Free Objective Test 01 Practice Test - 11th and 12th

Question 1

Let X = {1,2,3,4,5} and Y = {1,3,5,7,9}. Which of the following is/are relations from X to Y

$R_{1} = {(x, y) y = 2 + x, x \in X, y \in Y}$

$R_{2} = {(1, 1), (2, 1), (3, 3), (4, 3), (5, 5)}$

$R_{3} = {(1, 1), (1, 3), (3, 5), (3, 7), (5, 7)}$

$R_{4} = {(1, 3), (2, 5), (2, 4), (7, 9)}$

SOLUTION

Solution : A, B, and C

If 'R' is a relation from 'A' to 'B' , then 'R' is defined as {(x,y)| x $\in$ A and y $\in$ B}

In all the ordered pairs in the realtions of R $_{1}$ , R $_{2}$ , R $_{3}$ first component $\in$ X and second component $\in$ Y. So, R $_{1}$ , R $_{2}$ , R $_{3}$ are relations from X to Y.

$R_{4}$ is not a relation from X to Y , because in ordered pain (7,9) the first component 7 $\notin$ X.

Question 2

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

SOLUTION

Solution : D

Since A $\subseteq$ A . $∴$ Relation ' $\subseteq$ ' is relfexive

Since A $\subseteq$ B , B $\subseteq$ C $\Rightarrow$ A $\subseteq$ C

$∴$ Relation ' $\subseteq$ ' is transitive.

But A ' $\subseteq$ ' B , $\Rightarrow$ B $\subseteq$ A . $∴$ relation is not symmetric.

Question 3

If R be a relation $<$ from A={1,2,3,4} to B={1,3,5} i.e., (a,b) $\in R \Leftrightarrow a < b, t h e n R o R^{- 1}$ is

{(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}

{(3, 1) (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)}

{(3, 3), (3, 5), (5, 3), (5, 5)}

{(3, 3) (3, 4), (4, 5)}

SOLUTION

Solution : C

We have, $R$ ={(1,3);(1,5);(2,3);(2,5);(3,5);(4,5)}
$R^{- 1}$ = {(3,1);(5,1);(3,2);(5,2);(5,3);(5,4)}
Hence $R o R^{- 1}$ = {(3,3);(3,5);(5,3);(5,5)}

Question 4

The relation R is defined on the set of natural numbers as {(a,b) : a = 2b}. Then $R^{- 1}$ is given by

{(2, 1), (4, 2), (6, 3).....}

{(1, 2), (2, 4), (3, 6)....}

$R^{- 1}$ is not defined

None of these

SOLUTION

Solution : B

R = {(2,1),(4,2),(6,3),.....}.

So, $R^{- 1}$ = {(1,2),(2,4),(3,6),......}.

Question 5

The relation R defined on the set of natural numbers as {(a, b) : a differs from b by 3}, is given by

{(1, 4, (2, 5), (3, 6),.....}

{(4, 1), (5, 2), (6, 3),.....}

{(1, 3), (2, 6), (3, 9),..}

None of these

SOLUTION

Solution : B

$R = {(a, b) : a, b \in N, a - b = 3} = {((n + 3), n) : n \in N}$

={(4,1),(5,2),(6,3), .......} .

Question 6

Given two finite sets A and B such that n(A) = 3, n(B) = 3. Then total number of relations from A to B is

512

SOLUTION

Solution : C

Here n(A × B) = 3 × 3 = 9

Since every subset of A × B defines a relation from A to B, the number of relations from A to B is equal to the number of subsets of A $\times$ B = $2^{n(A \times B)}$
= $2^{9}$
= 512

Question 7

Let A = {1, 2, 3}. The total number of distinct relations that can be defined over A is

$2^{9}$

None of these

SOLUTION

Solution : A

n ( A × A) = n(A)n(A) = $3^{2}$ = 9

So, the total number of subsets of A × A is $2^{9}$

and a subset of A × A is a relation over the set A .

Question 8

Let A = {1,2,3}, B = {1,3,5}. A relation R:A $\to$ B is

defined by R = {(1,3),(1,5),(2,1)}. Then $R^{- 1}$ is defined by

{(1,2), (3,1), (1,3), (1,5)}

{(1, 2), (3, 1), (2, 1)}

{(1, 2), (5, 1), (3, 1)}

None of these

SOLUTION

Solution : C

( x,y) $\in$ R ⇔ (y,x) $\in R^{- 1},$ $∴$ $R^{- 1} = {(3, 1), (5, 1), (1, 2)}$ .

