Download App

MATHS 2 Marks Questions

PART - 1 CH - 2 Relations and Functions · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 12 questions.

Questions

2 Marks Questions

🎯

Test yourself on this topic

12 questions · timed · auto-graded

Question 12 Marks
If $f: R \rightarrow R$ is defined as-
$f(x)=\left\{\begin{array}{r}1 \text { if } x \in Q \\ -1 \text { if } x \notin Q \end{array}\right.$
then
(i) Find the value of the following-
$f\left(\frac{1}{2}\right), f(\pi), f(\sqrt{2})$
(ii) Find the set of images of set R under $f$.
(iii) Find the pre images of 1 and -1
Answer
(i)
$\begin{array}{ll}\because \frac{1}{2} \in Q, & \therefore f\left(\frac{1}{2}\right)=1 \\\because \quad \pi \notin Q, & \therefore f(\pi)=-1 \\\text { and because } \sqrt{2} \notin Q, & \therefore f(\sqrt{2})=-1\end{array}$
(ii) $\because R$ is a set of rational and irrational numbers
$\therefore \forall x \in Q, f(x)=1$
and $\forall x \notin Q , f(x)=-1$, while $x$ is a real number.
Hence, set of images of R under $f=\{1,-1\}$.
(iii) Since 1 is the image of all those real numbers which are the elements of Q , therefore the pre- image set of 1 is $Q$. Similarly -1 is the image of those real numbers which are elements of ( $R -$ $Q )$. Therefore pre image set of -1 is $( R - Q )$.
View full question & answer→
Question 22 Marks
A relation $R$ is defined on the set of integers such that $x R y \Leftrightarrow x ^2+ y ^2= 2 5$ then write $R$ and $R^{-1}$ as the set of ordered pairs and also find its domain.
Answer
Given relation R is defined as-
$\begin{array}{l}x R y \Leftrightarrow x^2+y^2=25 \\\Rightarrow \quad y= \pm \sqrt{25-x^2}\end{array}$
Here, $x=0, \Rightarrow y= \pm 5 \therefore(0,5) \in R$ and $(0,-5) \in R$
$x= \pm 3 \Rightarrow y= \pm 4$
$\begin{array}{ll}\therefore & (3,4) \in R,(-3,4) \in R,(3,-4) \in R \text { and }-3,, 4) \in R\ \\& x= \pm 4 \Rightarrow y= \pm 3 \\\therefore & (4,3) \in R,(4,-3) \in R,(-4,3) \in R \text { and }(-4,-3) \in R \\\text { and } & x= \pm 5 \Rightarrow y=0 \\\therefore & (5,0) \in R \text { and }(-5,0) \in R\end{array}$
We see here that for any other integer value of $x$, there is no integer value of $y$, which satisfies the given relation.
$\begin{aligned}\text { So, } \quad R= & \{(0,5),(0,-5),(3,4),(-3,4),(3,-4), \\& (-3,-4),(4,3),(4,-3),(-4,3), \\& (-4,-3),(5,0),(-5,0)\} \\R^{-1}= & \{(5,0),(-5,0),(4,3),(4,-3),(-4,3), \\& (-4,-3),(3,4),(3,-4),(-3,4), \\& (-3,-4),(0,5),(0,-5)\}\end{aligned}$
$\begin{array}{l}\text { Domain of } R=\{0,3,-3,4,-4,5,-5\} \\=\text { Domain of } R^{-1}\end{array}$
View full question & answer→
Question 32 Marks
If $x, y \in\{1,2,3\}$, then which of the relation are functions-
(i) $f_1=\{(x, y): x+y>3\}$
(ii) $f_2=\{(x, y): x>3\}$
(iii) $f_3=\{(x, y): y=x+1\}$
(iv) $f_4=\{(x, y): x+y=4\}$
Answer
(i) Relation $f_1$ can be written in the form of ordered pairs as follows-
$f_1=\{(1,3),(2,2),(2,3),(3,1),(3,2),(3,3)\}$
This relation is not a function because the first element of ordered pairs $(2,2)$ and $(2,3)$ are same. Similarly the first elements of ordered pairs $(3,1)$, $(3,2)$ and $(3,3)$ are same.
