Relations and Functions Extra Questions for Class 12 Mathematics

Chapter 1 Relations and Functions

1. In Z , the set of integers, inverse of – 7 , w.r.t. ‘ * ‘ defined by a * b = a + b + 7 for all {tex}a,b \in Z{/tex} ,is
   1. -7
   2. -14
   3. 14
   4. 7
2. If A = { 1, 2, 3}, then the relation R = {(1, 2), (2, 3), (1, 3)} in A is ___.
   1. transitive only
   2. reflexive only
   3. symmetric only
   4. symmetric and transitive only
3. A relation R on a set A is called an empty relation if
   1. no element of A is related to any element of A
   2. every element of A is related to one element of A
   3. one element of A is related to all the elements of A
   4. every element of A is related to any element of A
4. Let f and g be two functions from R to R defined as {tex}f(x) = \left\{ {\begin{array}{*{20}{c}} {0,x\;is\;rational} \\ {1,x\;is\;irrational} \end{array}} \right\}{/tex}, {tex}g(x) = \left\{ {\begin{array}{*{20}{c}} { – 1,x\;is\;rational} \\ {0,x\;is\;irrational} \end{array}} \right\}{/tex} then, (gof)(e) + (fog) {tex}(\pi ){/tex} =.
   1. 1
   2. 2
   3. -1
   4. 0
5. Let R be the relation on N defined as xRy if x + 2 y = 8. The domain of R is
   1. {2, 4, 6, 8}
   2. {2, 4, 8}
   3. {1, 2, 3, 4}
   4. {2, 4, 6}
6. If n(A) = p and n(B) = q, then the number of relations from set A to set B = ________.
7. A function is called an onto function, if its range is equal to ________.
8. A binary operation * on a set X is said to be ________, if a * b = b * a, where a, b {tex}\in{/tex} X.

9. Find gof f(x) = |x|, g(x) = |5x + 1|.

10. Show that function f: N {tex}\rightarrow{/tex} N, given by f(x) = 2x, is one – one.

11. Let S = {1, 2, 3} Determine whether the function f: S {tex}\rightarrow{/tex} S defined as below have inverse.
    f = {(1, 1), (2, 2), (3, 3)}

12. Let f : {1, 3, 4} {tex}\rightarrow{/tex} {1, 2, 5} and g : {1, 2, 5} {tex}\rightarrow{/tex} {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof.

13. Consider f: {1, 2, 3} {tex}\rightarrow{/tex} {a, b, c} given by f(1) = a, f(2) = b and f(3) = c find f^-1 and show that (f^-1)^-1 = f.

14. Is g = {(1, 1), (2, 3), (3, 5), (4, 7)} a function? If g is described by g(x) = ax + b, then what value should be assigned to a and b.

15. Show that the relation R defined by (a, b) R (c, d) {tex}\Rightarrow{/tex} a + d = b + c on the set N{tex}\times{/tex}N is an equivalence relation.

16. Let the function f : R {tex}\rightarrow{/tex} R be defined by f(x) = cosx, {tex}\forall{/tex}x {tex}\in{/tex} R. Show that f is neither one-one nor onto.

17. Let L be the set of all lines in plane and R be the relation in L define if R = {(l₁, L₂ ): L₁ is{tex} \bot {/tex} to L₂ } .Show that R is symmetric but neither reflexive nor transitive.

18. If the function f : R {tex}\to{/tex} R is given by f (x) = x² + 2 and g: R {tex}\to{/tex} R is given by {tex}g ( x ) = \frac { x } { x – 1 } ; x \neq 1{/tex} then find fog and gof and hence find fog(2) and gof(- 3).

