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MATHS MCQ

RELATIONS AND FUNCTIONS · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 165 questions.

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165 questions · auto-graded multiple-choice test.

MCQ 11 Mark
If R is a relation on a finite set having n elements, then the number of relations on A is:
  • A
    $2^{\text{n}}$
  • ✓
    $2^{\text{n}^2}$
  • C
    $\text{n}^2$
  • D
    $\text{n}^\text{n}$
Answer
Correct option: B.
$2^{\text{n}^2}$
  1. $2^{\text{n}^2}$
Solution:
Given, A finite set with n elements
Its Cartesian product with itself will have $\text{n}^2$ elements.
$\therefore$ Number of relations on $\text{A}=2^{\text{n}^2}$
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MCQ 21 Mark
Let a relation R be defined by R = {(4, 5), (1, 4), (4, 6), (7, 6), (3, 7)}, then ROR is equal to:
  • ✓
    {(1, 5), (1, 6), (3, 6)}
  • B
    {(1, 4), (1, 5), (3, 6)}
  • C
    {(1, 5), (1, 6), (3, 7)}
  • D
    {(1, 4), (1, 5), (3, 7)}
Answer
Correct option: A.
{(1, 5), (1, 6), (3, 6)}
  1. {(1, 5), (1, 6), (3, 6)}
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MCQ 31 Mark
Which function is shown in graph?
  • ✓
    Constant
  • B
    Modulus
  • C
    Identity
  • D
    Signum function
Answer
Correct option: A.
Constant
{(-1, 1), (1, 1), (-2, 2), (2, 2), (-3, 3), (3, 3), ……}. This function involves relation {(x, y), y = |x|} which is involved in modulus function.
So, above function is modulus function.
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MCQ 41 Mark
Which of the following is not a function?
  • A
    {(1, 2), (2, 4), (3, 6)}
  • B
    {(-1, 1), (-2, 4), (2, 4)}
  • ✓
    {(1, 2), (1, 4), (2, 5), (3, 8)}
  • D
    {(1, 1), (2, 2), (3, 3)}
Answer
Correct option: C.
{(1, 2), (1, 4), (2, 5), (3, 8)}
A relation from a set A to a set B is said to be a function if every element of set A has one and one image in set B.
In {(1, 2), (1, 4), (2, 5), (3, 8)}, since element 1 has two images 2 and 4 which is not possible in a function so, it is not a function. Rest all have
one and only one image so they can be called a function.
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MCQ 51 Mark
If n is the smallest natural number such that n + 2n + 3n + …. + 99n is a perfect square, then the number of digits in square of n is:
  • A
    1
  • B
    2
  • ✓
    3
  • D
    4
Answer
Correct option: C.
3
  1. 3
Solution:
Given thatn + 2n + 3n + …. + 99n
$ =\text{ n} × (1 + 2 + 3 + …….. + 99)$
$=\frac{\text{n} \times99 \times100}{2}$
$=\text{n} × 99 × 50$
$= \text{n} × 9 × 11 × 2 × 25$
To make it perfect square we need $2 \times 11$
So $n=2 \times 11=22$.
Now $n^2=22 \times 22=484$
So, the number of digit in $n^2=3$.
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MCQ 61 Mark
The domain of definition of the function $\text{f(x)}=\sqrt{\text{x}-1}+\sqrt{3-\text{x}}$ is:
  • A
    $[1,\infty)$
  • B
    $\big(-\infty,3\big)$
  • C
    $(1,3)$
  • ✓
    $\big[1,3\big]$
Answer
Correct option: D.
$\big[1,3\big]$
$\text{f(x)}=\sqrt{\text{x}-1}+\sqrt{3-\text{x}}$
For f(x) to be defined,
$(\text{x}-1)\geq0$
$\Rightarrow\text{x}\geq1\ ...(\text{i})$
and $(3-\text{x})\geq0$
From (i) and (ii),
$\text{x}\in[1,3]$
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MCQ 71 Mark
Let R be a relation from a set A to a set B, then:
  • A
    $\text{ R} = \text{A}∪\text{B}$
  • B
    $\text{ R} = \text{A}\cap\text{B}$
  • ✓
    $ ​​\text{R} \subseteq \text{A}\times{\text{B}}$
  • D
    $ ​​\text{R} \subseteq \text{B}\times{\text{A}}$
Answer
Correct option: C.
$ ​​\text{R} \subseteq \text{A}\times{\text{B}}$
  1. $ ​​\text{R} \subseteq \text{A}\times{\text{B}}$
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MCQ 81 Mark
If $\text{x}\neq1$ and $\text{f(x)}=\frac{\text{x}+1}{\text{x}-1}$ is a real function, then $\text{f}(\text{f}(\text{f(2)}))$ is:
  • A
    1
  • B
    2
  • ✓
    3
  • D
    4
Answer
Correct option: C.
3
$\text{f(x)}=\frac{\text{x}+1}{\text{x}-1}$
$\text{f}(\text{f}(\text{f(2)}))$
$=\text{f}\Big(\text{f}\Big(\frac{2+1}{2-1}\Big)\Big)$
$=\text{f}(\text{f}(3))$
$=\text{f}\Big(\frac{3+1}{3-1}\Big)$
$=\text{f}(2)=3$
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MCQ 91 Mark
If (a, b) = (x, y) then___________.
  • A
    a = x
  • B
    a = y
  • ✓
    a = y and b = x
  • D
    a = x and b = y
Answer
Correct option: C.
a = y and b = x
Two ordered pairs are said to be equal if and only if their corresponding elements are equal i.e. a = x and b = y.
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MCQ 101 Mark
If $f(x)=3 x^4-5 x^2+9$, then value of $f(x-1)$ is:
  • A
    $3 x^4+12 x+13 x+2 x+7$
  • B
    $3 x^4-12 x-13 x-2 x-7$
  • ✓
    $3 x^4-12 x+13 x-2 x+7$
  • D
    $3 x^4-12 x-13 x+2 x+7$
Answer
Correct option: C.
$3 x^4-12 x+13 x-2 x+7$
  1. $3 x^4-12 x+13 x-2 x+7$
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MCQ 111 Mark
The function f(x) = x - [x] has period of:
  • A
    1
  • ✓
    2
  • C
    3
Answer
Correct option: B.
2
Let T is a positive real number.
Let f(x) is periodic with period T.
Now, f(x + T) = f(x), for all $\text{x} \in \text{R}$
⇒ x + T - [x + T] = x - [x], for all $ \text{x} \in \text{R}$
⇒ [x + T] - [x] = T, for all $ \text{x} \in \text{R}$
Thus, there exist T > 0 such that f(x + T) = f(x) for all $ \text{x} \in \text{R}$
Now, the smallest value of T satisfying f(x + T) = f(x) for all $ \text{x} \in \text{R}$ is 1
So, f(x) = x - [x] has period 1
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MCQ 121 Mark
If f(x) = (x - 1)(x - 3)(x - 4)(x - 6) + 19 for all real value of x is:
  • ✓
    positive
  • B
    negative
  • C
    zero
  • D
    none of these
Answer
Correct option: A.
positive
  1. positive
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MCQ 131 Mark
If $f(x)=e x$ and $g(x)=$ loge $x$ then the value of fog(1) is:
  • A
    0
  • ✓
    1
  • C
    -1
  • D
    None of these
Answer
Correct option: B.
1
  1. 1
Solution:
Given, $f(x)=e^x$
and $g(x)=\log x$
$fog(x)=f(g(x))$
$=f(\log x)$
$=e^{\log x}$
$=\mathrm{x}$
So, fog $(1)=1$
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MCQ 141 Mark
If A × B = {(1, a), (1, b), (1, c), (2, a), (2, b), (2, c)} then find set A:
  • A
    {1}
  • ✓
    {1, 2}
  • C
    {1, a}
  • D
    {a, b, c}
Answer
Correct option: B.
{1, 2}
  1. {1, 2}
Solution:
In each ordered pair of A × B, first element belongs to set A and second element belongs to set B.
1, $ 2 ∈\text{A} $ so, $ \text{A} = {1, 2}.$
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MCQ 151 Mark
The domain of definition of $\text{f(x)}=\sqrt{\frac{\text{x}+3}{(2-\text{x})(\text{x}-5)}}$ is:
  • ✓
    $(-\infty,-3]\cup(2,5)$
  • B
    $(-\infty,-3]\cup(2,5)$
  • C
    $(-\infty,-3]\cup[2,5]$
  • D
    None of these.
Answer
Correct option: A.
$(-\infty,-3]\cup(2,5)$
$\text{f(x)}=\sqrt{\frac{\text{x}+3}{(2-\text{x})(\text{x}-5)}}$
For f(x) to be defined,
$(2-\text{x})(\text{x}-5)\neq0$
$\Rightarrow\text{x}\neq2,5\ ...(\text{i})$
Also, $\frac{(\text{x}+3)}{(2-\text{x})(\text{x}-5)}\geq0$
$\Rightarrow\frac{(\text{x}+3)(2-\text{x})(\text{x}-5)}{(2-\text{x})^2(\text{x}-5)^2}\geq0$
$\Rightarrow(\text{x}+3)(\text{x}-2)(\text{x}-5)\leq0$
$\Rightarrow\text{x}\in\big(-\infty,-3\big]\cap(2,5)\ ...(\text{ii})$
From (i) and (ii)
$\text{x}\in\big(-\infty,-3\big]\cup(2,5)$
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MCQ 161 Mark
Choose the correct answers: If f(x) = ax + b, where a and b are integers, f(–1) = –5 and f(3) = 3, then a and b are equal to.
  • A
    a = –3, b = –1
  • ✓
    a = 2, b = –3
  • C
    a = 0, b = 2
  • D
    a = 2, b = 3
Answer
Correct option: B.
a = 2, b = –3
Given that: f(x) = ax + b
⇒ f(-1) = a(-1) + b
⇒ -5 = -a + b
⇒ a - b = 5 ...........(i)
f(3) = 3a + b
⇒ 3 = 3a + b
⇒ 3a + b = 3 ........(ii)
On solving eqn. (i) and (ii), We get a = 2, b = -3
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MCQ 171 Mark
If $ \text{aN} = \frac{\text{ax}}{\text{x}\in\text{N}}$ and $\text{bN}\cap\text{cN}=\text{d}\text{N}$ Where $ \text{b}, \text{c }\in\text{ N}$
  • ✓
    d = bc
  • B
    c = bd
  • C
    b = cd
  • D
    None
Answer
Correct option: A.
d = bc
  1. d = bc
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MCQ 181 Mark
The range of the function $\text{f(x)}=\frac{\text{x}^2-\text{x}}{\text{x}^2+2\text{x}}$ is:
  • A
    $\text{R}$
  • B
    $\text{R}-\{1\}$
  • ✓
    $\text{R}-\Big\{\frac{1}{2},1\Big\}$
  • D
    None of these.
Answer
Correct option: C.
$\text{R}-\Big\{\frac{1}{2},1\Big\}$
$\text{f(x)}=\frac{\text{x}^2-\text{x}}{\text{x}^2+2\text{x}}$
Let, $\text{y}=\frac{\text{x}^2-\text{x}}{\text{x}^2+2\text{x}}$ $\big[\text{Also},\text{ x}\neq0\big]$
$\Rightarrow\text{y}=\frac{\text{x}(\text{x}-1)}{\text{x}(\text{x}+2)}$
$\Rightarrow\text{y}=\frac{(\text{x}-1)}{(\text{x}+2)}$
$\Rightarrow\text{xy}+2\text{y}=\text{x}-1$
$\Rightarrow\text{x}=\frac{2\text{y}+1}{1-\text{y}}$
Here, $1-\text{y}\neq0$
Or, $\text{y}\neq1$
Also, $\text{x}\neq0$
$\Rightarrow\frac{2\text{y}+1}{1-\text{y}}\neq0$
$\Rightarrow\text{y}\neq-\frac{1}{2}$
Thus, range $\text{(f)}=\text{R}-\Big\{-\frac{1}{2},1\Big\}$
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MCQ 191 Mark
The function $ \text{f}(\text{x}) = \sin \Big(\frac{\pi‎\text{x}}{2}\Big) +\cos \Big(\frac{\pi‎\text{x}}{2}\Big)$ is periodic with period:
  • ✓
    4
  • B
    6
  • C
    12
  • D
    24
Answer
Correct option: A.
4
Period of $\sin \Big(\frac{\pi‎\text{x}}{2}\Big) = 2\pi\Big(\frac{\pi}{2}\Big) = 4$
Period of $\cos \Big(\frac{\pi‎\text{x}}{2}\Big) = 2\pi\Big(\frac{\pi}{2}\Big) = 4$
So, period of f(x) = LCM (4, 4) = 4
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MCQ 201 Mark
If $R$ is a relation from a finite set $A$ having $m$ elements of a finite set $B$ having $n$ elements, then the number of relations from $A$ to $B$ is:
  • ✓
    $2^{\mathrm{mn}}$
  • B
    $2^{m n}-1$
  • C
    $2 mn$
  • D
    $m^n$
Answer
Correct option: A.
$2^{\mathrm{mn}}$
  1. $2^{\mathrm{mn}}$
Solution:
Given, n(A) = m
n(B) = n
$\therefore$ n(A × B) = mn
Then, the number of relations from A to is $2^{\mathrm{mn}}$​​​​​​​
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MCQ 211 Mark
Domain of $\sqrt{\text{a}^{2} -\text{x}^{2}} (\text{a} > 0)$ is:
  • A
    (-a, a)
  • ✓
    (-a, a)
  • C
    (0, a)
  • D
    (-a, 0)
Answer
Correct option: B.
(-a, a)
  1. (-a, a)
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MCQ 221 Mark
Let A = {1, 2, 3}. The total number of distinct relations that can be defined over A, is:
  • ✓
    29
  • B
    6
  • C
    8
  • D
    None of these
Answer
Correct option: A.
29
  1. 29
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MCQ 231 Mark
Two functions f and g are said to be equal if:
  • A
    The domain of f = the domain of g
  • B
    The co-domain of f = the co-domain of g
  • C
    F(x) = g(x) for all x
  • ✓
    All of above
Answer
Correct option: D.
All of above
Two functions f and g are said to be equal if
  1. The domain of f = the domain of g
  2. The co-domain of f = the co-domain of g
  3. F(x) = g(x) for all x
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MCQ 241 Mark
If the function$ \text{f}(\text{x})=\frac{\text{ax} -{\text{x}}}{2,(\text{a>2})}$, then $ \text{f}(\text{x + y}) + \text{f}(\text{x – y})$ is equal to:
  • ✓
    $ 2\text{f}(\text{x} ) \text{ f}(\text{y})$
  • B
    $ \text{f}(\text{x} ) \text{ f}(\text{y})$
  • C
    $ \frac{\text{f} (\text{x})}{\text{f}(\text{y})}$
  • D
    $ \text{None of these} $
Answer
Correct option: A.
$ 2\text{f}(\text{x} ) \text{ f}(\text{y})$
  1. $ 2\text{f}(\text{x} ) \text{ f}(\text{y})$
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MCQ 251 Mark
If set A has 2 elements and set B has 4 elements then how many relations are possible?
  • A
    32
  • B
    128
  • ✓
    256
  • D
    64
Answer
Correct option: C.
256
  1. 256
Solution:
We know, $\mathrm{A} \times \mathrm{B}$ has $2 \times 4$ i.e. 8 elements.
Number of subsets of $A \times B$ is $2^8$ i.e. 256 .
A relation is a subset of cartesian product so,
number of possible relations are 256 .
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MCQ 261 Mark
If the set A has p elements, B has q elements, then the number of elements in A × B is:
  • A
    p + q
  • B
    p + q + 1
  • ✓
    pq
  • D
    $p^2$
Answer
Correct option: C.
pq
  1. pq
Solution:
n(A × B) = n(A) × n(B)
n(A × B) = p × q = pq
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MCQ 271 Mark
If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x)=2 x+3$ and $g(x)=x^2+7$, then the values of $x$ such that $g(f(x))=8$ are:
  • A
    1, 2
  • B
    -1, 2
  • ✓
    -1, -2
  • D
    1, -2
Answer
Correct option: C.
-1, -2
  1. -1, -2
Solution:
$f(x)=2 x+3 \text { and } g(x)=x^2+7$
$g(f(x))=8$
$\Rightarrow(f(x))^2+7=8$
$\Rightarrow(2 x+3)^2+7=8$
$\Rightarrow x^2+3 x+2=0$
$\Rightarrow(x+2)(x+1)=0$
$\Rightarrow x=-1,-2$
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MCQ 281 Mark
Which of the following relation is a function?
