Relations And Functions(11th Grade > Mathematics ) Questions and answers for exam Preparation

11th Grade > Mathematics

RELATIONS AND FUNCTIONS MCQs

Total Questions : 30 | Page 3 of 3 pages

Question 21.

$Let f (x) = [x] c o s (\frac{π}{[x + 2]}) where denotes the greatest integer function. Then, the domain of f is$

$x ϵ R, x n o t a n i n t e g e r$
$(- \infty, - 2) \cup [- 1, \infty)$
$x ϵ R, x \neq - 2$
$(- \infty, - 1]$

Discuss Question

Answer: Option B. ->

(- \infty, - 2) \cup [- 1, \infty)

:
B

[x + 2] \neq 0 [x] + 2 \neq 0 [x] \neq - 2 x should not belong to [- 2, - 1) Domain of f is (\infty, - 2) \cup [- 1, \infty) .

Question 22.

$The domain of the function f (x) = l o g_{e} (x^{2} + x + 1) + s i n \sqrt{x - 1} i s$

(-2,1)
$(- 2, \infty)$
$(1, \infty)$
None of these

Discuss Question

Answer: Option C. ->

(1, \infty)

:
C

We must have x - 1 \geq 0. Note that (x^{2} + x + 1) is always positive combining , the domain is [1, \infty) .

Question 23.

The function $f (x) = l o g_{a} x ((a > 0 a n d a \neq 1))$

Odd
Even
Neither even nor odd
Both even and odd

Discuss Question

Answer: Option C. -> Neither even nor odd
:
C

∵

domain is

(0, \infty)

and is not symmetric about the origin

Question 24.

$The function$ $f : R + \to (1, e) defined by f (x) = \frac{X^{2} + e}{X^{2} + 1}$ is

One-one but not onto
Onto but not one-one
Both one-one and onto
Neither one-one nor onto

Discuss Question

Answer: Option C. -> Both one-one and onto
:
C

$f (x) = \frac{x^{2} + e}{x^{2} + 1} f^{^{'}} (x) = \frac{2 x (x^{2} + 1) - 2 x (x^{2} + e)}{(x^{2} + 1)^{2}} = \frac{2 x^{3} + 2 x - 2 x^{3} - 2 e x}{(x^{2} + 1)^{2}} = \frac{2 x - 2 x e}{(x^{2} + 1)^{2}} = \frac{2 x (1 - e)}{(x^{2} + 1)^{2}} < 0 f^{^{'}} (x) < 0, f (x) is decreasing$ $Hence f is one-one function .$ $x \to 0, f (x) \to e$ $x \to \infty, f (x) \to e$ $Hence range = (1, e) = co-domain$

Question 25.

If f(x+2y, x-2y)=xy, then f(x, y) equals

$\frac{x^{2} - y^{2}}{8}$
$\frac{x^{2} - y^{2}}{4}$
$\frac{x^{2} + y^{2}}{4}$
$\frac{x^{2} - y^{2}}{2}$

Discuss Question

Answer: Option A. ->

\frac{x^{2} - y^{2}}{8}

:
A

Let x + 2y = p
x – 2y = q
Solving we got $x = \frac{p + q}{2}$
$y = \frac{p - q}{2} ∴ f (p, q) = \frac{p^{2} - q^{2}}{8} F (x, y) = \frac{x^{2} - y^{2}}{8}$

Question 26.

The entire graphs of the equation $y = x^{2} + k x - x + 9$ is strictly above the x-axis if and only if

k<7
-5<k<7
k>-5
-7<k<5

Discuss Question

Answer: Option B. -> -5<k<7
:
B

$y = x 2 + (k - 1) x + 9 = {(x + \frac{k + 1}{2})}^{2} + 9 - {(\frac{k - 1}{2})}^{2}$
For entire graph to be above x-axis, we should have
$9 - {(\frac{x - 1}{2})}^{2}$ > 0
$\Rightarrow k^{2} - 2 k - 35 < 0 \Rightarrow (k - 7) (k + 5)$ < 0
$\Rightarrow$
i.e., -5<k<7

Question 27.

If the function $f : R \to$ A given by $f (x) = \frac{x^{2}}{x^{2} + 1}$ is a surjection, then A is

R
[0,1]
(0,1]
[0,1)

Discuss Question

Answer: Option D. -> [0,1)
:
D

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

x \to 0, f (x) \to 0

x \to \pm \infty, f (x) \to 1

∴ A ϵ [0, 1)

Question 28.

If $f : [1, \infty) \to [0, \infty) a n d f (x) = \frac{x}{1 + x}$ then f is

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

Discuss Question

Answer: Option A. -> one-one and into
:
A

Given that $f : [0, \infty) \to [0, \infty)$
$s . t . f (x) = \frac{x}{x + 1}$
then $f^{'} (x)$ = $\frac{1 + x - x}{(1 + x^{2})} = \frac{1}{(1 + x)^{2}}$ >0, $\forall x$
$∴$ f is an increasing function $\Rightarrow$ is one-one.
Also $D_{f} = [0, \infty)$
And for range let $\frac{x}{x + 1} = y \Rightarrow x = \frac{y}{y + 1}$

Question 29.

If $f (x) f (\frac{1}{x}) = f (x) + f (\frac{1}{x}) \forall x ϵ R - 0$ where f(x) be a polynomial function and f(5) =126 then f(3)=

Discuss Question

Answer: Option A. -> 28
:
A

f (x) = 1 \pm x^{n} o r, f (5) = 1 \pm 5^{n}

or,

126 = \pm 5^{n}

or,

\pm 5^{n} = 125 \Rightarrow \pm 5^{n} = 5^{3}

n=3

f (3) = 1 + 3^{3} = 28

Question 30.

Let X = {1,2,3,4,5} and Y = {1,3,5,7,9}. Which of the following is/are relations from X to Y

$R_{1} = {(x, y) y = 2 + x, x \in X, y \in Y}$
$R_{2} = {(1, 1), (2, 1), (3, 3), (4, 3), (5, 5)}$
$R_{3} = {(1, 1), (1, 3), (3, 5), (3, 7), (5, 7)}$
$R_{4} = {(1, 3), (2, 5), (2, 4), (7, 9)}$

Discuss Question

Answer: Option A. ->

R_{1} = {(x, y) y = 2 + x, x \in X, y \in Y}

:
A, B, and C

If 'R' is a relation from 'A' to 'B' , then 'R' is defined as {(x,y)| x $\in$ A and y $\in$ B}
In all the ordered pairs in the realtions of R $_{1}$ , R $_{2}$ , R $_{3}$ first component $\in$ X and second component $\in$ Y. So, R $_{1}$ , R $_{2}$ , R $_{3}$ are relations from X to Y.
$R_{4}$ is not a relation from X to Y , because in ordered pain (7,9) the first component 7 $\notin$ X.