Miscellaneous Exercise answers: Relations and Functions

Open NCERT p. 40Checked by computer

1. f is given by two rules: x² on [0,3], and 3x on [3,10]. The two pieces only overlap at the single point x = 3.
2. Check both rules at x = 3: the first gives 3² = 9, and the second gives 3×3 = 9 — they agree.
3. Since both pieces give the same value where they meet, every x in [0,10] gets exactly one output, so f is a function.
4. g is given by x² on [0,2], and 3x on [2,10]. The pieces overlap only at x = 2.
5. Check both rules at x = 2: the first gives 2² = 4, and the second gives 3×2 = 6 — they disagree.
6. Since x = 2 would be sent to two different values (4 and 6), g is not a function.

Answerf is a function, because its two pieces agree (both give 9) at the meeting point x = 3. g is not a function, because its two pieces disagree (4 vs 6) at the meeting point x = 2.

Watch this explained “Two rules glued together, and where they meet”, 9:43 into The one-output rule that promotes a relation to a function

Open NCERT p. 40Matches NCERT’s answer

1. f(x) = x², so f(1.1) = 1.1² = 1.21 and f(1) = 1² = 1.
2. f(1.1) − f(1) = 1.21 − 1 = 0.21.
3. 1.1 − 1 = 0.1.
4. Divide: 0.21 ÷ 0.1 = 2.1.

Answer2.1 (i.e. 21/10).

Watch this explained “Nine squares, five answers”, 5:56 into Each standard function is pinned down by its picture as much as by its rule

Open NCERT p. 40Matches NCERT’s answer

1. This function is a fraction, so it is undefined wherever the bottom (denominator) is zero.
2. Set x² − 8x + 12 = 0. Factorise: (x−2)(x−6) = 0.
3. So the denominator is zero at x = 2 and x = 6 — these two values must be left out.

AnswerDomain = all real numbers except 2 and 6, i.e. R − {2, 6}.

Watch this explained “One polynomial over another, and what it gives up”, 8:02 into Each standard function is pinned down by its picture as much as by its rule

Open NCERT p. 40Matches NCERT’s answer

1. A square root needs a non-negative value inside it, so x − 1 ≥ 0, i.e. x ≥ 1.
2. So the domain is [1, ∞).
3. As x runs from 1 upward, x − 1 runs from 0 upward, so √(x−1) runs from 0 upward too.

AnswerDomain = [1, ∞), Range = [0, ∞).

Watch the lesson Each standard function is pinned down by its picture as much as by its rule

Open NCERT p. 40Matches NCERT’s answer

1. |x−1| is defined for every real x, so the domain is all of R.
2. An absolute value is never negative, so |x−1| ≥ 0 always, and it can be made as large as we like by choosing x far from 1.

AnswerDomain = R, Range = [0, ∞).

Watch this explained “The corner at nought”, 10:48 into Each standard function is pinned down by its picture as much as by its rule

Open NCERT p. 40Matches NCERT’s answer

1. Given: f = {(x, x²/(1 + x²)) : x ∈ R}.
2. Write y = x²/(1 + x²). The bottom, 1 + x², is at least 1, so f is defined for every real x.
3. The top x² is 0 or more and the bottom is positive, so y ≥ 0. At x = 0, y = 0.
4. Also y = 1 − 1/(1 + x²), and 1/(1 + x²) is always positive, so y < 1: the value 1 is never reached.
5. Every y with 0 ≤ y < 1 is reached: from y(1 + x²) = x² we get x² = y/(1 − y), which is 0 or more, so x = √(y/(1 − y)) works. For example, y = 1/2 comes from x = 1.

AnswerRange of f = [0, 1), that is, every real number y with 0 ≤ y < 1.

Watch the lesson Each standard function is pinned down by its picture as much as by its rule

Open NCERT p. 40Matches NCERT’s answer

1. (f + g)(x) = f(x) + g(x) = (x+1) + (2x−3) = 3x − 2.
2. (f − g)(x) = f(x) − g(x) = (x+1) − (2x−3) = x + 1 − 2x + 3 = 4 − x.
3. (f/g)(x) = f(x)/g(x) = (x+1)/(2x−3), which is defined everywhere except where 2x−3 = 0, i.e. x = 3/2.

