MCQ 1011 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
$R – (3, –2)$
AnswerCorrect option: A. $R – \{3, –2\}$
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 $f(x) = R - \{-2, 3\}$
View full question & answer→MCQ 1021 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
$-2$
- ✓
AnswerGiven,
$\text{f(x)}=\cos(\log\text{x})$
$\therefore\ \text{f(y)}=\cos(\log\text{y})$
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
$\text{f(xy)}=\cos(\log\text{xy})=\cos(\log\text{x}+\log\text{y})$
$\Rightarrow\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f(xy)}=\cos(\log\text{x}-\log\text{y})+\cos(\log\text{x}+\log\text{y})$
$\Rightarrow\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f(xy)}=2\cos(\log\text{x})\cos(\log\text{y})$
$\Rightarrow\frac{1}{2}\Big[\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f(xy)}\Big]=\cos(\log\text{x})\cos(\log\text{y})$
$\Rightarrow\text{f(x})\text{f}(\text{y})-\frac{1}{2}\Big\{\text{f}\Big(\frac{\text{x}}{\text{y}}\Big)+\text{f}\big(\text{x}\text{y}\big)\Big\}\\=\cos(\log\text{x})\cos(\log\text{y})-\cos(\log\text{x})\cos(\log\text{y})=0$
View full question & answer→MCQ 1031 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
$f(3x)$
- B
${f(x)}3$
- ✓
$3f(x)$
- D
$-f(x)$
AnswerCorrect option: C. $3f(x)$
$\text{f(x)}=\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$ 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, $\text{f}(\text{g(x)})=\log=\log\Big(\frac{1+\text{g(x)}}{1-\text{g(x)}}\Big)$
$=\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big)^3$
$=3\text{f}(\text{x})$
View full question & answer→MCQ 1041 Mark
A relation $R$ is defined from $\{2, 3, 4, 5\}$ to $\{3, 6, 7, 10\}$ by $: x R y \Leftrightarrow x$ is relatively prime to $y.$ Then, domain of $R$ is:
- A
$\{2, 3, 5\}$
- B
$\{3, 5\}$
- C
$\{2, 3, 4\}$
- ✓
$\{2, 3, 4, 5\}$
AnswerCorrect option: D. $\{2, 3, 4, 5\}$
Given,
From $\{2, 3, 4, 5\}$ to $\{3, 6, 7, 10\}, x R y \Leftrightarrow x$ 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 $\{2, 3, 4, 5\}$
View full question & answer→MCQ 1051 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
$\{(\text{x, y}):\text{x},\text{y}\in\text{R},\text{y}=\text{x}^3\}$
AnswerCorrect option: B. $\{(\text{x, y}):\text{x},\text{ y}\in\text{R},\text{y}^2=\text{x}\}$
$y^2 = x$ 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.
View full question & answer→MCQ 1061 Mark
The range of the function: $\text{f}(\text{x})=\sqrt{(\text{x}-1)(3-\text{x})}$ :
- ✓
$(-1, 1)$
- B
$(-1, 1)$
- C
$(-3, 3)$
- D
$(-3, 1)$
AnswerCorrect option: A. $(-1, 1)$
View full question & answer→MCQ 1071 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 codomain of the relation:
- A
$\{1, 4, 9\}$
- B
$\{-1, 1, -2, 2, -3, 3\}$
- C
$\{1, 4, 8, 9\}$
- ✓
$\{-1, 1, -2, 2, -3, 3, 5\}$
AnswerCorrect option: D. $\{-1, 1, -2, 2, -3, 3, 5\}$
We know, codomain of a relation is the set to which relation is defined i.e. set $B.$
So, codomain $= \{-1, 1, -2, 2, -3, 3, 5\}.$
View full question & answer→MCQ 1081 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
$(-\infty,4)$
AnswerCorrect 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 $(\text{f(x)})=\big[4,\infty)$
View full question & answer→MCQ 1091 Mark
If $f : [-2, 2] \rightarrow R$ 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 $\{\text{x}\in[-2,2]:\text{x}\leq0\text{ and }\text{f}(\text{|x|})=\text{x}\}=$
AnswerCorrect 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 $(ii)]$
$=\Big\{\text{x}:2\text{x}=\frac{-1}{2}\Big\}$
$=\Big\{\text{x}:\text{x}=\frac{-1}{2}\Big\}$
$=\Big\{-\frac{1}{2}\Big\}$
View full question & answer→MCQ 1101 Mark
If $f : R \rightarrow R$ is defined by $f(x) = 3x + |x|,$ then $f(2x) - f(-x) - 6x =$
- ✓
$f(x)$
- B
$2f(x)$
- C
$-f(x)$
- D
$f(-x)$
AnswerCorrect option: A. $f(x)$
View full question & answer→MCQ 1111 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:
AnswerGiven, $\text{f(x)}=\cos(\log\text{x})$
Then, $\text{f(x)}\text{f(4)}-\frac{1}{2}\Big\{\text{f}\Big(\frac{\text{x}}{4}\Big)+\text{f}(4\text{x})\Big\}$
$=\cos(\log\text{x})\cos(\log4)+\frac{1}{2}\Big\{\cos\Big(\log\frac{\text{x}}{4}\Big)+\cos(\log4\text{x})\Big\}$
