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
$\text{n}^\text{n}$
AnswerCorrect option: B. $2^{\text{n}^2}$
- $2^{\text{n}^2}$
Solution:
Given, A finite set with n elements
Its Cartesian product with itself will have $\text{n}^2$ elements.
$\therefore$ Number of relations on $\text{A}=2^{\text{n}^2}$ View full question & answer→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:
MCQ 31 Mark
Which function is shown in graph?
Answer{(-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.
View full question & answer→MCQ 41 Mark
Which of the following is not a function?
- A
- B
{(-1, 1), (-2, 4), (2, 4)}
- ✓
{(1, 2), (1, 4), (2, 5), (3, 8)}
- D
AnswerCorrect 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.
View full question & answer→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:
Answer
- 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 $n^2=3$. View full question & answer→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
$(1,3)$
- ✓
$\big[1,3\big]$
AnswerCorrect 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),
$\text{x}\in[1,3]$
View full question & answer→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}}$
AnswerCorrect option: C. $ \text{R} \subseteq \text{A}\times{\text{B}}$
- $ \text{R} \subseteq \text{A}\times{\text{B}}$
View full question & answer→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:
Answer$\text{f(x)}=\frac{\text{x}+1}{\text{x}-1}$
$\text{f}(\text{f}(\text{f(2)}))$
$=\text{f}\Big(\text{f}\Big(\frac{2+1}{2-1}\Big)\Big)$
$=\text{f}(\text{f}(3))$
$=\text{f}\Big(\frac{3+1}{3-1}\Big)$
$=\text{f}(2)=3$
View full question & answer→MCQ 91 Mark
If (a, b) = (x, y) then___________.
AnswerTwo ordered pairs are said to be equal if and only if their corresponding elements are equal i.e. a = x and b = y.
View full question & answer→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
$3 x^4-12 x-13 x+2 x+7$
AnswerCorrect option: C. $3 x^4-12 x+13 x-2 x+7$
View full question & answer→MCQ 111 Mark
The function f(x) = x - [x] has period of:
AnswerLet 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
View full question & answer→MCQ 121 Mark
If f(x) = (x - 1)(x - 3)(x - 4)(x - 6) + 19 for all real value of x is:
MCQ 131 Mark
If $f(x)=e x$ and $g(x)=$ loge $x$ then the value of fog(1) is:
Answer
- 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 $(1)=1$ View full question & answer→MCQ 141 Mark
If A × B = {(1, a), (1, b), (1, c), (2, a), (2, b), (2, c)} then find set A:
Answer
- {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}.$ View full question & answer→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
AnswerCorrect 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)
$\text{x}\in\big(-\infty,-3\big]\cup(2,5)$
View full question & answer→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.
AnswerGiven 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
View full question & answer→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}$
View full question & answer→MCQ 181 Mark
The range of the function $\text{f(x)}=\frac{\text{x}^2-\text{x}}{\text{x}^2+2\text{x}}$ is:
AnswerCorrect 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 $\text{(f)}=\text{R}-\Big\{-\frac{1}{2},1\Big\}$
View full question & answer→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:
AnswerPeriod 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
View full question & answer→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
$m^n$
AnswerCorrect option: A. $2^{\mathrm{mn}}$
- $2^{\mathrm{mn}}$
Solution:
Given, n(A) = m
n(B) = n
$\therefore$ n(A × B) = mn
Then, the number of relations from A to is $2^{\mathrm{mn}}$ View full question & answer→MCQ 211 Mark
Domain of $\sqrt{\text{a}^{2} -\text{x}^{2}} (\text{a} > 0)$ is:
View full question & answer→MCQ 221 Mark
Let A = {1, 2, 3}. The total number of distinct relations that can be defined over A, is:
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
- ✓
AnswerTwo functions f and g are said to be equal if
- The domain of f = the domain of g
- The co-domain of f = the co-domain of g
- F(x) = g(x) for all x
View full question & answer→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} $
AnswerCorrect option: A. $ 2\text{f}(\text{x} ) \text{ f}(\text{y})$
- $ 2\text{f}(\text{x} ) \text{ f}(\text{y})$
View full question & answer→MCQ 251 Mark
If set A has 2 elements and set B has 4 elements then how many relations are possible?
Answer
- 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 . View full question & answer→MCQ 261 Mark
If the set A has p elements, B has q elements, then the number of elements in A × B is:
Answer
- pq
Solution:
n(A × B) = n(A) × n(B)
n(A × B) = p × q = pq
View full question & answer→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:
Answer
- -1, -2
Solution:
$f(x)=2 x+3 \text { and } g(x)=x^2+7$
$g(f(x))=8$
$\Rightarrow(f(x))^2+7=8$
$\Rightarrow(2 x+3)^2+7=8$
$\Rightarrow x^2+3 x+2=0$
$\Rightarrow(x+2)(x+1)=0$
$\Rightarrow x=-1,-2$ View full question & answer→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)
AnswerCorrect option: C. (a, b) (b, e) (c, e) (b, x)
- (a, b) (b, e) (c, e) (b, x)
View full question & answer→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
AnswerCorrect option: B. ((4, 1), (5, 2), (6, 3),.....)
- ((4, 1), (5, 2), (6, 3),.....)
View full question & answer→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
$\text{R}\subseteq\text{B}\times\text{A}$
AnswerCorrect 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.
