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Standard XII

Mathematics

Number of Elements in a Cartesian Product

Let A = 4, 6,...

Question

Let $A = {4, 6, 8, 10}, B = {3, 4, 5, 6, 7}$ , and $f : A \to B$ be defined by $f (x) = \frac{1}{2} x + 1$ , then which of the following options is correct, if we represent $f$ by a set of ordered pairs?

{(4, 3), (6, 5), (8, 4), (10, 7)}

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{(4, 3), (6, 4), (8, 5), (10, 6)}

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{(4, 3), (6, 4), (8, 5), (10, 7)}

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{(4, 3), (6, 5), (8, 4), (10, 6)}

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Solution

The correct option is B ${(4, 3), (6, 4), (8, 5), (10, 6)}$
Given,
$f (x) = \frac{1}{2} x + 1$
$f (4) = (\frac{1}{2} \times 4) + 1 = 2 + 1 = 3$
$f (6) = (\frac{1}{2} \times 6) + 1 = 3 + 1 = 4$
$f (8) = (\frac{1}{2} \times 8) + 1 = 4 + 1 = 5$
$f (10) = (\frac{1}{2} \times 10) + 1 = 5 + 1 = 6$
The set of ordered pairs are $f = {(4, 3), (6, 4), (8, 5), (10, 6)}$

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Similar questions

Q. Let A = {4, 6, 8, 10 } and B = { 3, 4, 5, 6, 7 }. If

f : A \to B

is defined by

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

then represent f(x) by a set of ordered pairs

Q. Let

A = {5, 6, 7, 8},

B = {- 11, 4, 7, - 10, - 7, - 9, - 13}

, and

f = {(x, y) : y = 3 - 2 x, x \in A, y \in B}

. What is the range of

f

Q. Let

A = {4, 6, 8, 10}

and

B = {3, 4, 5, 6, 7}

and

f : A \to B

be defined by

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

, then represent

f

by:
(i) An arrow diagram (ii) A set of ordered pairs (iii) A table

Match the following

$\begin{matrix} 1. P C l_{3} & a. 2 lone pairs & A. 5 Ligands 2. C l F_{3} & b. 3 lone pairs & B. 3 Ligands 3. X e F_{6} & c. 0 lone pairs & C. 6 Ligands 4. S b C l_{5} & d. 1 lone pairs & D. 4 Ligands \end{matrix}$

Q. Match

List I

with the

List II

and select the correct answer using the code given below the lists :

\begin{matrix} List I & List II (A) & Let f : [- 1, \infty) \to (0, \infty) be defined by f (x) = e^{x^{2} + | x |} . Then f is & (P) & one-one (B) & Let f : (1, \infty) \to [3, \infty) be defined by f (x) = \sqrt{10 - 2 x + x^{2}} . Then f is & (Q) & into (C) & Let f : R \to Z be defined by f (x) = {tan}^{5} π [x^{2} + 2 x + 3] where [. ] denotes & (R) & many-one the greatest integer function. Then f is (D) & Let f : [3, 4] \to [4, 6] be defined by f (x) = | x - 1 | + | x - 2 | + | x - 3 | + | x - 4 | . & (S) & onto Then f is (T) & periodic \end{matrix}

Which of the following is a CORRECT combination?

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Let A={4,6,8,10}, B={3,4,5,6,7}, and f:A→B be defined by f(x)=12x+1, then which of the following options is correct, if we represent f by a set of ordered pairs?

The correct option is B {(4,3),(6,4),(8,5),(10,6)}Given, f(x)=12x+1 f(4)=(12×4)+1=2+1=3 f(6)=(12×6)+1=3+1=4 f(8)=(12×8)+1=4+1=5 f(10)=(12×10)+1=5+1=6 The set of ordered pairs are f={(4,3),(6,4),(8,5),(10,6)}

Let $A = {4, 6, 8, 10}, B = {3, 4, 5, 6, 7}$ , and $f : A \to B$ be defined by $f (x) = \frac{1}{2} x + 1$ , then which of the following options is correct, if we represent $f$ by a set of ordered pairs?