CBSE Class 11 Mathematics Relations and Functions Competitive Exam MCQs Set 01

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

Explanation:
1. A × B = {(x, y) : x ∈ A, y ∈ B}, i.e. every element of A is paired with every element of B.
2. n(A × B) = 3 × 2 = 6 ordered pairs.
3. A × B = {(1, a), (1, b), (2, a), (2, b), (3, a), (3, b)}.

Teacher's Note:
1. Count first: n(A × B) = n(A) × n(B), which immediately rules out options with fewer pairs.
2. In each ordered pair of A × B the first entry comes from A and the second from B.

Answer: (b) x ∈ (5, ∞)

Explanation:
1. \( \frac{1}{\sqrt{x - 5}} \) needs x − 5 > 0, i.e. x > 5.
2. x² is defined for all real x.
3. \( \frac{1}{\sqrt{x + 7}} \) needs x + 7 > 0, i.e. x > −7.
4. Domain = (5, ∞) ∩ R ∩ (−7, ∞) = (5, ∞).

Teacher's Note:
1. A square root in the denominator needs a strictly positive expression, because it can be neither negative nor zero.
2. The domain of a sum of functions is the intersection of the individual domains.

Answer: (a) x ∈ φ

Explanation:
1. \( \sqrt{1 - x} \) needs 1 − x ≥ 0, i.e. x ≤ 1.
2. \( \sqrt{x - 5} \) needs x − 5 ≥ 0, i.e. x ≥ 5.
3. No real number is both ≤ 1 and ≥ 5, so the domain is φ.

Teacher's Note:
1. For a sum, both terms must be defined together, so take the intersection, not the union.
2. Option (c) is the union of the two conditions, a common trap.

Answer: (a) y ≠ 2

Explanation:
1. Domain: x ≠ 5.
2. y(x − 5) = 2x + 1 ⇒ x(y − 2) = 5y + 1 ⇒ \( x = \frac{5y + 1}{y - 2} \).
3. x exists for every real y except y = 2 (and x = 5 is never obtained, since that would need 5y − 10 = 5y + 1).
4. Range = R − {2}, i.e. y ≠ 2.

Teacher's Note:
1. For y = (ax + b)/(cx + d), the value a/c is never attained, so the range is R − {a/c}.
2. Option (b) describes the domain, not the range; read the question carefully.

Answer: (c) \( \left[ -\frac{1}{2}, \frac{1}{2} \right] \)

Explanation:
1. Put y = x/(1 + x²) ⇒ yx² − x + y = 0.
2. If y = 0, x = 0 works. If y ≠ 0, this quadratic in x has real roots only when 1 − 4y² ≥ 0.
3. So \( y^2 \le \frac{1}{4} \), i.e. \( -\frac{1}{2} \le y \le \frac{1}{2} \).
4. Both ends are attained (x = 1 gives \( \frac{1}{2} \), x = −1 gives \( -\frac{1}{2} \)); range = \( \left[ -\frac{1}{2}, \frac{1}{2} \right] \).

Teacher's Note:
1. Rewriting y = f(x) as a quadratic in x and demanding real x (discriminant ≥ 0) is a quick way to get such ranges.
2. Always check the case where the coefficient of x² becomes zero (here y = 0) separately.

Answer: (a) [1, ∞)

Explanation:
1. Let t = x² + 1, so t ≥ 1 and \( f = t - 1 + \frac{1}{t} \).
2. For t ≥ 1, \( t + \frac{1}{t} - 2 = \frac{(t - 1)^2}{t} \ge 0 \), so f ≥ 1, with f = 1 at t = 1 (x = 0).
3. As t increases beyond 1, \( t + \frac{1}{t} \) increases without bound, so f takes every value from 1 onwards.
4. Range = [1, ∞).

Teacher's Note:
1. Substituting t = x² + 1 turns the function into the simpler form t + 1/t − 1.
2. The result \( t + \frac{1}{t} \ge 2 \) for t > 0, with equality at t = 1, is worth remembering.

Answer: (c) \( \frac{1}{\sqrt{2}} \) or \( -\frac{1}{\sqrt{2}} \)

Explanation:
1. f(2α) = 4α² − 6α + 1 and 2f(α) = 2α² − 6α + 2.
2. 4α² − 6α + 1 = 2α² − 6α + 2 ⇒ 2α² = 1.
3. \( \alpha = \pm\frac{1}{\sqrt{2}} \).

Teacher's Note:
1. To find f(2α), replace every x in the formula by 2α, including inside the square.
2. The −6α terms cancel on both sides, which is why both signs of α work.

Answer: (b) f(x) = x + |x|

Explanation:
1. From the graph, f(x) = 0 for x < 0 and f(x) = 2x for x ≥ 0.
2. For x ≥ 0, |x| = x, so x + |x| = 2x; for x < 0, |x| = −x, so x + |x| = 0.
3. Hence f(x) = x + |x|.

