Miscellaneous Exercise — Relations and Functions

1. 2.M.13 marksNCERT Class 11 Mathematics, Relations and Functions, Reprint 2026-27

   f is defined by f(x)=x^2 for 0<=x<=3, f(x)=3x for 3<=x<=10. g is defined by g(x)=x^2 for 0<=x<=2, g(x)=3x for 2<=x<=10. Show f is a function and g is not.

   Hint. Check the single point where the two pieces of each definition overlap -- that is the only place a piecewise rule can go wrong.

   For f, the two pieces overlap only at x=3: the first piece gives f(3)=3^2=9, and the second piece gives f(3)=3(3)=9 -- they agree, so f assigns exactly one value everywhere and is a function. For g, the two pieces overlap at x=2: the first piece gives g(2)=2^2=4, but the second piece gives g(2)=3(2)=6 -- these disagree, so x=2 would need to map to both 4 and 6, which is not allowed for a function.

   ✦ Working through each part gives: f is a function since its pieces agree at x=3 (both give 9); g is not a function since its pieces disagree at x=2 (4 versus 6).

2. 2.M.22 marksNCERT Class 11 Mathematics, Relations and Functions, Reprint 2026-27

   If f(x) = x^2, find [f(1.1) - f(1)] / (1.1 - 1).

   Hint. Compute f(1.1) and f(1) separately first, then form the difference quotient.

   f(1.1)=1.21, f(1)=1, so the numerator is 1.21-1=0.21, and the denominator is 1.1-1=0.1.

   ✦ Working through each part gives: 0.21 / 0.1 = 2.1.

3. 2.M.32 marksNCERT Class 11 Mathematics, Relations and Functions, Reprint 2026-27

   Find the domain of f(x) = (x^2+2x+1) / (x^2-8x+12).

   Hint. A rational function is undefined exactly where its denominator is zero -- factorise the denominator to find those points.

   x^2-8x+12 factors as (x-6)(x-2), which is zero at x=6 and x=2, so the function is undefined at exactly those two points.

   ✦ Working through each part gives: domain = R minus {2, 6}.

4. 2.M.42 marksNCERT Class 11 Mathematics, Relations and Functions, Reprint 2026-27

   Find the domain and range of f(x) = sqrt(x - 1).

   Hint. The quantity under the square root must be non-negative.

   x-1>=0 means x>=1, fixing the domain; over that domain the square root itself takes every value from 0 upward.

   ✦ Working through each part gives: domain = [1, infinity). Range = [0, infinity).

5. 2.M.52 marksNCERT Class 11 Mathematics, Relations and Functions, Reprint 2026-27

   Find the domain and range of f(x) = |x - 1|.

   Hint. An absolute value expression is defined for every real input and never produces a negative output.

   There is no restriction on x, so the domain is all reals; the output |x-1| can be any non-negative number, achieved by choosing x on either side of 1.

   ✦ Working through each part gives: domain = R. Range = [0, infinity).

6. 2.M.63 marksNCERT Class 11 Mathematics, Relations and Functions, Reprint 2026-27

   Let f = {(x, x^2/(1+x^2)) : x in R} be a function from R into R. Determine the range of f.

   Hint. Since x^2 is always non-negative and strictly less than 1+x^2, think about what values the ratio x^2/(1+x^2) can and cannot reach.

   At x=0 the ratio is 0. As |x| grows large, x^2/(1+x^2) gets closer and closer to 1 but never actually reaches it, since the denominator is always slightly larger than the numerator.

   ✦ Working through each part gives: range = [0, 1), every value from 0 up to but not including 1.

7. 2.M.73 marksNCERT Class 11 Mathematics, Relations and Functions, Reprint 2026-27

   Let f, g : R to R be defined by f(x)=x+1, g(x)=2x-3. Find f+g, f-g, and f/g.

   Hint. Combine the two rules pointwise using the standard definitions of sum, difference and quotient of functions.

   (f+g)(x) = (x+1)+(2x-3) = 3x-2. (f-g)(x) = (x+1)-(2x-3) = 4-x. (f/g)(x) = (x+1)/(2x-3), defined wherever 2x-3 is not zero.

   ✦ Working through each part gives: f+g = 3x-2; f-g = 4-x; f/g = (x+1)/(2x-3), x not equal to 3/2.