Answer(f + g)(x) = 3x − 2, (f − g)(x) = 4 − x, (f/g)(x) = (x+1)/(2x−3), x ≠ 3/2.

Watch this explained “All four, built from one pair”, 7:17 into Combining two functions point by point, and the one case that fails

Open NCERT p. 40Matches NCERT’s answer

1. f(x) = ax + b, and from the pair (0, −1), f(0) = −1, so a(0) + b = −1, giving b = −1.
2. From the pair (1, 1), f(1) = 1, so a(1) + b = 1, i.e. a − 1 = 1, so a = 2.
3. Check with the remaining pairs: f(2) = 2(2) − 1 = 3 ✓, and f(−1) = 2(−1) − 1 = −3 ✓.

Answera = 2, b = −1.

Watch this explained “The form that gathers the first two”, 3:34 into Each standard function is pinned down by its picture as much as by its rule

Open NCERT p. 40Checked by computer

(i) (a,a) ∈ R, for all a ∈ N

1. A pair (a,b) is in R exactly when a = b².
2. (a,a) would need a = a², which is only true when a = 1 (since a is a natural number).
3. For any other a, such as a = 2 (2 ≠ 4), (a,a) is not in R.
4. So the claim 'for all a' is false.

AnswerFalse.

(ii) (a,b) ∈ R, implies (b,a) ∈ R

1. Take a = 4, b = 2: since 4 = 2², the pair (4,2) is in R.
2. For (2,4) to be in R we would need 2 = 4² = 16, which is false.
3. So (a,b) ∈ R does not force (b,a) ∈ R.

AnswerFalse.

(iii) (a,b) ∈ R, (b,c) ∈ R implies (a,c) ∈ R

1. Take c = 2, so b = c² = 4, and a = b² = 16.
2. (a,b) = (16,4) is in R since 16 = 4²; (b,c) = (4,2) is in R since 4 = 2².
3. For (a,c) = (16,2) to be in R we would need 16 = 2² = 4, which is false.
4. So the chain does not carry through.

AnswerFalse.

Watch this explained “A sentence, turned into pairs”, 3:49 into A relation is nothing more than a chosen part of the product

Open NCERT p. 40Matches NCERT’s answer

(i) f is a relation from A to B

1. Check every pair of f against A and B: each first entry (1,2,3,4) is in A, and each second entry (5,9,1,5,11) is in B.
2. Since every pair of f lies in A × B, f is a subset of A × B, so it is a relation from A to B.

AnswerYes, f is a relation from A to B.

(ii) f is a function from A to B

1. For f to be a function, every element of A must be sent to exactly one element of B.
2. But 2 appears twice: once paired with 9, and once paired with 11 — two different outputs for the same input.

AnswerNo, f is not a function from A to B, because 2 has two images (9 and 11).

Watch this explained “Two tests you can run on a list”, 0:42 into The one-output rule that promotes a relation to a function

Open NCERT p. 41Matches NCERT’s answer

1. f pairs up (product, sum) for integers a, b — the first number is a×b, the second is a+b.
2. Try a = 2, b = 3: product = 6, sum = 5, giving the pair (6, 5).
3. Try a = 1, b = 6: product is also 6, but sum = 7, giving the pair (6, 7).
4. The same first number, 6, is sent to two different second numbers, 5 and 7.

AnswerNo, f is not a function, because 6 (= 2×3 = 1×6) is paired with both 5 and 7.

Watch this explained “Find the counterexample yourself”, 11:44 into The one-output rule that promotes a relation to a function

Open NCERT p. 41Matches NCERT’s answer

1. Find the prime factors of each number in A: 9 = 3×3, 10 = 2×5, 11 is prime, 12 = 2×2×3, 13 is prime.
2. Take the highest (largest) prime factor for each: f(9)=3, f(10)=5, f(11)=11, f(12)=3, f(13)=13.
3. Collect the distinct outputs, since a set does not repeat a member.

AnswerRange = {3, 5, 11, 13}.

Watch this explained “Three lists, two tests each”, 5:29 into The one-output rule that promotes a relation to a function