sets relations and functions jee mains questions | JEE Main Mathematics Sets Relations And Functions

Q 31 :

The greatest integer function $f : R \to R$ , given by $f (x) = [x]$ is:

one-one
onto
both one-one and onto
neither one-one nor onto

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(4)

Ans. neither one-one nor onto

Explanation:

We know

[2.2] = 2

[2.5] = 2

$∵$ [2.2] = [2.5]

but 2.2 ≠ 2.5

$\Rightarrow f$ is not one-one function.

Now, consider $0.6 \in R$

It is known that $f (x) = [x]$ is always an integer. Thus, there does not exist any element $x \in R$ such that $f (x) = 0.6$ .

$∴$ $f$ is not onto.

Hence, greatest integer function is neither one-one nor onto.

Q 32 :

Let $f : R \to R$ be defined by $f (x) = \frac{x^{2} - 8}{x^{2} + 2},$ then $f$ is:

one-one but not onto
one-one and onto
onto but not one-one
Neither one-one nor onto

1

2

3

4

(4)

Ans. Neither one-one nor onto

Explanation:

Given, $f (x) = \frac{x^{2} - 8}{x^{2} + 2}$

$f (3) = \frac{9 - 8}{9 + 2} = \frac{1}{11}$

$f (- 3) = \frac{9 - 8}{9 + 2} = \frac{1}{11}$

Here, $f (3) = f (- 3)$ but $3 \neq - 3$ .

$\Rightarrow f$ is not one-one.

Now, $f (x) = \frac{x^{2} - 8}{x^{2} + 2} = 1 - \frac{10}{x^{2} + 2}$

So, minimum value of $f (x) = 1 - 5 = - 4$

$∴$ It is not onto.

Hence it is neither one-one nor onto.

Q 33 :

Let $f : R \to R$ be defined as $f (x) = x^{4}$ . Choose the correct answer:

$f$ is one-one onto
$f$ is many-one onto
$f$ is one-one but not onto
$f$ is neither one-one nor onto

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(4)

Ans. $f$ is neither one-one nor onto

Explanation:

$f : R \to R$ is defined as $f (x) = x^{4}$

Let $x, y \in R$ such that

$f (x) = f (y)$

$\Rightarrow$ $x^{4} = y^{4}$

$\Rightarrow$ $x^{4} - y^{4} = 0$

$\Rightarrow$ $(x^{2} + y^{2}) (x^{2} - y^{2}) = 0$

$\Rightarrow$ $x^{2} = y^{2}$

$\Rightarrow$ $x = \pm y$

$∴$ $f (x) = f (y)$ does not imply that $x = y$

For instance,

$f (1) = f (- 1) = 1$

Here $f (1) = f (- 1)$ but $- 1 \neq 1$ . Hence, $f$ is not one-one.

To check onto,

$f (x) = x^{4}$

Let $f (x) = y$ such that $y \in R$

$\Rightarrow$ $x^{4} = y$

$\Rightarrow$ $x = \pm y^{1 / 4}$

Note that $y$ is a real number, but it can also be negative.

For example, put $y = - 3$

$\Rightarrow$ $x = {(\pm 3)}^{1 / 4}$

$\Rightarrow$ $x = \pm {(- \sqrt{3})}^{1 / 2}$

which is not possible, as root of a negative number is not a real i.e., $f (x) = x^{4}$ is not a onto function.

Q 34 :

If $f (x + 1) = x^{2} - 3 x + 2,$ then $f (x)$ is equal to:

$x^{2} - 5 x - 6$
$x^{2} + 5 x - 6$
$x^{2} + 5 x + 6$
$x^{2} - 5 x + 6$

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(4)

Ans. $x^{2} - 5 x + 6$

Explanation:

$f (x + 1) = x^{2} - 3 x + 2$

This function is valid for all real values of $x$

Hence, put $x - 1$ in place of $x$ ,

$f (x) = {(x - 1)}^{2} - 3 (x - 1) + 2$

$= x^{2} + 1 - 2 x - 3 x + 3 + 2$

$= x^{2} - 5 x + 6$

Q 35 :

Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let $f$ = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Then, $f$ is:

one-one but not onto
one-one and onto
onto but not onto
neither one-one nor onto

1

2

3

4

(1)

Ans. one-one but not onto

Explanation:

Here, $f : A \to B$ is defined as {(1, 4),(2, 5),(3, 6)}.

Since, the image of distinct elements of A under $f$ are distinct as:

$f (1) = 4,$ $f (2) = 5$ and $f (3) = 6$

From above it is evident that

$x_{1} \neq x_{2}$ and $f (x_{1}) \neq f (x_{2})$

$∴$ $f : A \to B is one-one.$

Here, Co-domain of $f \neq$ Range of $f$

Hence, $f$ is not onto.

Topic Question Set

Q 31 :

The greatest integer function $f : R \to R$ , given by $f (x) = [x]$ is:

Q 32 :

Let $f : R \to R$ be defined by $f (x) = \frac{x^{2} - 8}{x^{2} + 2},$ then $f$ is:

Q 33 :

Let $f : R \to R$ be defined as $f (x) = x^{4}$ . Choose the correct answer:

Q 34 :

If $f (x + 1) = x^{2} - 3 x + 2,$ then $f (x)$ is equal to:

Q 35 :

Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let $f$ = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Then, $f$ is:

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Topic Question Set

Q 31 : The greatest integer function f:R→R, given by f(x)=[x] is:

Q 32 : Let f:R→R be defined by f(x)=x2-8x2+2, then f is:

Q 33 : Let f:R→R be defined as f(x)=x4. Choose the correct answer:

Q 34 : If f(x+1)=x2-3x+2, then f(x) is equal to:

Q 35 : Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Then, f is:

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Q 31 :

The greatest integer function $f : R \to R$ , given by $f (x) = [x]$ is:

Q 32 :

Let $f : R \to R$ be defined by $f (x) = \frac{x^{2} - 8}{x^{2} + 2},$ then $f$ is:

Q 33 :

Let $f : R \to R$ be defined as $f (x) = x^{4}$ . Choose the correct answer:

Q 34 :

If $f (x + 1) = x^{2} - 3 x + 2,$ then $f (x)$ is equal to:

Q 35 :

Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let $f$ = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Then, $f$ is: