The document NCERT Exemplar - Relations and Functions (Part - 1) Notes | EduRev is a part of the JEE Course Mathematics (Maths) Class 12.

All you need of JEE at this link: JEE

**SHORT ANSWER TYPE QUESTIONS**

**Q.1. ****Let A = {a, b, c} and the relation R be defined on A as follows: R = {(a, a), (b, c), (a, b)}.**

Here, R = {(a, a), (b, c), (a, b)}

for reflexivity; (b, b), (c, c) and for transitivity; (a, c)

Hence, the required ordered pairs are (b, b), (c, c) and (a, c)

For real value of f(x), 25 â€“ x

â‡’ â€“ x

Hence, D âˆˆ - 5 â‰¤ x â‰¤ 5 or [- 5, 5]

Here, f(x) = 2x + 1 and g(x) = x

âˆ´ g of = g[f(x)]

= [2x + 1]

Hence, g of = 4x

Here, f(x) = 2x â€“ 3

Let f(x) = y = 2x â€“ 3

â‡’ y + 3 = 2x

Let y = f(x) âˆ´ x = f

âˆ´ If f = {(a, b), (b, d), (c, a), (d, c)}

then f

Here, f(x) = x

âˆ´ f [f(x)] = [f(x)]

= (x

= x

= x

Hence, f [f(x)] = x

Yes, g = {(1, 1), (2, 3), (3, 5), (4, 7)} is a function.

Here, g(x) = Î±x + Î²

For (1, 1), g(1) = Î±.1 + Î²

1 = Î± + Î² ... (1)

For (2, 3), g(2) = Î±.2 + Î²

3 = 2Î± + Î² ... (2)

Solving eqs. (1) and (2) we get, Î± = 2, Î² = â€“ 1

fog = f [g(x)]

= f [g(2)] = f(3) = 5

= f [g(5)] = f(1) = 2

= f [g(1)] = f(3) = 5

Hence, fog = {(2, 5), (5, 2), (1, 5)}

Here, f(z) = |z| âˆ€ z âˆˆ C

f(1) = |1| = 1

f(- 1) = |- 1| = 1

f(1) = f(- 1)

But 1 â‰ - 1

Therefore, it is not one-one.

Now, let f(z) = y = |z|. Here, there is no pre-image of negative numbers. Hence, it is not onto.

Here, f(x) = cos x âˆ€ x âˆˆ R

Therefore, the given function is not one-one. Also, it is not onto function as no pre-image of any real number belongs to the range of cos x i.e., [â€“1, 1].

Here, given that X = {1, 2, 3}, Y = {4, 5}

âˆ´ X Ã— Y = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}

f is not a function because there is no unique image of each element of domain under f.

Yes, g is a function because each element of its domain has a unique image.

Yes, it is a function because each element of its domain has a unique image.

Clearly k is also a function.

Let x

gof {f(x

â‡’ g(x

âˆ´ x

Hence, f is one-one. But g is not onto as there is no pre-image of A in B under g.

Range of cos x is [â€“ 1, 1]

â‡’ 2y â€“ y cos x = 1 â‡’ y cos x = 2y â€“ 1

Now â€“ 1 â‰¤ cos x â‰¤ 1

Here, âˆ€ a, b âˆˆ Z and a R b if and only if a - b is divisible by n.

The given relation is an equivalence relation if it is reflexive, symmetric and transitive.

a R a â‡’ (a - a) = 0 divisible by n

So, R is reflexive.

a R b = b R a âˆ€ a, b âˆˆ Z

a - b is divisible by n (Given)

â‡’ - (b - a) is divisible by n

â‡’ b â€“ a is divisible by n

â‡’ b R a

Hence, R is symmetric.

a R b and b R c â‡” a R c âˆ€ a, b, c âˆˆ Z

a - b is divisible by n

b - c is also divisible by n

â‡’ (a - b) + (b - c) is divisible by n

â‡’ (a - c) is divisible by n

Hence, R is transitive.

So, R is an equivalence relation.

**LONG ANSWER TYPE QUESTIONS**

**Q.16. If A = {1, 2, 3, 4}, define relations on A which have properties of being: ****(a) Reflexive, transitive but not symmetric ****(b) Symmetric but neither reflexive nor transitive ****(c) Reflexive, symmetric and transitive.****Ans.**

Given that A = {1, 2, 3, 4}

âˆ´ ARA = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4), (2, 1), (3, 1), (4, 1), (3, 2), (4, 2), (4, 3)}**(a)** Let R_{1} = {(1, 1), (2, 2), (1, 2), (2, 3), (1, 3)}

So, R_{1} is reflexive and transitive but not symmetric.**(b)** Let R_{2} = {(2, 3), (3, 2)}

So, R_{2} is only symmetric.**(c)** Let R_{3} = {(1, 1), (1, 2), (2, 1), (2, 4), (1, 4)}

So, R_{3} is reflexive, symmetric and transitive.**Q.17. Let R be relation defined on the set of natural number N as follows: R = {(x, y) : x âˆˆN, y âˆˆ N, 2x + y = 41}. Find the domain and range of the relation R. Also verify whether R is reflexive, symmetric and transitive.****Ans.**

Given that x âˆˆ N, y âˆˆ N and 2x + y = 41

âˆ´ Domain of R = {1, 2, 3, 4, 5, ..., 20}

and Range = {39, 37, 35, 33, 31, ..., 1}

Here, (3, 3) âˆˆ R

as 2 Ã— 3 + 3 â‰ 41

So, R is not reflexive.

