Question Set B

Relations and Functions Important Questions

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Questions:

Generic 2

14 x 2 = 28

If f(x) = x³ -\(1\over x^3\) ,Then find f(x) + f\(({1\over x})\) = 0

A function f: R\(\rightarrow\) R is defined as f(x) = \(\begin{cases} 2x+3,x\ge3\\ 7,x<3 \end{cases}\)
find f(1)

Let A = {1, 3}, B = {2, 4} and C = {1, 5}. Find A x B x C

A function f: R \(\rightarrow\)R is defined as
\(f(x)=\begin{cases} x^2+1\quad x\le-2 \\ 2x+1 \quad -2<x\le3\\ 2x^3-3\quad 3<x<8 \end{cases}\)
Find f(-1)

Find the domain and range of the following functions:\(f(x)=1-|x-3|,x\in R\)

A function t is defined by f(x) = 2x - 5. Write down the values of f(- 3)

Let R be a relation on set N of natural numbers defined by R = {(x, y) : x, y\(\in\)N, x + 3y = 12} Find Range of R

Let A = {1, 2, 3}, B = {2, 3, 4, 5}, state as to which of the following sets R represents a relation of set A into set B?
If R represents a relation, write its domain and range.
R = {(1, 1), (1, 4), (1, 5), (2, 3)}

Let R be a relation on Q,defined by
R=, \(\{ (a,b):a,b\in Q\)and \(a-b\in Z\} \)show that
\((a,b)\in R\Rightarrow (b,a)\in R\)

10.

Determine the domain and range of the relations R, where R = \(\left( x,{ x }^{ 3 } \right) \) : x is a prime number less than 10}

11.

Find the domain for which the functions \(f\left( x \right) =2{ x }^{ 2 }-1\) and \(g\left( x \right) =1-3x\) are equal

12.

Let f(x)=x² and g(x)=2x+1 be two real functions.Find (fg) (x)

13.

If A x B = {a,1), (b,3), (a,3), (b,1), (a,2), (b,2)}. Then, find A and B.

14.

Which of the following relations are functions?
\(\left\{ (2,0),(4,8),(2,1),(3,6) \right\} \)

Generic 3

5 x 3 = 15

15.

Let A = {1, 2, 3, 4, 5, 6}. Define a relation R from A to A by R = {(x, y) : y = x + 1 }
(i) Depict this relation using an arrow diagram.
(ii) Write down the domain, codomain and range of R.

16.

If A = {1,3,6} and B = {X,Y}, then represent the following catesian products by an arrow diagrams \(B\times A\)

17.

Find the domain of the function f defined by \(f(x)=\sqrt { 4-x } +\frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } \)

18.

If f and g be two real function defined by \(f\left( x \right) =\sqrt { x+1 } \)and \(g\left( x \right) =\sqrt { 9-{ x }^{ 2 } } \) .Then, describe each of the following functions. \(\frac { g }{ f } \)

19.

If f and g be two real function defined by \(f\left( x \right) =\sqrt { x+1 } \)and \(g\left( x \right) =\sqrt { 9-{ x }^{ 2 } } \). Then, describe each of the following functions. f+g

Generic 5

4 x 5 = 20

20.

Let A, B, C and D be any non-empty sets.
Prove that (A xB)\(\cap\)(CxD) =(A\(\cap\)C) x (B\(\cap\)D).

21.

Let R be a relation from N to N defined by R = {(a, b): a, b \(\in\) N and a = b²}. Are the following true?
(i) (a, a) \(\in\)R for all a\(\in\) N
(ii) (a, b)\(\in\) R implies (b, a)\(\in\) R
(iii) (a, b)\(\in\)R, (b, c) \(\in\) R implies (a, c)\(\in\) R

22.

The relation f is defined by
\(f(x)=\begin{cases} { x }^{ 2 },\quad 0\le x \le 3 \\ 3x,\quad3\le x<10\end{cases}\)
The relation g is defined by
\(g(x)=\begin{cases} { x }^{ 2 },\quad 0\le x \le 2 \\ 3x,\quad2\le x\le10\end{cases}\)
Show that f is a function and g is not a function.