(ii) Relation $f_2$ can be written in the form of ordered pairs as follows-
$f_2=\{(2,1),(3,1),(3,2)\}$
This relation is not a function, because $1 \in\{1,2,3\}$, but 1 has no image in $\{1,2,3\}$
(iii) Relation $f_3$ can be written in the form of ordered pairs as follows-
$f_3=\{(1,2),(2,3)\}$
This relation is not a function, because $3 \in\{1,2,3\}$, but 3 has no image in $\{1,2,3\}$
(iv) Relation $f_4$ can be written in the form of ordered pairs as follows-
$f_4=\{(1,3),(2,2),(3,1)\}$
Thus, relation is a function because each element of set $\{1,2,3\}$ is associated with a unique element of $\{1,2,3\}$.
View full question & answer→
Question 42 Marks
If $f: R \rightarrow R , f(x)=e^x$, then find -
(i) Set of images of R under $f$.
(ii) $\{f / f(y)=1\}$
(iii) Is $f(x+y)=f(x) \cdot f(y)$ is true?
Answer
$\text {(i)}\quad \because e^x$ is a positive real number, $\forall ~x \in R$
$\therefore \quad f(x)=e^x$ is a positive real number $\forall~ x \in R$
For each positive real number $x,$
$ f(\log x)=e^{\log _e x}=x $
So, set $f$ images of R under $f$ is $R ^{+}$i.e. set of positive real numbers.
$\begin{aligned}\text {(ii)}\therefore\quad f(y)=1 & \quad \Rightarrow e^y=1=e^0 \\ \Rightarrow e^y=e^0 &\quad \Rightarrow y=0\end{aligned}$
$\quad\quad\therefore \quad\{y / f(y)=1\}=\{0\}$
$\begin{aligned}\text {(iii)} \because \quad f(x+y) & =e^{x+y}=e^x \cdot e^y \quad \forall x, y \in R \\ \quad f(x+y) & =f(x) \cdot f(y) \\ \therefore \quad f(x+y) & =f(x) \cdot f(y) \text { is true. }\end{aligned}$
View full question & answer→
Question 62 Marks
Is $g=\{(1,1),(2,3),(3,5),(4,7)\}$ is a function? If $g$ is represented by $g(x)=\alpha x+\beta$ then find the value of $\alpha$ and $\beta$.
Answer
Given $g=\{(1,1),(2,3),(3,5),(4,7)\}$
Here $g$ is a function because image of elements $\text {1, 2, 3, 4}$ are $1,3,5,7$ respectively. Here each element has unique image.
Hence, this is a function.
$ \begin{aligned}\text { Now }g(1) & =1, g(2)=3, g(3)=5, g(5)=7 \\g(x) & =\alpha x+\beta \\g(1) & =\alpha \cdot 1+\beta=\alpha+\beta=1\ldots\ldots \text {(i)} \\g(2) & =\alpha \cdot 2+\beta=2 \alpha+\beta=3\ldots\ldots\text {(ii)}\end{aligned}$
Subtracting (i) from (ii)
$2 \alpha-\alpha=3-1$
or $\quad \alpha=2$
putting the value of $\alpha$ in equation (i)
$\begin{array}{l}\alpha+\beta=1 \\2+\beta=1 \Rightarrow \beta=-1 \\\therefore \alpha=2, \beta=-1\end{array}$
View full question & answer→
Question 72 Marks
Find the domain of functions $f$ and $g$ for which $f(x)=2 x^2-1$ and $g(x)=1-3 x$ are equal. Find domain for which these functions are not equal.
Answer
Given, $f(x)=g(x)$
$\begin{array}{l}\therefore 2 x^2-1=1-3 x \\\Rightarrow 2 x^2+3 x-2=0 \Rightarrow(2 x-1)(x+2)=0 \\\Rightarrow x=\frac{1}{2} \text { or } x=-2 \text { or both. }\end{array}$
Functions $f$ and $g$ are equal for the domain $\left\{\frac{1}{2},-2\right\}$.
Functions $f$ and $g$ are not equal for the domain R -
$\left\{\frac{1}{2},-2\right\} .$
