MCQ 511 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
AnswerCorrect 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, $\text{domain(f)}=\big[-3,-2\big]\cup\big[2,3\big]$
View full question & answer→MCQ 521 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
$\{1, 2, 3, 4, 5\}$
AnswerCorrect 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 $\{3, 4, 5\}.$
View full question & answer→MCQ 531 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)$
AnswerCorrect option: A. $ (-2, \infty )$
View full question & answer→MCQ 541 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
AnswerCorrect 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 $\text{f}(\beta)=3\big((3)^2-12\big)=-9$
$\Rightarrow\text{f}(\alpha)=\text{f}(\beta)=-9$
View full question & answer→MCQ 551 Mark
Let a relation $R$ be defined by $R = ((4, 5), (1, 4), (4, 6), (7, 6), (3, 7)),$ then $\text{ROR}$ is equal to:
- ✓
$((1, 5), (1, 6), (3, 6))$
- B
$((1, 4), (1, 5), (3, 6))$
- C
$((1, 5), (1, 6), (3, 7))$
- D
$((1, 4), (1, 5), (3, 7))$
AnswerCorrect option: A. $((1, 5), (1, 6), (3, 6))$
View full question & answer→MCQ 561 Mark
If $A$ is the null set and $B$ is an infinite set, then what is $A \times B?$
View full question & answer→MCQ 571 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
$(0,4)$
- ✓
$[0,4]$
AnswerCorrect 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 $(\text{f})=[0,4]$
View full question & answer→MCQ 581 Mark
If the relation $R : A \rightarrow B,$ where $A = \{1, 2, 3, 4\}$ and $B = \{1, 3, 5\}$ is defined by $\text{R} =\{(\text{x, y}) : \text{x} < \text{y}, \text{x } \text{iA}, \text{y} \in \text{B}\},$ then $R^{-1}\text{OR}$ 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
AnswerCorrect option: C. $(3, 3), (3, 5), (5, 3), (5, 5)$
View full question & answer→MCQ 591 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
$f(c) = g(a)$
AnswerCorrect option: C. $f(d) = g(b)$
Given, f$(x) = ax + b$ and $g(x) = cx + d$ and
Now, $f{g(x)} = g{f(x)}$
$\Rightarrow f{cx + d} = g{ax + b}$
$\Rightarrow a(cx + d) + b = c(ax + b) + d$
$\Rightarrow ad + b = cb + d$
$\Rightarrow f(d) = g(b)$
View full question & answer→MCQ 601 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 $
AnswerCorrect option: B. $ -1– \sqrt{3} ≤ \times ≤ -1+\sqrt{3}$
View full question & answer→MCQ 611 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 $= \{-1, 1\}$
AnswerCorrect 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 $f(x) = R - \{-1\}$
View full question & answer→MCQ 621 Mark
The function $f : R \rightarrow R$ 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
$\Big(\frac{3}{4},1\Big)$
AnswerCorrect 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 $1$
$\therefore\ \text{f(R)}=\Big(\frac{3}{4},1\Big)$
View full question & answer→MCQ 631 Mark
If $A \times B = (5, 5), (5, 6), (5, 7), (8, 6), (8, 7), (8, 5),$ then the value $A:$
- A
$(5)$
- B
$(8)$
- ✓
$(5, 8)$
- D
$(5, 6, 7, 8)$
AnswerCorrect option: C. $(5, 8)$
View full question & answer→MCQ 641 Mark
The domain of the function $^{7-\text{x}}\text{p}_{\text{x}-3}$ is:
- ✓
$\{1, 2, 3\}$
- B
$\{3, 4, 5, 6\}$
- C
$\{3, 4, 5\}$
- D
$\{1, 2, 3, 4, 5\}$
AnswerCorrect option: A. $\{1, 2, 3\}$
$\frac{\text{ (x}^{2} + \ 2)}{(\text{x}^{2} + \ 1)}$
Given $f(x) = x^2 + 2$ and $\text{g}(\text{x})=\frac{\text{x}}{(\text{x}\ - \ 1)}$
Now, $gof(x) = g(x^2+ 2)$
$= \text{got}(\text{x})=\text{g}(\text{x}^2 +2)$
$=\frac{\text{ (x}^{2} +\ 2)}{(\text{x}^{2} \ +\ 2\ -\ 1)}$
$=\frac{(\text{x}^2\ +\ 2)}{(\text{x}^2\ +\ 1)}$
View full question & answer→MCQ 651 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
$\{3, 6, 9, 12, 15\}$
AnswerCorrect option: C. $\{3, 6, 9, 12\}$
View full question & answer→MCQ 661 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}$
AnswerCorrect option: C. $ (-1, 2)\ \cup \ (3, \infty)$
View full question & answer→MCQ 671 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
$1 + 2x^2$
- ✓
$2 + x^2$
- C
$1 + x$
- D
$2 + x$
AnswerCorrect option: B. $2 + x^2$
View full question & answer→MCQ 681 Mark
Let $R$ be the relation on $Z$ defined by $R = \{(a, b) : a, b Îz, a - b$ is an interger$\}.$ Find the domain and Range of $R:$
- ✓
$z, z$
- B
$z+ , z$
- C
$z, z-$
- D
AnswerCorrect option: A. $z, z$
View full question & answer→MCQ 691 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
$\{(1, 3), (1, 4), (2, 3), (2, 4)\}$
AnswerCorrect 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 $A \times B$ 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 $B.$
View full question & answer→MCQ 701 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
$\frac{1}{14}\Big(-\frac{3}{\text{x}}+5\text{x}+6\Big)$
- ✓
Answer$3\text{f(x)}+5\text{f}\Big(\frac{1}{\text{x}}\Big)=\frac{1}{\text{x}}-3\ ...(\text{i})$
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.