23.

Determine the domain and range of the relation defined as: R = {(a, b) : a, b\(\in\)N, a < 5, b = 3a + 1}

Multiple Choice Question

5 x 1 = 5

24.

The domain of the function \(f(x)=\sqrt{x-1}+\sqrt{3-x}\) is ______.

(a)

(1,\(\infty\))

(b)

(\(\infty\),5)

(c)

(1, 3)

(d)

[1, 3]

25.

If \(f(x)={2^x+2^{-x}\over 2}\) then f(x+y) f(x-y)is equal to ______.

(a)

\({1\over2}[f(2x)+f(2y)]\)

(b)

\({1\over2}[f(2x)-f(2y)]\)

(c)

\({1\over3}[f(2x)+f(2y)]\)

(d)

\({1\over3}[f(2x)-f(2y)]\)

26.

Let f(x) = Ix-1I then ______.

(a)

f(x²) = [f(x)]²

(b)

f(x +y) = f(x) f(y)

(c)

f(IxI) = I f(x) I

(d)

none of these

27.

If R is a relation from a finite set A having m elements to a finite set B having n elemen.ts, then the number of relations from A to B is ______.

(a)

2^mn

(b)

2^mn-1

(c)

2mn

(d)

mⁿ

28.

If the set A has m elements, B has n elements then the number of elements in A x B is ______.

(a)

m + n

(b)

m + n + 1

(c)

(d)

n²

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Answers Key:

Generic 2

14 x 2 = 28

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

-1

Domain = R, Range = (-\(\infty\),1].

Here f(x) = 2x - 5
putting x = - 3
\(\therefore\) f(-3) = 2 x -3 - 5 = - 11

Range of R = {3, 2, 1}

Let \(a,b\in R\quad then\)
\(a,b\in R\Rightarrow a-b\in Z,a,b\in Q\)
\(\Rightarrow b-a\in Z\therefore (b,a)\in R\)

10.

R = \(\left( x,{ x }^{ 3 } \right) \) : x is a prime number less than 10}
R = {(2,8),(3,27),(5,125),(7,343)}
[\(\therefore \) 2,3,5,7 are primes less than 10]
Domain(R) = {2,3,5,7}
Range(R) = {8,27,125,343}

11.

Given,
\(f\left( x \right) =2{ x }^{ 2 }-1\)
\(g\left( x \right) =1-3x\)
\(Sice,\ f\left( x \right) =g\left( x \right) \)
\(\therefore 2{ x }^{ 2 }-1=1-3x\)
\(\Rightarrow \ 2{ x }^{ 2 }+3x-2=0\)
\(\Rightarrow \ 2{ x }^{ 2 }+4x-x-2=0\)
\(\Rightarrow \ 2x(x+2)-1(x+2)=0\)
\(\Rightarrow \ (2x-1)(x+2)=0\)
\(\Rightarrow \ 2x-1=0\ or\ x+2=0\)
\(\Rightarrow \ x=\frac { 1 }{ 2 } orx=-2\)
Thus, domain for which the function \(f\left( x \right) =g\left( x \right) \ is \ \left\{ \frac { 1 }{ 2 } ,-2 \right\} \)

12.

\((f+g)(x)=x^{2}+2 x+1,(f-g)(x)=x^{2}-2 x-1,\)
\((\text { fg })(x)=x^{2}(2 x+1)=2 x^{3}+x^{2},\left(\frac{f}{g}\right)(x)=\frac{x^{2}}{2 x+1}, x \neq-\frac{1}{2} \)

13.

Here, first element of each ordered pair of A x B gives the elements of set A and corresponding second element gives the elements of set B
\(\therefore\) A = {a, b} and B = {l, 3, 2}

14.

\(\left\{ (2,0),(4,8),(2,1),(3,6) \right\} \)
It is not a function because first elements of (2,0) and (2,1) are same.

Generic 3

5 x 3 = 15

15.

(i) By the definition of the relation, R = {(1,2), (2,3), (3,4), (4,5), (5,6)}.
(ii) We can see that the domain ={1, 2, 3, 4, 5,} Similarly, the range = {2, 3, 4, 5, 6} and the codomain = {1, 2, 3, 4, 5, 6}.