  • A
    (a, b) (b, e) (c, e) (b, x)
  • B
    (a, d) (a, m) (b, e) (a, b)
  • ✓
    (a, b) (b, e) (c, e) (b, x)
  • D
    (a, d) (b, m) (b, y) (d, x)
Answer
Correct option: C.
(a, b) (b, e) (c, e) (b, x)
  1. (a, b) (b, e) (c, e) (b, x)
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MCQ 291 Mark
The relation R defined on the set of natural numbers as (a, b) : a differs from b by 3 is given:
  • A
    ((1, 4), (2, 5), (3, 6),.....)
  • ✓
    ((4, 1), (5, 2), (6, 3),.....)
  • C
    ((1, 3), (2, 6), (3, 9),.....)
  • D
    none of these
Answer
Correct option: B.
((4, 1), (5, 2), (6, 3),.....)
  1. ((4, 1), (5, 2), (6, 3),.....)
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MCQ 301 Mark
Let R be a relation from a set A to a set B, then:
  • A
    $\text{R}=\text{A}\cup\text{B}$
  • B
    $\text{R}=\text{A}\cap\text{B}$
  • ✓
    $\text{R}\subseteq\text{A}\times\text{B}$
  • D
    $\text{R}\subseteq\text{B}\times\text{A}$
Answer
Correct option: C.
$\text{R}\subseteq\text{A}\times\text{B}$
If R is a relation from set A to set B, then R is always a subset of A × B.
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MCQ 311 Mark
Let A = {1, 2} B = {1, 2, 3, 4} C = {5, 6} and D = {5, 6, 7, 8} Following statements are given below:
  1. $ \text{A }\times ({\text{B} \cap\text{C})} = (\text{A}\times \text{ B}) ∩ (\text{A}\times \text{ C})$
  2. A, C is a subset of $\text{ B }\times\text{ D}$
Which of the following statment is correct?
  • A
    Only I
  • B
    Only II
  • ✓
    Both I and II
  • D
    None of these
Answer
Correct option: C.
Both I and II
  1. Both I and II
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MCQ 321 Mark
If $\text{R}=\{(\text{x, y}):\text{x, y}\in\text{Z},\text{ x}^2+\text{y}^2\leq4\}$ is a relation on Z, then the domain of R is:
  • A
    {0, 1, 2}
  • B
    {0, -1, -2}
  • ✓
    {-2, -1, 0, 1, 2}
  • D
    none of these.
Answer
Correct option: C.
{-2, -1, 0, 1, 2}
$\text{R}=\{(\text{x, y}):\text{x, y}\in\text{Z},\text{ x}^2+\text{y}^2\leq4\}$
We know that,
$(-2)^2+0^2\leq4$
$\Rightarrow(2)^2+0^2\leq4$
$\Rightarrow(-1)^2+0^2\leq4$
$\Rightarrow(1)^2+0^2\leq4$
$\Rightarrow(-1)^2+(1)^2\leq4$
$\Rightarrow0^2+0^2\leq4$
$\Rightarrow(1)^2+(1)^2\leq4$
$\Rightarrow(-1)^2+(-1)^2\leq4$
Hence, domain(R) = {-2, -1, 0, 1, 2}
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MCQ 331 Mark
If A × B = {(1, a), (1, b), (1, c), (2, a), (2, b), (2, c)} then find set B:
  • A
    {1}
  • B
    {1, 2}
  • C
    {1, a}
  • ✓
    {a, b, c}
Answer
Correct option: D.
{a, b, c}
  1. {a, b, c}
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MCQ 341 Mark
If the function f : R → R be given by $ \text{f}(\text{x}) = \text{x}^2 + 2$ and g : R → R is given by $\text{g}(\text{x})=\frac{\text{x}}{\text{ x - 1}}$ The value of gof(x) is.
  • ✓
    $ \frac{(\text{x}^{2} + 2)}{(\text{x}^{2} + 1)}$
  • B
    $ \frac{\text{x}^{2}}{(\text{x}^{2} + 1)}$
  • C
    $ \frac{\text{x}^{ 2}}{(\text{x}^{2} + 2)}$
  • D
    $ \text{None of these}$
Answer
Correct option: A.
$ \frac{(\text{x}^{2} + 2)}{(\text{x}^{2} + 1)}$
Given $ \text{f}(\text{x}) = \text{x}^2 + 2$ and $\text{gof}(\text{x}) =\text{g}(\text{x}^{2} - 1)$
Now, $\text{gof}(\text{x}) =\text{g}(\text{x}^{2} + 2)$
$ = \frac{(\text{x}^{2} + 2)}{(\text{x}^{2} + 2 – 1)}$
$ = \frac{(\text{x}^{2} + 2)}{(\text{x}^{2} + 1)}$
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MCQ 351 Mark
Let A = {1, 2, 3}, B = {1, 3, 5}. If relation R from A to B is given by = {(1, 3), (2, 5), (3, 3)}, Then $R^{-1}$ is:
  • ✓
    {(3, 3), (3, 1), (5, 2)}
  • B
    {(1, 3), (2, 5), (3, 3)}
  • C
    {(1, 3), (5, 2)}
  • D
    none of these.
Answer
Correct option: A.
{(3, 3), (3, 1), (5, 2)}
  1. {(3, 3), (3, 1), (5, 2)}
Solution:
$A=\{1,2,3\}, B=\{1,3,5\}$
$R=\{(1,3),(2,5),(3,3)\}$
$\therefore R^{-1}=\{(3,3),(3,1),(5,2)\}$
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MCQ 361 Mark
If $\text{e}^{\text{f(x)}}=\frac{10+\text{x}}{10-\text{x}},\text{ x}\in(-10,10)$ and $\text{f(x)}=\text{kf}\Big(\frac{200\text{x}}{100+\text{x}^2}\Big),$ then k =
  • ✓
    0.5
  • B
    0.6
  • C
    0.7
  • D
    0.8
Answer
Correct option: A.
0.5
$\text{e}^{\text{f(x)}}=\frac{10+\text{x}}{10-\text{x}}$
$\Rightarrow\text{ f(x)}=\log_\text{e}\Big(\frac{10+\text{x}}{10-\text{x}}\Big)\ ...(\text{i})$
$\Rightarrow\ \text{f(x)}=\text{kf}\Big(\frac{200\text{x}}{100+\text{x}^2}\Big)$
$\Rightarrow\ \log_\text{e}\Big(\frac{10+\text{x}}{10-\text{x}}\Big)=\text{k}\log_\text{e}\Bigg(\frac{10+\frac{200\text{x}}{100+\text{x}^2}}{10-\frac{200\text{x}}{100+\text{x}^2}}\Bigg)$ {from (1)}
$\Rightarrow\ \log_\text{e}\Big(\frac{10+\text{x}}{10-\text{x}}\Big)=\text{k}\log_\text{e}\Big(\frac{1000+10\text{x}^2+200\text{x}}{1000+10\text{x}^2-200\text{x}}\Big)$
$\Rightarrow\ \log_\text{e}\Big(\frac{10+\text{x}}{10-\text{x}}\Big)=\text{k}\log_\text{e}\bigg(\frac{(\text{x}+10)^2}{(\text{x}-10)^2}\bigg)$
$\Rightarrow\ \log_\text{e}\Big(\frac{10+\text{x}}{10-\text{x}}\Big)=2\text{k}\log_\text{e}\frac{(\text{x}+10)}{(\text{x}+10)}$
$\Rightarrow\ 1=2\text{k}$
$\Rightarrow\ \text{k}=\frac{1}{2}=0.5$
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MCQ 371 Mark
Let f(x) = x, $\text{g(x)}=\frac{1}{\text{x}}$ and h(x) = f(x) g(x). Then, h(x) = 1
  • A
    $\text{x}\in\text{R}$
  • B
    $\text{x}\in\text{Q}$
  • C
    $\text{x}\in\text{R}-\text{Q}$
  • ✓
    $\text{x}\in\text{R},\text{ x}\neq0$
Answer
Correct option: D.
$\text{x}\in\text{R},\text{ x}\neq0$
Given,
$\text{f(x)}=\text{x},\text{ g(x)}=\frac{1}{\text{x}}$ and $\text{h(x)}=\text{f(x)}\text{g(x)}$
Now,
$\text{h(x)}=\text{x}\times\frac{1}{\text{x}}=1$
We observe that the domain of f is R and the domain of g is R - {0}
$\therefore\ \text{Domain of h}=\text{Domain of f }\cap\text{ Domain of g}\\\ \ \ =\text{R }\cap\big[\text{R}-\{0\}\big]=\text{R}-\{0\}$
$\Rightarrow\text{x}\in\text{R},\text{ x}\neq0$
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MCQ 381 Mark
If $f(x)=3 x^4-5 x^2+9$, then value of $(x-1)$ is:
  • ✓
    $3 x^4+12 x^2+13 x^2+2 x+7$
  • B
    $3 x^4-12 x^3-13 x^2-2 x-7$
  • C
    $3 x^4-12 x^3+13 x^2-2 x+7$
  • D
    $3 x^4-12 x^3-13 x^2+2 x+7$
Answer
Correct option: A.
$3 x^4+12 x^2+13 x^2+2 x+7$
  1. $3 x^4+12 x^2+13 x^2+2 x+7$
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MCQ 391 Mark
If $f: R \rightarrow R$ is defined by $f(x)=x^2-3 x+2$, the $f(y)$ is:
  • A
    $x^4+6 x^3+10 x^2+3 x$
  • B
    $x^4-6 x^3+10 x^2+3 x$
  • C
    $x^4+6 x^3+10 x^2-3 x$
  • ✓
    $x^4-6 x^3+10 x^2-3 x$
Answer
Correct option: D.
$x^4-6 x^3+10 x^2-3 x$
  1. $x^4-6 x^3+10 x^2-3 x$
Solution:
Given, $f(x)=x^2-3 x+2$
Now, $f(f(y))=f\left(x^2-3 x+2\right)$
$=\left(x^2-3 x+2\right)^2-3\left(x^2-3 x+2\right)+2$
$=x^4-6 x^3+10 x^2-3 x$
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MCQ 401 Mark
If A = {1, 2, 3}, B = {1, 2} and C = {2, 3}, which one of the following is correct?
  • A
    $(\text{A} \times\text{B})\ \cap\ (\text{B}\times\text{A})=(\text{A} \times\text{C})\ \cap\ (\text{B}\times\text{C})$
  • B
    $(\text{A} \times\text{B})\ \cap\ (\text{B}\times\text{A})=(\text{C} \times\text{A})\ \cap\ (\text{C}\times\text{B})$
  • ✓
    $(\text{A} \times\text{B})\ \cup\ (\text{B}\times\text{A})=(\text{A} \times\text{B})\ \cup\ (\text{B}\times\text{C})$
  • D
    $(\text{A} \times\text{B})\ \cup\ (\text{B}\times\text{A})=(\text{A} \times\text{B})\ \cup\ (\text{A}\times\text{C})$
Answer
Correct option: C.
$(\text{A} \times\text{B})\ \cup\ (\text{B}\times\text{A})=(\text{A} \times\text{B})\ \cup\ (\text{B}\times\text{C})$
  1. $(\text{A} \times\text{B})\ \cup\ (\text{B}\times\text{A})=(\text{A} \times\text{B})\ \cup\ (\text{B}\times\text{C})$
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MCQ 411 Mark
Choose the correct answers:The domain and range of real function f defined by $\text{f(x)}=\sqrt{\text{x}-1}$ is given by.
  • ✓
    Domain $= [1, \infty),$ Range $= [0, \infty)$
  • B
    Domain $= [1, \infty),$ Range $= [0, \infty)$
  • C
    Domain $= [1, \infty),$ Range $= [0, \infty)$
  • D
    Domain $= [1, \infty),$ Range $= [0, \infty)$
Answer
Correct option: A.
Domain $= [1, \infty),$ Range $= [0, \infty)$
We have, $\text{f(x)}=\sqrt{\text{x}-1}$
Clearly, f(x) is defined if $\text{x}-1\geq0$
$\Rightarrow\text{x}\geq1$
$\therefore$ Domain of $\text{f}=[1, \infty)$
Now for $\text{x}\geq1,\text{x}-1\geq0$
$\Rightarrow\sqrt{\text{x}-1}\geq1$
⇒ Range of $= [0, \infty)$
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MCQ 421 Mark
If (x + 2, y - 3) = (5, 7) then find values of x and y:
  • ✓
    x = 3 and y = 10
  • B
    x = 3 and y = 4
  • C
    x = 7 and y = 4
  • D
    x = 7 and y = 10
Answer
Correct option: A.
x = 3 and y = 10
Two ordered pairs are said to be equal if and only if their corresponding elements are equal.
x + 2 = 5 ⇒ x = 3
y - 3 = 7 ⇒ y = 10
Hence, x = 3 and y = 10.
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MCQ 431 Mark
The range of the function f(x) = |x - 1| is:
  • A
    $\big(-\infty,0\big)$
  • ✓
    $\big[0,\infty\big)$
  • C
    $\big(0,\infty\big)$
  • D
    $\text{R}$
Answer
Correct option: B.
$\big[0,\infty\big)$
$\text{f(x)}=|\text{x}-1|\geq0\ \forall\text{ x}\in\text{R}$
Thus, range $=\big[0,\infty\big)$
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MCQ 441 Mark
Let f : R → R be defined by f(x) = 2x + |x|. Then f(2x) + f(-x) - f(x) =
  • A
    2x
  • ✓
    2|x|
  • C
    -2x
  • D
    -2|x|
Answer
Correct option: B.
2|x|
f(x) = 2x + |x|
Then, f(2x) + f(-x) - f(x)
= 2(2x) + 2|x| + (-2x) + |-x| - 2x + |x|
= 4x - 2x - 2x + 2|x| + |-x| - |x|
= 0 + 2|x| + |x| - |x| = 2|x|
= 2|x|
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MCQ 451 Mark
Let A = {1, 2, 3, 4, 5} and R be a relation from A to A, R = {(x, y) : y = x + 1}. Find the range:
  • A
    {1, 2, 3, 4, 5}
  • ✓
    {2, 3, 4, 5}
  • C
    {1, 2, 3, 4}
  • D
    {1, 2, 3, 4, 5, 6}
Answer
Correct option: B.
{2, 3, 4, 5}
Range is the set of elements of codomain which have their preimage in domain.
Relation R = {(1, 2), (2, 3), (3, 4), (4, 5)}.
Range = {2, 3, 4, 5}.
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MCQ 461 Mark
The function $ \text{f}(\text{x})=\sin(\frac{\pi\text{x}}{2})+2\cos\Big(\frac{\pi\text{x}}{3}\Big) -\tan\Big(\frac{\pi\text{x}}{4}\Big) +2\cos\Big(\frac{\pi\text{x}}{3}\Big)-\tan\Big(\frac{\pi\text{x}}{4}\Big)$ is periodic with period:
  • A
    4
  • B
    6
  • C
    8
  • ✓
    12
Answer
Correct option: D.
12
Period of sin$ \sin\Big(\frac{\pi\text{x}}{2}\Big)=\frac{2\pi}{\big(\frac{\pi\text{x}}{2}\big) }=4=\frac{2\pi}{\big(\frac{\pi\text{x}}{2}\big)}=4$
Period of cos$ \cos\Big(\frac{\pi\text{x}}{3}\Big)=\frac{2\pi}{\big(\frac{\pi\text{x}}{3}\big) }=6 $
Period of tan$ \sin\Big(\frac{\pi\text{x}}{2}\Big)=\frac{2\pi}{\big(\frac{\pi\text{x}}{2}\big) }=4$
So, period of f(x) = LCM (4, 6, 4) = 12
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MCQ 471 Mark
Let R be a relation in the set of real numbers defined as a R b if $ |\text{a}-\text{b}| ≥ \frac{1}{2}$, Then the relation R is:
  • A
    An equivalence relation.
  • B
    Reflexive and symmetric but not transitive.
  • C
    Symmetric and transitive but not reflexive.
  • ✓
    Esymmetric but neither reflexive nor transitiv.
Answer
Correct option: D.
Esymmetric but neither reflexive nor transitiv.
