples in statement 2.

Answer: (b) An equivalence relation

Explanation:
1. Reflexive: a + b = b + a.
2. Symmetric: a + d = b + c gives c + b = d + a.
3. Transitive: a + d = b + c and c + f = d + e give a + f = b + e after adding and cancelling, so R is an equivalence relation.

Teacher's Note:
1. Add the two given equations for transitivity.
2. The relation says a - b = c - d.

Answer: (b) Symmetric only

Explanation:
1. R = {(1,4),(2,3),(3,2),(4,1)}.
2. Symmetric: every reverse pair is present. Not reflexive: (1,1) is missing.
3. Not transitive: (1,4) and (4,1) are in R but (1,1) is not.

Teacher's Note:
1. List the pairs explicitly.
2. Test (1,4) and (4,1) for transitivity.

Answer: (b) Is not an equivalence relation

Explanation:
1. Not reflexive: A A = O is false for A = I.
2. Not symmetric: for A = [[0,1],[0,0]] and B = [[1,0],[0,0]], AB = O but BA = [[0,1],[0,0]] is not O.
3. Not transitive either, so R is not an equivalence relation.

Teacher's Note:
1. Test A = identity for reflexivity.
2. Matrix multiplication is not commutative.

Answer: (b) 6

Explanation:
1. [e] = [2.718...] = 2.
2. [-pi] = [-3.14...] = -4.
3. So the value is 2 - (-4) = 6.

Teacher's Note:
1. The greatest integer function rounds down, also for negative numbers.
2. [-3.14] is -4, not -3.

Answer: (b) Infinite

Explanation:
1. Put t = {x}: 2t^2 - 5t + 2 = 0 gives t = 2 or t = 1/2.
2. Since 0 <= {x} < 1, only {x} = 1/2 is possible.
3. Every x of the form n + 1/2 (n an integer) works, so there are infinitely many solutions.

Teacher's Note:
1. Solve the quadratic in t = {x}.
2. Fractional part lies in [0, 1).

Answer: (b) 104

Explanation:
1. f(n) = 0 for n = 1 to 49, since 0.5 + n/100 < 1.
2. f(n) = 1 for n = 50 to 149 (100 values), contributing 100.
3. f(150) = [2.0] = 2 and f(151) = [2.01] = 2, contributing 4. Total 104.

Teacher's Note:
1. Find where 1/2 + n/100 crosses 1 and 2.
2. n = 50 gives exactly 1, n = 150 gives exactly 2.

Answer: (b) 3

Explanation:
1. Let u = 2x with [u] = k, {u} = u - k. Then k - 3{u} = 1 gives {u} = (k - 1)/3.
2. We need 0 <= (k - 1)/3 < 1, so k = 1, 2, 3.
3. These give u = 1, 7/3, 11/3, hence three values of x.

Teacher's Note:
1. Substitute u = 2x.
2. Fractional part must be in [0, 1).

Answer: (b) \( (1,2)\cup(2,3)\cup(3,4) \)

Explanation:
1. The fraction is negative when {x} > 0 and x^2 - 5x + 4 < 0.
2. x^2 - 5x + 4 < 0 gives 1 < x < 4; {x} > 0 excludes the integers 2 and 3.
3. So x is in (1,2) U (2,3) U (3,4).

Teacher's Note:
1. {x} is never negative, and it is zero at integers.
2. The sign depends only on the quadratic once {x} > 0.

Answer: (b) 7

Explanation:
1. x^3 - 4x^2 + 3x = x(x - 1)(x - 3).
2. It is positive for 0 < x < 1 and for x > 3; it is negative for x < 0 and for 1 < x < 3.
3. Integers in [-5, 10] with x > 3: 4, 5, ..., 10, which are 7 values.

Teacher's Note:
1. Factorise and study the sign on each interval.
2. No integer lies in (0, 1).

Answer: (b) 1

Explanation:
1. If {x} = 0, sgn = 0, so we need |1 - x| = 0, that is x = 1, and indeed {1} = 0: valid.
2. If {x} > 0, sgn = 1, so |1 - x| = 1 gives x = 0 or x = 2; but both have {x} = 0, a contradiction.
3. Hence only x = 1 works.

Teacher's Note:
1. Split into {x} = 0 and {x} > 0.
2. Check each candidate in the original equation.

Answer: (b) 4

Explanation:
1. f(1) = f(2) = 1 and f(pi) = -1, so 1 + 1 - 1 + f(p) = 0 gives f(p) = -1.
2. Therefore p must be irrational.
3. Among the options only 4 is rational, so p cannot be 4.

Teacher's Note:
1. Evaluate the known terms first.
2. f takes the value -1 only at irrational numbers.

Answer: (b) Reflexive

Explanation:
1. Reflexive: x - x + sqrt(2) = sqrt(2) is irrational.
2. Not symmetric: x = sqrt(2), y = 0 gives 2 sqrt(2) (irrational) but y - x + sqrt(2) = 0 is rational.
3. Not transitive: x = sqrt(2), y = 1, z = 2 sqrt(2) gives 2 sqrt(2) - 1 and 1 - 2 sqrt(2) + sqrt(2) = 1 - sqrt(2), both irrational, but x - z + sqrt(2) = 0 is rational.

Teacher's Note:
1. Test the relation with x = sqrt(2) and y = 0.
2. For transitivity choose z so that x - z + sqrt(2) = 0.

Answer: (b) \( R_1=R_2 \)

Explanation:
1. Modulo 5, 4 is congruent to -1 and 9 is congruent to -1.
2. So both conditions say m - n is divisible by 5.
3. Hence R1 = R2 (and both are symmetric).

Teacher's Note:
1. Reduce the coefficients 4 and 9 modulo 5.
2. Both are congruent to -1.

Answer: (b) The relation is neither reflexive, nor symmetric, nor transitive

Explanation:
1. Not reflexive: nobody is 5 years older than himself.
2. Not symmetric: if y is 5 years older than x, then x is not 5 years older than y.
3. Not transitive: if y is 5 years older than x and z is 5 years older than y, then z is 10 years older than x, not 5.

Teacher's Note:
1. Take people aged 10, 15 and 20.
2. Check each property separately.