These NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-1-miscellaneous-exercise/
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Miscellaneous Exercise
Question 1.
Let f : R → R be defined as f(x) = 10x + 7. Find the function g : R → R such that g o f = fog = 1R.
Solution:
Let y ∈R such that y = 10x + 7
⇒ 10x = y – 7
⇒ x = \(\frac { y -7 }{ 10 }\) ∈ R
Let g(x) = \(\frac { y -7 }{ 10 }\)
(fog)(x) = f(g(x)) = f(\(\frac { y -7 }{ 10 }\))
= 10(\(\frac { y -7 }{ 10 }\) ) + 7 = x
Similarly (gof)(x) = x
Thus (fog) = (gof) = IR
Question 2.
Let f : W → W be defined as f(n) = n – 1, if n is odd and f(n) = n + 1, if n is even. Show that/is invertible. Find the inverse off. Here, W is the set of all whole numbers.
Solution:
One-one
Let m and n be two distinct elements of W.
Case i :
When m and n are even
∴ f(m) = m + 1 and f(n) = n + 1
Since m ≠ n, then m + 1 ≠ n + 1
⇒ f(m) ≠ f(n)
Case ii :
When m and n are odd
f(m) = m – 1 and f(n) = n – 1
Since m ≠ n, m – 1 ≠ n – 1
⇒ f(m) ≠ f(n)
Case iii :
When m is odd and n is even
f(m) = m – 1 and f(n) = n + 1
Since m ≠ n, m – 1 ≠ n + 1
⇒ f(m) ≠ f(n)
Case iv :
When m is even and n is odd
f(m) = m – 1 and f(n) = n – 1
Since m ≠ n, m – 1 ≠ n – 1
⇒ f(m) ≠ f(n)
i.e., in any case if m ≠ n, then f(m) ≠ f(n)
Hence f is one-one.
Onto
Let y ∈ W, the co-domain of f
If y is odd, then y – 1 is even
f(y – 1) = y – 1 + 1 = y
If y is even, then y + 1 is odd
f(y + 1) = y + 1 – 1 = y
Hence every element in the co-domain has a preimage in the domain.
Hence f is onto.
i.e., f is one-one and onto, hence f is invertible.
We have proved that f(n – 1) = n, if n is odd and f(n + 1) = n, if n is even
∴ f-1 = \(\left\{\begin{array}{l}
n-1, \text { if } n \text { is odd } \\
n+1, \text { if } n \text { is even }
\end{array}\right.\)
i.e., f-1(n) = f(n)
Hence inverse of f is f itself.
Question 3.
If f: R → R is defined by f(x) = x² – 3x + 2, find f(f (x)).
Solution:
f(f(x)) = f(x² – 3x + 2)
= (x² – 3x + 2)² – 3(x² – 3x + 2) + 2
= x4 + 9x² + 4 – 6x³ – 12x + 4x² – 3x² + 9x – 6 + 2
= x4 – 6x³ + 10x² – 3x
Question 4.
Show that the function
f : R → {x ∈ R : – 1 < x < 1} defined by f(x) = \(\frac{x}{1+|x|}\), x ∈ R is one one and onto function.
Solution:
One-one
The domain of f is R
Let x1 x2 ∈ R
Case i :
x1 ≥ 0, x2 ≥ 0
Then 1 + x1 ≠ 1 + x2
Case ii :
since |x1| = – x1 and |x2| = – x2 as x1 ≤ 0 and x2 ≤ 0
⇒ f(x1) ≠ f(x2)
Case iii :
Thus in any case, f(x1) ≠ f(x2) when x1 ≠ x2
Hence f is one-one
Onto :
Let y ∈ R
i.e, when y ≥ 0, there is \(\frac{y}{1-y}\) ∈ R such that \(f\left(\frac{y}{1-y}\right)\)
case iii.
– 1 < y ≤ 0
i.e, when y < 0, there is \(\frac{y}{1+y}\) ∈ R such that \(f\left(\frac{y}{1-y}\right)\) = y
Question 5.
Show that the function f : R → R given by f(x) = x³ is injective.
Solution:
Let f(x1) = f(x2) for x1, x2 ∈ R
⇒ x³1 = x³2 ⇒ x³1 – x³2 = 0
⇒ (x1 – x2) (x²1 + x1x2 + x²2) = 0
⇒ x1 = x2
∴ f is one-one (injective)
Question 6.
Give examples of two functions f : N → Z and g : Z → Z such that gof is injective but g is not injective.