Teacher's Note:
1. Split any expression with |x| into the cases x ≥ 0 and x < 0 and compare each piece with the graph.
2. x − |x| gives the opposite picture: 0 for x ≥ 0 and 2x for x < 0.

Answer: (b) (2, 5]

Explanation:
1. (x + 2)(5 − x) ≥ 0 ⇒ (x + 2)(x − 5) ≤ 0 ⇒ −2 ≤ x ≤ 5.
2. x² − 4 > 0 ⇒ x < −2 or x > 2.
3. Taking both together: 2 < x ≤ 5, i.e. x ∈ (2, 5].

Teacher's Note:
1. (x − a)(x − b) ≤ 0 with a < b means x lies between a and b, ends included.
2. The second term has the root in the denominator, so x² − 4 must be strictly positive.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Let \( y = \frac{ax + b}{cx + d} \). Then cxy + dy = ax + b ⇒ x(cy − a) = b − dy ⇒ \( x = \frac{b - dy}{cy - a} = g(y) \).
2. g(y) is defined for all y except \( y = \frac{a}{c} \), so the Reason is true.
3. A value y is attained only if a real x = g(y) exists; so \( y = \frac{a}{c} \) is never attained and the Assertion is true.
4. The Reason is exactly the step that proves the Assertion, so the answer is (a).

Teacher's Note:
1. The range of y = f(x) is found from the values of y for which the expression for x in terms of y is defined.
2. The condition ad − bc ≠ 0 ensures the function is not a constant.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. The domain is the set of x for which f(x) is defined; if f(x) exists for every real x, the domain is R. So the Assertion is true.
2. The range is the set of values taken by f; if every real y equals f(x) for some x, the range is R. So the Reason is true.
3. The Reason speaks about the range while the Assertion is about the domain, so it does not explain the Assertion; the answer is (b).

Teacher's Note:
1. Domain questions ask 'for which x is f(x) defined?', while range questions ask 'which values does f(x) take?'.
2. Two true statements are linked as cause and explanation only when one actually proves the other.

Answer: (c) \( R - \left\{ \frac{3}{2} \right\} \)

Explanation:
1. Let \( y = \frac{3x + 1}{2x + 1} \) ⇒ 2xy + y = 3x + 1 ⇒ x(2y − 3) = 1 − y ⇒ \( x = \frac{1 - y}{2y - 3} \).
2. x exists for all y except \( y = \frac{3}{2} \); there is no common factor to cancel.
3. Range = \( R - \left\{ \frac{3}{2} \right\} \).

Teacher's Note:
1. For (ax + b)/(cx + d) with no common factor, the excluded value of the range is a/c.
2. Option (a) is the excluded value of the domain, \( x = -\frac{1}{2} \), not of the range.

Answer: (d) \( R - \left\{ 0, \frac{2}{3} \right\} \)

Explanation:
1. 2x² + 5x + 2 = (2x + 1)(x + 2), so \( f(x) = \frac{1}{x + 2} \), with x ≠ −2 and \( x \ne -\frac{1}{2} \).
2. \( \frac{1}{x + 2} \) takes every real value except 0.
3. At the cancelled point \( x = -\frac{1}{2} \), \( \frac{1}{x + 2} = \frac{2}{3} \), which must also be removed.
4. Range = \( R - \left\{ 0, \frac{2}{3} \right\} \).

Teacher's Note:
1. After cancelling a common factor, the cancelled x-value is still outside the domain, so its image must be excluded.
2. Always factorise the numerator and denominator before finding the range of a rational function.

Answer: (d) \( R - \left\{ \frac{3}{2} \right\} \)

Explanation:
1. 2x² + 5x + 2 = (2x + 1)(x + 2), so f(x) = x + 2 with \( x \ne -\frac{1}{2} \).
2. x + 2 takes every real value as x varies over R.
3. At \( x = -\frac{1}{2} \), x + 2 = \( \frac{3}{2} \), which must be removed.
4. Range = \( R - \left\{ \frac{3}{2} \right\} \).

Teacher's Note:
1. The graph of this function is the line y = x + 2 with one point (a 'hole') removed.
2. The hole's y-coordinate is found by substituting the excluded x into the simplified function.

Answer: 6

Explanation:
1. Domain: x − 6 ≥ 0 and 12 − x ≥ 0, so 6 ≤ x ≤ 12.
2. \( y^2 = (x - 6) + (12 - x) + 2\sqrt{(x - 6)(12 - x)} = 6 + 2\sqrt{(x - 6)(12 - x)} \).
3. (x − 6)(12 − x) = 9 − (x − 9)² varies from 0 (at x = 6, 12) to 9 (at x = 9).
4. So y² ranges over [6, 12] and y over \( [\sqrt{6}, 2\sqrt{3}] \), an interval of length \( 2\sqrt{3} - \sqrt{6} \).
5. Hence k = 6.