Chapter 1 Relations and Functions

Solution

1. 1. -7
      Explanation: If ‘ e ‘ is the identity ,then a*e = a {tex} \Rightarrow {/tex}a + e + 7 = a {tex} \Rightarrow {/tex} e = – 7 . Also,inverse of e is e itself. Hence , inverse of -7 is -7.
   1. transitive only
      Explanation: A relation R on a non-empty set A is said to be transitive if xRy and y Rz {tex} \Rightarrow {/tex} xRz, for all x {tex} \in {/tex} R. Here, (1, 2) and (2, 3) belongs to R implies that (1, 3) belongs to R.
   1. no element of A is related to any element of A
      Explanation: For any set A ,an empty relation may be defined on A as: there is no element exists in the relation set which satisfies the relation for a given set A i.e.
      let A={1,2,3,4,5} and R={(a,b): a,b {tex} \in {/tex} A and a+b= 10},so we get R={ } which is an empty relation.
   3. -1
      Explanation: (gof)(e) + (fog)({tex}\pi {/tex}) =g(f(e)) + f(g({tex}\pi {/tex}) = g(1) + f(0) = – 1 + 0 = -1.
   4. {2, 4, 6}
      Explanation: As xRy if x + 2 y = 8 , therefore, domain of the relation R is given by x = 8 – 2y ∈ N. When y = 1, {tex} \Rightarrow {/tex} x = 6, when y = 2, {tex} \Rightarrow {/tex} x =4 , when y =3 , {tex} \Rightarrow {/tex} x = 2. Therefore, domain is { 2, 4, 6 }.
2. 2^pq
3. codomain
4. commutative
5. gof (x) = g [f(x)] = g [|x|] =| 5 |x| + 1 |
6. For, f(x₁) = f(x₂)
   2x₁ = 2x₂
   x₁ = x₂
   So, The function f is one – one
7. Since different elements have different images. So, f is one – one. Also, every element of codomain has pre-image so, f is onto
   Now f is one – one and onto, so that f is invertible with inverse f^-1 = {(1, 1) (2, 2) (3, 3)}
8. f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}
   Now, f(1) = 2, f(3) = 5, f(4) = 1 and g(1) = 3, g(2) = 3, g(5) =1
   gof(x)=g(f(x))
   gof(1)=g(f(1))=g(2)=3
   g[f(3)] = g(5) = 1 and g[f(4)] = g(1) = 3
   Hence, gof = {(1, 3), (3, 1), (4, 3)}
9. f = {(1, a) (2, b) (3, c)}
   f^-1 = { (a, 1) (b, 2) (c, 3)}
   (f ^-1) ^-1 = {(1, a) (2, b) (3, c)}
   Hence (f^-1)^-1 = f.
10. Yes, g is a function since every element in domain has a unique image in range.
    Now, Let g(x) = ax + b Then Given,
    g(1) = a + b = 1 &
    g(2) = 2a +b = 3
    Subtracting g(1) from g(2) Gives
    (2a + b) – (a + b) = a = 2 & Substituting It into g(1)
    We have b = -1
11. we have,a + b = b + a for all (a, b) {tex}\in N×N{/tex}, which implies (a,b) R (a,b).Thus, R is reflexive.
    Let (a,b),(c,d) {tex}\in N×N{/tex} be such that
    (a, b) R (c, d)
    {tex}\Rightarrow{/tex} a + d= b + c
    {tex}\Rightarrow{/tex} d + a = c + b
    {tex}\Rightarrow{/tex} c + b = d + a
    {tex}\Rightarrow{/tex} (c, d) R (a, b) for all (a, b), (c, d) {tex}\in N×N{/tex}
    Hence R is symmetric.
    Let (a,b),(c,d),(e,f) {tex}\in N×N{/tex} such that (a,b) R (c,d) and
    (c,d) R (e,f).Then,
    (a, b) R (c, d) {tex}\Rightarrow{/tex} a + d = b + c……. (1)
    (c, d) R (e, f) {tex}\Rightarrow{/tex} c + f = d + e ………(2)
    Adding (1) and (2)
    (a + d) + (c+f) = (b + c) + (d + e)
    a + f = b+ e
    (a, b) R (e, f)
    Hence, R is transitive
    So, R is an equivalence relation.
12. Given function, f(x) = cosx, {tex}\forall{/tex}x {tex}\in{/tex} R
    Now, {tex}f\left( {\frac{\pi }{2}} \right) = \cos \frac{\pi }{2} = 0{/tex}
    {tex} \Rightarrow f\left( {\frac{{ – \pi }}{2}} \right) = \cos \frac{\pi }{2} = 0{/tex}
    {tex}\Rightarrow f\left( {\frac{\pi }{2}} \right) = f\left( {\frac{{ – \pi }}{2}} \right){/tex}
    But {tex}\frac{\pi }{2} \ne \frac{{ – \pi }}{2} = 0{/tex}
    So, f (x) is not one-one
    Now, f(x) = cosx, {tex}\forall{/tex}x {tex}\in{/tex} R is not onto as there is no pre-image for any real number. Which does not belong to the intervals [-1, 1], the range of cos x.
13. R is not reflexive, as a line L₁ cannot be {tex} \bot {/tex} to itself i.e (L₁, L₁ ){tex}\notin{/tex} R
    