  1. Esymmetric but neither reflexive nor transiti
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MCQ 481 Mark
Let R be a relation in the set of real numbers defined as a R b if |a - b| ≥ $\frac{1}{2}$, Then the relation R is:
  • A
    An equivalence relation
  • B
    Reflexive and symmetric but not transitive
  • C
    Symmetric and transitive but not reflexive
  • ✓
    Symmetric but neither reflexive nor transitive
Answer
Correct option: D.
Symmetric but neither reflexive nor transitive
d. Symmetric but neither reflexive nor transitive
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MCQ 491 Mark
Choose the correct answers: The domain of the function f defined by $\text{f(x)}=\sqrt{4-\text{x}}+\frac{1}{\sqrt{\text{x}^2-1}}$ is equal to.
  • ✓
    $(–\infty, –1) \cup (1, 4]$
  • B
    $(–\infty, –1] \cup (1, 4]$
  • C
    $(–\infty, –1) \cup [1, 4]$
  • D
    $(–\infty, –1) \cup [1, 4)$
Answer
Correct option: A.
$(–\infty, –1) \cup (1, 4]$
We have, $\text{f(x)}=\sqrt{4-\text{x}}+\frac{1}{\sqrt{\text{x}^2-1}}$
f(x) is defined if $4 - \text{x}\geq 0$and $\text{x}^2-1>0$
$\Rightarrow\text{x}\leq4$ and $(\text{x}+1)(\text{x}-1)>0$
$\Rightarrow\text{x}\leq4$ and $(\text{x}<-1 \ \text{or} \ \text{x}>1)$
$\therefore$ Domain of $\text{f}=(-\infty, -1)\cup(1, 4]$
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MCQ 501 Mark
Choose the correct answers: Domain of $\sqrt{\text{a}^2-\text{x}^2}(\text{a}>0)$ is.
  • A
    (-a, a)
  • ✓
    [-a, a]
  • C
    [0, a]
  • D
    (-a, 0]
Answer
Correct option: B.
[-a, a]
We have $\text{f(x)}\sqrt{\text{a}^2-\text{x}^2}$
Clearly f(x) is defined, if ${\text{a}^2-\text{x}^2}\geq0$
$\Rightarrow\text{x}^2\leq\text{a}^2$
$\Rightarrow-\text{a}\leq\text{x}\leq\text{a} \ [\therefore\text{a}>0]$
$\therefore$ Domain of f is [-a, a]
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MCQ 511 Mark
If A is the set of even natural numbers less than 8 and B is the set of prime numbers less than 7, then the number of relations from A to B is:
  • ✓
    29
  • B
    92
  • C
    32
  • D
    29 - 1
Answer
Correct option: A.
29
  1. 29
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MCQ 521 Mark
Let $n(A)=m$, and $n(B)=n$, Then the total number of non-empty relations that can be defined from $A$ to $B$ is:
  • A
    $\mathrm{m}^n$
  • B
    $\mathrm{n}^m-1$
  • C
    $m^n-1$
  • ✓
    $2^{m n}-1$
Answer
Correct option: D.
$2^{m n}-1$
  1. $2^{m n}-1$
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MCQ 531 Mark
If $\text{f(x)} = \sin^{2}\text{x}$ and the composite function $ \text{g}{\text{f(x)}} = | \sin\text{ x } |$, then the function g(x) is equal to:
  • A
    $\sqrt{\text{x} - 1}$
  • ✓
    $\sqrt{\text{x}}$
  • C
    $\sqrt{\text{x} + 1}$
  • D
    $-\sqrt{\text{x}}$
Answer
Correct option: B.
$\sqrt{\text{x}}$
  1. $\sqrt{\text{x}}$
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MCQ 541 Mark
The domain of definition of the function $\text{f(x)}=\sqrt{\frac{\text{x}-2}{\text{x}+2}}+\sqrt{\frac{1-\text{x}}{1+\text{x}}}$ is:
  • A
    $\big(-\infty,-2\big]\cap\big[2,-\infty\big)$
  • B
    $\big[-1,1\big]$
  • ✓
    $\phi$
  • D
    None of these.
Answer
Correct option: C.
$\phi$
$\text{f(x)}=\sqrt{\frac{\text{x}-2}{\text{x}+2}}+\sqrt{\frac{1-\text{x}}{1+\text{x}}}$
For f(x) to be defined,
$\text{x}+2\neq0$
$\Rightarrow\text{x}\neq-2\ ...(\text{i})$
And $1+\text{x}\neq0$
$\Rightarrow\text{x}\neq-1\ ...(\text{ii})$
Also, $\frac{\text{x}-2}{\text{x}+2}\geq0$
$\Rightarrow\frac{(\text{x}-2)(\text{x}-2)}{(\text{x}-2)^2}\ge0$
$\Rightarrow(\text{x}-2)(\text{x}+2)\geq0$
$\Rightarrow\text{x}\in(\infty,-2)\cup\big[2,\infty\big)\ ...(\text{iii})$
And $\frac{1-\text{x}}{1+\text{x}}\geq0$
$\Rightarrow\frac{(1-\text{x})(1+\text{x})}{(1+\text{x})^2}\geq0$
$\Rightarrow(1-\text{x})(1+\text{x})\geq0$
$\Rightarrow\text{ x}\in\big(-\infty,-1\big)\cup\big[1,\infty\big)\ ...(\text{iv})$
From (i), (ii), (iii) and (iv) we get
$\text{x }\in\phi$
Thus, domain $(\text{f(x)})=\phi$
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MCQ 551 Mark
If $\text{f(x)}=\cos(\log\text{x}),$ then value of $\text{f(x)}\text{f(4)}-\frac{1}{2}\Big\{\text{f}\Big(\frac{\text{x}}{4}\Big)+\text{f}(4\text{x})\Big\}$ is:
  • A
    1
  • B
    -1
  • ✓
    0
  • D
    $\pm1$
Answer
Correct option: C.
0
Given, $\text{f(x)}=\cos(\log\text{x})$
Then, $\text{f(x)}\text{f(4)}-\frac{1}{2}\Big\{\text{f}\Big(\frac{\text{x}}{4}\Big)+\text{f}(4\text{x})\Big\}$
$=\cos(\log\text{x})\cos(\log4)+\frac{1}{2}\Big\{\cos\Big(\log\frac{\text{x}}{4}\Big)+\cos(\log4\text{x})\Big\}$
$=\frac{1}{2}\big[\cos(\log\text{x}+\log4\big)+\cos(\log\text{x}-\log4)\big]\\-\frac{1}{2}\Big\{\cos\Big(\log\frac{\text{x}}{4}\Big)+\cos(\log4\text{x})\Big\}$
$=\frac{1}{2}\Big\{\cos(\log4\text{x})+\cos\Big(\log\frac{\text{x}}{4}\Big)-\cos\Big(\log\frac{\text{x}}{4}\Big)-\cos(\log4\text{x})\Big\}$
$=\frac{1}{2}\times0=0$
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MCQ 561 Mark
If f : R → R is defined by f(x) = 3x + |x|, then f(2x) - f(-x) - 6x =
  • ✓
    f(x)
  • B
    2f(x)
  • C
    -f(x)
  • D
    f(-x)
Answer
Correct option: A.
f(x)
  1. f(x)
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MCQ 571 Mark
If f : [-2, 2] → R is defined by $\text{f(x)}=\begin{cases}-1,&\text{for}-2\leq\text{x}\leq0\\\text{x}-1,&\text{for }0\leq\text{x}\leq2\end{cases}$ then $\{\text{x}\in[-2,2]:\text{x}\leq0\text{ and }\text{f}(\text{|x|})=\text{x}\}=$
  • A
    $\{-1\}$
  • B
    $\{0\}$
  • ✓
    $\Big\{-\frac{1}{2}\Big\}$
  • D
    $\phi$
Answer
Correct option: C.
$\Big\{-\frac{1}{2}\Big\}$
Given,
$\text{f(x)}=\begin{cases}-1,&\text{for}-2\leq\text{x}\leq0\\\text{x}-1,&\text{for }0\leq\text{x}\leq2\end{cases}$
We know,
$|\text{x}|\geq0$
$\Rightarrow\text{f(|x|)}=|\text{x}|-1\ ...(\text{i})$
Also,
If $\text{x}\leq0,$ then $|\text{x}|=-\text{x}\ ...(\text{ii})$
$\{\text{x}\in[-2,2]:\text{x}\leq0\text{ and }\text{f}(\text{|x|})=\text{x}\}$
$=\{\text{x}:|\text{x}|-1=\text{x}\}$ [Using(i)]
$=\{\text{x}:-\text{x}-1=\text{x}\}$ [Using (ii)]
$=\Big\{\text{x}:2\text{x}=\frac{-1}{2}\Big\}$
$=\Big\{\text{x}:\text{x}=\frac{-1}{2}\Big\}$
$=\Big\{-\frac{1}{2}\Big\}$
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MCQ 581 Mark
The domain of definition of $\text{f(x)}=\sqrt{\text{x}-3-2\sqrt{\text{x}-4}}-\sqrt{\text{x}-3+2\sqrt{\text{x}-4}}$ is:
  • ✓
    $\big[4,\infty\big)$
  • B
    $\big(-\infty,4\big]$
  • C
    $(4,\infty)$
  • D
    $(-\infty,4)$
Answer
Correct option: A.
$\big[4,\infty\big)$
$\text{f(x)}=\sqrt{\text{x}-3-2\sqrt{\text{x}-4}}-\sqrt{\text{x}-3+2\sqrt{\text{x}-4}}$
For f(x) to be defined, $\text{x}-4\geq0$
$\Rightarrow\text{x}-4\geq0$
$\Rightarrow\text{x}\geq4\ ...(\text{i})$
Also, $\text{x}-3-2\sqrt{\text{x}-4}\geq0$
$\Rightarrow\text{x}-3-2\sqrt{\text{x}-4}\geq0$
$\Rightarrow\text{x}-3\geq2\sqrt{\text{x}-4}$
$\Rightarrow(\text{x}-3)^2\geq\big(2\sqrt{\text{x}-4}\big)^2$
$\Rightarrow\text{x}^2+9-6\text{x}\geq4(\text{x}-4)$
$\Rightarrow\text{x}^2-10\text{x}+25\geq0$
$\Rightarrow\big(\text{x}-5\big)^2\geq0,$ which is always true.
Similarly, $\text{x}-3+2\sqrt{\text{x}-4}\geq0$ is always true.
Thus, domain $(\text{f(x)})=\big[4,\infty)$
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MCQ 591 Mark
If $A=\{1,4,8,9\}$ and $B=\{1,2,-1,-2,-3,3,5\}$ and $R$ is a relation from set $A$ to set $B\left\{(x, y) x=y^2\right\}$. Find codomain of the relation:
 
  • A
    $\{1,4,9\}$
  • B
    $\{-1,1,-2,2,-3,3\}$
  • C
    $\{1,4,8,9\}$
  • ✓
    $\{-1,1,-2,2,-3,3,5\}$
Answer
Correct option: D.
$\{-1,1,-2,2,-3,3,5\}$
  1. $\{-1,1,-2,2,-3,3,5\}$
Solution:
We know, codomain of a relation is the set to which relation is defined i.e. set B.
So, codomain $=\{-1,1,-2,2,-3,3,5\}$.
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MCQ 601 Mark
The range of the function: $\text{f}(\text{x})=\sqrt{(\text{x}-1)(3-\text{x})}$ :
  • ✓
    (-1, 1)
  • B
    (-1, 2)
  • C
    (-3, 3)
  • D
    (-3, 1)
Answer
Correct option: A.
(-1, 1)
  1. (-1, 1)
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MCQ 611 Mark
Which one of the following is not a function?
  • A
    $\{(\text{x, y}):\text{x},\text{ y}\in\text{R},\text{x}^2=\text{y}\}$
  • ✓
    $\{(\text{x, y}):\text{x},\text{ y}\in\text{R},\text{y}^2=\text{x}\}$
  • C
    $\{(\text{x, y}):\text{x},\text{ y}\in\text{R},\text{x}^2=\text{y}^3\}$
  • D
    $\{(\text{x, y}):\text{x},\text{y}\in\text{R},\text{y}=\text{x}^3\}$
Answer
Correct option: B.
$\{(\text{x, y}):\text{x},\text{ y}\in\text{R},\text{y}^2=\text{x}\}$
  1. $\{(\text{x, y}):\text{x},\text{ y}\in\text{R},\text{y}^2=\text{x}\}$
Solution:
$y^2$ = x gives two values of y for a value of x
i.e. there are two images for a value of x.
For example: $(2)^2 = 4$ and $(-2)^2 = 4$
Thus, it is not a function.
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MCQ 621 Mark
A relation R is defined from {2, 3, 4, 5} to {3, 6, 7, 10} by : x R y ⇔ x is relatively prime to y. Then, domain of R is:
  • A
    {2, 3, 5}
  • B
    {3, 5}
  • C
    {2, 3, 4}
  • ✓
    {2, 3, 4, 5}
Answer
Correct option: D.
{2, 3, 4, 5}
Given,
From {2, 3, 4, 5} to {3, 6, 7, 10}, x R y ⇔ x is relatively prime to y
2 is relatively prime to 3, 7
3 is relatively prime to 7, 10
4 is relatively prime to 3, 7
5 is relatively prime to 3, 6, 7
So, domain of R is {2, 3, 4, 5}
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MCQ 631 Mark
If $\text{f(x)}=\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$ and $\text{g(x)}=\frac{3\text{x}+\text{x}^3}{1+3\text{x}^2},$ then f(g(x)) is equal to:
  • A
    $f(3 x)$
  • B
    $\{f(x)\}^3$
  • ✓
    $3 \mathrm{f}(\mathrm{x})$
  • D
    $-f(x)$
Answer
Correct option: C.
$3 \mathrm{f}(\mathrm{x})$
  1. $3 \mathrm{f}(\mathrm{x})$
Solution:
$\text{f(x)}=\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$ and $\text{g(x)}=\frac{3\text{x}+\text{x}^3}{1+3\text{x}^2}$
Now,
$\frac{1+\text{g(x)}}{1-\text{g(x)}}=\frac{1+\frac{3\text{x}+\text{x}^3}{1+3\text{x}^2}}{1-\frac{3\text{x}+\text{x}^3}{1+3\text{x}^2}}$
$=\frac{1+3\text{x}^2+3\text{x}+\text{x}^3}{1+3\text{x}^2-3\text{x}-\text{x}^3}$
$=\frac{(1+\text{x})^3}{(1-\text{x})^3}$
Then, $\text{f}(\text{g(x)})=\log=\log\Big(\frac{1+\text{g(x)}}{1-\text{g(x)}}\Big)$
$=\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big)^3$
$=3\text{f}(\text{x})$
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MCQ 641 Mark
If $\text{f(x)}=\cos(\log\text{x}),$ then the value of $\text{f(x})\text{f}(\text{y})-\frac{1}{2}\Big\{\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f}\big(\text{x}\text{y}\big)\Big\}$ is:
  • A
    -1
  • B
    $\frac{1}{2}$
  • C
    -2
  • ✓
    None of these.
Answer
Correct option: D.
None of these.
Given,
$\text{f(x)}=\cos(\log\text{x})$
$\therefore\ \text{f(y)}=\cos(\log\text{y})$
Now,
$\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)=\cos\Big(\cos\Big(\frac{\text{x}}{\text{y}}\Big)\Big)=\cos(\log\text{x}-\log\text{y})$
and
$\text{f(xy)}=\cos(\log\text{xy})=\cos(\log\text{x}+\log\text{y})$
$\Rightarrow\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f(xy)}=\cos(\log\text{x}-\log\text{y})+\cos(\log\text{x}+\log\text{y})$
$\Rightarrow\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f(xy)}=2\cos(\log\text{x})\cos(\log\text{y})$
$\Rightarrow\frac{1}{2}\Big[\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f(xy)}\Big]=\cos(\log\text{x})\cos(\log\text{y})$
$\Rightarrow\text{f(x})\text{f}(\text{y})-\frac{1}{2}\Big\{\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f}\big(\text{x}\text{y}\big)\Big\}\\=\cos(\log\text{x})\cos(\log\text{y})-\cos(\log\text{x})\cos(\log\text{y})=0$
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MCQ 651 Mark
Choose the correct answers: The domain of the function f given by $\text{f(x)}=\frac{\text{x}^2+2\text{x}+1}{\text{x}^2-\text{x}-6}.$
  • ✓
    R – {3, –2}
  • B
    R – {–3, 2}
  • C
    R – [3, –2]
  • D
    R – (3, –2)
Answer
Correct option: A.