Teacher's Note:
1. Squaring a sum of two square roots often turns the problem into finding the range of a simple quadratic.
2. Completing the square, (x − 6)(12 − x) = 9 − (x − 9)², shows its greatest value 9 at x = 9.

Answer: 110

Explanation:
1. Let \( t = x + \frac{1}{x} \). Then \( t^3 = x^3 + \frac{1}{x^3} + 3\left( x + \frac{1}{x} \right) \).
2. So \( x^3 + \frac{1}{x^3} = t^3 - 3t \), i.e. f(t) = t³ − 3t.
3. f(5) = 125 − 15 = 110.

Teacher's Note:
1. Express the right side in terms of the bracket x + 1/x using the identity (a + b)³ = a³ + b³ + 3ab(a + b).
2. Here ab = \( x \cdot \frac{1}{x} = 1 \), which is why the identity becomes so simple.

Answer: (c) A = B

Explanation:
1. If A = B, then clearly A × B = B × A.
2. Conversely, let A × B = B × A (A, B non-empty). For a ∈ A and any b ∈ B, (a, b) ∈ A × B = B × A, so a ∈ B; thus A ⊆ B.
3. Similarly B ⊆ A, so A = B.

Teacher's Note:
1. The Cartesian product is not commutative: in general (a, b) ≠ (b, a).
2. This result assumes A and B are non-empty; if one of them is φ, both products are φ.

Answer: (b) {(1, 4)}

Explanation:
1. A − B = {1}
2. B − C = {4}
3. (A − B) × (B − C) = {(1, 4)}.

Teacher's Note:
1. Work out the set operations inside the brackets first, then form ordered pairs.
2. A Cartesian product is a set of ordered pairs, so (1, 4) without braces is not a set.

Answer: (a) (1, 5); (2, 3); (3, 5)

Explanation:
1. First components give 1, 2, 3 ∈ A; second components give 3, 5 ∈ B.
2. n(A) × n(B) = 6 with n(A) ≥ 3, n(B) ≥ 2 gives A = {1, 2, 3}, B = {3, 5}.
3. A × B = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 3), (3, 5)}; the remaining elements are (1, 5), (2, 3), (3, 5).

Teacher's Note:
1. In A × B, the first entry of every pair comes from A and the second from B.
2. Use n(A × B) = n(A) · n(B) to decide that no other elements can be in A or B.

Answer: (c) 4

Explanation:
1. (A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A).
2. A ∩ B = {2, 3}, which has 2 elements.
3. ∴ number of elements = 2 × 2 = 4.

Teacher's Note:
1. If A and B have n elements in common, then A × B and B × A have n² elements in common.
2. Here the common pairs are (2, 2), (2, 3), (3, 2), (3, 3).

Answer: (b) {(1, 3), (2, 3), (3, 3), (8, 3)}

Explanation:
1. A ∪ B = {1, 2, 3, 8}
2. A ∩ B = {3}
3. (A ∪ B) × (A ∩ B) = {(1, 3), (2, 3), (3, 3), (8, 3)}.

Teacher's Note:
1. The order matters: the first set in the product supplies the first entries of the pairs.
2. Option (a) is (A ∩ B) × (A ∪ B), obtained by reversing the order.

Answer: (c) {(3, 2), (3, 5)}

Explanation:
1. A − B = {3}
2. A ∩ B = {2, 5}
3. (A − B) × (A ∩ B) = {(3, 2), (3, 5)}.

Teacher's Note:
1. n((A − B) × (A ∩ B)) = 1 × 2 = 2, which matches only option (c).
2. Counting the elements first is a quick check before writing out the pairs.

Answer: (d) A subset of P × Q

Explanation:
1. A relation R from P to Q is a set of ordered pairs (p, q) with p ∈ P and q ∈ Q.
2. So R ⊆ P × Q.
3. Hence a relation from P to Q is a subset of P × Q.

Teacher's Note:
1. Every subset of P × Q, including φ and P × Q itself, is a relation from P to Q.
2. P × Q is only one particular relation (the universal relation), not the general definition.

Answer: (c) 2ⁿ²

Explanation:
1. A relation on A is a subset of A × A.
2. n(A × A) = n × n = n².
3. Number of subsets of A × A = 2ⁿ², so there are 2ⁿ² relations on A.

Teacher's Note:
1. Count the pairs in A × A first; each pair can be either in or out of the relation, giving 2 choices per pair.
2. For n = 2 this gives 2⁴ = 16 relations, a useful check.

Answer: (a) 2^mn

Explanation:
1. n(A × B) = mn.
2. Every subset of A × B is a relation from A to B.
3. Number of subsets of A × B = 2^mn.

