Free Objective Test 01 Practice Test - 11th and 12th

Question 1

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 2

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 3

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 4

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 5

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

Question 6

Let $f : R \to R, g : R \to R$ , be two functions, such that f(x) =2x – 3, g (x) = $x^{3}$ + 5.
The function $(f o g)^{- 1}$ (x) is equal to

{(\frac{x + 7}{2})}^{\frac{1}{3}}

{(x - \frac{7}{2})}^{\frac{1}{3}}

{(\frac{x - 2}{7})}^{\frac{1}{3}}

{(\frac{x - 7}{2})}^{\frac{1}{3}}

SOLUTION

Solution : D

We have, f : R $\to$ R, g: R $\to$ R defined by f(x) = 2x - 3 and g (x) = $x^{3}$ + 5
It can be checked that f(x) and g(x) are bijective functions
$∴$ fo g is also bijective and (fog) = f(g(x)) = f $(x^{3} + 5) = 2 (x^{3} + 5) - 3 = 2 x^{3} + 7$
$(f o g) (x) = y \Rightarrow 2 x^{3} + 7 = y \Rightarrow x = {(\frac{y - 7}{2})}^{\frac{1}{3}}$
$∴ (f o g)^{- 1} (x) = {(\frac{x - 7}{2})}^{\frac{1}{3}}, x ϵ R$
$∴$ The correct answer is (d).

Question 7

If f: R $\to$ R ,g : R $\to$ R and h: R $\to$ R are such that $f (x) = x^{2}$ , $g (x) = t a n x$ and $h (x) = l o g x$ , then the value of $(h (g (f (x))))$
if $x = \sqrt{\frac{π}{4}}$ will be

A. 0

B. 1

C. -1

π

SOLUTION

Solution : A

$(h o (g o f)) (x) = h {g {f (x)}}$
$= h {t a n x^{2}} = l o g {t a n x^{2}}$
$∴ A t x = \sqrt{\frac{π}{4}} \Rightarrow (h o (g o f)) (x) = l o g t a n \frac{π}{4}$
= log 1 = 0

Question 8

If the function $f : [1, \infty) \to [1, \infty)$ is defined by $f (x) = 2^{x (x - 1)}$ , then $f^{- 1} (x)$ is

{(\frac{1}{2})}^{x (x - 1)}

\frac{1}{2} (1 + \sqrt{1 + 4 l o g_{2} x})

(\frac{1}{2}) (1 - \sqrt{1 + 4 l o g_{2} x})

N o n e o f t h e s e

SOLUTION

Solution : B

$f (x) = y \Rightarrow 2^{x (x - 1)} = y \Rightarrow x (x - 1) l o g_{2} 2 = l o g_{2} y$
$\Rightarrow x (x - 1) = l o g_{2} y \Rightarrow x^{2} - x - l o g_{2} y = 0$
$\Rightarrow x = \frac{1 \pm \sqrt{1 + 4 l o g_{2} y}}{2}$
$∴ x = \frac{1 + \sqrt{1 + 4 l o g_{2} y}}{2}$
$∴ f^{- 1} (x) = \frac{1}{2} (1 + \sqrt{1 + 4 l o g_{2} x})$
$∴$ The correct answer is (b).

Question 9

$f : R \times R \to R$ such that f(x + iy) = $\sqrt{x^{2} + y^{2}}$ . Then, f is

A. many – one and into

B. one-one and onto

C. many – one and onto

D. one – one and into

SOLUTION

Solution : A

Since, f (x + iy) = f (x - iy) $∴$ f is many – one
Also, Range $= R^{+}$ and codomain = R.
$∴$ Range $⊏$ codomain $\Rightarrow$ f is into.
Hence, f is many – one into.

