OP Malhotra Class 12 Maths Solutions Chapter 2 Functions Ex 2(a)

Accessing OP Malhotra Class 12 Solutions Chapter 2 Functions Ex 2(a) can be a valuable tool for students seeking extra practice.

S Chand Class 12 ICSE Maths Solutions Chapter 2 Functions Ex 2(a)

Question 1.
If A = {1, 2, 3, 4} and B = {1, 2, 3, 4, 5, 6} are two sets and function f : A → B is defined by f (x) = x + 2, ∀x ∈ A, then the function f is
(a) bijective
(b) onto
(c) one-one
(d) many-one
Solution:
Given A = {1, 2, 3, 4} and B = {1, 2, 3, 4, 5, 6} and f : A → B is defined by
f (x) = x + 2 ∀x ∈ A
Here f(1) = 1 + 2 = 3;
f(2) = 2 + 2 = 4;
f(3) = 3 + 2 = 5;
f(4) = 4 + 2 = 6;
So different elements in A have different images in B
∴ f is one – one
Clearly 2 ∈ B and let x ∈ A s.t f(x) = 2
⇒ x + 2 = 2
⇒ x = 0 ∉ A
So element 2 in B has no pre-image 0 in A.
Thus, f is into (i.e., not onto)
Hence f is 1 – 1, into

Question 2.
Show that a function f : R → R given by f (x) = ax + b, a, b ∈ R, a ≠ 0 is a bijective.
Solution:
Given a function f : R → R defined by
f(x) = ax + b, a, b ∈ R, a ≠ 0
∀x, y ∈ R s.t f(x) = f(y)
⇒ ax + b = ay + b
⇒ ax = ay
⇒ x = y (∵ a ≠ 0)
Thus, f is one – one
Let yeR be any arbitrary element
and let y = ax + b is x = \(\frac { y – b }{ a }\) ∈ R, as
y ∈ R a, b ∈ R
∀ y∈R∃ x = \(\frac { y – b }{ a }\) ∈ R,
s.t. f(x) = f\(\left(\frac{y-b}{a}\right)=a\left(\frac{y-b}{a}\right)\) + b = y
Hence f is onto.
Thus f is one-one and onto and hence
f is bijective.

Question 3.
Let f : N → N given by f (x) = 2x ∀x ∈ N. Show that/is one-one and into.
Solution:
Given f : N → N defined by
f(x) = 2x ∀x ∈ N
∀x, y ∈ N s.t. f(x) = f(y)
⇒ 2x = 2y ⇒ x = y
Thus, f is one – one
Since 3 ∈ N and let x ∈ N (domain of f)
s.t. f(x) = 3
2x = 3 ⇒ x = \(\frac {3}{2}\) ∉ N
Hence 3 (codomain of f) has no pre-image in N (domain of f)
Thus, f is into.
Hence f is one-one, into.

Question 4.
Let f: R → R be defined a
f(x) = 3x. Choose the correct answer.
(a) f is one-one onto
(b) f is many-one onto
(c) f is one-one but not onto
(d) f is neither one-one nor onto
Solution:
Given f : R → R defined by
f(x) = 3x ∀x ∈ R
one – one : ∀x, y ∈ R s.t f(x) = f(y)
⇒ 3x = 3y ⇒ x = y
∴ f is one – one.
onto : Let y ∈ R be any arbitrary element
and let y = 3x ⇒ x = \(\frac {y}{3}\)
Since y ∈ R ⇒ \(\frac {y}{3}\) ∈ R ⇒ x ∈ R
∀ y ∈ R∃ y ∈ R
s.t f(x) = f(\(\frac {y}{3}\)) = 3 x \(\frac {y}{3}\) = y
∴ f is onto.
Thus, f is one – one and onto.

Question 5.
Show that f : R → R, defined by f(x) = x³ is a bijection.
Solution:
Given R → R, defined by
f(x) = x³ ∀x ∈ R
one – one : ∀x, y ∈ R s.t. f(x) = f(y)
⇒ x³ = y³
⇒ (x – y) (x² + xy + y²) = 0
⇒ x – y = 0
[Since x² + xy + y² ≠ 0 ∀x, y ∈ R]
⇒ x = y
∴ f is one – one.
onto : Let y∈R be any arbitrary element
and let y = x³ = f(x) ⇒ x = 3\(\sqrt{y}\)
as y ∈ R ⇒ 3\(\sqrt{y}\) ∈ R ⇒ x ∈ R
Thus, ∀ y ∈ R ∃ x ∈ R s.t. y = f(x)
∴ f is onto
Thus f is one – one and onto
∴ f is bijective.

