Free Relations and Functions 03 Practice Test - 12th Grade - Commerce

Question 1

Which one of the following function is not invertible?

f : R \to R, f (x) = 3 x + 1

f : R \to [0, \infty), f (x) = x^{2}

f : R^{+} \to R^{+}, f (x) = \frac{1}{x^{3}}

N o n e o f t h e a b o v e

SOLUTION

Solution : B

The function f(x) = $x^{2}$ , x $ϵ$ R is not one – one because
f(-4)=f (4) = 16
$∴$ It is not invertible

Question 2

The range of the function $f (x) = \frac{2 + x}{2 - x}, x \neq 2$ is

R

R - {- 1}

R - {1}

R - {- 2}

SOLUTION

Solution : B

$y = \frac{2 + x}{2 - x} \Rightarrow 2 y - y x = 2 + x \Rightarrow x (y + 1) = 2 y - 2 \Rightarrow x = \frac{2 y - 2}{y + 1} \Rightarrow f^{- 1} (x) = \frac{2 x - 2}{x + 1}$
$∴$ Range of $f =$ Domain of $f^{- 1} = R - {- 1}$

Question 3

If $2 f (s i n x) + f (c o s x) = x \forall x ϵ$ R then range of f(x) is

[\frac{- π}{3}, \frac{π}{3}]

[\frac{- 2 π}{3}, \frac{π}{3}]

[\frac{- 2 π}{3}, \frac{π}{6}]

[\frac{- π}{6}, \frac{π}{6}]

SOLUTION

Solution : B

Put $x = s i n^{- 1} x$
$2 f (x) + f (\sqrt{1 - x^{2}}) = s i n^{- 1} x \to (1)$
On Putting $x = c o s^{- 1} x$
$\Rightarrow 2 f (\sqrt{1 - x^{2}}) + f (x) = c o s^{- 1} x \to (2)$
Eq. $(1) \times 2 \Rightarrow 4 f (x) + 2 f (\sqrt{1 - x^{2}}) = 2 s i n^{- 1} x \to (3)$
On subtracting Eq. 2 from Eq. 3 we get -
$3 f (x) = 2 s i n^{- 1} x - c o s^{- 1} x$
$f (x) = \frac{2}{3} s i n^{- 1} x - \frac{1}{3} (\frac{π}{2} - s i n^{- 1} x)$
$= s i n^{- 1} x - \frac{π}{6}$
$f_{m a x} = \frac{π}{2} - \frac{π}{6} = \frac{π}{3}, f_{m i n} = - \frac{π}{2} - \frac{π}{6} = \frac{- 4 π}{6} = \frac{- 2 π}{3}$
$= [\frac{- 2 π}{3}, \frac{π}{3}]$

Question 4

The range of $f (x) = t a n^{- 1} (x^{2} + x + a)$ $\forall x ϵ$ R is a subset of $[0, \frac{π}{2})$ then the range of a is -

[- \sqrt{3}, \frac{1}{4}]

(\frac{- π}{2}, \frac{π}{2})

[- \sqrt{3}, - 1]

[\frac{1}{4}, \infty)

SOLUTION

Solution : D

$t a n^{- 1} (x^{2} + x + a) \geq 0 \Rightarrow x^{2} + x + a \geq 0$
$\Rightarrow D \leq 0 \Rightarrow 1 - 4 a \leq 0 \Rightarrow a \geq \frac{1}{4}$ ; D is discriminant of quadratic equation.
$\Rightarrow a ϵ [\frac{1}{4}, \infty)$

Question 5

The range of the function $f (x) = \frac{x + 3}{| x + 3 |}, x \neq - 3$ is

{3, - 3}

R - {- 3}

C. All positive integers

{- 1, 1}

SOLUTION

Solution : D

$f (x) = 1$ when $x + 3 > 0$
$f (x) = - 1$ when $x + 3 < 0$
Range $= {- 1, 1}$

Question 6

Let P = {(x,y) $x^{2} + y^{2} = 1, x, y \in R}$ . Then P is.

Reflexive

Symmetric

Transitive

Anti-symmetric

SOLUTION

Solution : B

Here we can see that the relation is neither reflexive nor transitive but it is symmetric,

because $x^{2} + y^{2} = 1 \Rightarrow y^{2} + x^{2} = 1$

Question 7

R is relation over the set of integers and it is given by (x, y) $ϵ$ R $\Leftrightarrow$ R |x - y| $\leq$ 1. Then, R is

A. Reflexive and transitive

B. reflexive and symmetric

C. Symmetric and transitive

D. an equivalence relation

SOLUTION

Solution : B

As (x,x) $ϵ$ R   $\Rightarrow$ $| x - x | \leq$ 1
$\Rightarrow$ 0 $\leq$ 1 (True),
Thus, reflexive.
As (x,y) $ϵ$ R   $\Rightarrow$ $| x - y | \leq$ 1
$\Rightarrow | y - x | | \leq 1 \Rightarrow$ (y,x) $ϵ$ R,
Thus, symmetric.
Again, (x, y) $ϵ$ R and (y, z) $ϵ$ R
$\Rightarrow | x - y | \leq$ 1   and $| y - z | 1 / \Rightarrow | x - z | \leq$ 1
$∴$ Not transitive

Question 8

Let R be a relation over the set $N \times n$ and it is defined by (a, b) R (c, d) $\Rightarrow$ a+ d = b + c. Then, R is

A. reflexive only

B. symmetric only

C. transitive only

D. an equivalence relation

SOLUTION

Solution : D

(a, b) R (a, b) because a + b = b + a. So, r is reflexive.
(a, b)R (c, d) $\Rightarrow$ a+d = b+c $\Rightarrow$ c+b = d+a
$\Rightarrow$ (c,d) R (a,b)
So, R is symmetric.
(a, b) R (c, d) and (c, d) R (e, f)
$\Rightarrow$                    a + d = b + c, c + f = d + e
       Adding,         a + d + c + f = b + c + d +e
$\Rightarrow$                   a + f = b + e
$\Rightarrow$                       (a, b) R (e, f).
$∴$ R is transitive.

Question 9

The period of the function $| s i n x | + | c o s x |$ is

\frac{π}{2}

2 π

4 π

π

SOLUTION

Solution : A

The smallest of $\frac{π}{2}, 2 π, 4 π, π$ is $\frac{π}{2}$
Let $f (x) = | s i n x | + | c o s x | .$
$∴ f (x + \frac{π}{2}) = ∣ ∣ s i n (x + \frac{π}{2}) ∣ ∣ + ∣ ∣ c o s (x + \frac{π}{2}) ∣ ∣$
$= | c o s x | + | - s i n x |$
$= | c o s x | + | s i n x |$
=f (x)
$∴$ The period of given function is $\frac{π}{2}$

Question 10

Which of the following functions are periodic?

A. f(x) = log x, x > 0

B. f(x) =

e^{x}

, x

ϵ

C. f(x) = x - [x], x

ϵ

D. f(x) = x + [x], x

ϵ

SOLUTION

Solution : C

f(x) = log x, is not periodic.
f(x) = $e^{x}$ , is not periodic.
f(x) = x - [x] = {x}, has period 1
f(x) = x + [x], is not periodic

Free Relations and Functions 03 Practice Test - 12th Grade - Commerce

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

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