Relations and Functions - Test Papers

 CBSE Test Paper 01

CH-02 Relations and Functions

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1. Two finite sets have m and n elements. The number o elements in the power set of the first is 48 more than the total number of elements in the power set of the second. Then the values of m and n are
   1. 6, 4
   2. 6, 3
   3. 3, 7
   4. 7, 6
2. Let f(x+1x)=x2+1x2,x≠0,$f\left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2},x\ne 0,$ then f(x) =
   1. x2−2$x^2-2$
   2. x2−1$x^2-1$
   3. x2$x^2$
   4. x2+1$x^2+1$
3. The function f (x) = log (x+x2+1−−−−−√)$(x+\sqrt{x^2+1})$ is
   1. a periodic function
   2. neither an even nor an odd function
   3. an odd function
   4. an even function
4. If A is the set of even natural numbers less than 8 and B is the set of prime numbers less than 7, then the number of relations from A to B is
   1. 32$3^2$
   2. 92$9^2$
   3. 29−1$2^9-1$
   4. 29$2^9$
5. The relation R = {1, 1), (2, 2), (3, 3)} on the set {1, 2, 3) is
   1. an equivalence relation
   2. reflexive only
   3. symmetric only
   4. transitive only
6. If f(1 + x) = x2 + 1, then f(2 - h) is ________.
7. Fill in the blanks: Let A and B be any two non-empty finite sets containing m and n elements respectively, then, the total number of subsets of (A ×$\times$ B) is ________.
8. If A ×$\times$ B = {(a, 1),(a, 5),(a, 2),(b, 2),(b, 5),(b, 1)}, then find A, B and B ×$\times$ A.
9. Find the domain of the function f(x)=x2+3x+5x2+x−6$f(x)=\frac{x^2+3x+5}{x^2+x-6}$.
10. Let f, g: R→R$R\to R$ be defined, respectively by f(x) = x + 1, g(x) = 2x – 3. Find f + g, f – g and fg$\frac{f}{g}$.
11. If A = (1, 2, 3), B = {4}, C = {5}, then verify that A×(B∪C)=(A×B)∪(A×C)$A\times (B\cup C)=(A\times B)\cup (A\times C)$.
12. Let f: R →$\to$ R and g: C →$\to$ C be two functions defined as f(x) = x2 and g(x) = 2x. Are they equal functions?
13. If A = {2, 3}, B = {4, 5}, C = {5, 6}, find A×(B∪C),A×(B∩C)$A\times (B\cup C),A\times (B\cap C)$,(A×B)∪(A×C)$(A\times B)\cup (A\times C)$
14. If (x3+1,y−23)=(53,13)$\left(\frac{x}{3}+1,y-\frac{2}{3}\right)=\left(\frac{5}{3},\frac{1}{3}\right)$find the values of x and y.
15. If A = {a,d}, B = {b, c, e} and C = {b, c, f}, then verify that
    1. A×$\times$(B ∪$\cup$ C) = (A×$\times$B) ∪$\cup$ (A×$\times$C)
    2. A×$\times$(B ∩$\cap$ C) = (A×$\times$B) ∩$\cap$ (A×$\times$C)

CBSE Test Paper 01
CH-02 Relations and Functions

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Solution

1. (a) 6, 4
   Explanation:

   Let A has m elements and B gas n elements. Then, no. of elements in

   P(A) = 2m and no. of elements in P(B) = 2n.]

   By the question,

   2m = 2n + 48

   ⇒$\Rightarrow$ 2m - 2n = 48

   This is possible, if 2m = 64, 2n = 16. (As 64 - 16 = 48)

   ∴2m=64⇒2m=26$\therefore 2^m=64\Rightarrow 2^m=2^6$

   ⇒m=6.$\Rightarrow m=6.$

   Also, 24=16⇒24=24$2^4=16\Rightarrow 2^4=2^4$

   ⇒n=4$\Rightarrow n=4$

2. (a) x2−2$x^2-2$

   Explanation:

   f(x+1x)=x2+1x2=(x+1x)2−2$\text{f}\left(\text{x}+\frac{1}{\text{x}}\right)={\text{x}}^2+\frac{1}{{\text{x}}^2}={\left(\text{x}+\frac{1}{\text{x}}\right)}^2-2$

   ∴f(x)=x2−2$\therefore \text{f}(\text{x})={\text{x}}^2-2$

3. (c) an odd function
   Explanation:

   f(−x)=log(−x+(−x)2+1−−−−−−−−√)=log(−x+x2+1−−−−−−√$\text{f}(-\text{x})=\log \left(-\text{x}+\sqrt{{(-\text{x})}^2+1}\right)=\log (-\text{x}+\sqrt{{\text{x}}^2+1}$