View full question & answer→MCQ 711 Mark
If $A = \{1, 2, 3\}$ and $B = \{x, y\},$ then the number of functions that can be defined from $A$ into $B$ is:
AnswerGiven,
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 $= 2^3 = 8$
View full question & answer→MCQ 721 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
$\{1, 2, 3, 4, 5, 6\}$
AnswerCorrect option: C. $\{1, 2, 3\}$
View full question & answer→MCQ 731 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
$\{1, 2, 3, 4, 5, 6\}$
AnswerCorrect 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 $= \{1,2,3,4,5\}.$
View full question & answer→MCQ 741 Mark
$\text{f}(\text{x})={\frac{\text{|x|}}{\text{x}}}$ for $ \text{x} ≠ 0$ and $ 0$ for $ \text{x} = 0$ Which function is this?
Answer$\text{f}(\text{x})={\frac{\text{|x|}}{\text{x}}} \{$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.
View full question & answer→MCQ 751 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
- D
AnswerCorrect option: B. range $= \{0, 1, 2\}$
View full question & answer→MCQ 761 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
$mn$
- B
$nm - 1 $
- C
$mn - 1$
- ✓
$2^{mn} - 1$
AnswerCorrect option: D. $2^{mn} - 1$
We have, $n(A) = m$ and $n(B) = n$
$n(A \times B) = n(A).n(B) = mn$
Total number of relation from $A$ to $B =$ Number of subsets of $A \times B = 2^{mn}$
So, total number if non$-$empty relations $= 2^{mn} - 1$
View full question & answer→MCQ 771 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
$x \in [2, 4)$
AnswerCorrect option: C. $x \in [2, 3]$
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, $x \in [2, 3]$
View full question & answer→MCQ 781 Mark
If $n(A) = 3, n(B) = 4,$ then $n (A \times A \times B)$ is equal to:
View full question & answer→MCQ 791 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
$[(3, 1), (2, 6), (3, 9)]$
- ✓
Answer$A = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$
$x R y ⇔ y = 3x$
For $x = 1, y = 3$
For $x = 2, y = 6$
For $x = 3, y = 9$
Thus, $R = \{(1, 3), (2, 6), (3, 9)\}$
View full question & answer→MCQ 801 Mark
If two sets $A$ and $B$ are having $99$ elements in common, then the number of elements common to each of the sets $A \times B$ and $B \times A$ are:
View full question & answer→MCQ 811 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
$\{1, 2, 3, 4\}$
AnswerCorrect option: C. $\{2, 4, 6\}$
$x + 2y = 8$
$\Rightarrow x = 8 - 2y$
For $y = 1, x = 6$
$y = 2, x = 4$
$y = 3, x = 2$
Then $R = \{(2, 3), (4, 2), (6, 1)\}$
$\therefore$ Domain of $R = \{2, 4, 6\}$
View full question & answer→MCQ 821 Mark
Domain of the function $ \text{ f}(\text{x}) =\sqrt{( 2 -\text{ 2x} - \text{x })} $ is:
AnswerCorrect option: B. -1-$\sqrt{3≤x≤-1}$
View full question & answer→MCQ 831 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
$(0,\infty)$
AnswerCorrect 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 $f(x) = \{-1, 1\}$
View full question & answer→MCQ 841 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)}]$
AnswerCorrect option: A. $ \frac{1}{2} \big[\text{}f(\text{2x}) +\text{ f} \text{(2y)}\big]$
View full question & answer→MCQ 851 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?
