JEE Mathematics Relations and Functions MCQs Set 16

Answer: (c) {−2, −1, 0, 1, 2}

Explanation:
1. \( x^{2} + y^{2} \le 4 \) gives \( x^{2} \le 4 \), so \( -2 \le x \le 2 \).
2. The integer values of \( x \) are −2, −1, 0, 1, 2, and each has a valid integer \( y \) (for example \( y = 0 \)).

Teacher's Note:
1. Domain is the set of first components that actually occur.
2. Restrict to integers before listing.

Answer: (c) 45

Explanation:
1. \( n(A \times A \times B) = n(A) \cdot n(A) \cdot n(B) = 3 \times 3 \times 5 = 45 \).

Teacher's Note:
1. The size of a Cartesian product is the product of the sizes.
2. Repeated sets are counted each time they appear.

Answer: (d) None of these

Explanation:
1. Checking all 25 pairs, \( R \) contains (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4), (4,5), (5,4), (5,5).
2. For example (1, 2) ∈ \( R \) since \( |1 - 4| = 3 \) < 16, but it is missing from options \( (a), (b), (c) \); also (2, 4) is in \( (b) \) but (2,1) is missing there.
3. So none of the three listed sets equals \( R \).

Teacher's Note:
1. Test each pair against the defining condition instead of guessing.
2. One missing or extra pair is enough to reject an option.

Answer: (c) {(6, 0), (0, 6)}

Explanation:
1. \( x \) and \( y \) must be non-negative integers with \( x^{2} + y^{2} = 36 \).
2. Only \( 0^{2} + 6^{2} = 36 \) and \( 6^{2} + 0^{2} = 36 \) work, as \( 36 - x^{2} \) is a perfect square only for \( x = 0 \) and \( x = 6 \).
3. So \( R = {(6, 0), (0, 6)} \).

Teacher's Note:
1. Surds are not integers, so options with roots can be rejected.
2. Check each \( x \) from 0 to 6.

Answer: (a) {1, 2, 3}, {7, 5}

Explanation:
1. \( x = 1 \) gives \( y = 7, x = 2 \) gives \( y = 5, x = 3 \) gives \( y = 5 \).
2. \( x = 4 \) gives 5.5, and \( x = 5 \) gives 6.2, which are not natural numbers.
3. So \( R = {(1, 7), (2, 5), (3, 5)} \), domain {1, 2, 3} and range {7, 5}.

Teacher's Note:
1. Keep only the \( x \) values for which \( y \) is also a natural number.
2. Range lists each value once.

Answer: (a) {(1, 3), (2, 3), (3, 1), (3, 2), (1, 1), (2, 1), (1, 2)}

Explanation:
1. A × \( B = {(1,1), (1,3), (2,1), (2,3)} \) and \( B \times A = {(1,1), (1,2), (3,1), (3,2)} \).
2. Their union has the seven distinct pairs (1,1), (1,3), (2,1), (2,3), (1,2), (3,1), (3,2), which is option \( (a) \).

Teacher's Note:
1. List both products fully and then merge, removing the repeated pair (1, 1).
2. Ordered pairs are different when the order is reversed.

Answer: (c) {(a, 2), (b, 3), (3, b)}

Explanation:
1. A relation from \( A \) to \( B \) is a subset of \( A \times B \), so each pair must have its first element in \( A \) and its second in \( B \).
2. In \( (c) \) the pair \( (3, b) \) has 3 ∉ \( A \), so it is not in \( A \times B \).
3. The other three sets are subsets of \( A \times B \).

Teacher's Note:
1. Check the first and second components of every pair against \( A \) and \( B \) separately.

Answer: (d) 16

Explanation:
1. \( n(A \times B) = 2 \times 2 = 4 \).
2. The number of subsets is \( 2^{4} = 16 \).

Teacher's Note:
1. A set with \( n \) elements has 2ⁿ subsets.
2. First find the number of elements in the product.

Answer: (a) 36

Explanation:
1. \( (R \times S) \) ∩ (\( S \times R) = (R \) ∩ \( S) \times (S \) ∩ \( R \)).
2. \( R \) ∩ \( S \) is the set of multiples of 15 up to 100: {15, 30, 45, 60, 75, 90}, which has 6 elements.
3. So the number of elements is \( 6 \times 6 = 36 \).

Teacher's Note:
1. Use \( (A \times B) \) ∩ (\( C \times D) = (A \) ∩ \( C) \times (B \) ∩ \( D \)).
2. Common multiples of 3 and 5 are multiples of the LCM 15.

Answer: (c) 50

Explanation:
1. pq = \( 7, a \) prime number, so \( {p, q} = {1, 7} \).
2. Hence \( p^{2} + q^{2} = 1 + 49 = 50 \).