8. 2.M.82 marksNCERT Class 11 Mathematics, Relations and Functions, Reprint 2026-27

   Let f = {(1,1),(2,3),(0,-1),(-1,-3)} be a function from Z to Z with f(x)=ax+b. Find a and b.

   Hint. Use two of the given points to set up two simultaneous equations in a and b.

   From (0,-1): f(0)=b=-1. From (1,1): f(1)=a+b=1, so a=1-b=1-(-1)=2. Checking against (2,3): f(2)=2(2)-1=3, matches; and (-1,-3): f(-1)=2(-1)-1=-3, matches.

   ✦ Working through each part gives: a = 2, b = -1.

9. 2.M.94 marksNCERT Class 11 Mathematics, Relations and Functions, Reprint 2026-27

   R is defined on N by R={(a,b): a,b in N, a=b^2}. Are the following true, with justification? (i) (a,a) in R for all a in N. (ii) (a,b) in R implies (b,a) in R. (iii) (a,b) in R and (b,c) in R implies (a,c) in R.

   Hint. Test each claim with a specific number before deciding -- a single counterexample disproves a 'for all' claim completely.

   (i) (a,a) in R needs a=a^2, true only for a=1, not for every natural number (e.g. a=2 fails, since 2 is not 4), so this is false in general. (ii) take a=4,b=2 (since 4=2^2): (4,2) is in R, but (2,4) would need 2=4^2=16, which is false, so this is false. (iii) take b=4,c=2 (b=c^2) and a=16 (a=b^2): (16,4) and (4,2) are both in R, but (16,2) would need 16=2^2=4, which is false, so this is also false.

   ✦ Working through each part gives: all three statements are false in general; each is disproved by a specific counterexample rather than holding for every choice of a, b, c.

10. 2.M.103 marksNCERT Class 11 Mathematics, Relations and Functions, Reprint 2026-27

    Let A={1,2,3,4}, B={1,5,9,11,15,16}, f={(1,5),(2,9),(3,1),(4,5),(2,11)}. Are the following true? (i) f is a relation from A to B. (ii) f is a function from A to B.

    Hint. For (i), check every pair has its first element in A and second in B. For (ii), check whether any first element repeats with a different second element.

    (i) every pair's first element is in A={1,2,3,4} and second element is in B, so f is indeed a subset of AxB and qualifies as a relation. (ii) the element 2 appears twice, once paired with 9 and once with 11 -- two different images for the same input -- so f fails the function requirement.

    ✦ Working through each part gives: (i) True, f is a relation from A to B. (ii) False, f is not a function, since 2 has two different images (9 and 11).

11. 2.M.113 marksNCERT Class 11 Mathematics, Relations and Functions, Reprint 2026-27

    Let f be the subset of Z x Z defined by f = {(ab, a+b) : a, b in Z}. Is f a function from Z to Z? Justify.

    Hint. Try to find two different pairs (a,b) that give the same product ab but different sums a+b -- that would break the function requirement.

    Take a=6,b=1: ab=6, a+b=7. Take a=2,b=3: ab=6, a+b=5. Both pairs give the same first element, 6, but different second elements, 7 and 5.

    ✦ Working through each part gives: f is not a function, since the same first element 6 arises from two different (a,b) choices that give different sums (7 and 5).

12. 2.M.123 marksNCERT Class 11 Mathematics, Relations and Functions, Reprint 2026-27

    Let A={9,10,11,12,13}, and f:A to N be defined by f(n) = the highest prime factor of n. Find the range of f.

    Hint. Factorise each of the five numbers in A and pick out its largest prime factor.

    9=3x3 has highest prime factor 3. 10=2x5 has highest prime factor 5. 11 is itself prime, so its highest prime factor is 11. 12=2x2x3 has highest prime factor 3. 13 is prime, so its highest prime factor is 13.

    ✦ Working through each part gives: range = {3, 5, 11, 13}.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Mathematics textbook, Reprint 2026-27 (kemh102.pdf) — three numbered exercises (2.1-2.3, 24 questions) plus the chapter's Miscellaneous Exercise (12 questions); Exercise 2.2 Q4's arrow-diagram figure (Fig 2.7) was rendered directly from the PDF at 250dpi and read visually to confirm the y=x-2 mapping, and the piecewise function boundaries in Miscellaneous Exercise Q1 were re-rendered at high resolution after the raw text extraction garbled the fraction/subscript layout, confirming f's pieces agree at x=3 (function) while g's pieces disagree at x=2 (not a function). Questions are referenced from the NCERT textbook for identification.