16.

\(B\times A=\{ x,y\} \times \{ 1,3,6\} \\ =\{ (x,1),(x,3),(x,6),(y,1),(y,3),(y,6)\} \)
Required arrow diagram is

17.

Domain = \((-\infty ,-1)\cup (1,4)\)

18.

Domain \((f)\cap \) Domain \((g)=\left[ -1,3 \right] \)
\(\left( \frac { g }{ f } \right) (x)=\sqrt { \frac { 9-{ x }^{ 2 } }{ x+1 } } ,x\neq -1\)

19.

Domain \((f)\cap \) Domain \((g)=\left[ -1,3 \right] \)
\((f+g)(x)=\sqrt { x+1 } +\sqrt { 9-{ x }^{ 2 } } \)

Generic 5

4 x 5 = 20

20.

Let (x, y) \(\in\) (A x B) \(\cap\)(C x D).
\(\Rightarrow\)(x, y)\(\in\) A x B and (x, y) \(\in\) C x D
\(\Rightarrow\) [X\(\in\)A and y\(\in\)B] and [X\(\in\)CandY\(\in\)D]
\(\Rightarrow\) [x \(\in\)A and x\(\in\)C] and [y \(\in\) Band Y\(\in\) D]
\(\Rightarrow\)x \(\in\)(A \(\cap\) C) and Y\(\in\) (B\(\cap\)D)
\(\Rightarrow\) (x, y)\(\in\)(A\(\cap\)C) x (B\(\cap\)D)
\(\therefore\) (A x B)\(\cap\)(C x D)\(\subset\)(A \(\cap\)C) x (B\(\cap\)D) ...(i)
Let (x, y)\(\in\) (A\(\cap\)C) x (B\(\cap\)D)
\(\Rightarrow\) x\(\in\) A\(\cap\)C and Y \(\in\) B\(\cap\)D
\(\Rightarrow\) [x \(\in\) A and x \(\in\) C] and [y\(\in\) Band Y\(\in\) D]
\(\Rightarrow\) [x \(\in\) A and Y \(\in\) B] and [x\(\in\)C and Y\(\in\) D]
\(\Rightarrow\)(x, y) \(\in\) A x B and (x, y)\(\in\) C x D
\(\Rightarrow\)(x, y)\(\in\) (A x B)\(\cap\)(C x D)
\(\therefore\)(A\(\cap\)C) x (B\(\cap\)D)\(\subset\)(A x B)\(\cap\)(C x D) ...(ii)
From (i) and (ii), we have
(A x B)\(\cap\)(C x D) = (A\(\cap\)B) x (B\(\cap\)D)

21.

R = {(a, b): a, b ∈ N and a = b²}
(i) It can be seen that 2 ∈ N;however, 2 ≠ 2² = 4.
Therefore, the statement “(a, a) ∈ R, for all a ∈ N” is not true.
(ii) It can be seen that (9, 3) ∈ N because 9, 3 ∈ N and 9 = 3².
Now, 3 ≠ 9² = 81; therefore, (3, 9) ∉ N
Therefore, the statement “(a, b) ∈ R, implies (b, a) ∈ R” is not true.
(iii) It can be seen that (16, 4) ∈ R, (4, 2) ∈ R because 16, 4, 2 ∈ N and 16 = 4² and 4 = 2².
Now, 16 ≠ 2² = 4; therefore, (16, 2) ∉ N
Therefore, the statement “(a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R” is not true.

22.

Here
f(x) = x² 0\(\le\)x \(\le\)3
f(x) = 3x 3 \(\le\) x \(\le\)10
At x=3
f(3) = (3)² = 9 and f(3) = 3 x 3 = 9.
We observe that f(x) takes unique value at each point in its domain [0,10]. So f is a function.
Now g(x) = x² 0\(\le\)x \(\le\)2
g(x) = 3x 2\(\le\)x \(\le\)10
At x=2
g(2) = (2)² = 4 and g(2) = 3 x 2 = 6
So g(x) does not have unique value at x = 2.
Hence g(x) is not a function.