Question 9

If A = { $x : x^{2} - 5 x + 6 = 0$ }, B = {2,4} , C = {4,5}, then A×(B $\cap$ C) is ___

{(2, 4), (3, 4)}

{(4, 2), (4, 3)}

{(2, 4), (3, 4), (4, 4)}

{(2,2), (3,3), (4,4), (5,5)}

SOLUTION

Solution : A

Given, A = { $x : x^{2} - 5 x + 6 = 0$ }
$∴$ The elements of A are the roots of $x^{2} - 5 x + 6 = 0$
$x^{2} - 5 x + 6 = 0 \Rightarrow (x - 3) (x - 2) = 0 \Rightarrow x = 3 and 2$
$∴$ A = {2,3} , B = {2,4}, C = {4,5}

$\Rightarrow B \cap C$ = {4}

$∴$ A × (B $\cap$ C) = {2, 3} $\times$ {4}
={(2,4), (3,4)}

Question 10

A relation from P to Q is

A universal set of P $\times$ Q

P $\times$ Q

An equivalent set of P $\times$ Q

A subset of P $\times$ Q

SOLUTION

Solution : D

A relation from P to Q is a subset of P $\times$ Q.

Question 11

$R a n g e o f t h e f u n c t i o n f (x) = \frac{x^{2} + x + 2}{x^{2} + x + 1}; x ϵ R i s$

$(1, \infty$ )

$(1, \frac{11}{7}]$

$(1, \frac{7}{3}]$

$(1, \frac{7}{5}]$

SOLUTION

Solution : C

$W e h a v e, f (x) = \frac{x^{2} + x + 2}{x^{2} + x + 1} = \frac{(x^{2} + x + 1)}{x^{2} + x + 1} = 1 + \frac{1}{{(x + \frac{1}{2})}^{2} + \frac{3}{4}} W e c a n s e e h e r e t h a t a s x \to \infty, f (x) \to 1 w h i c h i s t h e m i n v a l u e o f f (x) . A l s o f (x) i s m a x w h e n {(x + \frac{1}{2})}^{2} + \frac{3}{4} i s m i n w h i c h i s s o w h e n x = - \frac{1}{2} a n d t h e n \frac{3}{4} . ∴ f_{m a x} = 1 + \frac{1}{\frac{3}{4}} = \frac{7}{3} ∴ R_{i} = (1, \frac{7}{3}]$

Question 12

If f(x) is a function whose domain is symmetric about the origin, then f(x) + f(–x) is

One-one

Even

Odd

Both even and odd

SOLUTION

Solution : B

(a, b)
g(x) = f(x) + f(–x)
g(–x) = f(–x) + f(x) = g(x)
therefore g(x) is even

Question 13

Product of two odd functions is

Even function

Odd function

Neither even nor odd

Cannot be determined

SOLUTION

Solution : A

Let f(x), g(x) be odd
Let F(x) = f(x)g(x)
F(–x) = f(–x)g(–x) = F(x)
therefore F(x) is even

Question 14

If $f (x) + 2 f (1 - x) = x^{2} + 1 \forall x ϵ R$ then f(x) is

$\frac{1}{3} (x^{2} + 4 x - 3)$

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

$\frac{1}{3} (x^{2} - 4 x + 3)$

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

SOLUTION

Solution : C

$f (x) + 2 f (1 - x) = x^{2} + 1$ .........(i)
Replacing x by 1 – x
$f (1 - x) + 2 f (x) = (1 - x^{2}) + 1$ .......(ii)
multiplying (ii) by 2 and subtracting it from (1), we get
$- 3 f (x) = x^{2} - 2 (1 - x)^{2} - 1$
$3 f (x) = 2 (1 - x)^{2} + 1 - x^{2} = (1 - x) (2 - 2 x + 1 + x) = (1 - x) (3 - x) = x^{2} - 4 x + 3$
$f (x) = \frac{1}{3} (x^{2} - 4 x + 3)$

Question 15

Let f be a function satisfying $2 f (x) - 3 f (\frac{1}{x}) = x^{2}$ for any $x \neq 0$ , then the value of f(2) is

A. -2

\frac{- 7}{4}

- \frac{7}{8}

D. 4

SOLUTION

Solution : B

$2 f (x) - 3 f (\frac{1}{x}) = x^{2} \dots \dots (i) R e p l a c i n g x b y \frac{1}{x} 2 f (\frac{1}{x}) - 3 f (x) = \frac{1}{x^{2}} \dots \dots (i i)$
Solving (i) and (ii) we get
$- 5 f (x) = 2 x^{2} + \frac{3}{x^{2}} f (x) = \frac{- 1}{5} (2 x^{2} + \frac{3}{x^{2}}) ∴ f (2) = \frac{- 1}{5} (8 + \frac{3}{4}) = \frac{- 7}{4}$

Free Objective Test 01 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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