    Now ( L_1, L₂){tex}\in{/tex}R
    {tex}\Rightarrow{/tex}L₁ {tex}\bot {/tex} L₂
    {tex}\Rightarrow{/tex}L₂ {tex}\bot {/tex} L₁
    {tex}\Rightarrow{/tex}(L₂, L₁){tex} \in {/tex}R
    R is symmetric
    Now (L₁, L₂) and (L_2, L₃) {tex}\in{/tex} R
    i.e L₁ {tex}\bot{/tex} L₂ and L₂ {tex}\bot{/tex} L₃
    Then L₁ can never be {tex}\bot {/tex} to L₃ in fact L₁ || L₃
    i.e (L₁, L₂) {tex}\in{/tex} R, (L₂, L₃) {tex} \in{/tex} R.
    But (L₁, L₃) {tex}\notin{/tex} R
    R is not transitive.
14. We are given that, f : R {tex}\to{/tex} R and g : R {tex}\to{/tex} R defined as f(x) = x² + 2 and g(x) = {tex}\frac { x } { x – 1 } ; x \neq 1{/tex}.
    First we see whether fog and gof exist for the given functions.
    Since, range f {tex} \subseteq {/tex} domain g and range g {tex} \subseteq {/tex} domain f
    Hence,fog and gof exist for the given functions.
    Now, for any x {tex}\in{/tex} R – {1}, we have (fog)(x) = f[g(x)] = {tex}f \left[ \frac { x } { x – 1 } \right] = \left( \frac { x } { x – 1 } \right) ^ { 2 } + 2{/tex}
    {tex}= \frac { x ^ { 2 } + 2 ( x – 1 ) ^ { 2 } } { ( x – 1 ) ^ { 2 } }{/tex}= {tex}\frac { x ^ { 2 } + 2 \left( x ^ { 2 } + 1 – 2 x \right) } { ( x – 1 ) ^ { 2 } }{/tex}
    ={tex}\frac { 3 x^2 + 2 – 4 x } { ( x – 1 ) ^ 2 }{/tex}
    {tex}\therefore{/tex} fog: R {tex}\to{/tex} R is defined by
    (fog)(x)= {tex}\frac { 3 x ^ { 2 } – 4 x + 2 } { ( x – 1 ) ^ { 2 } } , x \neq 1{/tex} ……(i)
    For any x {tex}\in{/tex} R we have
    gof(x) = g[f(x)] = g(x² + 2) = {tex}\frac { x ^ { 2 } + 2 } { \left( x ^ { 2 } + 2 \right) – 1 } = \frac { x ^ { 2 } + 2 } { x ^ { 2 } + 1 }{/tex}
    {tex}\therefore{/tex} gof: R {tex}\to{/tex} R is defined by
    (gof)(x) = {tex}\frac { x ^ { 2 } + 2 } { x ^ { 2 } + 1 }{/tex} ……..(ii)
    On putting x = 2 in Eq. (i), we get
    fog(2) = {tex}\frac { 3 \times ( 2 ) ^ { 2 } – 4 ( 2 ) + 2 } { ( 2 – 1 ) ^ { 2 } }{/tex}= {tex}\frac { 3 \times 4 – 8 + 2 } { ( 1 ) ^ { 2 } }{/tex}
    = 12 – 8 + 2 = 6
    On putting x = – 3 in Eq. (ii), we get
    gof(-3) = {tex}\frac { ( – 3 ) ^ { 2 } + 2 } { ( – 3 ) ^ { 2 } + 1 } = \frac { 9+2 } { 9+1 }=\frac{11}{10}{/tex}