R – {3, –2}
  1. R – {3, –2}
Solution:
Given that: $\text{f(x)}=\frac{\text{x}^2+2\text{x}+1}{\text{x}^2-\text{x}-6}$
f(x) is defined if $\text{x}^2-\text{x}-6\neq0$
$\Rightarrow\text{x}^2-3\text{x}+2\text{x}-6\neq0$
$\Rightarrow(\text{x}-3)(\text{x}+2)\neq0$
$\Rightarrow\text{x}\neq-2,\text{x}\neq3$
So, the domain of f(x) = R - {-2, 3}
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MCQ 661 Mark
The range of the function $\text{f(x)}=\frac{\text{x}}{|\text{x}|}$ is:
  • A
    R - {0}
  • B
    R - {-1, 1}
  • ✓
    {-1, 1}
  • D
    None of these.
Answer
Correct option: C.
{-1, 1}
$\text{f(x)}=\frac{\text{x}}{|\text{x}|}$
Let $\text{y}=​​\frac{\text{x}}{|\text{x}|}$
For x > 0, |x| = x
$\Rightarrow\text{y}=\frac{\text{x}}{\text{x}}=1$
For x < 0, = -x
$\Rightarrow\text{y}=\frac{\text{x}}{-\text{x}}=-1$
Thus, range of f(x) is {-1, 1}
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MCQ 671 Mark
The range of $\text{f(x)}=\cos[\text{x}],$ for $-\frac{\pi}{2}<\text{x}<\frac{\pi}{2}$ is:
  • A
    $\{-1,1,0\}$
  • ✓
    $\{\cos1,\cos2,1\}$
  • C
    $\{\cos1,-\cos1,1\}$
  • D
    $[-1,1]$
Answer
Correct option: B.
$\{\cos1,\cos2,1\}$
Since, $\text{f(x)}=\cos[\text{x}],$ where $\frac{-\pi}{2}<\text{x}<\frac{\pi}{2}$
$-\frac{\pi}{2}<\text{x}<\frac{\pi}{2}$
$\Rightarrow-1.57<\text{x}<1.57$
$\Rightarrow[\text{x}]\ \in\ \{-1,0,1,2\}$
Thus, $\cos[\text{x}]=\{\cos(-1),\cos0,\cos1,\cos2\}$
Range of $\text{f(x)}=\{\cos1,1,\cos2\}$
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MCQ 681 Mark
Find the range of the function $f(x)=x^2+2$
  • A
    $(-2, 2)$
  • ✓
    $(2, \infty)$
  • C
    $(3, \infty)$
  • D
    None of these
Answer
Correct option: B.
$(2, \infty)$
  1. $(2, \infty)$
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MCQ 691 Mark
Given g(1) = 1 and g(2) = 3. If g(x) is described by the formula g(x) = ax + b, then the value of a and b is:
  • A
    2, 1
  • B
    -2, 1
  • ✓
    2, -1
  • D
    -2, -1
Answer
Correct option: C.
2, -1
Given, g(x) = ax + b
Again, g(1) = 1
⇒ a × 1 + b = 1
⇒ a + b = 1 ……… 1
and g(2) = 3
⇒ a × 2 + b = 3
⇒ 2a + b = 3 …….. 2
Solve equation 1 and 2, we get
a = 2, b = -1
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MCQ 701 Mark
Let A = {1, 2, 3} and B = {2, 3, 4}. Then which of the following is a function from A to B?
  • A
    {(1, 2), (1, 3), (2, 3), (3, 3)}
  • B
    {(1, 3), (2, 4)}
  • ✓
    {(1, 3), (2, 2), (3, 3)}
  • D
    {(1, 2), (2, 3), (3, 2), (3, 4)}
Answer
Correct option: C.
{(1, 3), (2, 2), (3, 3)}
We have,
R = {(1, 3), (2, 2), (3, 3)}
We observe that each element of the given set has appeared as first component in one and only one ordered pair of R.
So, R = {(1, 3), (2, 2), (3, 3)} is a function.
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MCQ 711 Mark
The relation R defined on the set $A=\{1,2,3,4,5\}$ by $R=\left\{(x, y):\left|x^2-y^2\right|<16\right\}$ is given by:
  • A
    {(1, 1), (2, 1), (3, 1), (4, 1), (2, 3)}
  • B
    {(2, 2), (3, 2), (4, 2), (2, 4)}
  • C
    {(3, 3), (4, 3), (5, 4), (3, 4)}
  • ✓
    None of the above
Answer
Correct option: D.
None of the above
  1. None of the above
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MCQ 721 Mark
If $2\text{f(x)}-3\text{f}\Big(\frac{1}{\text{x}}\Big)=\text{x}^2(\text{x}\neq0),$ then f(2) is equal to:
  • ✓
    $-\frac{7}{4}$
  • B
    $\frac{5}{2}$
  • C
    $-1$
  • D
    None of these.
Answer
Correct option: A.
$-\frac{7}{4}$
$2\text{f(x)}-3\text{f}\Big(\frac{1}{\text{x}}\Big)=\text{x}^2\ ...(\text{i})$ $(\text{x}\neq0)$
Replacing x by $\frac{1}{\text{x}}$
$2\text{f}\Big(\frac{1}{\text{x}}\Big)-3\text{f(x)}=\frac{1}{\text{x}^2}\ ...(\text{ii})$
Solving equations (i) & (ii)
$-5\text{f(x)}=\frac{3}{\text{x}^2}+2\text{x}^2$
$\Rightarrow\text{f(x)}=\frac{-1}{5}\Big(\frac{3}{\text{x}^2}+2\text{x}^2\Big)$
Thus, $\text{f(2)}=\frac{-1}{5}\Big(\frac{3}{4}+2\times4\Big)$
$=\frac{-1}{5}\Big(\frac{3+32}{4}\Big)$
$=-\frac{7}{4}$
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MCQ 731 Mark
In a function from set A to set B, every element of set A has___________ image in set B:
  • ✓
    one and only one
  • B
    different
  • C
    same
  • D
    many
Answer
Correct option: A.
one and only one
A relation from a set A to a set B is said to be a function if every element of set A has one and one image in set B.
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MCQ 741 Mark
If $\text{f(x)}=\frac{2^{\text{x}}+2^{-\text{x}}}{2},$ then f(x + y)f(x - y) is equal to:
  • ✓
    $\frac{1}{2}\big[\text{f(2}\text{x})+\text{f}(2\text{y})\big]$
  • B
    $\frac{1}{2}\big[\text{f(2}\text{x})-\text{f}(2\text{y})\big]$
  • C
    $\frac{1}{4}\big[\text{f(2}\text{x})+\text{f}(2\text{y})\big]$
  • D
    $\frac{1}{4}\big[\text{f(2}\text{x})-\text{f}(2\text{y})\big]$
Answer
Correct option: A.
$\frac{1}{2}\big[\text{f(2}\text{x})+\text{f}(2\text{y})\big]$
Given,
$\text{f(x)}=\frac{2^{\text{x}}+2^{-\text{x}}}{2}$
Now,
$\text{f}(\text{x}+\text{y})\text{f}(\text{x}-\text{y})=\Big(\frac{2^{\text{x}+\text{y}}+2^{-\text{x}-\text{y}}}{2}\Big)\Big(\frac{2^{\text{x}-\text{y}}+2^{-\text{x}+\text{y}}}{2}\Big)$
$\Rightarrow\ \text{f}(\text{x}+\text{y})\text{f}(\text{x}-\text{y})=\frac{1}{4}\big(2^{2\text{x}}+2^{-2\text{y}}+2^{2\text{y}}+2^{-2\text{x}}\big)$
$\Rightarrow\ \text{f}(\text{x}+\text{y})\text{f}(\text{x}-\text{y})=\frac{1}{2}\Big(\frac{2^{2\text{x}}+2^{-2\text{x}}}{2}+\frac{2^{2\text{y}}+2^{-2\text{y}}}{2}\Big)$
$\Rightarrow\text{f}(\text{x}+\text{y})\text{f}(\text{x}-\text{y})=\frac{1}{2}\big[\text{f}(2\text{x})+\text{f}(2\text{y})\big]$
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MCQ 751 Mark
If set $P$ has 4 elements and set $Q$ has 5 elements then find the number of elements in $P \times Q$ :
  • A
    $9$
  • B
    $4^5$
  • ✓
    $20$
  • D
    $5^4$
Answer
Correct option: C.
$20$
  1. $20$
Solution:
If set P has m elements and set Q has n elements then P × Q has m × n elements.
Here, m = 4 and n = 5 therefore P × Q has 4 × 5 = 20 element.
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MCQ 761 Mark
If $[\text{x}^2]-5[\text{x}]+6=0,$ where [.] denotes the greatest integer function, then:
  • A
    $\text{x}\in[3,4]$
  • B
    $\text{x}\in\big(2,3\big]$
  • C
    $\text{x}\in\big[2,3\big]$
  • ✓
    $\text{x}\in\big[2,4\big)$
Answer
Correct option: D.
$\text{x}\in\big[2,4\big)$
The given equation is $[\text{x}^2]-5[\text{x}]+6=0$
$[\text{x}^2]-5[\text{x}]+6=0$
$\Rightarrow[\text{x}^2\big]-3\big[\text{x}\big]-2\big[\text{x}\big]+6=0$
$\Rightarrow\big[\text{x}\big]\big([\text{x}]-3\big)-2\big([\text{x]}-3\big)=0$
$\Rightarrow\big([\text{x}]-2)\big([\text{x}]-3)=0$
$\Rightarrow\big[\text{x}\big]-2=0$ or $[\text{x}]-3=0$
$\Rightarrow[\text{x}]=2$ or $[\text{x}]=3$
$\Rightarrow\text{x}\in\big[2,3\big)$ or $\big[3,4\big)$
$\Rightarrow\text{x}\in\big[2,4\big)$
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MCQ 771 Mark
If $A=\{1,4,8,9\}$ and $B=\{1,2,-1,-2,-3,3,5\}$ and $R$ is a relation from set $A$ to set $B\left\{(x, y) x=y^2\right\}$. Find range of the relation:
  • A
    {1, 4, 9}
  • ✓
    {-1, 1, -2, 2, -3, 3}
  • C
    {1, 4, 8, 9}
  • D
    {-1, 1, -2, 2, -3, 3, 5}
Answer
Correct option: B.
{-1, 1, -2, 2, -3, 3}
  1. {-1, 1, -2, 2, -3, 3}
Solution:
Range is the set of elements of codomain which have their preimage in domain.
Relation R = {(1, 1), (1, -1), (4, 2), (4, -2), (9, 3), (9, -3)}.
Range = {-1, 1, -2, 2, -3, 3}.
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MCQ 781 Mark
If $\text{f(x)}=\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big),$ then $\text{f}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ is equal to:
  • A
    $\{\mathrm{f}(\mathrm{x})\}^2$
  • B
    $\{\mathrm{f}(\mathrm{x})\}^3$
  • ✓
    $2 f(x)$
  • D
    $3 f(x)$
Answer
Correct option: C.
$2 f(x)$
  1. $2 f(x)$
Solution:
$\text{f(x)}=\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$
Then, $\text{f}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)=\log\Bigg(\frac{1+\frac{2\text{x}}{1+\text{x}^2}}{1-\frac{2\text{x}}{1+\text{x}^2}}\Bigg)$
$=\log\Bigg(\frac{\frac{1+\text{x}^2+2\text{x}}{1+\text{x}^2}}{1-\frac{2\text{x}}{1+\text{x}^2}}\Bigg)$
$=\log\bigg(\frac{(1+\text{x})^2}{(1-\text{x})^2}\bigg)$
$=2\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$
$=2(\text{f(x)})$
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MCQ 791 Mark
Let R = {(2, 3), (3, 4)} be relation defined on the set of natural numbers. The minimum number of ordered pairs required to be added in R so that enlarged relation becomes an equivalence relation is:
  • A
    3
  • B
    5
  • C
    7
  • ✓
    9
Answer
Correct option: D.
9
  1. 9
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MCQ 801 Mark
The domain of the function $\text{f}(\text{x})=\frac{\text{x}}{(1+\text{x}^2)}$ is:
  • A
    R - {1}
  • B
    R - {-1}
  • ✓
    R
  • D
    None of these
Answer
Correct option: C.
R
Given, function $\text{f}(\text{x})=\frac{\text{x}}{(1+\text{x}^2)}$
Since f(x) is defined for all real values of x
So, domain(f) = R.
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MCQ 811 Mark
A relation $\phi$ from C to R is defined by $\text{x }\phi\text{ y}\Leftrightarrow|\text{x}|=\text{y}.$ Which one is correct?
  • A
    $(2+3\text{i})\ \phi\ 13$
  • B
    $3\phi\ (-3)$
  • C
    $(1+\text{i})\ \phi\ 2$
  • ✓
    $\text{i}\ \phi\ 1$
Answer
Correct option: D.
$\text{i}\ \phi\ 1$
We have,
$|\text{i}|=\sqrt{1^2+0^2}=1$
Thus, $\text{i }\phi\ 1$ satisfies $\text{x}\ \phi\text{ y}\Leftrightarrow|\text{x}|=\text{y}$
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MCQ 821 Mark
A relation $R$ is defined from the set of integers to the set of real numbers as $(x, y)=R$ if $x^2+y^2=16$ then the domain of $R$ is:
  • A
    $(0,4,4)$
  • ✓
    $1$
  • C
    $(0,-4,-4)$
  • D
    None of these
Answer
Correct option: B.
$1$
  1. $1$
Solution:
Given, $f(x)=e^x$
and $g(x)=\log x$
$f \circ g(x)=f(g(x))$
$=f(\log x)$
$=e^{\log x}$
$=\mathrm{x}$
So, fog $(1)=1$
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MCQ 831 Mark
If $ \text{f}(\text{x}) = 2\text{x} + 2-\sqrt{\frac{\text{x}}{2}}$, then $\text{f}(\text{x + y}), \text{f}(\text{x - y}) =$
  • ✓
    $ \frac{1}{2} \big[\text{}f(\text{2x}) +\text{ f} \text{(2y)}\big]$
  • B
    $ \frac{1}{4} \big[\text{[}f(\text{2x}) +\text{ f} \text{(2y)}\big]$
  • C
    $ \frac{1}{2} \big[\text{}f(\text{2x}) \text{ f} \text{(2y)}\big]$
  • D
    $ \frac{1}{4} \text{[}f(\text{2x}) \text{ f} \text{(2y)}]$
Answer
Correct option: A.
$ \frac{1}{2} \big[\text{}f(\text{2x}) +\text{ f} \text{(2y)}\big]$
  1. $ \frac{1}{2} \big[\text{}f(\text{2x}) +\text{ f} \text{(2y)}\big]$
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MCQ 841 Mark
The range of the function $\text{f(x)}=\frac{\text{x}+2}{|\text{x}+2|},\text{ x}\neq-2$ is:
  • ✓
    {-1, 1}
  • B
    {-1, 0, 1}
  • C
    {1}
  • D
    $(0,\infty)$
Answer
Correct option: A.
{-1, 1}
$\text{f(x)}=\frac{\text{x}+2}{|\text{x}+2|},\text{ x}\neq-2$
Let $\text{y}=\frac{\text{x}+2}{|\text{x}+2|}$
For |x + 2| > 0
Or x > -2
$\text{y}=\frac{\text{x}+2}{\text{x}+2}=1$
For |x + 2| < 0
Or x < -2
$\text{y}=\frac{\text{x}+2}{-(\text{x}+2)}=-1$
Thus, y = {-1, 1}
Or range f(x) = {-1, 1}
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MCQ 851 Mark
Domain of the function $ \text{ f}(\text{x}) =\sqrt{( 2 -\text{ 2x} - \text{x })} $ is:
  • A
    $ - \sqrt{3 ≤\text{ x} ≤} + \sqrt{3}$
  • ✓
    $ -1- \sqrt{3 ≤ \text{x} ≤ -1}$
  • C
    $ + \sqrt{3}$
  • D
    $ -2 ≤ \text{x} ≤ 2$
Answer
Correct option: B.
$ -1- \sqrt{3 ≤ \text{x} ≤ -1}$
  1. $ -1- \sqrt{3 ≤ \text{x} ≤ -1}$
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MCQ 861 Mark
Let R be a relation on N defined by x + 2y = 8. The domain of R is:
  • A
    {2, 4, 8}
  • B
    {2, 4, 6, 8}
  • ✓
    {2, 4, 6}
  • D
    {1, 2, 3, 4}
Answer
Correct option: C.