R is not symmetric as (2, 37) âˆˆ R but (37, 2) âˆ‰R

R is not transitive as (11, 19) âˆˆ R and (19, 3) âˆˆ R but (11, 3) âˆ‰ R.

Hence, R is neither reflexive, nor symmetric and nor transitive.**Q.18. Given A = {2, 3, 4}, B = {2, 5, 6, 7}. Construct an example of each of the following: ****(a) An injective mapping from A to B ****(b) A mapping from A to B which is not injective ****(c) A mapping from B to A.****Ans.**

Here, A = {2, 3, 4} and B = {2, 5, 6, 7}**(a)** Let f: A â†’ B be the mapping from A to B

f = {(x, y): y = x + 3}

âˆ´ f = {(2, 5), (3, 6), (4, 7)} which is an injective mapping.**(b)** Let g: A â†’ B be the mapping from A â†’ B such that

g = {(2, 5), (3, 5), (4, 2)} which is not an injective mapping.**(c)** Let h: B â†’ A be the mapping from B to A

h = {(y, x): x = y - 2}

h = {(5, 3), (6, 4), (7, 3)} which is the mapping from B to A.**Q.19. Give an example of a map ****(i) Which is one-one but not onto ****(ii) Which is not one-one but onto ****(iii) Which is neither one-one nor onto.****Ans.****(i)** Let f: N â†’ N given by f(x) = x_{2}

Let x_{1}, x_{2} âˆˆ N then f(x_{1}) = x_{1}^{2} and f(x_{2}) = x_{2}^{2}

Now, f (x_{1}) = f(x_{2}) â‡’ x_{1}^{2} =x_{2}^{2} â‡’ x_{1}^{2} -x_{2}^{2}= 0 â‡’ (x_{1} + x_{2}) (x_{1} - x_{2}) = 0

Since x_{1}, x_{2} âˆˆ N, so x_{1} + x_{2} = 0 is not possible.

âˆ´ x_{1} â€“ x_{2} = 0 â‡’ x_{1} = x_{2}

âˆ´ f(x_{1}) = f(x_{2}) â‡’ x_{1} = x_{2}

So, f (x) is one to one function.

Now, Let f(x) = 5 âˆˆ N

then x^{2} = 5 â‡’ x = Â± âˆš5 âˆ‰ N

So, f is not onto.

Hence, f (x) = x^{2} is one-one but not onto.

Since f (1) = f(2) but 1 â‰ 2,

So, f is not one-one.

Now, let y âˆˆ N be any element.

Then f (n) = y

â‡’ n = 2y â€“ 1 if y is even

n = 2y if y is odd or even

â‡’

âˆ´ Every y âˆˆ N has pre-image

âˆ´ f is onto.

Hence, f is not one-one but onto.**(iii)** Let f: R â†’ R be defined as f(x) = x_{2}

Let x_{1} = 2 and x_{2} = - 2

f(x_{1}) = x_{1}^{2} = (2)^{2}= 4

f(x_{2}) = x_{2}^{2} = (- 2)^{2} = 4

f(2) = f(- 2) but 2 â‰ - 2

So, it is not one-one function.

Let f (x) = â€“ 2 â‡’ x^{2} = â€“ 2 âˆ´ x = Â± âˆš-2 âˆ‰ R

Which is not possible, so f is not onto.

Hence, f is neither one-one nor onto.**Q.20. Let A = R â€“ {3}, B = R â€“ {1}. Let f: A â†’ B be defined by f (x) ****= (x-2)/(x-3) ****âˆ€ x âˆˆ A . Then show that f is bijective.****Ans.**

Here, A âˆˆ R - {3}, B = R - {1}

Let x_{1}, x_{2} âˆˆ f(x)

âˆ´ f(x_{1}) = f(x_{2})

So, it is injective function.

â‡’ xy â€“ 3y = x â€“ 2

â‡’ xy â€“ x = 3y â€“ 2

â‡’ x(y â€“ 1) = 3y â€“ 2

â‡’ f(x) = y âˆˆ B.

So, f (x) is surjective function.

Hence, f (x) is a bijective function.**Q.21. ****Let A = [â€“1, 1]. Then, discuss whether the following functions defined on A are one-one, onto or bijective:****(i) f(x) = 2/x (ii) g(x) = |x| (iii) h(x) = x |x| (iv) k(x) = x**

Let x

So, f(x) is one-one function.

Let f(x) = y = x/2 â‡’ x = 2y

For y = 1, x = 2 âˆ‰ [-1, 1]

So, f (x) is not onto. Hence, f(x) is not bijective function.

g(x

So, g (x) is not one-one function.