Question 6.
Check the injectivity and surjectivity of the following functions:
(a) f : N → N given by f(x) = x²
(b) f : Z → Z given by f(x) = x²
(c) f : R → R given by f (x) = x²
(d) f : N → N given by f(x) = x³
(e) f : Z → Z given by f(x) = x³
Solution:
(a) Given f : N →N given by f(x) = x²
∀ y ∈ N
Injectivity : ∀x, y ∈ N s.t f(x) = f(y)
⇒ x² = y²
⇒ (x – y) (x + y) = 0
⇒ x – y = 0
[∵ Since x + y ≠ 0 ∀x, y ∈ N]
⇒ x = y
∴ f is one – one or injective

Surjectivity : Now 2 ∈ N (codomain of f)
Let x ∈ N (domain of f) s.t. f(x) = 2
⇒ x² = 2
⇒ x = ± \(\sqrt{2}\) ∈ N
Hence 2 ∈ N (codomain off) has no pre-image in N (domain of f)
∴ f is not surjective.
Thus,/is injective but not surjective.

(b) Given f : Z → Z defined by
f (x) = x² ∀X∈Z
Since elements 1 and – 1 ∈ Z (domain of f)
has same image 1 ∈ Z (codomain of f)
∴ f is many – one
Thus, f is not injective.
Further 2 ∈ Z (codomain of f)
let X ∈ Z (domain of f) s.t. f(x) = 2
⇒ x² = 2
⇒ x = ± \(\sqrt{2}\) ∈ z
Hence 2 has no pre image in Z (domain of f)
∴ f is not surjective
Thus, f is neither surjective nor injective.

(c) Given f : R → R defined by
f (x) = x² ∀ X ∈ R
Since f(1) = 1² = 1; f(-1) = (-1)² = 1
∴ different elements 1 and -1 have same image I∈Z (codomain of J)
∴ f is many – one and hence f is not injective.
onto : since – 1 ∈R and let x ∈R (domain of f) such that f(x) = -1
⇒ x² = -1
it does not gives real values of JCGR
i.e., negative real numbers has no (pre image in R)
Thus, f is not surjective (onto)
Hence, f is neither injective nor surjective.

(d) Given f : N → N defined by
f (x) = x³ ∀ x ∈ N
one-one : ∀x, y ∈ N s.t f(x) = f(y)
⇒ (x – y) (x² + xy + y²) = 0
⇒ x – y = 0 [∵ x² + xy + y² ≠ 0 ∀x, y ∈N]
⇒ x = y
∴ f is one – one i.e. injective
Now 2∈N and let X∈N (domain of f)
s.t. f(x) = 2 ⇒ x³ = 2
⇒ x = ± 3\(\sqrt{2}\) ∈ N
Thus, 2∈N (codomain of f) has no pre image in N (domain of f)
∴ f is not onto i.e., surjective.
Hence, f is injective but not surjective,

(e) Given a function
f : Z → Z defined by f (x) = x³
Injectivity: ∀x, y∈Z s.t f(x) = f(y)
⇒ x³ = y³
⇒ x³ – y³ = 0
⇒ (x – y) (x² + xy + y²) = 0
⇒ x – y = 0
[∵ x² + xy + y² ≠ 0 ∀x, y ∈Z]
⇒ x = y
∴ f is one – one i.e. injective.
Surjectivity : since 2∈Z and let X∈Z (domain of f) be s.t. f(x) = 2
⇒ x³ = 2
⇒ x = 3\(\sqrt{2}\) ∉ Z
Hence 2 has no pre-image in Z (domain of f)
∴ f is not surjective

Question 7.
Show that the function f : W → W defined by
f(x) = \(\left\{\begin{array}{l}
n+1, \text { if } n \text { is even } \\
n-1, \text { if } n \text { is odd }
\end{array}\right.\)
is a bijective function.
Solution: Given function f : W → W defined by
f(x) = \(\left\{\begin{array}{l}
n+1, \text { if } n \text { is even } \\
n-1, \text { if } n \text { is odd }
\end{array}\right.\)
one – one : Case-I :
∀ x, y ∈ W and x, y are even
s.t f(x) = f(y)
⇒ x + 1 = y + 1 ⇒ x = y

Case-II :
∀x, y∈W and x, y both are odd
s-t f(x) = f(y)
⇒ x – 1 = y – 1 ⇒ x = y

Case-III :
if x is odd and y is even
f(x) = x – 1 & f(y) = x + 1
i.e. x ≠ y ⇒ f(x) ≠ f(y) ∴ f is 1 – 1.

Case-IV :
if x is even and y is odd
∴ f(x) = x + 1 and f(y) = y – 1
so x ≠ y ⇒ f(x) ≠ f(y)
∴ f is one – one.
so combining all four cases, f is one – one.

onto : If n be odd natural number ∃ an even natural number
n – 1 ∈ N s.t. f(n – 1) = n – 1 + 1 = n
∴ f is onto.
Thus, f is one – one and onto.
Hence f is bijective function.