{2, 4, 6}
x + 2y = 8
⇒ x = 8 - 2y
For y = 1, x = 6
y = 2, x = 4
y = 3, x = 2
Then R = {(2, 3), (4, 2), (6, 1)}
$\therefore$ Domain of R = {2, 4, 6}
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MCQ 871 Mark
If two sets A and B are having 99 elements in common, then the number of elements common to each of the sets A × B and B × A are:
  • A
    $2^{99}$
  • ✓
    $99^2$
  • C
    100
  • D
    18
Answer
Correct option: B.
$99^2$
  1. $99^2$
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MCQ 881 Mark
If R is a relation on the set A = {1, 2, 3, 4, 5, 6, 7, 8, 9} given by x R y ⇔ y = 3x, then R =
  • A
    [(3, 1), (6, 2), (8, 2), (9, 3)]
  • B
    [(3, 1), (6, 2), (9, 3)]
  • C
    [(3, 1), (2, 6), (3, 9)]
  • ✓
    none of these.
Answer
Correct option: D.
none of these.
A = {1, 2, 3, 4, 5, 6, 7, 8, 9}
x R y ⇔ y = 3x
For x = 1, y = 3
For x = 2, y = 6
For x = 3, y = 9
Thus, R = {(1, 3), (2, 6), (3, 9)}
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MCQ 891 Mark
If n(A) = 3, n(B) = 4, then n (A × A × B) is equal to:
  • ✓
    36
  • B
    12
  • C
    108
  • D
    None of these
Answer
Correct option: A.
36
  1. 36
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MCQ 901 Mark
Choose the correct answers: If $[x]^2-5[x]+6=0$, where [.] denote the greatest integer function, then.
  • A
    $x \in[3,4]$
  • B
    $x \in(2,3]$
  • ✓
    $x \in[2,3]$
  • D
    $x \in[2,4)$
Answer
Correct option: C.
$x \in[2,3]$
  1. $x \in[2,3]$
Solution:
We have $[x]^2-5[x]+6=0$
$\Rightarrow[x]^2-3[x] 2[x]+6=0$
$\Rightarrow[x]([x]-3)-2([x]-3)=0$
$\Rightarrow([x]-3)([x]-2)=0 \Rightarrow[x]=2,3$
So, $x \in[2,3]$
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MCQ 911 Mark
Choose the correct answers: Let n(A) = m, and n(B) = n. Then the total number of non-empty relations that can be defined from A to B is.
  • A
    $m^n$
  • B
    $\mathrm{n}^m-1$
  • C
    $m n-1$
  • ✓
    $2^{m n}-1$
Answer
Correct option: D.
$2^{m n}-1$
  1. $2^{m n}-1$
Solution:
We have, $n(A)=m$ and $n(B)=n$
$n(A \times B)=n(A) \cdot n(B)=m n$
Total number of relation from $A$ to $B=$ Number of subsets of $A \times B=2^{m n}$
So, total number if non-empty relations $=2^{\mathrm{mn}}-1$
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MCQ 921 Mark
For the following relation R = {(0, 0), (0, 1), (1, 1), (2, 1), (2, 2), (2, 0), (1, 0), (0, 2), (0, 1)}:
  • A
    domain = {0, 1}
  • ✓
    range = {0, 1, 2}
  • C
    both correct
  • D
    none of these
Answer
Correct option: B.
range = {0, 1, 2}
  1. range = {0, 1, 2}
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MCQ 931 Mark
$\text{f}(\text{x})={\frac{\text{|x|}}{\text{x}}}$ for $ \text{x} ≠ 0$ and $ 0$ for $ \text{x} = 0$ Which function is this?
  • A
    Constant
  • B
    Modulus
  • C
    Identity
  • ✓
    Signum function
Answer
Correct option: D.
Signum function
$\text{f}(\text{x})={\frac{\text{|x|}}{\text{x}}}$ {for $ \text{x} ≠ 0$ and 0 for $ \text{x} = 0$}.
Function is {(-3, -1), (-2, -1), (-1, 1), (0, 0), (1, 1), (2, 1), (3, 1), …….}
This is signum function.
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MCQ 941 Mark
Let A = {1, 2, 3, 4, 5} and R be a relation from A to A, R = {(x, y) : y = x + 1}. Find the domain:
  • ✓
    {1, 2, 3, 4, 5}
  • B
    {2, 3, 4, 5}
  • C
    {1, 2, 3 ,4}
  • D
    {1, 2, 3, 4, 5, 6}
Answer
Correct option: A.
{1, 2, 3, 4, 5}
We know, codomain of a relation is the set to which relation is defined i.e. set A.
So, codomain = {1,2,3,4,5}.
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MCQ 951 Mark
The range of the function f(x) = P is:
  • A
    {1, 2, 3, 4}
  • B
    {1, 2, 3, 4, 5}
  • ✓
    {1, 2, 3}
  • D
    {1, 2, 3, 4, 5, 6}
Answer
Correct option: C.
{1, 2, 3}
  1. {1, 2, 3}
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MCQ 961 Mark
If $A=\{1,2,3\}$ and $B=\{x, y\}$, then the number of functions that can be defined from $A$ into $B$ is:
  • A
    12
  • ✓
    8
  • C
    6
  • D
    3
Answer
Correct option: B.
8
  1. 8
Solution:
Given,
Number of elements in set $A=3$
Number of elements in set $B=2$
Therefore, the number of functions that can be defined from $A$ into $B$ is $=2^3=8$
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MCQ 971 Mark
If $3\text{f(x)}+5\text{f}\Big(\frac{1}{\text{x}}\Big)=\frac{1}{\text{x}}-3$ for all non-zero x, then f(x) =
  • A
    $\frac{1}{14}\Big(\frac{3}{\text{x}}+5\text{x}-6\Big)$
  • B
    $\frac{1}{14}\Big(-\frac{3}{\text{x}}+5\text{x}-6\Big)$
  • C
    $\frac{1}{14}\Big(-\frac{3}{\text{x}}+5\text{x}+6\Big)$
  • ✓
    None os these.
Answer
Correct option: D.
None os these.
$3\text{f(x)}+5\text{f}\Big(\frac{1}{\text{x}}\Big)=\frac{1}{\text{x}}-3\ ...(\text{i})$
Multiplying (1) by 3,
$15\text{f}\Big(\frac{1}{\text{x}}\Big)+9\text{f(x)}=\frac{3}{\text{x}}-9\ ...(\text{ii})$
Replacing x by $\frac{1}{\text{x}}$ in (i)
$3\text{f}\Big(\frac{1}{\text{x}}\Big)+5\text{f(x)}=\text{x}-3$
Multiplying by 5
$15\text{f}\Big(\frac{1}{\text{x}}\Big)+25\text{f(x)}=5\text{x}-15\ ...(\text{iii})$
Solving (ii) and (iii),
$-16\text{f(x)}=\frac{3}{\text{x}}-5\text{x}+6$
$\Rightarrow\text{f(x)}=\frac{1}{16}\Big(-\frac{3}{\text{x}}+5\text{x}-6\Big)$
Disclaimer: The question in the book has some error, so, none of the options are matching with the solution. The solution is created according to the question given in the book.
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MCQ 981 Mark
Let A = {1, 2} and B = {3, 4}. Which of the following cannot be relation from set A to set B?
  • ✓
    {(1, 1), (1, 2), (1, 3), (1, 4)}
  • B
    {(1, 3), (1, 4)}
  • C
    {(2, 3), (2 ,4)}
  • D
    {(1, 3), (1, 4), (2, 3), (2, 4)}
Answer
Correct option: A.
{(1, 1), (1, 2), (1, 3), (1, 4)}
A relation from set A to set B is a subset of cartesian product of A × B In ordered pair, first element should belong to set A and secondelement should belongs to set B.
In {(1, 1), (1, 2), (1, 3), (1, 4)}, 1 and 2 should also be in the set B which is not so as given in question.
Hence, {(1, 1), (1, 2), (1, 3), (1, 4)} is not a relation from set A to set B.
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MCQ 991 Mark
Let $R$ be the relation on $Z$ defined by $R=\{(a, b): a, b$ $\hat{l} z, a-b$ is an interger $\}$. Find the domain and Range of $R$ :
  • ✓
    $z, z$
  • B
    $\mathrm{z}^{+}, \mathrm{z}$
  • C
    $z, z-$
  • D
    None of these
Answer
Correct option: A.
$z, z$
  1. $z, z$
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MCQ 1001 Mark
If $ \text{g}(\text{x}) = 1 +\sqrt{\text{x}}$ and $ \text{fg} (\text{x}) = 3 + 2\sqrt{\text{x} +\text{ x}},$ then $\text{f}(\text{x})=$
  • A
    $1+2 x^2$
  • ✓
    $2+x^2$
  • C
    $1+\mathrm{x}$
  • D
    $2+x$
Answer
Correct option: B.
$2+x^2$
  1. $2+x^2$
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MCQ 1011 Mark
$ \text{f} (\text{x}) = \frac{\sqrt{(\text{x} 1) (\text{x} 3)}} {\text{(x} 2)}=$ is a real valued function in the domain:
  • A
    $ (-∞, -1 ) \ \cup\ ( 3, ∞)$
  • B
    $ (-∞, -1) \ \cup\ (2, 3)$
  • ✓
    $ (-1, 2)\ \cup \ (3, \infty)$
  • D
    $ \text{none of these}$
Answer
Correct option: C.
$ (-1, 2)\ \cup \ (3, \infty)$
  1. $ (-1, 2) \ \cup\ (3, \infty)$
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MCQ 1021 Mark
Which one of the following is the domain of the relation R defined on the set N of natural numbers as R = {(m, n) : 2m + 3n = 30m, in}.
  • A
    {2, 4, 6, 8}
  • B
    {3, 7, 11, 15}
  • ✓
    {3, 6, 9, 12}
  • D
    {3, 6, 9, 12, 15}
Answer
Correct option: C.
{3, 6, 9, 12}
  1. {3, 6, 9, 12}
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MCQ 1031 Mark
If A × B = (5, 5), (5, 6), (5, 7), (8, 6), (8, 7), (8, 5), then the value A:
  • A
    (5)
  • B
    (8)
  • ✓
    (5, 8)
  • D
    (5, 6, 7, 8)
Answer
Correct option: C.
(5, 8)
  1. (5, 8)
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MCQ 1041 Mark
The function f : R → R is defined by $\text{f(x)}=\cos^2\text{x}+\sin^4\text{x}.$ Then, f(R) =
  • A
    $\Big[\frac{3}{4},1\Big]$
  • B
    $\Big(\frac{3}{4},1\Big]$
  • ✓
    $\Big[\frac{3}{4},1\Big]$
  • D
    $\Big(\frac{3}{4},1\Big)$
Answer
Correct option: C.
$\Big[\frac{3}{4},1\Big]$
Given,
$\text{f(x)}=\cos^2\text{x}+\sin^4\text{x}$
$\Rightarrow\text{f(x)}=1-\sin^2\text{x}+\sin^4\text{x}$
$\Rightarrow\text{f(x)}=\Big(\sin^2\text{x}-\frac{1}{2}\Big)^2+\frac{3}{4}$
The minimum value of $\text{f(x)}$ is $\frac{3}{4}$
Also,
$\sin^2\text{x}\leq1$
$\Rightarrow\ \sin^2\text{x}-\frac{1}{2}\leq\frac{1}{2}$
$\Rightarrow\ \Big(\sin^2\text{x}-\frac{1}{2}\Big)^2\leq\frac{1}{4}$
$\Rightarrow\ \Big(\sin^2\text{x}-\frac{1}{2}\Big)^2+\frac{3}{4}\leq\frac{1}{4}+\frac{3}{4}$
$\Rightarrow\ \text{f(x)}\leq1$
The maximum value of f(x) is 1
$\therefore\ \text{f(R)}=\Big(\frac{3}{4},1\Big)$
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MCQ 1051 Mark
Choose the correct answers: The domain and range of the real function f defined by $\text{f(x)}=\frac{4-\text{x}}{\text{x}-4}$ is given by.
  • A
    Domain = R, Range = {-1, 1}
  • B
    Domain = R - {1}, Range = R
  • ✓
    Domain = R - {4}, Range = {-1}
  • D
    Domain = R - {-4}, Range = {-1, 1}
Answer
Correct option: C.
Domain = R - {4}, Range = {-1}
Given that: $\text{f(x)}=\frac{4-\text{x}}{\text{x}-4}$
We know that f(x) is defined if $\text{x}-4\neq0 \Rightarrow \text{x}\neq4$
So, the domain of f(x) is = R - {4}
Let $\text{f(x)}=\text{y}=\frac{4-\text{x}}{\text{x}-4}$
$\Rightarrow\text{yx}-4\text{y}=4-\text{x}\Rightarrow\text{yx}+\text{x}=4\text{y}+4$
$\Rightarrow\text{x}(\text{y}+1)=4\text{y}+4\Rightarrow\text{x}=\frac{4(1+\text{y})}{1+\text{y}}$
If x is real number, then $1+\text{y}\neq0\Rightarrow\text{x}\neq1$
$\therefore$ Range of f(x) = R - {-1)
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MCQ 1061 Mark
The domain of the function $ \text{f}(\text{x}) = \sqrt{(2-2\text{x}-\text{x2})}$ is:
  • A
    $ – \sqrt{3} ≤ \times ≤ \sqrt{3}$
  • ✓
    $ -1– \sqrt{3} ≤ \times ≤ -1+\sqrt{3}$
  • C
    $ -2 ≤ \times ≤ 2$
  • D
    $ -2 – \sqrt{3} ≤\times ≤ – 2 + \sqrt3 $
Answer
Correct option: B.
$ -1– \sqrt{3} ≤ \times ≤ -1+\sqrt{3}$
  1. $ -1– \sqrt{3} ≤ \times ≤ -1+\sqrt{3}$
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MCQ 1071 Mark
If f(x) = ax + b and g(x) = cx + d and f{g(x)} = g{f(x)} then:
  • A
    f(a) = g(c)
  • B
    f(b) = g(b)
  • ✓
    f(d) = g(b)
  • D
    f(c) = g(a)
Answer
Correct option: C.
f(d) = g(b)
Given, f(x) = ax + b and g(x) = cx + d and
Now, f{g(x)} = g{f(x)}
⇒ f{cx + d} = g{ax + b}
⇒ a(cx + d) + b = c(ax + b) + d
⇒ ad + b = cb + d
⇒ f(d) = g(b)
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MCQ 1081 Mark
If the relation R : A → B, where A = {1, 2, 3, 4} and B = {1, 3, 5} is defined by $ \text{R} = {(\text{x, y}) : \text{x} < \text{y}, \text{x } \text{iA}, \text{y} ∈ \text{B}},$ then $R^{-1}$ OR is:
  • A
    (1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)
  • B
    (3, 1), (5, 1), (5, 2), (5, 3), (5, 4)
  • ✓
    (3, 3), (3, 5), (5, 3), (5, 5)
  • D
    None of the above
Answer
Correct option: C.
(3, 3), (3, 5), (5, 3), (5, 5)
  1. (3, 3), (3, 5), (5, 3), (5, 5)
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MCQ 1091 Mark
The domain of definition of $\text{f(x)}=\sqrt{4\text{x}-\text{x}^2}$ is:
  • A
    $\text{R}-[0,4]$
  • B
    $\text{R}-(0,4)$
  • C
    $(0,4)$
  • ✓
    $[0,4]$
Answer
Correct option: D.
$[0,4]$
$\text{f(x)}=\sqrt{4\text{x}-\text{x}^2}$
Clearly, f(x) assumes real values if
$4\text{x}-\text{x}^2\geq0$
$\Rightarrow\text{x}(4-\text{x})\geq0$
$\Rightarrow-\text{x}(\text{x}-4)\geq0$
$\Rightarrow\text{x}(\text{x}-4)\leq0$
$\Rightarrow\text{x}\in[0,4]$
Hence, domain $(\text{f})=[0,4]$
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MCQ 1101 Mark
If A is the null set and B is an infinite set, then what is A × B?
  • A
    Infinite set
  • B
    f
  • ✓
    Undefined
  • D
    A singleton set
Answer
Correct option: C.
Undefined
  1. Undefined
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MCQ 1111 Mark
Let a relation $R$ be defined by $R = ((4, 5), (1, 4), (4, 6), (7, 6), (3, 7)),$ then $\ce{ROR}$ is equal to:
  • A
    $((1, 5), (1, 6), (3, 6))$
  • B
    $((1, 4), (1, 5), (3, 6))$
  • C
    $((1, 5), (1, 6), (3, 7))$
  • ✓
    $((1, 4), (1, 5), (3, 7))$
Answer
Correct option: D.