Let g(x) =y= |x| â‡’ x =

So, g (x) is not onto function.

Hence, g (x) is not bijective function.

h(x

â‡’ x

So, h (x) is one-one function.

Now, let h(x) = y = x|x| = x

âˆ´ h(x) is not onto function.

Hence, h (x) is not bijective function.

â‡’ x

So, k (x) is not one-one function.

Now, let k(x) = y = x

If y = â€“ 1 â‡’ x = Â± âˆš- 1 âˆ‰ A âˆ€ y âˆˆ A

âˆ´ k(x) is not onto function.

Hence, k (x) is not a bijective function.

(iv) x + 4y = 10 x, y âˆˆ N.

For reflexivity x > x âˆ€ x âˆˆ N which is not true

So, it is not reflexive relation.

So, it is not symmetric relation.

For transitivity, x R y, y R z â‡’ x R z âˆ€x, y, z âˆˆ N â‡’ x > y, y > z â‡’ x > z

So, it is transitive relation.

R = {(1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)}

For reflexive: 5 + 5 = 10, 5 R 5 â‡’ (x, x) âˆˆ R

So, R is reflexive.

For symmetric: (1, 9) âˆˆ R and (9, 1) âˆˆ R

So, R is symmetric.

For transitive: (3, 7) âˆˆ R, (7, 3) âˆˆ R but (3, 3) âˆ‰ R

So, R is not transitive.

For reflexive: x R x = x . x = x

[âˆ´ Square of an integer is also an integer]

So, R is reflexive.

For symmetric: x R y = y R x âˆ€ x, y âˆˆ N

âˆ´ xy = yx (integer)

So, it is symmetric.

For transitive: x R y and y R z â‡’ x R z

Let xy = k

which is again a square of an integer.

So, R is transitive.

R = {(2, 2), (6, 1)}

For reflexivity: (2, 2) âˆˆ R

So, R is reflexive.

For symmetric: (x, y) âˆˆ R but (y, x) âˆ‰ R (6, 1) âˆˆ R but (1, 6) âˆ‰ R

So, R is not symmetric.

For transitive: (x, y) âˆˆ R but (y, z) âˆ‰ R and (x, z) âˆˆ R

So, R is not transitive.

Here, A = {1, 2, 3, ..., 9}

and R â†’ A Ã— A defined by (a, b) R (c, d) â‡’ a + d = b + c

âˆ€ (a, b), (c, d) âˆˆ A Ã— A

For reflexive: (a, b) R (a, b) = a + b = b + a âˆ€ a, b âˆˆ A which is true. So, R is reflexive.

For symmetric: (a, b) R (c, d) = (c, d) R (a, b)

L.H.S. a + d = b + c

R.H.S. c + b = d + a

L.H.S. = R.H.S. So, R is symmetric.

For transitive: (a, b) R (c, d) and (c, d) R (e, f) â‡” (a, b) R (e, f)

â‡’ a + d = b + c and c + f = d + e

â‡’ a + d = b + c and d + e = c + f

â‡’ (a + d) - (d + e) = (b + c) - (c + f)

â‡’ a - e = b - f

â‡’ a + f = b + e

â‡’ (a, b) R (e, f)

So, R is transitive.

Hence, R is an equivalence relation.

Equivalent class of {(2, 5)} is {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}

A function f : X â†’ Y is said to be invertible if there exists a

function g : Y â†’ X such that gof = I

f

A function f : X â†’ Y is said to be invertible if f is a bijective function.

= (2x â€“ 3)

= (x

= x

= x

So, * is binary operation.

a * b = a â€“ b and b * a = b â€“ a âˆ€ a, b âˆˆ Q

a - b â‰ b - a

So, * is not commutative.

so * is a binary operation. a * b = b * a

â‡’ a

So, * is commutative.

a * b = a + ab and b * a = b + ba

a + ab â‰ b + ba â‡’ a * b â‰ b * a âˆ€ a, b âˆˆ Q.

So, * is not commutative.

a * b = (a - b)

â‡’ (a - b)

So, * is commutative.

Q.27. Let * be binary operation defined on R by a * b = 1 + ab, âˆ€ a, b âˆˆ R. Then the operation * is **(i) Commutative but not associative ****(ii) Associative but not commutative ****(iii) Neither commutative nor associative ****(iv) Both commutative and associative****Ans.****(i)**: Given that

a * b = 1 + ab âˆ€ a, b âˆˆ R

and b * a = 1 + ba âˆ€ a, b âˆˆ R

a * b = b * a = 1 + ab

So, * is commutative.

Now a * (b * c) = (a * b) * c âˆ€ a, b, c âˆˆ R

L.H.S. a * (b * c) = a * (1 + bc) = 1 + a(1 + bc) = 1 + a + abc

R.H.S. (a * b) * c = (1 + ab) * c = 1 + (1 + ab) . c = 1 + c + abc

L.H.S. â‰ R.H.S.

So, * is not associative.

Hence, * is commutative but not associative.

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!