$=\frac{1}{2}\big[\cos(\log\text{x}+\log4\big)+\cos(\log\text{x}-\log4)\big]-\frac{1}{2}\Big\{\cos\Big(\log\frac{\text{x}}{4}\Big)+\cos(\log4\text{x})\Big\}$
$=\frac{1}{2}\Big\{\cos(\log4\text{x})+\cos\Big(\log\frac{\text{x}}{4}\Big)-\cos\Big(\log\frac{\text{x}}{4}\Big)-\cos(\log4\text{x})\Big\}$
$=\frac{1}{2}\times0$
$=0$
View full question & answer→MCQ 1121 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:
AnswerCorrect 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 $(\text{f(x)})=\phi$
View full question & answer→MCQ 1131 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
$-\sqrt{\text{x}}$
AnswerCorrect option: B. $\sqrt{\text{x}}$
View full question & answer→MCQ 1141 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
$mn$
- B
$nm - 1$
- C
$mn - 1$
- ✓
$2mn - 1$
AnswerCorrect option: D. $2mn - 1$
View full question & answer→MCQ 1151 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:
View full question & answer→MCQ 1161 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
$(-a, 0]$
AnswerCorrect 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 $[-a, a]$
View full question & answer→MCQ 1171 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 1181 Mark
Let $R$ be a relation in the set of real numbers defined as $\text{a R b}$ if $|a - b| ≥ \frac{1}{2}$, Then the relation $R$ is:
- ✓
- B
Reflexive and symmetric but not transitive
- C
Symmetric and transitive but not reflexive
- D
Symmetric but neither reflexive nor transitive
View full question & answer→MCQ 1191 Mark
Let $R$ be a relation in the set of real numbers defined as $\text{a R b}$ 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.
View full question & answer→MCQ 1201 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:
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 $f(x) = \text{LCM} (4, 6, 4) = 12$
View full question & answer→MCQ 1211 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
$\{1, 2, 3, 4, 5, 6\}$
AnswerCorrect 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 $= \{2, 3, 4, 5\}.$
View full question & answer→MCQ 1221 Mark
Let $f : R \rightarrow R$ be defined by $f(x) = 2x + |x|.$ Then $f(2x) + f(-x) - f(x) =$
- A
$2x$
- ✓
$2|x|$
- C
$-2x$
- D
$-2|x|$
AnswerCorrect option: B. $2|x|$
$f(x) = 2x + |x|$
Then, $f(2x) + f(-x) - f(x)$
$= 2(2x) + 2|x| + (-2x) + |-x| - 2x + |x|$
$= 4x - 2x - 2x + 2|x| + |-x| - |x|$
$= 0 + 2|x| + |x| - |x| = 2|x|$
$= 2|x|$
View full question & answer→MCQ 1231 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 1241 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 $y = 10$
AnswerCorrect option: A. $x = 3$ and $y = 10$
Two ordered pairs are said to be equal if and only if their corresponding elements are equal.
$x + 2 = 5 \Rightarrow x = 3$
$y - 3 = 7 \Rightarrow y = 10$
Hence, $x = 3$ and $y = 10.$
View full question & answer→MCQ 1251 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 $\text{x}\geq1,\text{x}-1\geq0$
$\Rightarrow\sqrt{\text{x}-1}\geq1$
$\Rightarrow$ Range of $= [0, \infty)$
View full question & answer→MCQ 1261 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. $(\text{A} \times\text{B})\ \cup\ (\text{B}\times\text{A})=(\text{A} \times\text{B})\ \cup\ (\text{B}\times\text{C})$
View full question & answer→MCQ 1271 Mark
If $f : R \rightarrow R$ is defined by $f(x) = x^2 – 3x + 2,$ the $f(y)$ is:
- A
$x^4 + 6x^3 + 10x^2 + 3x$
- B
$x^4 - 6x^3 + 10x^2 + 3x$
- C
$x^4 + 6x^3 + 10x^2 - 3x$
- ✓
$x^4 - 6x^3 + 10x^2 - 3x$
AnswerCorrect option: D. $x^4 - 6x^3 + 10x^2 - 3x$
Given, $f(x) = x^2 - 3x + 2$
Now, $f(f(y)) = f(x^2- 3x + 2)$
$= (x^2 - 3x + 2)^2 - 3(x^2 - 3x + 2) + 2$
$= x^4 - 6x^3 + 10x^2 - 3x.$
View full question & answer→MCQ 1281 Mark
If $f(x) = 3x^4 - 5x^2 + 9$, then value of $(x - 1)$ is:
- ✓
$3x^4 - 12x^3 + 13x^2- 2x + 7$
- B
$3x^4 - 12x^3 - 13x^2 - 2x - 7$
- C
$3x^4 - 12x^3 + 13x^2 - 2x + 7$
- D
$3x^4 - 12x^3 - 13x^2 + 2x + 7$
AnswerCorrect option: A. $3x^4 - 12x^3 + 13x^2- 2x + 7$
View full question & answer→MCQ 1291 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 1301 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 1311 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