$((1, 4), (1, 5), (3, 7))$
$((1, 4), (1, 5), (3, 7))$
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MCQ 1121 Mark
If $\text{f(x)}=64\text{x}^3+\frac{1}{\text{x}^3}$ and $\alpha,\beta$ are the roots of $4\text{x}+\frac{1}{\text{x}}=3.$ Then,
  • ✓
    $\text{f}(\alpha)=\text{f}(\beta)=-9$
  • B
    $\text{f}(\alpha)=\text{f}(\beta)=63$
  • C
    $\text{f}(\alpha)\neq\text{f}(\beta)$
  • D
    None of these.
Answer
Correct option: A.
$\text{f}(\alpha)=\text{f}(\beta)=-9$
$\text{f(x)}=64\text{x}^3+\frac{1}{\text{x}^3}$
$\Rightarrow\text{f(x)}=\Big(4\text{x}+\frac{1}{\text{x}}\Big)\Big(16\text{x}^2+\frac{1}{\text{x}^2}-4\Big)$
$\Rightarrow\text{f(x)}=\Big(4\text{x}+\frac{1}{\text{x}}\Big)\Big(\Big(4\text{x}+\frac{1}{\text{x}}\Big)^2-12\Big)$
$\Rightarrow\text{f}(\text{a})=\Big(4\alpha+\frac{1}{\alpha}\Big)\Big(\Big(4\alpha+\frac{1}{\alpha}\Big)^2-12\Big)$ and $\text{f}(\beta)=\Big(4\beta+\frac{1}{\beta}\Big)\Big(\Big(4\beta+\frac{1}{\beta}\Big)^2-12\Big)$
Since $\alpha$ and $\beta$ are the roots of $4\text{x}+\frac{1}{\text{x}}=3,$
$4\alpha+\frac{1}{\alpha}=3$ and $4\beta+\frac{1}{\beta}=3$
$\Rightarrow\text{f}(\alpha)=3\big((3)^2-12\big)=-9$ and $\text{f}(\beta)=3\big((3)^2-12\big)=-9$
$\Rightarrow\text{f}(\alpha)=\text{f}(\beta)=-9$
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MCQ 1131 Mark
If the domain of the function $\text{f}(\text{x})=\text{x}^2$ then the range of function is:
  • ✓
    $ (-2, \infty )$
  • B
    $ ( ​-\infty, \infty )$
  • C
    $ (-2, +1)$
  • D
    $ (-\infty, -2)$
Answer
Correct option: A.
$ (-2, \infty )$
  1. $ (-2, \infty )$
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MCQ 1141 Mark
The domain of the function $\text{f}(\text{x})= ^{7-\text{x}}\text{P}_\text{x-3}$ is:
  • A
    {1, 2, 3}
  • B
    {3, 4, 5, 6}
  • ✓
    {3, 4, 5}
  • D
    {1, 2, 3, 4, 5}
Answer
Correct option: C.
{3, 4, 5}
The function $\text{f}(\text{x})= ^{7-\text{x}}\text{P}_\text{x-3}$ is defined only if x is an integer satisfying the following inequalities:
$ 7 - \text{x} ≥ 0.....(1)$
$ \text{x} - 3 ≥ 0......(2)$
$ 7 - \text{x} ≥ \text{x} - 3........(3)$
Now, from 1, we get $ \text{x} ≤ 7 ……… (4)$
from 2, we get $\text{x} ≥ 3 ……………. (5)$
and from 2, we get $\text{x} ≤ 5 ………. (6)$
From 4, 5 and 6, we get
$ 3 ≤ \text{x} ≤ 5$
So, the domain is {3, 4, 5}.
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MCQ 1151 Mark
The domain of the function $\text{f(x)}=\sqrt{5|\text{x}|-\text{x}^2-16}$ is:
  • A
    $(-3,-2)\cup(2,3)$
  • B
    $\big[-3,-2\big]\cup\big[2,3\big) $
  • ✓
    $\big[-3,-2\big]\cup\big[2,3\big] $
  • D
    None os these.
Answer
Correct option: C.
$\big[-3,-2\big]\cup\big[2,3\big] $
$\text{f(x)}=\sqrt{5|\text{x}|-\text{x}^2-16}$
For f(x) to be defined, $5|\text{x}|-\text{x}^2-6\geq0$
$\Rightarrow5|\text{x}|-\text{x}^2-6\geq0$
$\Rightarrow\text{x}^2-5|\text{x}|+6\leq0$
For x > 0, |x| = x
$\Rightarrow\text{x}^2-5\text{x}+6\leq0$
$\Rightarrow(\text{x}-2)(\text{x}-3)\leq0$
$\Rightarrow\text{x}\in[2,3]\ ...(\text{i})$
For x < 0, |x| = -x
$\Rightarrow\text{x}^2+5\text{x}+6\leq0$
$\Rightarrow(\text{x}+2)(\text{x}+3)\leq0$
$\Rightarrow\text{x}\in\big[-3,-2\big]\ ...(\text{ii})$
From (i) and (ii),
$\text{x}\in\big[-3,-2\big]\cup\big[2,3\big]$
Or, $\text{domain(f)}=\big[-3,-2\big]\cup\big[2,3\big]$
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MCQ 1161 Mark
If $A = \{1, 2, 3\}, B = \{3, 4\}, C = \{4, 5, 6\},$ then:
  • A
    $ (\text{A}\times\text{B}) ∩ (\text{B}\times\text{C})$
  • B
    $ \{(1, 4)\} \{(3, 4)\}$
  • ✓
    $ \{(1, 4), (3, 4)\}$
  • D
    None of these
Answer
Correct option: C.
$ \{(1, 4), (3, 4)\}$
$ \{(1, 4), (3, 4)\}$
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MCQ 1171 Mark
f is a real valued function given by $\text{f(x)}=27\text{x}^3+\frac{1}{\text{x}^3}$ and $\alpha,\beta$ are roots of $3\text{x}+\frac{1}{\text{x}}=12$ Then,
  • A
    $\text{f}(\alpha)\neq\text{f}(\beta)$
  • B
    $\text{f}(\alpha)=10$
  • C
    $\text{f}(\beta)=-10$
  • ✓
    None of these.
Answer
Correct option: D.
None of these.
$\text{f(x)}=27\text{x}^3+\frac{1}{\text{x}^3}$
$\Rightarrow\text{f(x)}=\Big(3\text{x}+\frac{1}{\text{x}}\Big)\Big(9\text{x}^2+\frac{1}{\text{x}^2}-3\Big)$
$\Rightarrow\text{f(x)}=\Big(3\text{x}+\frac{1}{\text{x}}\Big)\Big(\Big(3\text{x}+\frac{1}{\text{x}}\Big)^2-9\Big)$
$\Rightarrow\text{f}(\alpha)=\Big(3\alpha+\frac{1}{\alpha}\Big)\Big(\Big(3\alpha+\frac{1}{\alpha}\Big)^2-9\Big)$
Since $\alpha$ and $\beta$ are the roots of $3\text{x}+\frac{1}{\text{x}}=12,$
$3\alpha+\frac{1}{\alpha}=12$ and $3\beta+\frac{1}{\beta}=12$
$\Rightarrow\text{f}(\alpha)=12\big((12)^2-9\big)$ and $\text{f}(\beta)=12\big((12)^2-9\big)$
$\Rightarrow\text{f}(\alpha)=\text{f}(\beta)=\big((12)^2-9\big)$
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MCQ 1181 Mark
If the function f : R → R be given by $f(x)=x^2+2$ and g : R → R is given by $ \text{g}(\text{x})=\frac{\text{x}}{(\text{x} - 1)}$. The value of gof(x) is:
  • ✓
    $\frac{\text{ (x}^{2} + 2)}{(\text{x}^{2} + 1)}$
  • B
    $ \frac{\text{x}^{2}}{(\text{x} - 1)}$
  • C
    $ \frac{\text{x}^{2}}{(\text{x} +2)}$
  • D
    $\text{none of these}$
Answer
Correct option: A.
$\frac{\text{ (x}^{2} + 2)}{(\text{x}^{2} + 1)}$
  1. $\frac{\text{ (x}^{2} + 2)}{(\text{x}^{2} + 1)}$
Solution:
$\frac{\text{ (x}^{2} + \ 2)}{(\text{x}^{2} + \ 1)}$
Given $f(x)=x^2+2$ and $ \text{g}(\text{x})=\frac{\text{x}}{(\text{x}\ - \ 1)}$
Now, $\text{gof(x)} = \text{g}(\text{x}^2 + 2) = \text{got}(\text{x})=\text{g}(\text{x}^2 +2)=\frac{\text{ (x}^{2} +\ 2)}{(\text{x}^{2} \ +\ 2\ -\ 1)}=\frac{(\text{x}^2\ +\ 2)}{(\text{x}^2\ +\ 1)}$
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MCQ 1191 Mark
If the set A has 3 elements and the set B = {1, 3, 4, 5}, then the number of elements in (A × B) is:
  • A
    11
  • ✓
    12
  • C
    13
  • D
    15
Answer
Correct option: B.
12
  1. 12
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MCQ 1201 Mark
Choose the correct answers: Range of $\text{f(x)}=\frac{1}{1-2\cos\text{x}}$ is.
  • A
    $\Big[\frac{1}{3}, 1\Big]$
  • B
    $\Big[-1, \frac{1}{3}\Big]$
  • ✓
    $(-\infty, -1]\cup\Big[\frac{1}{3},\infty\Big)$
  • D
    $\Big[-\frac{1}{3}, 1\Big]$
Answer
Correct option: C.
$(-\infty, -1]\cup\Big[\frac{1}{3},\infty\Big)$
We know that, $-1\leq-\cos\text{x}\leq1$
$\Rightarrow-1\leq-\cos\text{x}\leq1$
$\Rightarrow-2\leq-2\cos\text{x}\leq2$
$\Rightarrow-1\leq-2\cos\text{x}\leq3$
Now $\text{f(x)}=\frac{1}{1-2\cos\text{x}}$ is defined if
$-1\leq-2\cos\text{x}\leq0$ or $0<1-2\cos\text{x}\leq3$
$\Rightarrow-1\geq\frac{1}{1-2\cos\text{x}}>-\infty$ or $\infty>\frac{1}{1-2\cos\text{x}}\geq\frac{1}{3}$
$\Rightarrow\frac{1}{1-2\cos\text{x}}\in(-\infty, -1]\cup\Big[\frac{1}{3},\infty\Big)$
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MCQ 1211 Mark
The domain of the function $ \text {f} (\text{x}) = \frac{1}{(2 -\cos 3\text{x})}$ is:
  • A
    $ \Big (\frac{1}{3}, 1\Big)$
  • B
    $ \Big (\frac{1}{3}, 1\Big)$
  • C
    $ \Big (\frac{1}{3}, 1\Big)$
  • ✓
    $ \text{R}$
Answer
Correct option: D.
$ \text{R}$
Given,
function is $ \text{f}(\text{x}) = \frac{1}{(2 -\cos 3\text{x})}$
Since $ -1 \leq \cos \text{3x} \leq1$ for all$\text{ x }∈\text{R}$
So,$ -1 \leq 2 \cos \text{3x} \leq1$for all$\text{ x }∈\text{R}$
⇒ $\text{f}(\text{x})$ is defined for all$\text{ x }∈\text{R}$
So, domain of f(x) is R.
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MCQ 1221 Mark
The domain of the function $\text{f(x)}=\sqrt{\frac{(\text{x}+1)(\text{x}-3)}{\text{x}-2}}$ is:
  • ✓
    $\big[-1,2\big)\cap\big[3,\infty\big)$
  • B
    $(-1,2)\cap[3,\infty)$
  • C
    $[-1,2]\cap[3,\infty]$
  • D
    None of these.
Answer
Correct option: A.
$\big[-1,2\big)\cap\big[3,\infty\big)$
$\text{f(x)}=\sqrt{\frac{(\text{x}+1)(\text{x}-3)}{\text{x}-2}}$
For f(x) to be defined,
$(\text{x}-2)\neq0$
$\Rightarrow\text{x}\neq2\ ...(\text{i})$
Also,
$\frac{(\text{x}+1)(\text{x}-3)}{\text{x}-2}\geq0$
$\Rightarrow\frac{(\text{x}+1)(\text{x}-3)(\text{x}-2)}{(\text{x}-2)^2}\geq0$
$\Rightarrow(\text{x}+1)(\text{x}-3)(\text{x}-2)\geq0$
$\Rightarrow\text{x}\in\big[-1,2\big)\cup\big[3,\infty\big)\ ...(\text{ii})$
From (i) and (ii),
$\text{x}\in\big[-1,2\big)\cap\big[3,\infty\big)$
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MCQ 1231 Mark
Let $\text{f(x)}=\sqrt{\text{x}^2+1}$ Then which of the following is correct?
  • A
    $\text{f(xy)}=\text{f(x)}\text{f(y)}$
  • B
    $\text{f(xy)}\geq\text{f(x)}\text{f(y)}$
  • ✓
    $\text{f(xy)}\leq\text{f(x)}\text{f(y)}$
  • D
    None os these.
Answer
Correct option: C.
$\text{f(xy)}\leq\text{f(x)}\text{f(y)}$
Given, $\text{f(x)}=\sqrt{\text{x}^2+1}\ ...(\text{i})$
Replacing x by y in (i), we get
$\text{f(y)}=\sqrt{\text{y}^2+1}$
$\therefore\ \text{f(x)}\text{f(y)}=\sqrt{\text{x}^2+1}\sqrt{\text{y}^2+1}$
$=\sqrt{(\text{x}^2+1)(\text{y}^2+1)}$
$=\sqrt{\text{x}^2\text{y}^2+\text{x}^2+\text{y}^2+1}$
Also, replacing x by xy in (i), we get
$\text{f(xy)}=\sqrt{\text{x}^2\text{y}^2+1}$
Now,
$\text{x}^2\text{y}^2+1\leq\text{x}^2\text{y}^2+\text{x}^2+\text{y}^2+1$
$\Rightarrow\sqrt{\text{x}^2\text{y}^2+1}\leq\sqrt{\text{x}^2\text{y}^2+\text{x}^2+\text{y}^2+1}$
$\Rightarrow\text{f}(\text{xy})\leq\text{f(x)}\text{f(y)}$
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MCQ 1241 Mark
The period of the function $\text{f(x)}=\sin \big(\frac{2\pi\text{x}}{3}\big)+\cos\big(\frac{\pi\text{x}}{3}\big)$:
  • A
    3
  • B
    4
  • ✓
    12
  • D
    None of these
Answer
Correct option: C.
12
Given, function $ \text{f}(\text{x})=\sin \big(\frac{2\pi\text{x}}{3}\big)+\cos \big(\frac{\pi\text{x}}{2}\big)$
Now, period of $ \text{f}(\text{x})=\big(\frac{2\pi\text{x}\times{3}}{2\pi}\big)=3$
and period of $\cos \Big(\frac{\text{n}\pi}{2}\Big)=\frac{2\text{n}}{\frac{\text{n}}{2}} = \big(\frac{2\pi\text{}\times{2}}{ \pi}\big)= 2 × 2 = 4$ $$
Now, period of f(x) = LCM(3, 4) = 12 Hence, period of function $\text{f(x)} = \sin\frac{2\pi\text{x}}{3} + \cos \big(\frac{\pi\text{x}}{2}\big) + \cos$ $$is 12
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MCQ 1251 Mark
$R$ is a relation from $\{11,12,13\}$ to $\{8,10,12\}$ defined by $y=x-3$. Then, $R^{-1}$ is:
  • ✓
    $\{(8,11),(10,13)\}$
  • B
    $\{(11,8),(13,10)\}$
  • C
    $\{(10,13),(8,11),(12,10)\}$
  • D
    none of these.
Answer
Correct option: A.
$\{(8,11),(10,13)\}$
  1. {(8, 11), (10, 13)}
Solution:
$R$ is a relation from $\{11,12,13\}$ to $\{8,10,12\}$, defined by $y=x-3$
Now, we have,
$11-3=8$
$13-3=10$
So, $R=\{(13,10),(11,8)\}$
$\therefore R^{-1}=\{(10,13),(8,11)\}$
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MCQ 1261 Mark
If $\text{f(x)}=\cos(\log\text{x}),$ then the value of $\text{f(x}^2)\text{f}(\text{y}^2)-\frac{1}{2}\Big\{\text{f}\Big(\frac{\text{x}^2}{\text{y}^2}\Big)+\text{f}\big(\text{x}^2\text{y}^2\big)\Big\}$ is:
  • A
    -2
  • B
    -1
  • C
    $\frac{1}{2}$
  • ✓
    None of these.