Solution:
Let f : N → Z defined by f(x) = x and
g : Z → Z defined by g(x) = |x|
We have 1 ≠ – 1, but |1| = |- 1|
∴ g(1) = g(- 1)
Hence g is not injective
gof: N → Z
Let x1, x2 ∈ N
such that (gof)(x1) = (gof)(x2)
⇒ g(f(x1)) = g(f(x2))
⇒ g(x1) = g(x2)
⇒ |x1| = |x2|
⇒ x1 = x2
since x1, x2 ∈ N
∴ gof is injective
Question 7.
Give examples of two functions f : N → N and g : N → N such that gof is onto but f is not onto.
Solution:
Let f, g : N → N defined by f(x) = x + 1
g(x) = \(\left\{\begin{array}{l}
x-1 \text { if } x>1 \\
1 \quad \text { if } x=1
\end{array}\right.\)
Let x ∈ N, x ≥ 1 ⇒ x + 1 ≥ 2
⇒ f(x) ≥ 2 for all x ∈ N
Range of f ≠ N
∴f is not onto
gof : N → N
(gof)(x) = g(f(x)) = g(x + 1)
= (x + 1) – 1 since x + 1 > 1 = x
i.e., (gof)x = x for all x ∈ N
i.e., gof is an identity function and hence onto.
Question 8.
Given a non empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows: For subsets A, B in P(X), ARB if and only if A ⊂ B. Is R an equivalence relation on P(X)? Justify your answer.
Solution:
ARB if A ⊂ B
i. ARA since A ⊂ A, for all A ∈ P(X) Hence R is reflexive.
ii. ARB need not imply BRA since A ⊂ B
⇒ B ⊄ A . Hence R is not symmetric.
Hence R is not an equivalence relation.
Question 9.
Given a non-empty set X, consider the binary operation * : P(X) x P(X) → P(X) given by A * B = A ∩B for all A, B in P(X), where P(X) is the power set of X. Show that X is the identity element for this operation and X is the only invertible element in P(X) with respect to the operation *.
Solution:
X ∈ P(X)
For any A e P(X), A * X = A ∩ X = A and
X * A= X ∩ A = A
∴ X is the identity element in P(X).
Let S be an invertible element in P(X).
Then S * X = X * S = X
S ∩ X = X ∩ S = X
S = X
∴ X is the only invertible element in P(X).
Question 10.
Find the number of all onto functions from the set {1, 2, 3,…, n} to itself.
Solution:
Since the functions are onto, every element of {1, 2,3 , n} has a pre-image. The element I can have pre-images in n ways. The element 2 can have pre-images in (n – 1) ways and so on. The last element n can have pre-iniage in only one way. Therefore number of ways we can have pre images is n(n – 1)(n – 2) …… 2.1 n! The total number of all onto functions from the set {1, 2, 3, …….. n} to itself is n!.
Question 11.
Let S = {a, b, c} and T = {1, 2, 3}. Find F-1 of the following functions F from S to T, if it exists.
i. F = {(a, 3), (b, 2), (c, 1)}
ii. F = {(a, 2), (b, 1), (c, 1)}
Solution:
i.
F-1 : T → S is defined as
F-1 = {(3, a), (2, b),(1, c)}
ii.
As 1 has 2 preimages, F is not bijective. F-1 is not defined.
Question 12.
Consider the binary operations *: R x R → R and ° : R x R → R defined as a * b = |a – b| and a ° b = a, for all a, b ∈ R. Show that * is commutative but not associative, ° is associative but not commutative. Further, I show that for all a, b, c ∈ R, a * (b ° c) = (a * b) o (a * b). [If it is so, we say that the j operation * distributes over the operation °]. Does o distribute over *? Justify your answer.
Solution:
Let a, b ∈ R, then
a * b = |a – b| = |b – a| = b * a
i.e., a * b = b * a for all a, b ∈ R
Hence * is commutative.
Let 1, 2, 3 ∈ R
(1 * 2) * 3 = |1 – 2| * 3
= 1 * 3
= |1 – 3| = 2
1 * (2 * 3) = 1 * |2 – 3| = 1 * 1
= |1 – 1| = 0
i.e., (1 * 2) * 3 ≠ 1 * (2 * 3)
∴ * is not associative
Thus * is commutative but not associative.
Let a, b ∈ R
a ° b – a and boa = b
∴ a ° b ≠ b ° a
Hence ° is not commutative.
Now (a ° b) ° c = a ° c = a
a ° (b ° c) = a ° b = a
i.e., a ° (b ° c) = (a ° b) ° c
i.e., o is associative
Hence ° is associative but not commutative.