   =log(x2+1−−−−−−√−x)=log((x2+1√−x)(x2+1√+x)(x2+1√+x))$=\log (\sqrt{{\text{x}}^2+1}-\text{x})=\log \left(\frac{(\sqrt{{\text{x}}^2+1}-\text{x})(\sqrt{{\text{x}}^2+1}+\text{x})}{(\sqrt{{\text{x}}^2+1}+\text{x})}\right)$

   =log(1(x2+1√+x))=log(1)−log(x+x2+1−−−−−−√)$=\log \left(\frac{1}{(\sqrt{{\text{x}}^2+1}+\text{x})}\right)=\log \left(1\right)-\log \left(\text{x}+\sqrt{{\text{x}}^2+1}\right)$

   =0−log(x+x2+1−−−−−−√)$=0-\log \left(\text{x}+\sqrt{{\text{x}}^2+1}\right)$

   ⇒f(−x)=−f(x)$\Rightarrow \text{f}(-\text{x})=-\text{f}(\text{x})$

   ⇒$\Rightarrow$ f is an odd fucntion

4. (d) 29$2^9$

   Explanation:

   Here, A = {2,3,4}; B={2,3,5}

   n(A) = 3, n)B) = 3

   ∴$\therefore$no. of relations from A to B =2n(A)×n(B)=23×3=29$=2^{n(A)\times n(B)}=2^{3\times 3}=2^9$

5. (a) an equivalence relation

6. h2 - 2h + 2

7. 2mn

8. A ×$\times$ B = {(a,1),(a,5),(a,2),(b,2),(b,5),(b,1)$(a,1),(a,5),(a,2),(b,2),(b,5),(b,1)$}. Clearly,
   A is the set of first elements of all ordered pairs in A ×$\times$ B and B is set of second elements of all ordered pairs in A ×$\times$ B.
   ∴$\therefore$ A=$A=$ {a, b}, B = {1,5,2$1,5,2$}
   and B ×$\times$ A = { 1,5,2}×$\times${ a , b }
   = {(1,a),(1,b),(5,a),(5,b),(2,a),(2,b)$(1,a),(1,b),(5,a),(5,b),(2,a),(2,b)$}

9. Here f(x)=x2+3x+5x2+x−6=x2+3x+5(x+3)(x−2)$f(x)=\frac{x^2+3x+5}{x^2+x-6}=\frac{x^2+3x+5}{(x+3)(x-2)}$
   The function f(x) is defined for all values of x except
   x + 3 = 0, x  - 2 = 0 i.e. x = -3 and x = 2
   Thus domain of f(x) = R - {-3, 2}

10. Here f (x) = x + 1 and g (x) = 2x – 3
    Now (f + g) (x) =f (x) + g(x) = x + 1 + 2x - 3 = 3x - 2
    (f - g) (x) = f(x) - g(x) = x + 1- (2x - 3) = x + 1 - 2x + 3 = -x + 4
    (f)(g)(x)=f(x)g(x)=x+12x−3,x≠32$\frac{(f)}{(g)}(x)=\frac{f(x)}{g(x)}=\frac{x+1}{2x-3},x\ne \frac{3}{2}$

11. As given in the question,

    A = {1, 2, 3}, B = {4} and C = {5}

    ∴B∪C={4}∪{5}$\therefore B\cup C=\{4\}\cup \{5\}$ = {4, 5}

    ∴A×(B∪C)$\therefore A\times (B\cup C)$= {1, 2, 3} ×$\times$ {4, 5}
    ⇒A×(B∪C)$\Rightarrow A\times (B\cup C)$= {(1, 4), (1, 5), (2, 4) , (2, 5), (3, 4), (3, 5)}.......(a)

    Now,

    (A×B)$(A\times B)$ = {1, 2, 3} ×$\times$ {4} = {(1, 4), (2, 4), (3, 4)}

    and, (A×C)$(A\times C)$ = {1, 2, 3}×$\times$  {5} = {(1, 5), (2, 5), (3, 5)}

    ∴(A×B)∪(A×C)$\therefore (A\times B)\cup (A\times C)$= {(1, 4), (2, 4), (3, 4)} ∪$\cup$ {(1, 5), (2, 5), (3, 5)}
    ⇒(A×B)∪(A×C)$\Rightarrow (A\times B)\cup (A\times C)$= {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}......(b)

    From equations (a) and (b), we get
    A×(B∪C)=(A×B)∪(A×C)$A\times (B\cup C)=(A\times B)\cup (A\times C)$

    Hence verified.

