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Let \( \Large f:N \rightarrow R:f \left(x\right)=\frac{ \left(2x-1\right) }{2} \) and \( \Large g:Q \rightarrow R:g \left(x\right)=x+2 \) be two functions then \( \Large \left(gof\right) \left(\frac{3}{2}\right) \)

A) 3
B) 1
C) \( \Large \frac{7}{2} \)
D) None of these

Correct Answer:

A) 3

Description for Correct answer:
\( \Large f \left(x\right)=\frac{2x-1}{2}\ and\ g \left(x\right)=x+2 \)

\( \Large \left(gof \left(x\right) \right)=g \left(\frac{1}{2} \left(2x-1\right) \right) \)

= \( \Large x-\frac{1}{2}+2 = x+\frac{3}{2} \)

\( \Large \therefore \left(gof\right) \left(\frac{3}2{}\right)=\frac{3}{2}+\frac{3}{2}=3 \)

But, \( \Large \frac{3}{2} \notin N\)

Part of solved Set theory questions and answers : >> Elementary Mathematics >> Set theory

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Similar Questions

1). If N be the set of all natural numbers, consider \( \Large f:N \rightarrow N:f \left(x\right)=2x \forall x \epsilon N \), then f is:

A). one-one onto

B). one-one into

C). many-one

D). one of these

2). N is the set of natural numbers. The relation R is defined on \( \Large N \times N \) as follows: \( \Large \left(a,\ b\right)R \left(c,\ d\right) \Leftrightarrow a+d=b+c \) is:

A). reflexive

B). symmetric

C). transitive

D). all of these

3). Let \( \Large A = \{ 2,\ 3,\ 4,\ 5 \} \) and
\( \Large R = \{ \left(2,\ 2\right),\ \left(3,\ 3\right),\ \left(4,\ 4\right),\ \left(5,\ 5\right),\
\left(2,\ 3\right),\ \left(3,\ 2\right),\) \( \Large \ \left(3,\ 5\right),\ \left(5,\ 3\right) \} \) be a relation in A, Then R is:

A). reflexive and transitive

B). reflexive and symmetric

C). reflexive and anti-symmetric

D). none of the above

4). For real numbers x and y, we write
\( \Large x R y \Leftrightarrow x^{2}-y^{2}+\sqrt{3} \)
is an irrational number. Then the relation R is:

A). reflexive

B). symmetric

C). transitive

D). none of these

5). \( \Large f \left(x\right)=\frac{1}{2}-\tan \frac{ \pi x}{2},\ -1 < x < 1\ and\ g \left(x\right) \) \( \Large =\sqrt{ \left(3+4x-4x^{2}\right) } \) then dom \( \Large \left(f + g\right) \) is given by:

A). \( \Large \left[ \frac{1}{2}, 1 \right] \)

B). \( \Large \left[ \frac{1}{2}, -1 \right] \)

C). \( \Large \left[ -\frac{1}{2}, 1 \right] \)

D). \( \Large \left[ -\frac{1}{2}, -1 \right] \)

6). If \( \Large R \subset A \times B\ and\ S \subset B \times C \) be two relations, then \( \Large \left(SOR\right)^{-1} \) is equal to:

A). \( \Large S^{-1}OR^{-1} \)

B). \( \Large R^{-1}OS^{-1} \)

C). SOR

D). ROS

7). If \( \Large A = \{ x:x\ is\ multiple\ of\ 4 \} \) and \( \Large B = \{ x:x\ is\ multiples\ of 6 \} \) then \( \Large A \subset B \) consists of all multiples of:

A). 16

B). 12

C). 8

D). 4

8). The relation "less than" in the set of natural numbers is:

A). only symmetric

B). only transitive

C). only reflexive

D). equivalence relation

9). The solution set of \( \Large 8x \equiv 6 \left(mod\ 14\right),\ x \epsilon z \) are:

A). \( \Large \left[ 8 \right] \cup \left[ 6 \right] \)

B). \( \Large \left[ 8 \right] \cup \left[ 14 \right] \)

C). \( \Large \left[ 6 \right] \cup \left[ 13 \right] \)

D). \( \Large \left[ 8 \right] \cup \left[ 6 \right] \cup \left[ 13 \

10). Consider the following relations:
\( \Large 1.\ A-B=A- \left(A \cap B\right) \)
\( \Large 2.\ A= \left(A \cap B\right) \cup \left(A-B\right) \)
\( \Large 3.\ A- \left(B \cup C\right) = \left(A-B\right) \cup \left(A-C\right) \)
which of these is/are correct?

A). (1) and (3)

B). (2) only

C). (2) and (3)

D). (1) and (2)