Answer
Correct option: D.
None of these.
Given,
$\text{f(x)}=\cos(\log\text{x})$
$\Rightarrow\ \text{f(x}^2)=\cos(\log(\text{x}^2))$
$\Rightarrow\ \text{f(x}^2)=\cos(2\log(\text{x}))$
Similarly,
$\text{f}(\text{y}^2)=\cos(2\log(\text{y}))$
Now,
$\text{f}\Big(\frac{\text{x}^2}{\text{y}^2}\Big)=\cos\Big(\log\Big(\frac{\text{x}^2}{\text{y}^2}\Big)\Big)=\cos\big(\log\text{x}^2-\log\text{y}^2\big)$
and $\text{f}(\text{x}^2\text{y}^2)=\cos(\log\text{x}^2\text{y}^2)=\cos\big(\log\text{x}^2+\log\text{y}^2\big)$
$\Rightarrow\ \text{f}\Big(\frac{\text{x}^2}{\text{y}^2}\Big)+\text{f}(\text{x}^2\text{y}^2)=\cos\big((2\log\text{x}-2\log\text{y})\big)+\cos\big((2\log\text{x}-2\log\text{y})\big)$
$\Rightarrow\ \text{f}\Big(\frac{\text{x}^2}{\text{y}^2}\Big)+\text{f}(\text{x}^2\text{y}^2)=2\cos(2\log\text{x})\cos(2\log\text{y})$
$\Rightarrow\frac{1}{2}\bigg[\Big(\text{f}\Big(\frac{\text{x}^2}{\text{y}^2}\Big)+\text{f}(\text{x}^2\text{y}^2)\bigg]=\cos(2\log)\cos(2\log\text{y})$
$\Rightarrow\ \text{f(x}^2)\text{f}(\text{y}^2)-\frac{1}{2}\Big\{\text{f}\Big(\frac{\text{x}^2}{\text{y}^2}\Big)+\text{f}\big(\text{x}^2\text{y}^2\big)\Big\}\\\ =\cos(2\log)\cos(2\log\text{y})-\cos(2\log)\cos(2\log\text{y})=0$
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MCQ 1271 Mark
Choose the correct answers: The domain for which the functions defined by $f(x)=3 x^2-1$ and $g(x)=3+x$ are equal is.
  • ✓
    $\Big\{-1, \frac{4}{3}\Big\}$
  • B
    $\Big[-1, \frac{4}{3}\Big]$
  • C
    $\Big(-1, -\frac{4}{3}\Big)$
  • D
    $\Big[-1, -\frac{4}{3}\Big)$
Answer
Correct option: A.
$\Big\{-1, \frac{4}{3}\Big\}$
  1. $\Big\{-1, \frac{4}{3}\Big\}$
Solution:
We have, $f(x)=3 x^2-1$ and $g(x)=3+x$
$f(x)=g(x)$
$\Rightarrow 3 x^2-1=3+x$
$\Rightarrow 3 x^2-x-4=0$
$\Rightarrow(3 x-4)(x+1)=0$
$\therefore x=-1, \frac{4}{3}$
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MCQ 1281 Mark
If $f: Q \rightarrow Q$ is defined as $f(x)=x^2$, then $f^{-1}(9)$ is equal to:
  • A
    3
  • B
    -3
  • ✓
    {-3, 3}
  • D
    $\phi$
Answer
Correct option: C.
{-3, 3}
  1. {-3, 3}
Solution:
If f : A → B, such that $\text{y}\in\text{B},$ then $\text{f}^{-1}(\text{y})=\{\text{x}\in\text{A}:\text{f(x)}=\text{y}\}$
In other words, $\text{f}^{-1}{y}$ is the set of pre-images of y.
Let $\text{f}^{-1}\{9\}=\text{x}$
Then, $\text{f(x)}=9$
$\Rightarrow\text{x}^2=9$
$\Rightarrow\text{x}=\pm3$
$\therefore\ \text{f}^{-1}\{9\}=\{-3,3\}$
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MCQ 1291 Mark
If $A=\{1,4,8,9\}$ and $B=\{1,2,-1,-2,-3,3,5\}$ and $R$ is a relation from set $A$ to set $B\left\{(x, y): x=y^2\right\}$. Find domain of the relation:
  • A
    {1, 4, 9}
  • B
    {-1, 1, -2, 2, -3, 3}
  • ✓
    {1 , 4, 8, 9}
  • D
    {-1, 1, -2, 2, -3, 3, 5}
Answer
Correct option: C.
{1 , 4, 8, 9}
  1. {1, 4, 8, 9}

Solution:
We know, domain of a relation is the set from which relation is defined i.e. set A.
So, domain = {1, 4, 8, 9}.
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MCQ 1301 Mark
Find range of function |x|:
  • A
    Set of real numbers
  • ✓
    Set of positive real numbers
  • C
    Set of integers
  • D
    Set of natural numbers
Answer
Correct option: B.
Set of positive real numbers
Since the above function can have positive real value of y for all real values of x. So, range is set of positive real numbers.
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MCQ 1311 Mark
The domain of definition of the function $\text{f(x)}=\log|\text{x}|$ is:
  • A
    $\text{R}$
  • B
    $\big(-\infty,0\big)$
  • C
    $(0,\infty)$
  • ✓
    $\text{R}-\{0\}$
Answer
Correct option: D.
$\text{R}-\{0\}$
$\text{f(x)}=\log|\text{x}|$
For f(x) to be defined,
$|\text{x}|>0,$ which is always true.
But $|\text{x}|\neq0$
$\Rightarrow\text{x}\neq0$
Thus, $\text{domain(f)}=\text{R}-\{0\}$
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MCQ 1321 Mark
If set A has 2 elements and set B has 3 elements then how many subsets does A × B have?
  1. 6
  2. 8
  3. 32
  4. 64
  • A
    6
  • B
    8
  • C
    32
  • ✓
    64
Answer
Correct option: D.
64
  1. 64
Solution:
If set $A$ has $m$ elements and set $B$ has $n$ elements then $A \times B$ has $m \times n$ elements. We know, a set has $2^r$ subsets if it has r number of elements.
Here, $A \times B$ has $2 \times 3=6$ elements. So, number of subsets of $A \times B$ will be $2^6$ i.e. 64 .
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MCQ 1331 Mark
$\text{f}(\text{x})=\sqrt{9-\text{x}^2}$. Find the domain of the function:
  • A
    (0, 3)
  • B
    (0, 3)
  • ✓
    (-3, 3)
  • D
    (-3, 3)
Answer
Correct option: C.
(-3, 3)
We know radical cannot be negative.
So, $9-\text{x}^2,\geq 0$
$ (3 - \text{x}) (3 + \text{x}) ≥$
$\Rightarrow (\text{x} - 3) (\text{x} + 3) ≤ 0$
$ \Rightarrow \text{x} ∈ [-3,3].$
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MCQ 1341 Mark
Let $\text{A}=\{\text{x}\in\text{R}:\text{x}\neq0-4\leq\text{x}\leq4\}$ and $\text{f}:\text{A}\in\text{R}$ be defined by $\text{f(x)}=\frac{|\text{x}|}{\text{x}}$ for $\text{x}\in\text{A}$ Then A:
  • ✓
    $[1,-1]$
  • B
    $\big[\text{x}:0\leq\text{x}\leq4\big]$
  • C
    $\{1\}$
  • D
    $\{\text{x}:-4\leq\text{x}\leq0\}$
Answer
Correct option: A.
$[1,-1]$
As, $\text{|x|}=\begin{cases}\text{x},\ \text{x}\geq0\\-\text{x}<0\end{cases}$
So, $\text{f(x)}=\frac{\text{x}}{|\text{x}|}$
When, $\text{x}<0\text{ i.e.,}\text{ x}\in\big[-4,0\big)$
$\text{f(x)}=\frac{\text{x}}{-\text{x}}=-1$
and when, $\text{x}>0\text{ i.e., x}\in\big(0,4\big]$
$\text{f(x)}=\frac{\text{x}}{\text{x}}=1$
So, range $\text{f}=\{-1,1\}$
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MCQ 1351 Mark
If f(x) = (x - 1), (x - 3), (x - 4), (x - 6) + 19 for all real value of x is:
  • ✓
    positive
  • B
    negative
  • C
    zero
  • D
    none of thes
Answer
Correct option: A.
positive
  1. positive
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MCQ 1361 Mark
Let $f: R \rightarrow R$ be a function given by $f(x)=x^2+1$ then the value of $f^{-1}(26)$ is:
  • A
    $ ​​5$
  • B
    $ -5$
  • ✓
    $ ±5$
  • D
    $ \text{None of these}$
Answer
Correct option: C.
$ ±5$
  1. $ ±5$
Solution:
Let $\text{y}=\text{f}(\text{x})=\text{x}^{2}+1$
⇒ $\text{y}=\text{x}^{2}+1$
⇒ $ \text{y}-1=\text{x}^{2}$
⇒ $ \text{x} = ±\sqrt{(\text{y} – 1)}$
⇒ $\text{f}^{-1} \text{x} = ±\sqrt{(\text{x} – 1)}$
Now, $\text{f}^{-1} (26) = ±\sqrt{(\text{26} – 1)}$
⇒ $\text{f}^{-1} (26) = ±\sqrt{(\text{25} )}$
⇒ $\text{f}^{-1} (26) = ± {5 }$
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MCQ 1371 Mark
Find domain of function |x|:
  • ✓
    Set of real numbers
  • B
    Set of positive real numbers
  • C
    Set of integers
  • D
    Set of natural numbers
Answer
Correct option: A.
Set of real numbers
Since the above function can have all real values of x. So, domain is set of real numbers.
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MCQ 1381 Mark
Let f(x) = |x - 1|. Then:
  • A
    $(\text{x}^2) = [\text{f(x)}]^2$
  • B
    $\text{f(x + y) = f(x)f(y)}$
  • C
    $\text{f(|x|) = |f(x)|}$
  • ✓
    None of these.
Answer
Correct option: D.
None of these.
$\text{f(x)}=|\text{x}-1|$
Since, $|\text{x}^2-1|\neq|\text{x}-1|^2$
$\text{f(x)}^2\neq(\text{f(x)})^2$
Thus, (i) is wrong.
Since, $|\text{x}+\text{y}-1|\neq|\text{x}-1||\text{y}-1|$
$\text{f}(\text{x}+\text{y})\neq\text{f(x)}\text{f(y)}$
Thus, (ii) is wrong.
Since, $|\text{|x|}-1\neq||\text{x}-1||=|\text{x}-1|$
$\text{f(|x|)}\neq|\text{f(x)}|$
Thus, (iii) is wrong.
Hence, none of the given options is the answer.
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MCQ 1391 Mark
If $\text{f(x)}=\cos(\log_\text{e}),$ then $\text{f}\Big(\frac{1}{\text{x}}\Big)\text{f}\Big(\frac{1}{\text{y}}\Big)-\frac{1}{2}\Big\{\text{f(xy)}+\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)\Big\}$ is equal to:
  • A
    $\cos(\text{x}-\text{y})$
  • B
    $\log(\cos(\text{x}-\text{y}))$
  • C
    $1$
  • ✓
    $\cos(\text{x}+\text{y})$
Answer
Correct option: D.
$\cos(\text{x}+\text{y})$
$\text{f(x)}=\cos(\log_\text{e}\text{x})$
$\Rightarrow\text{f}\Big(\frac{1}{\text{x}}\Big)=\cos\Big(\log_\text{e}\Big(\frac{1}{\text{x}}\Big)\Big)$
$\Rightarrow\text{f}\Big(\frac{1}{\text{x}}\Big)=\cos\Big(-\log_\text{e}(\text{x})\Big)$
$\Rightarrow\text{f}\Big(\frac{1}{\text{x}}\Big)=\cos\Big(\log_\text{e}(\text{x})\Big)$
Similarly,
$\text{f}\Big(\frac{1}{\text{y}}\Big)=\cos(\log_\text{e}\text{y})$
Now,
$\text{f(xy)}=\cos(\log_\text{e}\text{xy})=\cos\big(\log_\text{e}\text{x}+\log_\text{e}\text{y}\big)$
and
$\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)=\cos\Big(\log_\text{e}\frac{\text{x}}{\text{y}}\Big)=\cos\big(\log_\text{e}\text{x}-\log_\text{e}\text{y}\big)$
$\Rightarrow\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f(xy)}=\cos\big(\log_\text{e}\text{x}+\log_\text{e}\text{y}\big)+\cos\big(\log_\text{e}\text{x}-\log_\text{e}\text{y}\big)$
$\Rightarrow\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f(xy)}=2\cos\big(\log_\text{e}\text{x}\big)\cos(\log_\text{e}\text{y})$
$\Rightarrow\frac{1}{2}\Big[\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f(xy)}\Big]=\cos\big(\log_\text{e}\text{x}\big)\cos\big(\log_\text{e}\text{y}\big)$
$\Rightarrow\text{f}\Big(\frac{1}{\text{x}}\Big)\text{f}\Big(\frac{1}{\text{y}}\Big)-\frac{1}{2}\Big\{\text{f(xy)}+\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)\Big\}=\cos\big(\log_\text{e}\text{x}\big)\cos\big(\log_\text{e}\text{y}\big)\\\ \ -\cos\big(\log_\text{e}\text{x}\big)\cos\big(\log_\text{e}\text{y}\big)=0$
Disclaimer: The question in the book has some error, so none of the options are matching with the solution. The solution is created according to the question given in the book.
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MCQ 1401 Mark
If $\text{f(x)}=\sin[\pi^2]\text{x}+\sin[-\pi^2]\text{x},$ where [x] denotes the greatest integer less than or equal to x, then:
  • ✓
    $\text{f}\Big(\frac{\pi}{2}\Big)=1$
  • B
    $\text{f}(\pi)=2$
  • C
    $\text{f}\Big(\frac{\pi}{4}\Big)=-1$
  • D
    None of these.
Answer
Correct option: A.
$\text{f}\Big(\frac{\pi}{2}\Big)=1$
$\text{f(x)}=\sin[\pi^2]\text{x}+\sin[-\pi^2]\text{x}$
$\Rightarrow\text{f(x)}=\sin\big[9.8\big]\text{x}+\sin\big[-9.8\big]\text{x}$
$\Rightarrow\text{f(x)}=\sin9\text{x}-\sin10\text{x}$
$\Rightarrow\text{f}\Big(\frac{\pi}{2}\Big)=\sin9\times\frac{\pi}{2}-\sin10\times\frac{\pi}{2}$
$\Rightarrow\text{f}\Big(\frac{\pi}{2}\Big)=1-0=1$
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MCQ 1411 Mark
Choose the correct answers: Let $\text{f(x)}=\sqrt{1+\text{x}^2}$ then.
  • A
    $\text{f(xy)} = \text{f(x)}.\text{f(y)}$
  • B
    $\text{f(xy)} \geq \text{f(x)}.\text{f(y)}$
  • ✓
    $\text{f(xy)} \leq \text{f(x)}.\text{f(y)}$
  • D
    None of these
Answer
Correct option: C.
$\text{f(xy)} \leq \text{f(x)}.\text{f(y)}$
Given that: $\text{f(x)}=\sqrt{1+\text{x}^2}$
$\Rightarrow\text{f(xy)}=\sqrt{1+\text{x}^2\text{y}^2}$
and $\text{f(x)}.\text{f(y)}=\sqrt{1+\text{x}^2}.\sqrt{1+\text{x}^2}=\sqrt{1+\text{x}^2 +\text{y}^2+\text{x}^2\text{y}^2}$
$\therefore\sqrt{1+\text{x}^2\text{y}^2}\leq\sqrt{1+\text{x}^2 +\text{y}^2+\text{x}^2\text{y}^2}$
$\text{f(xy)} \leq \text{f(x)}.\text{f(y)}$
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MCQ 1421 Mark
If P × Q is an empty set then which of the following is a null set?
  • A
    only P
  • B
    only Q
  • ✓
    either P or Q
  • D
    both P and Q
Answer
Correct option: C.
either P or Q
If either set P or set Q is a null set then $\text{P}\times\text{Q}$ is an empty set.
i.e. if P is $ \phi$ or Q is $ \phi$ then $ \text{P} ×\text{Q}=\phi $.