Teacher's Note:
1. The empty relation is also counted, so do not subtract 1.
2. Do not confuse this with the number of pairs, mn, which is the size of A × B itself.

Answer: (d) None of these

Explanation:
1. (x, y) ∈ R iff |x² − y²| < 16. The squares are 1, 4, 9, 16, 25.
2. Pairs not in R are those with |x² − y²| ≥ 16: (1, 5), (2, 5), (3, 5), (5, 1), (5, 2), (5, 3); all other 19 pairs, e.g. (1, 1), (1, 2), (4, 4), (4, 5), are in R.
3. Each of options (a), (b), (c) lists only a few of these 19 pairs, so none of them is R. Hence (d).

Teacher's Note:
1. A relation ‘given by’ a rule must contain every pair satisfying the rule, not just a few of them.
2. List the squares 1, 4, 9, 16, 25 and check the differences; only pairs involving 5 with 1, 2 or 3 fail.

Answer: (d) {2, 3, 4, 5}

Explanation:
1. 2 is coprime to 3 and 7; 3 is coprime to 7 and 10; 4 is coprime to 3 and 7; 5 is coprime to 3, 6 and 7.
2. So each of 2, 3, 4, 5 is related to at least one element.
3. ∴ Domain of R = {2, 3, 4, 5}.

Teacher's Note:
1. The domain is the set of first elements that actually appear in some pair of R, not simply the whole first set.
2. Two numbers are relatively prime when their HCF is 1.

Answer: (c) {2, 4, 6}

Explanation:
1. x = 8 − 2y with x, y ∈ N.
2. y = 1, 2, 3 give x = 6, 4, 2; y = 4 gives x = 0 ∉ N.
3. R = {(6, 1), (4, 2), (2, 3)}, so the domain is {2, 4, 6}.

Teacher's Note:
1. Both x and y must be natural numbers, so stop as soon as x becomes 0 or negative.
2. Here the range is {1, 2, 3}, which is option (d) — a common trap.

Answer: (c) {−2, −1, 0, 1, 2}

Explanation:
1. x² ≤ x² + y² ≤ 4, so x ∈ {−2, −1, 0, 1, 2}.
2. Each of these occurs: (±2, 0), (±1, 0), (0, 0) ∈ R.
3. ∴ Domain of R = {−2, −1, 0, 1, 2}.

Teacher's Note:
1. Do not forget negative integers: x² ≤ 4 gives −2 ≤ x ≤ 2.
2. Taking y = 0 shows that every such x really appears in R.

Answer: (c) {1}

Explanation:
1. x > y with x ∈ {1, 2, 3}, y ∈ {1, 4, 6, 9}.
2. Only (2, 1) and (3, 1) satisfy this, so R = {(2, 1), (3, 1)}.
3. Range of R = {1}.

Teacher's Note:
1. The range is the set of second elements of the pairs in R.
2. Every element of B except 1 is at least 4, which is bigger than every element of A.

Answer: (b) (−∞, 0)

Explanation:
1. We need |x| − x > 0.
2. For x ≥ 0, |x| − x = x − x = 0, which is not allowed.
3. For x < 0, |x| − x = −x − x = −2x > 0, which works.
4. Domain = (−∞, 0).

Teacher's Note:
1. Expressions containing |x| should be split into the cases x ≥ 0 and x < 0.
2. |x| − x is never negative; it is zero for all x ≥ 0 and equals −2x for x < 0.

Answer: (a) 18

Explanation:
1. p(x) = (a − a₁)x² + (b − b₁)x + (c − c₁) is a quadratic, since a ≠ a₁.
2. A quadratic that vanishes only at x = −1 has −1 as a repeated root, so p(x) = k(x + 1)².
3. p(−2) = k(−1)² = k = 2, so p(x) = 2(x + 1)².
4. p(2) = 2 × 9 = 18.

Teacher's Note:
1. The difference of two functions is found term by term: (f − g)(x) = f(x) − g(x).
2. A quadratic with exactly one root must have equal roots, so it is a constant times a perfect square.

Answer: (c) \( \left( 1, \frac{7}{3} \right] \)

Explanation:
1. \( f(x) = 1 + \frac{1}{x^2 + x + 1} \).
2. \( x^2 + x + 1 = \left( x + \frac{1}{2} \right)^2 + \frac{3}{4} \) takes all values in \( \left[ \frac{3}{4}, \infty \right) \).
3. So \( \frac{1}{x^2 + x + 1} \) takes all values in \( \left( 0, \frac{4}{3} \right] \).
4. Range = \( \left( 1, \frac{7}{3} \right] \).

Teacher's Note:
1. Writing the function as 1 + (something) by division makes its range easy to read off.
2. Completing the square gives the least value of a quadratic with positive leading coefficient.