Teacher's Note:
1. A prime product forces the factors 1 and the prime itself.
2. The sum of squares is the same in either order.

Answer: (b) 8

Explanation:
1. A = \( {0, 1, 2}, B = {2, 3} \), so \( A \) ∪ \( B = {0, 1, 2, 3} \) with 4 elements.
2. \( C \) has 2 elements, so (\( A \) ∪ B) × C has \( 4 \times 2 = 8 \) elements.

Teacher's Note:
1. Write each set out before taking the union.
2. Whole numbers start from 0, natural numbers from 1.

Answer: (d) None of these

Explanation:
1. For integers, \( x^{2} + y^{2} = 9 \) only for (0, ±3) and (±3, 0), since \( 9 - x^{2} \) is not a perfect square for \( x = \pm 1, \pm 2 \).
2. So \( R \) is as in \( (a) \), the domain is {−3, 0, 3} and the range is {−3, 0, 3}.
3. Statements \( (a), (b), (c) \) are all true, so the false-statement choice is 'None of these'.

Teacher's Note:
1. Read the question stem carefully: it asks for the false statement.
2. Try \( x = 0, \pm 1, \pm 2, \pm 3 \) to list all integer solutions.

Answer: (c) [−2, 2] ∪ [3, 4]

Explanation:
1. \( x^{2} - 5x + 6 \ge 0 \) gives \( x \le 2 \) or \( x \ge 3 \).
2. \( 2x + 8 - x^{2} \ge 0 \) gives \( x^{2} - 2x - 8 \le 0 \), so \( -2 \le x \le 4 \).
3. The intersection is [−2, 2] ∪ [3, 4].

Teacher's Note:
1. Every square root needs a non-negative radicand.
2. Take the intersection of the conditions.

Answer: (d) [−4, −1]

Explanation:
1. Since \( 0.625 = 5/8 \) and \( 1.6 = 8/5 \), we need \( (5/8)^{4-3x} \ge (8/5)^(x(x+8)) = (5/8)^(-x(x+8)) \).
2. The base is less than 1, so the inequality reverses: \( 4 - 3x \le -x^{2} - 8x \).
3. This gives \( x^{2} + 5x + 4 \le 0 \), i.e. \( (x + 1)(x + 4) \le 0 \), so \( x \) ∈ [−4, −1].

Teacher's Note:
1. Notice that 0.625 and 1.6 are reciprocals.
2. For a base between 0 and 1, an exponential inequality reverses direction.

Answer: (d) [1, ∞)

Explanation:
1. \( f(x) = (x + 1)^{2} + 1 \ge 1 \), and the value 1 is attained at \( x = -1 \).
2. So the range is [1, ∞.)

Teacher's Note:
1. Complete the square to find the minimum value.
2. The minimum is attained, so the bracket is closed.

Answer: (a) (−∞, 2] − {−1}

Explanation:
1. Using \( x^{3} + 1 = (x + 1)(x^{2} - x + 1) \), the expression under the root becomes \( [2(x + 1) - (x^{2} - x + 1) - (2x - 1)] / (x^{3} + 1) = (-x^{2} + x + 2)/(x^{3} + 1) \).
2. This equals \( -(x - 2)(x + 1) / [(x + 1)(x^{2} - x + 1)] = -(x - 2)/(x^{2} - x + 1) \) for \( x \ne -1 \).
3. Since \( x^{2} - x + 1 \) > 0, we need \( x \le 2 \), with \( x = -1 \) excluded. The domain is (−∞, 2] − {−1}.

Teacher's Note:
1. Factorise \( x^{3} + 1 \) and combine into one fraction.
2. The cancelled factor \( (x + 1) \) is still excluded from the domain.

Answer: (a) (1/2, 1]

Explanation:
1. The first term needs \( 2x - 1 \) > 0, so \( x \) > 1/2.
2. The second term needs \( 1 - x^{2} \ge 0 \), so \( -1 \le x \le 1 \).
3. The common domain is (1/2, 1].

Teacher's Note:
1. Strict inequality for a denominator, non-strict for a plain root.
2. Intersect the two intervals.

Answer: (c) h = {(n, 1/n) | n ∈ I}

Explanation:
1. For \( h, n = 0 \) belongs to I, but 1/0 is not defined, so 0 has no image.
2. \( f, g \) and \( t \) assign exactly one image to every element of their domains.

Teacher's Note:
1. A function needs a unique image for every element of the domain.
2. Look for domain values where the rule breaks.

Answer: (d) −∞ < x < 1

Explanation:
1. 2ʸ = 2 − 2ˣ must be positive, so 2ˣ < 2, which gives \( x \) < 1.
2. Every \( x \) < 1 gives y = log₂(2 − 2ˣ), so the domain is (−∞, 1).