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MCQ 1431 Mark
Let f : R × R be a function defined by $ \text{f}(\text{x}) = \cos(2\text{x} + 5)$, then f is:
  • A
    injective
  • B
    surjective
  • C
    bijective
  • ✓
    None of these
Answer
Correct option: D.
None of these
Given, $ \text{f}(\text{x}) = \cos(2\text{x} + 5)$
Period of $\text{f}(\text{x})=\frac{2\pi}{5}$
Since f(x) is a periodic function with period $\text{f}(\text{x})=\frac{2\pi}{5}$ so it is not injective.
The function f is not surjective also as its range [-1, 1] is a proper subset of its co-domain R.
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MCQ 1441 Mark
If g= {(1, 1), (2, 3), (3, 5), (4, 7)} is a function described by the formula, g(x) = ax + b then what values should be assigned to a and b?
  • A
    a = 1, b = 1
  • ✓
    a = 2, b = -1
  • C
    a = 1, b = -2
  • D
    a = -2, b = -1
Answer
Correct option: B.
a = 2, b = -1
  1. a = 2, b = -1
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MCQ 1451 Mark
If $\text{f(x)}=\frac{\sin^{4}\text{x}+\cos^2\text{x}}{\sin^2\text{x}+\cos^4\text{x}}$ for $\text{x}\in\text{R},$ then f(2002) =
  • ✓
    1
  • B
    2
  • C
    3
  • D
    4
Answer
Correct option: A.
1
Given,
$\text{f(x)}=\frac{\sin^{4}\text{x}+\cos^2\text{x}}{\sin^2\text{x}+\cos^4\text{x}}$
On dividing the numerator and denominator by $\cos^4\text{x},$ we get
$\text{f(x)}=\frac{\tan^4\text{x}+\sec^2\text{x}}{1+\tan^2\text{x}\sec^2\text{x}}$
$=\frac{1+\tan^4\text{x}+\tan^2\text{x}}{1+\tan^2\text{x}(1+\tan^2\text{x})}$
$=\frac{1+\tan^{4}\text{x}+\tan^{2}\text{x}}{1+\tan^{4}\text{x}+\tan^{2}\text{x}}=1$ $(\text{For every x}\in\text{R})$
$\text{For x}=2002,$
We have,
$\text{f}(2002)=1$
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MCQ 1461 Mark
If f : R → R be given by for all $\text{f(x)}=\frac{4^{\text{x}}}{4^{\text{x}}+2}\text{ x }\in\text{R},$ then:
  • A
    f(x) = f(1 - x)
  • B
    f(x) + f(1 - x) = 0
  • ✓
    f(x) + f(1 - x) = 1
  • D
    f(x) + f(x - 1) = 1
Answer
Correct option: C.
f(x) + f(1 - x) = 1
$\text{f(x)}=\frac{4^{\text{x}}}{4^{\text{x}}+2}\text{ x }\in\text{R},$
$\text{f}\big(\text{1}-\text{x}\big)=\frac{4^{1-\text{x}}}{4^{1-\text{x}}+2}$
$=\frac{4}{2\times4^{\text{x}}+4}$
$=\frac{2}{4^{\text{x}}+2}$
$\text{f(x)}+\text{f}(1-\text{x})=\frac{4^{\text{x}}}{4^{\text{x}}+2}+\frac{2}{4^{\text{x}}+2}$
$=\frac{4^{\text{x}}+2}{4^{\text{x}}+2}=1$
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MCQ 1471 Mark
If $\text{f}(\text{x})=\frac{\text{x - 1}^{3}}{\text{x}^3}$ then $\text{f}(\text{x}) +\text{f}\big(\frac{1}{\text{x}}\big)$is equal to:
  • A
    $2\text{x}^{3}$
  • B
    $\frac{1}{\text{x}^{3}}$
  • ✓
    $0$
  • D
    $1$
Answer
Correct option: C.
$0$
  1. $0$
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MCQ 1481 Mark
Domain and range of $ \text{f}(\text{x})=\frac{|\text{x - 3}|}{\text{x - 3}}$ are respectively:
  • A
    R, (-1, 1)
  • ✓
    R - (3), (1, -1)
  • C
    RT, R
  • D
    None of these
Answer
Correct option: B.
R - (3), (1, -1)
  1. R - (3), (1, -1)
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MCQ 1491 Mark
The range of $\text{f(x)}=\frac{1}{1-2\cos\text{x}}$ is:
  • A
    $\Big[\frac{1}{3},1\Big]$
  • ✓
    $\Big[-1,\frac{1}{3}\Big]$
  • C
    $\big(-\infty,-1\big)\cup\Big[\frac{1}{3},\infty\Big)$
  • D
    $\Big[-\frac{1}{3},1\Big]$
Answer
Correct option: B.
$\Big[-1,\frac{1}{3}\Big]$
We know that $-1\leq\cos\text{x}\leq1$ for all $\text{x}\in\text{R}$
Now,
$-1\leq\cos\text{x}\leq1$
$\Rightarrow-1\leq\cos\text{x}\leq1$
$\Rightarrow-2\leq-2\cos\text{x}\leq2$
$\Rightarrow-1\leq1-2\cos\text{x}\leq3$ (Adding 1 ro each term)
But,
$\cos\text{x}\neq\frac{1}{2}$
$\Rightarrow1-2\cos\text{x}\in\big[-1,3\big]-\{0\}$
$\Rightarrow\frac{1}{1-2\cos\text{x}}\in\big(-\infty,-1\big]\cap\Big[\frac{1}{3},\infty\Big)$
$\therefore\ \text{Range of }\text{f(x)}=\big(-\infty,-1\big]\cap\Big[\frac{1}{3},\infty\Big)$
Disclaimer: The range of the function does not matches with either of the given options. The range matches with option (c) if it is given as $\big(-\infty,-1\big]\cap\Big[\frac{1}{3},\infty\Big)$
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MCQ 1501 Mark
If : R → R is defined by f(x) = 3x + |x|, then f(2x) - f(-x) -6x =:
  • A
    f(x)
  • B
    2f(x)
  • C
    -f(x)
  • ✓
    f(-x)
Answer
Correct option: D.
f(-x)
  1. f(-x)
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MCQ 1511 Mark
Choose the correct answers:The domain and range of the function f given by f(x) = 2 - x - 5| is.
  • A
    Domain = $\mathrm{R}^{+}$, Range $= ( –\infty, 1]$
  • ✓
    Domain = R, Range $= ( –\infty, 2]$
  • C
    Domain = R, Range $= ( –\infty, 2]$
  • D
    Domain = $\mathrm{R}^{+}$, Range $= ( –\infty, 2]$
Answer
Correct option: B.
Domain = R, Range $= ( –\infty, 2]$
  1. Domain = R, Range $= ( –\infty, 2]$
Solution:
We have, f(x) = 2 - |x - 5|
Clearly, f(x) is defined for all $\text{x}\in\text{R}.$
$\therefore$ Domain of f = R
Now, $|\text{x}-5|\geq0,\forall\text{x}\in\text{R}$
$\Rightarrow-|\text{x}-5|\leq0$
$\Rightarrow2-|\text{x}-5|\leq2$
$\therefore\text{f(x)}\leq2$
$\therefore$ Range of $\text{f}=(-\infty, 2]$
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MCQ 1521 Mark
If P, Q and R are subsets of set A, then $\text{R }\times(\text{p}^\text{c} \cup\text{Q}^{\text{c}})^\text{c}=$
  • ✓
    $(\text{R}\times\text{P}) \cap(\text{R}\times\text{Q})$
  • B
    $(\text{R}\times\text{Q}) \cap(\text{R}\times\text{P})$
  • C
    $(\text{R}\times\text{P}) \cup(\text{R}\times\text{Q})$
  • D
    $\text{ None of these}$
Answer
Correct option: A.
$(\text{R}\times\text{P}) \cap(\text{R}\times\text{Q})$
  1. $(\text{R}\times\text{P}) \cap(\text{R}\times\text{Q})$
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MCQ 1531 Mark
Let $n(A)=m$ and $n(B)=n$, Then, the total number of non-empty relations that can be defined from $A$ to $B$ is:
  • A
    $m^n$
  • B
    $-1{ }^m n$
  • C
    $2 m n-1$
  • ✓
    $2^{m n}-1$
Answer
Correct option: D.
$2^{m n}-1$
  1. $2^{m n}-1$
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MCQ 1541 Mark
If $A=\left\{x: x^2-5 x+6=0\right\} B=\{2,4\}, C=\{4,5\}$, then $A \times(B \cap C)$ is:
  • ✓
    {(2, 4), (3, 4)}
  • B
    {(4, 2), (4, 3)}
  • C
    {(2, 4), (3, 4), (4, 4)}
  • D
    {(2, 2), (3, 3), (4, 4), (5, 5)}
Answer
Correct option: A.
{(2, 4), (3, 4)}
  1. {(2, 4), (3, 4)}
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MCQ 1551 Mark
The domain of the function $\text{f(x)}=\sqrt{2-2\text{x}-\text{x}^2}$ is:
  • A
    $\big[-\sqrt{3},\sqrt{3}\big]$
  • ✓
    $\big[-1,-\sqrt{3},-1+\sqrt{3}\big]$
  • C
    $\big[-2,2\big]$
  • D
    $\big[-2-\sqrt{3},-2+\sqrt{3}\big]$
Answer
Correct option: B.
$\big[-1,-\sqrt{3},-1+\sqrt{3}\big]$
$\text{f(x)}=\sqrt{2-2\text{x}-\text{x}^2}$
Since, $2-2\text{x}-\text{x}^2\geq0$
$\text{x}^2+2\text{x}-2\leq0$
$\Rightarrow\text{x}^2-2\text{x}-2+1-1\leq0$
$\Rightarrow(\text{x}-1)^2-\big(\sqrt{3}\big)^2\leq0$
$\Rightarrow\big[\text{x}-\big(1-\sqrt{3}\big)\big]\big[\text{x}-\big(1+\sqrt{3}\big)\big]\leq0$
$\Rightarrow\big(-1-\sqrt{3}\big)\leq\text{x}\leq(-1+\sqrt{3})$
Thus, domain $(\text{f})=\big[-1-\sqrt{3},-1+\sqrt{3}\big]$
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MCQ 1561 Mark
$\text{f}(\text{x})=\sqrt{9-\text{x}^2}$. Find the range of the function:
  • A
    $R$
  • B
    $R^{+}$
  • C
    $[-3,3]$
  • ✓
    $[0,3]$
Answer
Correct option: D.
$[0,3]$
  1. $[0,3]$
Solution:
We know, square root is always non-negative.$\sqrt{9-\text{x}^2}>0$.
So, the range of the function is set of positive real numbers from 0 to 3.
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MCQ 1571 Mark
If A = {1, 2, 4}, B = {2, 4, 5}, C = {2, 5}, then (A - B) × (B - C) is:
  • A
    {(1, 2), (1, 5), (2, 5)}
  • ✓
    {(1, 4)}
  • C
    (1, 4)
  • D
    none of these.
Answer
Correct option: B.
{(1, 4)}
A = {1, 2, 4}, B = {2, 4, 5} and C = {2, 5}
(A - B) = {1}
(B - C) = {4}
So, (A - B) × (B - C) = {(1, 4)}
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MCQ 1581 Mark
Consider the following statements:
  1. If n(A) = p and n(B) = q then n(A × B) = pq
  2. A × f = f
  3. In general, A × B¹ B × A
Which of the above statements are true?
  • A
    only (i)
  • B
    only (ii)
  • C
    only (iii)
  • ✓
    All the above
Answer
Correct option: D.
All the above
  1. All the above
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MCQ 1591 Mark
Consider the following statements:
  1. If $\text{A } \cap \text{B}=\phi$ then either $ \text{A}=\phi$ or $ \text{B}=\phi$
  2. For $ \text{a} ≠ \text{b}, \ ({\text{a}},\text{b})=(\text{b}, \text{a})$ and $ \text{a} ≠ \text{b}$
  3. If $\text{A}\subseteq \text{B}$, then $ \text{A } \times\text{A }\subseteq ( \text{A } \times\text{B })\ \cap ( \text{B} \times\text{A })$
  4. If $\text{A}\subseteq \text{B}$ and $\text{C}\subseteq \text{D}$, then $ \text{A } \times\text{C }\subseteq ( \text{B} \times\text{D }) $ Which of these is/are correct?
  • A
    Only $(II)$
  • B
    Only $(I)$
  • C
    Only $(IV)$
  • ✓
    $(II), (III)$ and $(IV)$
Answer
Correct option: D.
$(II), (III)$ and $(IV)$
$(II), (III)$ and $(IV)$
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MCQ 1601 Mark
If $P \times Q$ has $10$ elements then which is not possible?
  • A
    $n(P) = 1$ and $n(Q ) = 10$
  • B
    $n(P) = 10$ and $n(Q) = 1$
  • C
    $n(P) = 2$ and $n(Q) = 5$
  • ✓
    $n(P) = 5$ and $n(Q) = 4$
Answer
Correct option: D.
$n(P) = 5$ and $n(Q) = 4$
$n(P) = 5$ and $n(Q) = 4$
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MCQ 1611 Mark
If A and B are two sets, then $ \text{A}\times\text{ B }=\text{ B}\times\text{A}$ If and only if:
  • A
    $\text{A}\subset \text{ B}$
  • B
    $\text{B}\subset\text{C}$
  • ✓
    $\text{A}= \text{ B}$
  • D
    $\text{None of the above} $
Answer
Correct option: C.
$\text{A}= \text{ B}$
  1. $\text{A}= \text{ B}$
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MCQ 1621 Mark
The domain of $ \tan^{-1}(2\text{x}+1)$ is:
  • ✓
    $ \text{R}$
  • B
    $ \text{R}-\frac{1}{2}$
  • C
    $ \text{R}-\frac{-1}{2}$
  • D
    $\text{None of these}$
Answer
Correct option: A.
$ \text{R}$
Since $ \tan^1$ x exists if $\text{x}\in(-\infty,\infty)$
So, (2x + 1) is defined if
$ -\infty < 2\text{x} + 1 <\infty$
$\Rightarrow-\infty < \times <\infty$
$\Rightarrow \text{x}\in (-\infty,\infty)$
$\Rightarrow \text{x}\in \text{R}$
So, domain of $ \tan^-1(2\text{x}+1)$ is R.
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MCQ 1631 Mark
Which of the following are functions?
  • A
    $\{(\text{x, y}):\text{y}^2=\text{x, x, y}\in\text{R}\}$
  • ✓
    $\{(\text{x, y}):\text{y}=\text{|x|},\text{x, y}\in\text{R}\}$
  • C
    $\{(\text{x, y}):\text{x}^2+\text{y}^2=1,\text{x, y}\in\text{R}\}$
  • D
    $\{(\text{x, y}):\text{x}^2-\text{y}^2=1\text{x, y}\in\text{R}\}$
Answer
Correct option: B.
$\{(\text{x, y}):\text{y}=\text{|x|},\text{x, y}\in\text{R}\}$
For every value of $\text{x}\in\text{R},$ there is a unique value $\text{y} \in\text{R}$
i.e., there is a unique image for all values of $\text{x}\in\text{R},$
Also, values of x occur only once in the ordered pairs.
Thus, it is a function.
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MCQ 1641 Mark
If $9(x)=3 x^4-5 x^2+$, then value of $f(x-1)$ is:
  • ✓
    $3 x^4+12 x+13 x+2 x+7$
  • B
    $3 x^4-12 x-13 x-2 x-7$
  • C
    $3 x^4-12 x+13 x-2 x+7$
  • D
    $3 x^4-12 x-13 x+2 x+7$
Answer
Correct option: A.
$3 x^4+12 x+13 x+2 x+7$
  1. $3 x^4+12 x+13 x+2 x+7$
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MCQ 1651 Mark
If A = {1, 2, 3}, B = {1, 4, 6, 9} and R is a relation from A to B defined by 'x' is greater than y. The range of R is
  • A
    {1, 4, 6, 9}
  • B
    {4, 6, 9}
  • ✓
    {1}
  • D
    none of these.
Answer
Correct option: C.
{1}
A = {1, 2, 3} and B = {1, 4, 6, 9}
R is a relation from A to B defined by: x is greater than y.
Then R = {(2, 1), (3, 1)}
$\therefore$ Range (R) = {1}
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