MCQ 11 Mark
If $f: R \rightarrow R , f(x)=\sin x$ and $g: R \rightarrow R , g(x)=x^2$ then $(f o g)(x)$ is equal to :
- ✓
$\sin x^2$
- B
$\sin x$
- C
$\sin ^2 x^2$
- D
$\sin ^2 x$.
AnswerCorrect option: A. $\sin x^2$
(A) $\sin x^2$.
View full question & answer→MCQ 21 Mark
A function $f: R \rightarrow R$ is defined such that $f(x)=2+$ $x^2$ then $f$ is :
- A
- B
- C
- ✓
Neither one-one nor onto.
AnswerCorrect option: D. Neither one-one nor onto.
(D) Neither one-one nor onto.
because $-1 \neq 1$
but $\quad f(-1)=f(1)=3$
and $-1 \in R$ (co-domain) then - 1 has not any pre image in R (domain).
View full question & answer→MCQ 31 Mark
Suppose $X =\left\{x^2, x \in N\right\}$ and $f: N \rightarrow X$ defined such that $f(x)=x^2, x \in N$ then function is:
Answer(D) one-one onto.
$\because \quad f=\{(1,1),(2,4),(3,9),(4,16), \ldots\}$
View full question & answer→MCQ 41 Mark
If $R =\left\{(x, y): x, y \in Z , x^2+y^2 \leq 4\right\}$ is a relation in Z then domain of R is :
- A
$\{0,1,2\}$
- ✓
$\{-2,-1,0,1,2\}$
- C
$\{0,-1,-2\}$
- D
$\{-1,0,1\}$.
AnswerCorrect option: B. $\{-2,-1,0,1,2\}$
(B) $\{-2,-1,0,1,2\}$.
View full question & answer→MCQ 51 Mark
If $A=\{1,2,3\}$ and a relation $R$ is such that$R =\{(1,3),(2,2),(3,2)\}$ then for making R reflexive and symmetric set of minimum ordered pair is :
- A
$\{(1,1),(2,3),(1,2)\}$
- B
$\{(3,3),(3,1),(1,2)\}$
- ✓
$\{(1,1),(3,3),(3,1),(2,3)\}$
- D
$\{(1,1),(3,3),(3,1),(1,2)\}$
AnswerCorrect option: C. $\{(1,1),(3,3),(3,1),(2,3)\}$
(C) $\{(1,1),(3,3),(3,1),(2,3)\}$
because for reflexivity $(1,1),(3,3)$ is necessary and for symmetricity $(3,1)$ and $(2,3)$ is necessary.
View full question & answer→MCQ 61 Mark
Let $f: R \rightarrow R$ be defined as $f(x)=x^3$. Choose the correct answer.
AnswerCorrect option: A. $f$ is one-one onto
(A) $f$ is one-one onto.
View full question & answer→MCQ 71 Mark
A relation R is defined as $R =\{(1,2)\}$ on set $A =$ $\{1,2,3\}$, then R is :
Answer(C) - Transitive.
$\because(1,1),(2,2),(3,3) \notin R \Rightarrow R$ is not reflexive.
$\because(1,2) \in R$ but $(2,1) \notin R \Rightarrow R$ is not symmetric. $R$ is transitive because $(1,2) \notin R$ there is not any ordered pair starting from 2 .
View full question & answer→MCQ 81 Mark
In which of the function is onto defined in $R \rightarrow R$.
- A
$f(x)=|x|$
- B
$f(x)=e^{-x}$
- ✓
$f(x)=x^3$
- D
$f(x)=\sin x$.
AnswerCorrect option: C. $f(x)=x^3$
(C) $
f(x)=x^3
$
Here $f: R \rightarrow R$
and$
f(x)=x^3
$
Suppose $y \in R$ co-domain if possible then pre image of $y$ in domain R is $x$.$
\begin{array}{c}
f(x)=y \\
\Rightarrow y=x^3 \therefore x=(y)^{1 / 3} \in R \forall y \in R
\end{array}
$
So, every value $y$ has exist pre image in domain R .
So, $f$ is onto function.
View full question & answer→MCQ 91 Mark
In which of the following function is onto :
- A
$f: Z \rightarrow Z$,$
f(x)=|x|
$
- B
$f: N \rightarrow N$,$
f(x)=|x|
$
- ✓
$f: R _0 \rightarrow R ^{+}, \quad f(x)=|x|$
- D
$f: C \rightarrow R$,$
f(x)=|x|
$
AnswerCorrect option: C. $f: R _0 \rightarrow R ^{+}, \quad f(x)=|x|$
(C) $f: R _0 \rightarrow R ^{+}, \quad f(x)=|x|$
$\because$ Every positive real number has exist the pre-image in domain $R _0$.
View full question & answer→MCQ 101 Mark
In which of the function is one-one defined in $R \rightarrow$ R .
- A
$f(x)=|x|$
- B
$f(x)=\cos x$
- ✓
$f(x)=e^x$
- D
$f(x)=x^2$.