Teacher's Note:
1. Exponentials are always positive.
2. Express \( y \) in terms of \( x \) and demand the logarithm's argument be positive.

Answer: (b) (−∞, 0)

Explanation:
1. We need \( |x| - x \) > 0, i.e. \( |x| \) > \( x \), which holds only when \( x \) < 0.
2. So the domain is (−∞, 0).

Teacher's Note:
1. \( |x| = x \) for \( x \ge 0 \), so those values make the denominator zero.
2. The expression under the root must be strictly positive.

Answer: (b) 4(2 − a)

Explanation:
1. \( f(a + 1) - f(a - 1) = 4[(a + 1) - (a - 1)] - [(a + 1)^{2} - (a - 1)^{2}] = 8 - 4a \).
2. This equals \( 4(2 - a) \).

Teacher's Note:
1. Use \( (a + 1)^{2} - (a - 1)^{2} = 4a \).
2. Substitute into each bracket separately.

Answer: (d) [1/√2, 1]

Explanation:
1. \( \sqrt{1 - x^{2}} \) needs \( -1 \le x \le 1 \).
2. The outer root needs \( x \ge \sqrt{1 - x^{2}} \ge 0 \), so \( x \ge 0 \) and \( x^{2} \ge 1 - x^{2} \), i.e. \( x \ge 1/\sqrt{2} \).
3. The domain is \( [1/\sqrt{2}, 1] \).

Teacher's Note:
1. Two nested roots give two conditions.
2. The right side is non-negative, so \( x \) itself must be non-negative.

Answer: (a) (0, 1)

Explanation:
1. We need \( x \) > 0 for \( x^{1/4} \). Then \( -\log_{1/2}(y) = \log_{2} y \), with \( y = 1 + x^{-1/4} \).
2. The outer logarithm needs \( \log_{2}(1 + x^{-1/4}) - 1 \) > 0, i.e. \( 1 + x^{-1/4} \) > 2, so \( x^{-1/4} \) > 1.
3. This gives 0 < \( x \) < 1.

Teacher's Note:
1. Convert \( \log \) to base 2 using \( \log_{1/2} y = -\log_{2} y \).
2. Work from the outside in.

Answer: (c) (−1, 0]

Explanation:
1. For \( x \ge 0, e^|x| \) = eˣ, so \( f(x) = 0 \).
2. For \( x \lt 0, f(x) \) = (eˣ − e⁻ˣ)/(eˣ + e⁻ˣ) = \( \tanh x \), which takes every value in (−1, 0).
3. The range is (−1, 0].

Teacher's Note:
1. Split the function into the cases \( x \ge 0 \) and \( x \) < 0.
2. \( \tanh x \) lies strictly between −1 and 1.

Answer: (c) E ∪ F ⊂ f(E) ∪ f(F)

Explanation:
1. \( f(E) = [0, 1] \) and \( f(F) = [0, 1] \), so \( (a) \) is true.
2. \( E \) ∩ \( F = {0} \) ⊂ [0, 1], so \( (b) \) is true, and f(E ∩ F) = {0}, so \( (d) \) is true.
3. \( E \) ∪ \( F = [-1, 1] \) is not a subset of \( f(E) \) ∪ \( f(F) = [0, 1] \) because it contains negative numbers, so \( (c) \) is false.

Teacher's Note:
1. Find the images \( f(E) \) and \( f(F) \) first.
2. Negative numbers can never lie in the image of \( x^{2} \).

Answer: (d) −x

Explanation:
1. Replace \( x \) by \( 1/x \): \( 3f(1/x) - f(x) = -4 \ln x \).
2. Substituting \( f(1/x) = 3f(x) - 4 \ln x \) from the original equation gives \( 9f(x) - 12 \ln x - f(x) = -4 \ln x \), so \( f(x) = \ln x \).
3. Therefore f(e⁻ˣ) = ln(e⁻ˣ) = \( -x \).

Teacher's Note:
1. Form a second equation by replacing \( x \) with \( 1/x \), then eliminate \( f(1/x) \).
2. Treat \( f(x) \) and \( f(1/x) \) as two unknowns.

Answer: (b) ½[f(2x) + f(2y)]

Explanation:
1. \( f(x + y) f(x - y) = \frac{1}{4}(2^{x+y} + 2^{-x-y})(2^{x-y} + 2^{-x+y}) \).
2. Expanding gives \( \frac{1}{4}(2^{2x} + 2^{2y} + 2^{-2y} + 2^{-2x}) \).
3. This equals \( \frac{1}{2}[f(2x) + f(2y)] \).

Teacher's Note:
1. Multiply out the four terms and regroup.
2. Recognise each pair as \( 2\cdot f \) of a doubled argument.