Relations And Functions Ii(12th Grade > Mathematics ) Questions and answers for exam Preparation

12th Grade > Mathematics

RELATIONS AND FUNCTIONS II MCQs

Relations And Functions

Total Questions : 89 | Page 1 of 9 pages

Question 1. The function

f (x) = c o s \sqrt{x}

Periodic with period 2√π
Periodic with period √π
Periodic with period 4π2
Not a periodic function

Discuss Question

Answer: Option D. -> Not a periodic function
:
D
Try drawing

c o s \sqrt{x}

graph. It’s not periodic.

Question 2. Range of

f (x) = \frac{t a n (π [x^{2} - x])}{1 + s i n (c o s x)}

is where

[x]

denotes the greatest integer function

(−∞,∞)∼[0,tan1]
(−∞,∞)∼[tan2,0)
[tan2,tan1]
{0}

Discuss Question

Answer: Option D. -> {0}
:
D

f (x) = \frac{t a n (π [x^{2} - x])}{1 + s i n (c o s x)} = {0} because of [x^{2} - x] i s integer .

Question 3. If f: R

\to

R ,g : R

\to

R and h: R

\to

R are such that

f (x) = x^{2}

g (x) = t a n x

and

h (x) = l o g x

, then the value of

(h (g (f (x))))

x = \sqrt{\frac{π}{4}}

will be

0
1
-1
π

Discuss Question

Answer: Option A. -> 0
:
A

(h o (g o f)) (x) = h {g {f (x)}}

= h {t a n x^{2}} = l o g {t a n x^{2}}

∴ A t x = \sqrt{\frac{π}{4}} \Rightarrow (h o (g o f)) (x) = l o g t a n \frac{π}{4}

= log 1 = 0

Question 4. Let X be a family of sets and R be a relation on X defined by 'A is disjoint from B'. Then R is

Reflexive
Symmetric
Anti-symmetric
Transitive

Discuss Question

Answer: Option B. -> Symmetric
:
B
Clearly, the relation is symmetric but it is neither reflexive nor transitive.

Question 5.

T h e i n v e r s e o f f (x) = {(5 - {(x - 8)}^{5})}^{\frac{1}{3}} i s

5−(x−8)5
8+(5−x3)15
8−(5−x3)15
(5−(x−8)15)3

Discuss Question

Answer: Option B. -> 8+(5−x3)15
:
B

y = f (x) = {(5 - {(x - 8)}^{5})}^{\frac{1}{3}}

then

y^{3} = 5 - (x - 8)^{5} \Rightarrow (x - 8)^{5} = 5 - y^{3}

\Rightarrow x = 8 + (5 - y^{3})^{\frac{1}{5}}

Let,

z = g (x) = 8 + (5 - x^{3})^{\frac{1}{5}}

To check,

f (g (x)) = [5 - {(x - 8)}^{5}]^{\frac{1}{3}}

= {(5 - {[{(5 - x^{3})}^{\frac{1}{5}}]}^{5})}^{\frac{1}{3}} = (5 - 5 + x^{3})^{\frac{1}{3}} = x

similarly , we can show that

g (f (x)) = x

Hence,

g (x) = 8 + (5 - x^{3})^{\frac{1}{5}}

is the inverse of

f (x)

Question 6. Let

f : [0, \infty) \to

[0,2] be defined by

f (x) = \frac{2 x}{1 + x}

, then f is

one – one but not onto
onto but not one – one
both one – one and onto
neither one – one nor onto

Discuss Question

Answer: Option A. -> one – one but not onto
:
A
We can draw the graph and see that the given function is one - one.
Since, Range

⊏

Codomain

\Rightarrow

f is into

∴

f is only injective.

Question 7. Inverse exists for a function which is

Injective
Surjective
Bijective
Many-one

Discuss Question

Answer: Option C. -> Bijective
:
C
We have seen that inverse exists only when function is one-one and onto, i.e. Bijective.

Question 8. With reference to a universal set, the inclusion of a subset in another, is relation, which is

Symmetric only
Equivalence relation
Reflexive only
None of these

Discuss Question

Answer: Option D. -> None of these
:
D
Since

A \subseteq A

∴

Relation

^{'} \subseteq^{'}

is reflexive.
Since

A \subseteq B, B \subseteq C \Rightarrow A \subseteq C

∴

Relation

^{'} \subseteq^{'}

is transitive.
But If

A \subseteq B

, Doesn't imply

B \subseteq A

∴

Relation is not symmetric

Question 9.

f (x) = \frac{x^{2} - 3 x + 4}{x^{2} + 3 x + 4}

the range of f(x) is

[0,17]
(−∞,17)∪(7,∞)
(−∞,7)
[17,7]

Discuss Question

Answer: Option D. -> [17,7]
:
D

y = \frac{x^{2} - 3 x + 4}{x^{2} + 3 x + 4}

y x^{2} + 3 x y + 4 y = x^{2} - 3 x + 4

x^{2} (y - 1) + 3 x (y + 1) + 4 (y - 1) = 0

\geq 0 \Rightarrow 9 (y + 1)^{2} - 4.4 (y - 1)^{2} \geq 0

(3 (y + 1) - 4 (y - 1))

(3 (y + 1) + 4 (y - 1)) \geq 0

(- y + 7) (7 y - 1) \geq 0

(y - 7) (y - \frac{1}{7}) \leq 0

\frac{1}{7} \leq y \leq 7

Question 10. If f: R

\to

R ,g : R

\to

R and h: R

\to

R are such that f(x) =

x^{2}

, g(x) = tan x and h (x) = log x, then the value of (ho (gof)) (x)
if x =

\sqrt{\frac{π}{4}}

will be

0
1
-1
π

Discuss Question

Answer: Option A. -> 0
:
A

(h o (g o f)) (x) = h {g {f (x)}}

= h {t a n x^{2}} = l o g {t a n x^{2}}

∴ A t x = \sqrt{\frac{π}{4}} \Rightarrow (h o (g o f)) (x) = l o g t a n \frac{π}{4}

= log 1 = 0

Latest Videos

Chapter 1 - GLOBAL STEEL SCENARIO & INDIAN STEEL INDUSTRY Part 1 : (13-04-2024) INDUSTRY AND COMPANY AWARENESS (ICA)

Direction Sense Test Part 1 Reasoning (Hindi)

Chapter 1 - RMHP / OHP / OB & BP Part 1 : (14-02-2024) GPOE

Cube & Cuboid Part 1 Reasoning (Hindi)

Data Interpretation (DI) Basic Concept Reasoning (Hindi)

Counting Figures Part 1 Counting Of Straight Lines Reasoning (Hindi)

Sail E0 free webinar Webinars

Real Numbers Part 7 Class 10 Maths

Real Numbers Part 1 Class 10 Maths

Polynomials Part 1 Class 10 Maths

Latest Test Papers

Pie Chart - Test Paper 1 Data Interpretation Reasoning & DI

Tabulation - Test Paper 2 Data Interpretation Reasoning & DI

Test Paper 1 Direction Sense Test Reasoning & DI

ICA , RDIC,GFM Free Modal Test Paper - Non - Technical Branch Old Syllabus 2022 - SAIL Junior Officer E0

GFM Combined 1 Free Revision Test SAIL JO E-0