Question 8.
Show that the function f : R → R given by f(x) = cos x for all x ∈ R is neither one-one nor onto.
Solution:
Given a function f : R → R defined by f(x) = cos x ∀ x ∈R
one – one : ∀x, y∈R s.t f(x) = f(y)
⇒ cos x = cos y
⇒ x = 2nn ± y ∀n∈I
⇒ x ≠ y
i.e. f(0) = cos 0 = 1; f(2π) = cos2π = 1
i.e. 0 and 2π∈R (domain off) have same image 1 ∈ R (codomain of f)
∴ f is many one and hence not one-one.
onto : since 2∈R
and let r∈R (domain off) s.t. f(x) = 2
⇒ cos x = 2 which does not have any real solution
since |cos x| ≤ 1
Thus, 2 has no pre image in R domain of f
∴ f is not onto.
Hence f is neither one-one nor onto.

Question 9.
Show that the function f:
[0, ∞ ) → [0, ∞) defined by
f(x) = \(\frac {2x}{1+2x}\) is
(a) one-one and onto
(b) one-one but not onto
(c) not one-one but onto
(d) neither one-one nor onto.
Solution:
Given a function f : [0, ∞) → [0, ∞)
defined by f(x) = \(\frac {2x}{1+2x}\)
one – one : ∀x, y∈[0, ∞) s.t f(x) = f(y)
⇒ \(\frac{2 x}{1+2 x}=\frac{2 y}{1+2 y}\)
⇒ 2x(1 + 2y) = 2y(1 + 2x)
⇒ 2x + 4xy = 2y + 4xy
⇒ 2x = 2y ⇒ x = y
∴ f is one – one.
onto : Let y ∈ [0, ∞) be any arbitrary element
Then f(x) = y ⇒ \(\frac{2 x}{1+2 x}\) = y
⇒ 2x = y + 2xy
⇒ 2x (1 – y) = y ⇒ x = \(\frac{y}{2(1-y)}\)
which does not exists at y = 1 ∈[0, ∞) so 1 ∈ [0, ∞) (codomain of f) has no (pre image) in [0, ∞) (domain of f) f is not onto.
Hence f is one-one but not onto.

Question 10.
If f : R → R be a function defined by f(x) = 2x³ – 5, show that the function f is a bijective function.
Solution:
Given f : R → R be a function defined
by f(x) = 2x³ – 5 ∀ x ∈ R
one – one : ∀x, y ∈ R s.t f(x) = f(y)
⇒ 2x³ – 5 = 2y³ – 5
⇒ x³ = y³
⇒ (x – y) (x² + xy + y²) = 0
⇒ x – y = 0
⇒ x = y
[∵ x² + xy + y² = x² + xy + \(\frac{y^2}{4}+\frac{3}{4}\)y² = \(\left(x+\frac{y}{2}\right)^2+\frac{3}{4}\)y² ≠ 0]
onto : Let y ∈ R be any arbitrary element
Then f(x) = y
⇒ 2x³ – 5

∴ f is onto.
∴ f is one-one and onto and hence
f is bijective.

Question 11.
Let f : R → R be defined as f(x) = x⁵. Show that it is a bijective function.
Solution:
Given f : R → R be defined as
f(x) = x⁵ ∀ x ∈ R
∀x, y ∈ R s.t f(x) = f(y) ⇒ x⁵ + y⁵
⇒ x⁵ – y⁵ = 0
⇒(x – y) (x⁴ + x³y + x²y² + xy³ + y⁴) = 0
⇒ x – y = 0
[∵ x⁴ + x³y + x²y² + xy³ + y⁴ ≠ 0 ∀x, y ∈ R]
⇒ x – y
∴ f is one-one.

onto : Let y∈R (co-domain of f) be any orbitrary element Then f(x) = y
⇒ x⁵ = y ⇒ x = (y)^1/5
as y ∈ R ⇒ y^1/5 ∈ R ⇒ y ∈ R
∀ y ∈ R ∃ x ∈ R
S.t f(x) = f(y^1/5) = (y^1/5)⁵ = y
∴ f is onto.
Thus, f is one – one and onto and hence
f is bijective function.

Question 12.
A mapping f : N → N, where N is the set of natural numbers is defined as
f(n) = \(\left\{\begin{array}{l}
n^2, \text { for } n \text { odd } \\
2 n+1, \text { for } n \text { even }
\end{array}\right.\)
for neN. Show that/is neither injective nor surjective.
Solution:
Given a mapping f : N → N defined
by \(\begin{cases}n^2, & n \text { is odd } \\ 2 n+1, & n \text { is even }\end{cases}\)
since f(3) = 3² = 9 and
since f(4) = 2 x 4 + 1 = 9 .
Thus elements 3 and 4 has same image 9. so different elements in N (domain of f) has same image 9 ∈ N (codomain of f).
∴ f is many one.
Thus, f is not injective i.e. one – one.
Since 2 ∈ N (codomain of f).
Let x ∈ N (domain of f) s.t f(x) = 2
if x is odd then f(x) = x² = 2
⇒ x = ±\(\sqrt{2}\) ∉ N
If x is even then f(x) = 2
⇒ 2x + 1 = 2 ⇒ x = \(\frac { 1 }{ 2 }\) ∉ N
so in both cases, 2 ∈ N (codomain of f)
has no pre-image in N (domain off)
∴ f is not onto i.e. surjective.
Hence, f is neither injective nor surjective.