AnswerCorrect option: D. $\text{i}\ \phi\ 1$
We have,
$|\text{i}|=\sqrt{1^2+0^2}=1$
Thus, $\text{i }\phi\ 1$ satisfies $\text{x}\ \phi\text{ y}\Leftrightarrow|\text{x}|=\text{y}$
View full question & answer→MCQ 861 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
AnswerGiven, 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 $(f) = R.$
View full question & answer→MCQ 871 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:
View full question & answer→MCQ 881 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
${f(x)}2$
- B
${f(x)}3$
- ✓
$2f(x)$
- D
$3f(x)$
AnswerCorrect option: C. $2f(x)$
$\text{f(x)}=\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$
Then, $\text{f}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)=\log\Bigg(\frac{1+\frac{2\text{x}}{1+\text{x}^2}}{1-\frac{2\text{x}}{1+\text{x}^2}}\Bigg)$
$=\log\Bigg(\frac{\frac{1+\text{x}^2+2\text{x}}{1+\text{x}^2}}{1-\frac{2\text{x}}{1+\text{x}^2}}\Bigg)$
$=\log\bigg(\frac{(1+\text{x})^2}{(1-\text{x})^2}\bigg)$
$=2\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$
$=2(\text{f(x)})$
View full question & answer→MCQ 891 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 $B \{(x, y) x = y^2\}.$ Find range of the relation:
- A
$\{1, 4, 9\}$
- ✓
$\{-1, 1, -2, 2, -3, 3\}$
- C
$\{1, 4, 8, 9\}$
- D
$\{-1, 1, -2, 2, -3, 3, 5\}$
AnswerCorrect option: B. $\{-1, 1, -2, 2, -3, 3\}$
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 $= \{-1, 1, -2, 2, -3, 3\}.$
View full question & answer→MCQ 901 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
$\text{x}\in\big[2,3\big]$
- ✓
$\text{x}\in\big[2,4\big)$
AnswerCorrect 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 $\big[3,4\big)$
$\Rightarrow\text{x}\in\big[2,4\big)$
View full question & answer→MCQ 911 Mark
If set $P$ has $4$ elements and set $Q$ has $5$ elements then find the number of elements in $P \times Q:$
AnswerIf set $P$ has $m$ elements and set $Q$ has $n$ elements then $P \times Q$ has $m \times n$ elements.
Here, $m = 4$ and $n = 5$ therefore $P \times Q$ has $4 \times 5 = 20$ element.
View full question & answer→MCQ 921 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
$\frac{1}{4}\big[\text{f(2}\text{x})-\text{f}(2\text{y})\big]$
AnswerCorrect 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,
$\text{f}(\text{x}+\text{y})\text{f}(\text{x}-\text{y})=\Big(\frac{2^{\text{x}+\text{y}}+2^{-\text{x}-\text{y}}}{2}\Big)\Big(\frac{2^{\text{x}-\text{y}}+2^{-\text{x}+\text{y}}}{2}\Big)$
$\Rightarrow\ \text{f}(\text{x}+\text{y})\text{f}(\text{x}-\text{y})=\frac{1}{4}\big(2^{2\text{x}}+2^{-2\text{y}}+2^{2\text{y}}+2^{-2\text{x}}\big)$
$\Rightarrow\ \text{f}(\text{x}+\text{y})\text{f}(\text{x}-\text{y})=\frac{1}{2}\Big(\frac{2^{2\text{x}}+2^{-2\text{x}}}{2}+\frac{2^{2\text{y}}+2^{-2\text{y}}}{2}\Big)$
$\Rightarrow\text{f}(\text{x}+\text{y})\text{f}(\text{x}-\text{y})=\frac{1}{2}\big[\text{f}(2\text{x})+\text{f}(2\text{y})\big]$
View full question & answer→MCQ 931 Mark
In a function from set $A$ to set $B,$ every element of set $A$ has $............$ image in set $B:$
AnswerA 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 $B.$
View full question & answer→MCQ 941 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
AnswerCorrect 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, $\text{f(2)}=\frac{-1}{5}\Big(\frac{3}{4}+2\times4\Big)$
$=\frac{-1}{5}\Big(\frac{3+32}{4}\Big)$
$=-\frac{7}{4}$
View full question & answer→MCQ 951 Mark
The relation $R$ defined on the set $A = \{1, 2, 3, 4, 5\}$ by $R = \{(x, y) : | x^2 – y^2| <16\}$ is given by:
- A
$\{(1, 1), (2, 1), (3, 1), (4, 1), (2, 3)\}$
- B
$\{(2, 2), (3, 2), (4, 2), (2, 4)\}$
- C
$\{(3, 3), (4, 3), (5, 4), (3, 4)\}$
- ✓
View full question & answer→MCQ 961 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
$\{(1, 2), (2, 3), (3, 2), (3, 4)\}$
AnswerCorrect 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.
View full question & answer→MCQ 971 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
$-2, -1$
AnswerCorrect option: C. $2, -1$
Given, $g(x) = ax + b$
Again, $g(1) = 1$
$\Rightarrow a \times 1 + b = 1$
$\Rightarrow a + b = 1 ……… 1$
and $g(2) = 3$
$\Rightarrow a \times 2 + b = 3$
$\Rightarrow 2a + b = 3 …….. 2$
Solve equation $1$ and $2,$ we get
$a = 2, b = -1$
View full question & answer→MCQ 981 Mark
Find the range of the function $f(x) = x^2 + 2:$
- A
$(-2, 2)$
- ✓
$(2, \infty)$
- C
$(3, \infty)$
- D
AnswerCorrect option: B. $(2, \infty)$
View full question & answer→MCQ 991 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
$[-1,1]$
AnswerCorrect 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 $\text{f(x)}=\{\cos1,1,\cos2\}$
View full question & answer→MCQ 1001 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
AnswerCorrect 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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