View full question & answer→
Question 82 Marks
A relation $R$ is defined from the set of real numbers $R$ to the set of complex numbers $C$ such that $x R y \Leftrightarrow|x|=y$, then tell with reason which of the following are true or false-
(i) $(1+i) R 3$
(ii) $3 R (-3)$
(iii) $(2+3 i) R 13$
(iv) $(1+i) R 1$
Answer
Given that $x R y$ is true if $|x|=y$
(i) Here, $x=1+i$ and $y=3$
Now
$\begin{aligned}|x|=|1+i|=\sqrt{(1)^2+(1)^2} & =\sqrt{1+1}  =\sqrt{2}\end{aligned}$
$\therefore \quad|x| \neq y So ,(1+i) \text R 3$ is false.
(ii) $H \quad|x| \neq y So ,(1+i) R 3$ is false.
Here, $\quad x=3$ and $y=-3$
$\therefore \quad|x|=3 \neq y \text { So, }$
$3 R (-3)$ is false.
(iii) Here,
$x=2+3 i \text { and } y=13$
$\begin{aligned}\therefore \quad|x|=\sqrt{(2)^2+(3)^2} & =\sqrt{4+9}  =\sqrt{13}\end{aligned}
$
$\therefore \quad|x| \neq y \quad \therefore(2+3 i) R 13$ is false.
(iv) Here,
$x=1+i \text { and } y=1$
$\begin{aligned}\therefore \quad|x|=\sqrt{(1)^2+(1)^2} & =\sqrt{1+1} =\sqrt{2}\end{aligned}$
$\therefore \quad|x| \neq y \quad \therefore(1+i) R 1$ is false.
View full question & answer→
Question 92 Marks
Write the following functions in the form of ordered pairs and also write their range.
(i) $f: A \rightarrow R , f(x)=x^2+1$
Where $A=\{-3,-2,-1,0,1,2,3\}$
(ii) $g: B \rightarrow I , g(x)=x^2-x$
Where $B =\{x: x \in N , x \leq 5\}$
Answer
(i) $f(x)=x^2+1$
By keeping all the values of $x$ here, the function $f$ can be expressed in the form of ordered pairs as follows-
$
\begin{array}{r}
f=\{(-3,10),(-2,5),(-1,2),(0,1),(1,2),(2,5),(3,10)\}
\end{array}
$
$\therefore \quad$ Range of $f=\{1,2,5,10\}$
(ii) By keeping all the values of $x$ here, the function $g$ can be expressed in the form of ordered pairs as follows-
$
\begin{array}{ll} 
& g=\{(1,0),(2,2),(3,6),(4,12),(5,20)\} \\
\therefore \quad & \text { Range of } g=\{0,2,6,12,20\}
\end{array}
$
View full question & answer→
Question 102 Marks
If $f: R \rightarrow R , f(x)=x^2$ then find-
(i) Range of $f$
(ii) $\{x / f(x)=4\}$
(iii) $\{y / f(y)=-1\}$
Answer
(i) Range of $f=\{x \in R / 0 \leq x<\infty\}$
(ii) $\because f(x)=x^2$
putting $x=2, f(2)=(2)^2=4$
putting $x=-2, f(-2)=(-2)^2=4$
$
\therefore\{2,-2\}
$
(iii) $\phi$ (It is not possible)
View full question & answer→
Question 122 Marks
If $A =\{2,4,9\}$ and $B =\{-3,-2,1,2\}$ and a relation R from set A to set B is defined such that $a R b \Leftrightarrow$ $b= \pm \sqrt{a}$ then write the relation R in the form of ordered pairs.
Answer

$\begin{array}{l}A=\{2,4,9\}, B=\{-3,-2,1,2\} \\\text { and } a R b \Leftrightarrow b= \pm \sqrt{a}\end{array}$
So, when $a=2 \in A$ and $b=\sqrt{2} \notin B$
$\begin{array}{l}a=4 \in A \text { then } b=\sqrt{4}= \pm 2 \in B \\\Rightarrow 4 R 2 \quad \text { 4R }(-2)\end{array}$
$a=9 \in A \text { then } b=\sqrt{9}= \pm 3,3 \notin B,-3 \in B$
$\Rightarrow 9 R-3$
Hence, $R=\{(4,2),(4,-2),(9,-3)\}$
View full question & answer→
Generate Paper FreeAll PART - 1 CH - 2 Relations and Functions topics
2 Marks Questions - MATHS STD 11 Science Questions - Vidyadip