AnswerCorrect option: A. $\{(3, 3), (3, 1), (5, 2)\}$
$A = \{1, 2, 3\}, B = \{1, 3, 5\}$
$R = \{(1, 3), (2, 5), (3, 3)\}$
$\therefore R^{-1} = \{(3, 3), (3, 1), (5, 2)\}$
View full question & answer→MCQ 1321 Mark
If the function $f : R \rightarrow R$ be given by $ \text{f}(\text{x}) = \text{x}^2 + 2$ and $g : R \rightarrow R$ 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 1331 Mark
If $A \times B = \{(1, a), (1, b), (1, c), (2, a), (2, b), (2, c)\}$ then find set $B:$
- A
$\{1\}$
- B
$\{1, 2\}$
- C
$\{1, a\}$
- ✓
$\{a, b, c\}$
AnswerCorrect option: D. $\{a, b, c\}$
View full question & answer→MCQ 1341 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
AnswerCorrect 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$(R) = \{-2, -1, 0, 1, 2\}$
View full question & answer→MCQ 1351 Mark
Let $A = \{1, 2\}, B = \{1, 2, 3, 4\}, C = \{5, 6\}$ and $D = \{5, 6, 7, 8\}$ Following statements are given below:
$i. \text{A }\times ({\text{B} \cap\text{C})} = (\text{A}\times \text{ B}) ∩ (\text{A}\times \text{ C})$
$ii. A, C$ 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
AnswerCorrect option: C. Both $I$ and $II$
View full question & answer→MCQ 1361 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 $A \times B.$
View full question & answer→MCQ 1371 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. $((4, 1), (5, 2), (6, 3),.....)$
View full question & answer→MCQ 1381 Mark
Which of the following relation is a function?
- A
$\text{(a, b) (b, e) (c, e) (b, x)}$
- B
$\text{(a, d) (a, m) (b, e) (a, b)}$
- ✓
$\text{(a, b) (b, e) (c, e) (b, x)}$
- D
$\text{(a, d) (b, m) (b, y) (d, x)}$
AnswerCorrect option: C. $\text{(a, b) (b, e) (c, e) (b, x)}$
View full question & answer→MCQ 1391 Mark
If $f : R \rightarrow R$ and $g : R \rightarrow R$ are defined by $f(x) = 2x + 3$ 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
$1, -2$
AnswerCorrect option: C. $-1, -2$
$f(x) = 2x + 3$ and $g(x) = x^2 + 7$
$g(f(x)) = 8$
$\Rightarrow (f(x))^2 + 7 = 8$
$\Rightarrow (2x + 3)^2 + 7 = 8$
$\Rightarrow x^2 + 3x + 2 = 0$
$\Rightarrow (x + 2)(x + 1) = 0$
$\Rightarrow x = -1, -2$
View full question & answer→MCQ 1401 Mark
If the set $A$ has $p$ elements, $B$ has $q$ elements, then the number of elements in $A \times B$ is:
- A
$p + q$
- B
$p + q + 1$
- ✓
$pq$
- D
$p2$
Answer$n(A \times B) = n(A) \times n(B)$
$n(A \times B) = p \times q = pq$
View full question & answer→MCQ 1411 Mark
If set $A$ has $2$ elements and set $B$ has $4$ elements then how many relations are possible?
AnswerWe know, $A \times B$ 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 1421 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})$
View full question & answer→MCQ 1431 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 $x$
- ✓
View full question & answer→MCQ 1441 Mark
Let $A = \{1, 2, 3\}.$ The total number of distinct relations that can be defined over $A,$ is:
View full question & answer→MCQ 1451 Mark
Domain of $\sqrt{\text{a}^{2} -\text{x}^{2}} (\text{a} > 0)$ is:
- A
$(-a, a)$
- ✓
$(-a, a)$
- C
$(0, a)$
- D
$(-a, 0)$
AnswerCorrect option: B. $(-a, a)$
View full question & answer→MCQ 1461 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:
- ✓
$2^{mn}$
- B
$2mn - 1$
- C
$2mn$
- D
$mn$
AnswerCorrect option: A. $2^{mn}$
Given, $n(A) = m$
$n(B) = n$
$\therefore n(A \times B) = mn$
Then, the number of relations from $A$ to is $2^{mn}$
View full question & answer→MCQ 1471 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 $f(x) = \text{LCM} (4, 4) = 4$
View full question & answer→MCQ 1481 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 1491 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
AnswerCorrect option: A. $d = bc$
View full question & answer→MCQ 1501 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
$a = 2, b = 3$
AnswerCorrect option: B. $a = 2, b = –3$
Given that: $f(x) = ax + b$
$\Rightarrow f(-1) = a(-1) + b$
$\Rightarrow -5 = -a + b$
$\Rightarrow a - b = 5 ...........(i)$
$f(3) = 3a + b$
$\Rightarrow 3 = 3a + b$
$\Rightarrow 3a + b = 3 ........(ii)$
On solving eqn. $(i)$ and $(ii),$ We get $a = 2, b = -3$
View full question & answer→