Let a, b, c ∈ R
a * (b°c) = a * b =|a – b|
(a * b) o (a * c) = |a – b| ° |a – c| = |a – b|
i.e., a * (b°c) = (a * b) ° (a * c)
i.e., * is distributive over °
Now a° (b * c) = a ° |b-c| = a
(a°b) * (a°c) = a * a = |a – a| = 0
i.e., in general a° (b * c) * (a ° b) * (a ° c)
i.e., ° is not distributive over *.
Question 13.
Given a non-empty set X,
let *: P(X) x P(X) → P(X) be defined as A * B = (A – B) ∪ (B – A), for all A, B ∈ P(X). Show that the empty set (Φ) is the identity for the operation * and all the elements A of P(X) are invertible with A-1 = A.
Solution:
Hence (Φ) is the identity element in P(X)
Now A * A = (A – A) ∪ (A – A)
= Φ∪Φ = Φ
∴ A is the inverse of itself, i.e., A-1 = A. Hence all the elements of P(X) are invertible with A-1 = A.
Question 14.
Define a binary operation * on the set {0,1,2,3,4, 5} as a * b = \(\left\{\begin{array}{l}
a+b, \quad \text { if } a+b<6 \\
a+b-6, \text { if } a+b \geq 6
\end{array}\right.\) Show that zero is the identity for this operation and each element a of the set is invertible with 6 – a being the inverse of a.
Solution:
The operation table for * is given below
From the table we observe that a * 0 = 0 * a = a for all a ∈ {0,1,2,3.4,5}. Hence 0 is the identity element for the operation *.
Let a ≠ 0 be any element of {0, 1, 2, 3, 4, 5}.
Then 6 – a ∈ {0, 1, 2, 3, 4, 5}
Now a * (6 – a) = a + 6 – a – 6 = 0
and (6 – a) * a = 6 – a + a – 6 = 0
i.e., a * (6 – a) = (6 – a) * a = 0, identity element of *
∴ 6 – a is the inverse of a ≠ 0.
Hence a ≠ 0 is invertible.
Also the inverse of 0 is 0.
Question 15.
Let A = {- 1, 0, 1, 2}, B = {- 4, – 2, 0, 2} and f, g : A → B be functions defined by f(x) = x² – x, x ∈ A and g(x) = 2|x – \(\frac { 1 }{ 2 }\)| – 1, x ∈ A. Are f and g equal? Justify your answer.
Solution:
Hence fa) = g(a) for all a ∈ A.
∴ f and g are equal functions.
Question 16.
Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is
a. 1
b. 2
c. 3
d. 4
Solution:
a. 1
Let R be the relation containing (1, 2) and 0,3)
∴ (1, 2), (1, 3) ∈ R
Since R is reflexive, R contains (1, 1), (2, 2), (3, 3)
Since R is symmetric, R contains (2, 1) and (3, 1)
∴ R = {(1, 1), (1, 2), (1, 3), (2, 2), (2, 1), (3, 1), (3, 3)}
Since R is not transitive (2, 1) and (1, 3) ∈ R but (2, 3) ∉ R.
If (3, 2) ∈ R, then (2, 3) ∈ R since R is symmetric.
∴(3, 2) ∉ R
∴ There is only one relation which is reflexive, symmetric and not transitive.
Question 17.
Let A = {1, 2, 3}. Then number of equivalence relations containing (1,2) is
a. 1
b. 2
c. 3
d. 4
Solution:
b. 2
Given (1, 2) i.e., 1 is related to 2.
Then there are two possibilities.
Case i :
1 is related to 3
A x A= {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3,1), (3, 2), (3, 3)}
Then A x A is reflexive, symmetric and transitive.
∴ A x A is an equivalence relation Containing (1, 2)
Case ii :
1 is not related to 3
Let R= {(1, 1), (1, 2), (2, 1), (2, 2), (3, 3)}
Then R is reflexive, symmetric and transitive, i.e., R is an equivalence relation containing (1 2).
Thus there are two equivalence relations containing (1, 2).
Question 18.
Let f : R → R be the Signum Function defined as
f(x) = \(\left\{\begin{array}{c}
1, x>0 \\
0, x=0 \\
-1, x<0
\end{array}\right.\)
and g : R → R be the Greatest Integer Function given by g (x) = [x], where [x] is greatest integer less than or equal fox. Then, does fog and gof coincide in (0, 1]?
Solution:
(fog)(x) = f(g(x)) = f([x]) x ∈ (0, 1]
= f(0) = 0
(gof)(x) = g(Ax)) = g(1) x ∈ (0,1]
= 1
∴ fog and gof do not coincide in (0, 1]
Question 19.
Number of binary operations on the set {a, b} are
a. 10
b. 16
c. 20
d. 8
Solution:
b. 16
{a, b) x {a, b) contains 4 elements.
Number of relations from {a, b} x {a, b) to {a, b) is 24 = 16