The domain of R = {1, 2, 3, 4} = A, The co-domain is B, Range of R is given by {3, 6, 10, 9}.

{(−5, −5), (−5, 5), (5, −5), (5, 5)}

$R = {(x, y); x^2 + y^2 = 13^2; x \in P, y \in Q}$

Domain of R = {−1, 0, 1, 2, 3}, Range of R = $\bigg\{ 6, 4, 3, \large\frac{12}{5}$$,2 \bigg\}$ or $

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x = 2 and y = 1

x = 3, y = -1

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

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

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

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

(i) f is a relation from A to B(ii) f is not a function

(i) not true (ii) not true (iii) not true

a and b are 2 and –1.

(f + g) (x) = 3x – 2, (f – g) (x) = –x + 4,$ \bigg( \large\frac{f}{g} \bigg)$$ (x) \large\frac{x+1}

Range of f = [0, 1 ]

Domain of f = R, Range of f is the set of all non-negative real numbers.

Here, f(x) = 1 – x, x < 0, this gives f(– 4) = 1 – (– 4)= 5; f(– 3) =1 – (– 3) = 4, f(– 2) = 1 –

Since $x^2 –5x + 4 = (x – 4) (x –1)$, the function f is defined for all real numbers except at x = 4

Since f is a linear function, f (x) = mx + c. Also, since (1, 1), (0, – 1) $\in $ R, f (1) = m + c =

(i) Since, a – a = 0 $ \in $ Z, if follows that (a, a) $ \in $ R. (ii) (a,b) $ \in $ R implies

Here f(0) = 10, f(1) = 11, f(2) = 12, ..., f(10) = 20, etc., and f(–1) = 9, f(–2) = 8, ..., f(–10)

0 ≤ y ≤ 5Hence (B) is the correct answer.

$ \bigg[ \large\frac{1}{4}$$, \large\frac{1}{2} \bigg]$Hence (C) is the correct answer.

f: R $ \rightarrow $ R defined by \[f(x) = \left\{ \begin{array}{l l} \large\frac{|x|}{x}, &...

hence (B) is the correct answer.

$9x^2 – 6x + 1$hence (B) is the correct answer.

$f_3 = \bigg\{ (x,y) : x^2+y^2=1$ and $ x,y \in R \bigg\}$Hence (B) is the correct answer.

Hence (D) is the correct answer.

Relations $R_2$ and $R_4$ are functions.Hence (B) is the correct answer.

[1, 2]Hence (C) is the correct answer.

$ \bigg[ \large\frac{-41}{8} \infty \bigg)$Hence (B) is the correct answer.

R − (−1, 0]hence (B) is the correct answer.

Hence (C) is the correct answer.

$f(x) = 2x^3− 5, x \in R$Hence (A) is the correct answer.

\[f(x) = \left\{ \begin{array}{l l} -x, & \quad x < 0 \\ [x] & \quad x \g...

$ (f-g)(y) = \large\frac{4-y}{(y-1)(y+2)}$Hence (D) is the correct answer.

\[(-2f)(x) = \left\{ \begin{array}{l l} -2, & \quad if \: x > 0 \\ 0, & \qua...

R − {−7, 3}Hence (C) is the correct answer.

28Hence (D) is the correct answer.

\[f(g)(x) = \left\{ \begin{array}{l l} \large\frac{1}{x}x \: if \: x > 0 \\ -\large\...

\[\bigg(\large\frac{f}{g} \bigg) (x) = \left\{ \begin{array}{l l} 1-\large\frac{2}{x}, &...

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Dom (f) = R{0} and range (f) R – {0}https://clay6.com/mpaimg/18.jpg

Dom (f) = R and range (f) = R.https://clay6.com/mpaimg/17.jpg

Dom (f) = R and range (f) = {x $ \in $ R : x $ \geq $ 0}.https://clay6.com/mpaimg/16.jpg

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