View full question & answer→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:
- $ \text{A }\times ({\text{B} \cap\text{C})} = (\text{A}\times \text{ B}) ∩ (\text{A}\times \text{ C})$
- A, C is a subset of $\text{ B }\times\text{ D}$
Which of the following statment is correct? View full question & answer→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:
Answer$\text{R}=\{(\text{x, y}):\text{x, y}\in\text{Z},\text{ x}^2+\text{y}^2\leq4\}$
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}
View full question & answer→MCQ 331 Mark
If A × B = {(1, a), (1, b), (1, c), (2, a), (2, b), (2, c)} then find set B:
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}$
AnswerCorrect 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)}$
View full question & answer→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:
Answer
- {(3, 3), (3, 1), (5, 2)}
Solution:
$A=\{1,2,3\}, B=\{1,3,5\}$
$R=\{(1,3),(2,5),(3,3)\}$
$\therefore R^{-1}=\{(3,3),(3,1),(5,2)\}$ View full question & answer→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 =
Answer$\text{e}^{\text{f(x)}}=\frac{10+\text{x}}{10-\text{x}}$
$\Rightarrow\text{ f(x)}=\log_\text{e}\Big(\frac{10+\text{x}}{10-\text{x}}\Big)\ ...(\text{i})$
$\Rightarrow\ \text{f(x)}=\text{kf}\Big(\frac{200\text{x}}{100+\text{x}^2}\Big)$
$\Rightarrow\ \log_\text{e}\Big(\frac{10+\text{x}}{10-\text{x}}\Big)=\text{k}\log_\text{e}\Bigg(\frac{10+\frac{200\text{x}}{100+\text{x}^2}}{10-\frac{200\text{x}}{100+\text{x}^2}}\Bigg)$ {from (1)}
$\Rightarrow\ \log_\text{e}\Big(\frac{10+\text{x}}{10-\text{x}}\Big)=\text{k}\log_\text{e}\Big(\frac{1000+10\text{x}^2+200\text{x}}{1000+10\text{x}^2-200\text{x}}\Big)$
$\Rightarrow\ \log_\text{e}\Big(\frac{10+\text{x}}{10-\text{x}}\Big)=\text{k}\log_\text{e}\bigg(\frac{(\text{x}+10)^2}{(\text{x}-10)^2}\bigg)$
$\Rightarrow\ \log_\text{e}\Big(\frac{10+\text{x}}{10-\text{x}}\Big)=2\text{k}\log_\text{e}\frac{(\text{x}+10)}{(\text{x}+10)}$
$\Rightarrow\ 1=2\text{k}$
$\Rightarrow\ \text{k}=\frac{1}{2}=0.5$
View full question & answer→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
AnswerCorrect 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}
$\therefore\ \text{Domain of h}=\text{Domain of f }\cap\text{ Domain of g}\\\ \ \ =\text{R }\cap\big[\text{R}-\{0\}\big]=\text{R}-\{0\}$
$\Rightarrow\text{x}\in\text{R},\text{ x}\neq0$
View full question & answer→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
$3 x^4-12 x^3-13 x^2+2 x+7$
AnswerCorrect option: A. $3 x^4+12 x^2+13 x^2+2 x+7$
- $3 x^4+12 x^2+13 x^2+2 x+7$
View full question & answer→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
$x^4+6 x^3+10 x^2-3 x$
- ✓
$x^4-6 x^3+10 x^2-3 x$
AnswerCorrect option: D. $x^4-6 x^3+10 x^2-3 x$
- $x^4-6 x^3+10 x^2-3 x$
Solution:
Given, $f(x)=x^2-3 x+2$
Now, $f(f(y))=f\left(x^2-3 x+2\right)$
$=\left(x^2-3 x+2\right)^2-3\left(x^2-3 x+2\right)+2$
$=x^4-6 x^3+10 x^2-3 x$ View full question & answer→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
$(\text{A} \times\text{B})\ \cup\ (\text{B}\times\text{A})=(\text{A} \times\text{B})\ \cup\ (\text{A}\times\text{C})$
AnswerCorrect option: C. $(\text{A} \times\text{B})\ \cup\ (\text{B}\times\text{A})=(\text{A} \times\text{B})\ \cup\ (\text{B}\times\text{C})$
- $(\text{A} \times\text{B})\ \cup\ (\text{B}\times\text{A})=(\text{A} \times\text{B})\ \cup\ (\text{B}\times\text{C})$
View full question & answer→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 $= [0, \infty)$
AnswerCorrect 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 $= [0, \infty)$
View full question & answer→MCQ 421 Mark
If (x + 2, y - 3) = (5, 7) then find values of x and y:
AnswerTwo 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.
View full question & answer→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
$\text{R}$
AnswerCorrect option: B. $\big[0,\infty\big)$
$\text{f(x)}=|\text{x}-1|\geq0\ \forall\text{ x}\in\text{R}$
Thus, range $=\big[0,\infty\big)$
View full question & answer→MCQ 441 Mark
Let f : R → R be defined by f(x) = 2x + |x|. Then f(2x) + f(-x) - f(x) =
Answerf(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|
View full question & answer→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:
AnswerRange 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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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:
AnswerPeriod 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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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
- B
Reflexive and symmetric but not transitive.
- C
Symmetric and transitive but not reflexive.
- ✓
Esymmetric but neither reflexive nor transitiv.
AnswerCorrect option: D. Esymmetric but neither reflexive nor transitiv.
- Esymmetric but neither reflexive nor transiti
View full question & answer→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
- B
Reflexive and symmetric but not transitive
- C
Symmetric and transitive but not reflexive
- ✓
Symmetric but neither reflexive nor transitive
AnswerCorrect option: D. Symmetric but neither reflexive nor transitive
d. Symmetric but neither reflexive nor transitive
View full question & answer→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
$(–\infty, –1) \cup [1, 4)$
AnswerCorrect 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 $\text{f}=(-\infty, -1)\cup(1, 4]$
View full question & answer→MCQ 501 Mark
Choose the correct answers: Domain of $\sqrt{\text{a}^2-\text{x}^2}(\text{a}>0)$ is.
AnswerWe 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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