AnswerCorrect option: C. $f(x)=e^x$
(C)$f(x)=e^x$
$x_1, x_2 \in R$ such that
$f\left(x_1\right)=f\left(x_2\right)$
$\Rightarrow \quad e^{x_1}=e^{x_2}$
$\Rightarrow \quad \log _e e^{x_1}=\log _e e^{x_2}$
$\Rightarrow \quad x_1 \log _e e=x_2 \log _e e$
$\Rightarrow \quad x_1=x_2 \quad\left[\because \log _e e=1\right]$
so, $\quad f\left(x_1\right)=f\left(x_2\right)$
$\Rightarrow \quad x_1=x_2 \quad \forall x_1, x_2 \quad R$
$\therefore f$ is one-one function
View full question & answer→MCQ 111 Mark
In which of the function is one-one-
- A
$f: R \rightarrow R$,$
f(x)=x^2
$
- ✓
$f: R \rightarrow R$,$
f(x)=x+1
$
- C
$f: R \rightarrow R$,$
f(x)=e^x+e^{-x}
$
- D
$f: R \rightarrow R$,$
f(x)=3 x^2
$
AnswerCorrect option: B. $f: R \rightarrow R$,$
f(x)=x+1
$
(B) $f: R \rightarrow R , \quad f(x)=x+1$
Suppose $x_1, x_2 \in R$ such that
$\begin{aligned} & & f\left(x_1\right) & =f\left(x_2\right) \\ \Rightarrow & & x_1+1 & =1 \\ \Rightarrow & & x_1 & =x_2+1\end{aligned}$
hence $f\left(x_1\right)=f\left(x_2\right) \Rightarrow x_1=x_2 \forall x_1, x_2 \in R$
$\therefore f$ is one-one function.
View full question & answer→MCQ 121 Mark
If $f(x)=2|x-2|-3|x-3|$ then $f(x)$ equals to, for 2 $< x< 3$.
- A
$5-x$
- B
$x-5$
- ✓
$5 x-13$
- D
$5+x$.
AnswerCorrect option: C. $5 x-13$
(C) $
\begin{aligned}
f(x) & =2|x-2|-3|x-3| \\
& =2\{x-2\}-3\{-(x-3)\} \\
& =2 x-4+3 x-9=5 x-13
\end{aligned}
$
View full question & answer→MCQ 131 Mark
If $f(x)=\frac{x}{1-x}=\frac{1}{y}$, then $f(y)=$
AnswerCorrect option: D. $\left(\frac{1-x}{2 x-1}\right)$.
(D) $f(x)=\frac{x}{1-x}=\frac{1}{y}$
$\Rightarrow \quad \frac{1}{y}=\frac{x}{1-x} \Rightarrow y=\frac{1-x}{x}$
then$
\begin{aligned}
f(y) & =\frac{y}{1-y}=\frac{\frac{1-x}{x}}{1-\left(\frac{1-x}{x}\right)}=\frac{\frac{1-x}{x}}{\frac{x-1+x}{x}} \\
& =\frac{1-x}{2 x-1}
\end{aligned}$
hence correct option is
View full question & answer→MCQ 141 Mark
In set of real numbers " $x$ is smaller than $y$ " will :
- A
- B
- ✓
- D
Reflexive and Anti symmetric.
Answer(C) This relation is only transitive.
If " $x$ is smaller and equal to $y$ " then this relation has reflexive and Anti symmetric.
View full question & answer→MCQ 151 Mark
If a relation $R =\{(a, b),(b, a),(a, a)\}$ in $A =\{a, b$, $c, d\}$ is defined as follows. Then R is :
Answer(A) - Symmetric and Transitive.
but it is not reflexive because $b, c \in A$ and $(b, b) \notin$ R and $(c, c) \notin R$.
View full question & answer→MCQ 161 Mark
A relation R is defined in set N of natural numbers such as $m R n \Leftrightarrow m$ is divisible of $n . \forall m, n \in N$, Then $R$ is :
- A
- B
- C
- ✓
Reflexive and transitive.
AnswerCorrect option: D. Reflexive and transitive.
(D) Reflexive and transitive because every number has divisible of itself here $m$ is divisible of $n$ but $n$ cannot be divisible of $m$. Hence relation is not symmetric. Suppose $x_1, x_2, x_3 \in N$. If $x_1$ is divisible of $x_2$ and $x_2$ is divisible of $x_1$ then $x_1$ will also divisible of $x_3$. Relation is transitive. Hence relation is reflexive and transitive. Correct option is .
View full question & answer→MCQ 171 Mark
If a relation $R$ is defined on $A=\{1,2,3\}$, then $R$ is, where $R =\{(1,1),(2,2),(3,3),(1,2),(2,3)$, $(1,3)\}$ :
- ✓
Reflexive but not symmetric
- B
Reflexive but not transitive
- C
- D
Neither symmetric nor reflexive.
AnswerCorrect option: A. Reflexive but not symmetric
(A) Reflexive but not symmetric because (1, 1), (2, 2), $(3,3) \in R \Rightarrow R$ in reflexive but $R$ is not symmetric because $(1,2) \in R$ but $(2,1) \notin R$.
View full question & answer→