Special Types of Real Functions Class 12 OP Malhotra Exe-2A ISC Maths Solutions Ch-2. In this article you would learn how to solve questions on special type of function and real function easily. These solutions are strictly as per latest syllabus of council and solved according to latest prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ISC Board Class-12 Mathematics.

Special Types of Real Functions Class 12 OP Malhotra Exe-2A ISC Maths Solutions Ch-2

Board	ISC
Publications	S Chand
Subject	Maths
Class	12th
Chapter-2	Functions
Writer	OP Malhotra
Exe-2(A)	Special Types of Functions and Real Functions.

Special Types of Functions and Real Functions

Class 12 OP Malhotra Exe-2A ISC Maths Solutions Ch-2

Que-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

Sol: 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

Que-2: Show that a function f : R → R given by f (x) = ax + b, a, b ∈ R, a ≠ 0 is a bijective.

Sol: 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 y ∈ R be any arbitrary element
and let y = ax + b is x = (y–b)/a ∈ R, as
y ∈ R a, b ∈ R
∀ y ∈ R ∃ x = (y–b)/a ∈ R,
s.t. f(x) = f(y−b/a) = a(y−b/a) + b = y
Hence f is onto.
Thus f is one-one and onto and hence
f is bijective.

Que-3: Let f : N → N given by f (x) = 2x ∀x ∈ N. Show that/is one-one and into.

Sol: 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 = 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.

Que-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

Sol: 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 = y/3
Since y ∈ R ⇒ y/3 ∈ R ⇒ x ∈ R
∀ y ∈ R∃ y ∈ R
s.t f(x) = f(y/3) = 3 x y/3 = y
∴ f is onto.
Thus, f is one – one and onto.

Que-5: Show that f : R → R, defined by f(x) = x³ is a bijection.

Sol: 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√y
as y ∈ R ⇒ 3√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.

Que-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

Sol: (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 = ± √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 = ± √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√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√2 ∉ Z
Hence 2 has no pre-image in Z (domain of f)
∴ f is not surjective

Que-7: Show that the function f : W → W defined by f(x) = [{n+1, if n is even} {n−1, if n is odd}]
is a bijective function.

Sol: Given function f : W → W defined by
f(x) = {n+1, if n is even n−1, if n is odd
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.

Que-8: Show that the function f : R → R given by f(x) = cos x for all x ∈ R is neither one-one nor onto.

Sol: 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.

Que-9: Show that the function f:
[0, ∞ ) → [0, ∞) defined by
f(x) = 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.

Sol: Given a function f : [0, ∞) → [0, ∞)
defined by f(x) = 2×1+2x
one – one : ∀x, y∈[0, ∞) s.t f(x) = f(y)
⇒ 2x/1+2x=2y/1+2y
⇒ 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 ⇒ 2×1+2x = y
⇒ 2x = y + 2xy
⇒ 2x (1 – y) = y ⇒ x = 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.

Que-10: If f : R → R be a function defined by f(x) = 2x³ – 5, show that the function f is a bijective function.

Sol: 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 + y24+34y² = (x+y/2)2+34y² ≠ 0]
onto : Let y ∈ R be any arbitrary element
Then f(x) = y
⇒ 2x³ – 5
x = (y+5/2)^1/3
Since y ∈ R => (y+5/2)^1/3 ∈ R => x ∈ R
Thus ∀y ∈ R ∃ x ∈ R
f(x) = f ((y+5/2)^1/3)
=>2(y+5/2)-5 = y
∴ f is onto.
∴ f is one-one and onto and hence
f is bijective.

Que-11: Let f : R → R be defined as f(x) = x⁵. Show that it is a bijective function.

Sol: 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 orbiratry 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.

Que-12: A mapping f : N → N, where N is the set of natural numbers is defined as f(n) = {n², for n odd 2n+1, for n even
for n∈N. Show that/is neither injective nor surjective.

Sol: Given a mapping f : N → N defined
by {n²,2n+1,n is odd n is even
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 = ±√2 ∉ N
If x is even then f(x) = 2
⇒ 2x + 1 = 2 ⇒ x = 12 ∉ 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.

–: End Special Types of Real Functions Class 12 OP Malhotra Exe-2A ISC Maths Solutions :–

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