Let A -1-2-3 and B -1-4-7-12-15-18- A function f - A - B is defined by f x - x 2-3- ThenA- range of f is 4-7-12B- range of f is 1-4-7-12 C- co-domain of f is 4-7-12D- co-domain of f is 1-4-7-12-15-18

The correct options are A range of f is { 4,7,12 } D co domain of f is { 1,4,7,12 ,15,18} nbsp; nbsp;Domain of f is {1,2,3} Co domain of f is { 1,4,7,12,15, ...

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Standard XII

Mathematics

Codomain

Let A =1,2,3 ...

Question

Let $A = {1, 2, 3}$ and $B = {1, 4, 7, 12, 15, 18}$ . A function $f : A \to B$ is defined by $f (x) = x^{2} + 3$ . Then

range of

f

{4, 7, 12}

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range of

f

{1, 4, 7, 12}

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co-domain of

f

{4, 7, 12}

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co-domain of

f

{1, 4, 7, 12, 15, 18}

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Solution

The correct options are
A range of $f$ is ${4, 7, 12}$
D co-domain of $f$ is ${1, 4, 7, 12, 15, 18}$
Domain of $f$ is ${1, 2, 3}$
Co-domain of $f$ is ${1, 4, 7, 12, 15, 18}$

$f (x) = x^{2} + 3$
$f (1) = 1 + 3 = 4$
$f (2) = 4 + 3 = 7$
$f (3) = 9 + 3 = 12$
$∴$ Range of $f$ is ${4, 7, 12}$

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Similar questions

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A = {1, 2, 3}

and

B = {1, 4, 7, 12, 15, 18}

. A function

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f (x) = x^{2} + 3

. Then

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f (x) = \sqrt{x - 1}

is given by
(a) Domain = (1, ∞), Range = (0, ∞)
(b) Domain = [1, ∞), Range =(0, ∞)
(c) Domain = [1, ∞), Range = [0, ∞)
(d) Domain = [1, ∞), Range = [0, ∞)

Q. Let

f

be a function defined as

f (x) = \frac{3 + 2 sin x}{9 - 4 {sin}^{2} x} .

Then which of the following option(s) is (are) CORRECT?

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Let A={1,2,3} and B={1,4,7,12,15,18}. A function f:A→B is defined by f(x)=x2+3. Then

The correct options are A range of f is {4,7,12} D co-domain of f is {1,4,7,12,15,18} Domain of f is {1,2,3} Co-domain of f is {1,4,7,12,15,18} f(x)=x2+3 f(1)=1+3=4 f(2)=4+3=7 f(3)=9+3=12 ∴ Range of f is {4,7,12}

Let $A = {1, 2, 3}$ and $B = {1, 4, 7, 12, 15, 18}$ . A function $f : A \to B$ is defined by $f (x) = x^{2} + 3$ . Then

The correct options are
A range of $f$ is ${4, 7, 12}$
D co-domain of $f$ is ${1, 4, 7, 12, 15, 18}$
Domain of $f$ is ${1, 2, 3}$
Co-domain of $f$ is ${1, 4, 7, 12, 15, 18}$

$f (x) = x^{2} + 3$
$f (1) = 1 + 3 = 4$
$f (2) = 4 + 3 = 7$
$f (3) = 9 + 3 = 12$
$∴$ Range of $f$ is ${4, 7, 12}$