Question 10

With reference to a universal set, the inclusion of a subset in another, is relation, which is

A. Symmetric only

B. Equivalence relation

C. Reflexive only

D. None of these

SOLUTION

Solution : D

Since $A \subseteq A$ .
$∴$ Relation $^{'} \subseteq^{'}$ is reflexive.
Since $A \subseteq B, B \subseteq C \Rightarrow A \subseteq C$
$∴$ Relation $^{'} \subseteq^{'}$ is transitive.
But If $A \subseteq B$ , Doesn't imply $B \subseteq A$ ,
$∴$ Relation is not symmetric

Question 11

If $f : R \to R$ and $g : R \to R$ are given by f(x) = |x| and g(x) = [x], then $g (f (x)) \leq f (g (x)$ is true for -

Z \cup (- \infty, 0)

(- \infty, 0)

Z

R

SOLUTION

Solution : D

$g (f (x)) \leq f (g (x)) \Rightarrow g (| x |) \leq f [x] \Rightarrow [| x |] \leq | [x] |$
This is true for $x ϵ R .$ .

Question 12

$f (x) = \frac{x^{2} - 3 x + 4}{x^{2} + 3 x + 4}$ the range of f(x) is

[0, \frac{1}{7}]

(- \infty, \frac{1}{7}) \cup (7, \infty)

(- \infty, 7)

[\frac{1}{7}, 7]

SOLUTION

Solution : D

$y = \frac{x^{2} - 3 x + 4}{x^{2} + 3 x + 4}$
$y x^{2} + 3 x y + 4 y = x^{2} - 3 x + 4$
$x^{2} (y - 1) + 3 x (y + 1) + 4 (y - 1) = 0$
D $\geq 0 \Rightarrow 9 (y + 1)^{2} - 4.4 (y - 1)^{2} \geq 0$
$(3 (y + 1) - 4 (y - 1))$ $(3 (y + 1) + 4 (y - 1)) \geq 0$
$(- y + 7) (7 y - 1) \geq 0$
$(y - 7) (y - \frac{1}{7}) \leq 0$
$\frac{1}{7} \leq y \leq 7$

Question 13

The range of the function $f (x) = c o s^{2} \frac{x}{4} + s i n \frac{x}{4}, x ϵ R$ is

[0, \frac{5}{4}]

[1, \frac{5}{4}]

(- 1, \frac{5}{4})

[- 1, \frac{5}{4}]

SOLUTION

Solution : D

$f (x) = 1 - s i n^{2} \frac{x}{4} + s i n \frac{x}{4} = - {s i n^{2} \frac{x}{4} - s i n \frac{x}{4}} + 1 = - {{(s i n \frac{x}{4} - \frac{1}{2})}^{2} - \frac{1}{4}} + 1$
$= \frac{5}{4} - {(s i n \frac{x}{4} - \frac{1}{2})}^{2}$
Maximun $f (x) = \frac{5}{4}$
Minimum $f (x) = \frac{5}{4} - {(- 1 - \frac{1}{2})}^{2} = \frac{5}{4} - \frac{9}{4} = - 1$
Range of $f (x) = [- 1, \frac{5}{4}]$

Question 14

Let $f (x) = \frac{x - [x]}{1 + x - [x]}, x ϵ R$ , where [ x] denotes the greatest integer function. Then, the range of f is

(0, 1)

[0, \frac{1}{2})

[0, 1]

[0, \frac{1}{2}]

SOLUTION

Solution : B

The graph of $y = x - [x]$ is as shown below

When x is an integer, $x - [x] = 0$
Hence, f(x) = 0 when x is an integer
$x \to [x]$ as x tends to an integer.
Let X = $x - [x]$
So, $f (x) = \frac{X}{1 + X}, X ϵ [0, 1)$
As $X \to 1, \frac{X}{1 + X} \to \frac{1}{2}$
Hence, the range of f(x) is $[0, \frac{1}{2})$ .

Question 15

Range of $f (x) = \frac{t a n (π [x^{2} - x])}{1 + s i n (c o s x)}$ is (where [x] denotes the greatest integer function)

(- \infty, \infty) \sim [0, t a n 1]

(- \infty, \infty) \sim [t a n 2, 0]

[t a n 2, t a n 1]

{0}

SOLUTION

Solution : D

$f (x) = \frac{t a n (π [x^{2} - x])}{1 + s i n (c o s x)} = {0}$ because of $[x^{2} - x]$ is integer

Free Objective Test 01 Practice Test - 11th and 12th

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

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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