12. We have,
    f : R →$\to$ R and g : C →$\to$ C

    where R is set of Real numbers and C is set of Complex numbers.

    From definitions as given,

    Domain of f = R and

    Domain of g = C

    Now, Two functions are said to be equal when domain and co-domain of both the functions are equal.

    As, Domain of f ≠ Domain of g,
    ∴$\therefore$ f(x) and g(x) are not equal functions.

13. We have,
    A = {2, 3}, B = {4, 5}, C = {5, 6}
    ∴B∪C={4,5}∪{5,6}$\therefore B\cup C=\{4,5\}\cup \{5,6\}$
    = {4, 5, 6}
    ∴A×(B∪C)={2,3}×{4,5,6}$\therefore A\times (B\cup C)=\{2,3\}\times \{4,5,6\}$
    = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
    Now,
    B∩C={4,5}∩{5,6}={5}$B\cap C=\{4,5\}\cap \{5,6\}=\{5\}$
    ∴A×(B∩C)={2,3}×{5}$\therefore A\times (B\cap C)=\{2,3\}\times \{5\}$
    ={(2, 5), (3, 5)}
    Now,
    A×B={2,3}×{4,5}$A\times B=\{2,3\}\times \{4,5\}$
    ={(2, 4), (2, 5), (3, 4),(3, 5)}
    and, A×C={2,3}×{5,6}$A\times C=\{2,3\}\times \{5,6\}$
    = {(2, 5), (2, 6), (3, 5), (3, 6)}
    ∴(A×B)∪(A×C)$\therefore (A\times B)\cup (A\times C)$= {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

14. Here (x3+1,y−23)=(53,13)$\left(\frac{x}{3}+1,y-\frac{2}{3}\right)=\left(\frac{5}{3},\frac{1}{3}\right)$
    ∴x3+1=53$\therefore \frac{x}{3}+1=\frac{5}{3}$and y−23=13$y-\frac{2}{3}=\frac{1}{3}$
    
    ⇒x3=53−1$\Rightarrow \frac{x}{3}=\frac{5}{3}-1$ and y=13+23$y=\frac{1}{3}+\frac{2}{3}$
    ⇒x3=23$\Rightarrow \frac{x}{3}=\frac{2}{3}$ and y=33$y=\frac{3}{3}$
    ⇒$\Rightarrow$ x = 2 and y = 1

15. 1. To determine A ×$\times$ (B ∪$\cup$ C)
       B ∪$\cup$ C = {b, c, e} ∪$\cup$ {b, c, f} = {b, c, e, f}
       ∴$\therefore$ A×$\times$(B ∪$\cup$ C) = {a, d} ×$\times$ {b, c, e, f}
       = {(a, b), (a, c), (a, e), (a, f), (d, b), (d, c), (d, e), (d, f)} ...(i)
       To determine (A ×$\times$ B) ∪$\cup$ (A ×$\times$ C)
       A ×$\times$ B = {a, d} ×$\times$ {b, c, e}
       = {(a, b), (a, c), (a, e), (d, b), (d, c), (d, e)}
       A ​​​​​​×$\times$ C = {a, d} ×$\times$ {b, c, f}
       = {(a, b), (a, c), (a, f), (d, b), (d, c), (d, f)}
       ∴$\therefore$ (A×$\times$B) ∪$\cup$ (A×$\times$C)
       = {(a, b), (a, c), (a, e), (a, f), (d, b), (d,c), (d,e),(d,f)} ...(ii)
       From Eqs. (i) and (ii), we get
       A×$\times$(B∪$\cup$C) = (A×$\times$B)∪$\cup$(A×$\times$C)
       Hence verified.
    2. To determine A ×$\times$ (B ∩$\cap$ C)
       (B ∩$\cap$ C) = {b, c, e} ∩$\cap$ {b, c, f} = {b, c}
       ∴$\therefore$ A ×$\times$ (B ∩$\cap$ C) = {a, d} ×$\times$ {b, c}
       = {(a, b), (a, c), (d, b), (d, c)} ...(iii)
       To determine (A×$\times$B)∩$\cap$(A×$\times$C)
       A ×$\times$ B = {(a, b), (a, c), (a, e), (d, b), (d, c), (d, e)}
       A ×$\times$ C = {(a, b), (a, c), (a, f), (d, b), (d, c), (d, f)}
       ∴$\therefore$ (A×$\times$B)∩$\cap$(A×$\times$C) = {(a, b), (a, c), (d, b), (d, c)} ...(iv)
       From Eqs. (iii) and (iv), we get
       A×$\times$(B∩$\cap$C) = (A×$\times$B)∩$\cap$(A×$\times$C)
       Hence verified.