Question 11 Mark
Assertion (A): The relation $R=\{(x, y):(x+y)$ is a prime number and $x, y \in N\}$ is not a reflexive relation.
Reason (R) : The number ' $2 n$ ' is composite for all natural numbers $n$.
AnswerNow, $\forall x \in R,(x, x) \notin R$ as $x+x=2 x$ is composite number i.e., not a prime number. So, $R$ is not a reflexive relation.
Reason is false as for $n=1,2 n$ is a prime number.
Hence, assertion (A) is true but reason (R) is false.
View full question & answer→Question 21 Mark
Assertion $( A )$ : The relation $f:\{1,2,3,4\} \rightarrow\{x, y$, $z, p\}$ defined by $f=\{(1, x),(2, y),(3, z)\}$ is a bijective function.
Reason $( R )$ : The function $f:\{1,2,3\} \rightarrow\{x, y, z, p\}$ such that $f=\{(1, x),(2, y),(3, z)\}$ is one-one.
AnswerThe element 4 has no image under $f \Rightarrow$ relation $f$ is not a function. So, Assertion is false.
The given function $f:\{1,2,3\} \rightarrow\{x, y, z, p\}$ is one - one as for each element of $\{1,2,3\}$, there is different image in $\{x, y, z, p\}$ under $f$.
$\therefore \quad$ Reason is true.
View full question & answer→Question 31 Mark
Assertion (A) : If $f: R \rightarrow R$ defined by $f(x)=7 x-[7 x]$, where [.] denotes greatest integer $\leq x \forall x \in R$, then $f$ is not one-one function.
Reason (R) : Fractional part functions are always many-one.
Answer(a) : $f(x)=7 x-[7 x]$
Let $7 x=y$
Then $f\left(\frac{y}{7}\right)=y-[y]=\{y\}$
$\Rightarrow f(x)$ is many-one.
$\therefore \quad$ Reason is correct & many-one function cannot be one-one function, so Assertion is also correct.
View full question & answer→Question 41 Mark
Assertion $(A):$ If set $A$ contains $7$ elements and set $B$ contains $6$ elements, then the number of one$-$one onto mapping from $A$ to $B$ is $420 $.
Reason $(R):$ If $A$ and $B$ are two non$-$empty sets containing $m$ and $n$ elements respectively, then number of one$-$one onto functions from $A$ to $B =\left\{\begin{array}{l}n !, \text { if } m=n \\0, \text { if } m \neq n\end{array}\right. \text {. }$
AnswerClearly, reason is true.
Now, $m=7$ and $n=6$ i.e., $m \neq n$
$\therefore$ Number of one$-$one onto mapping from $A$ to $B$ is $0$ .
$\therefore$ Assertion is false and Reason is true.
View full question & answer→Question 51 Mark
Assertion (A) : Let $f:(e, \infty) \rightarrow R$ defined by $f(x)=\log (\log (\log x))$ is bijective.
Reason (R) : A function $f$ will be bijective if $f$ is both one-one and onto.
Answer(a) : As $x \in(e, \infty)$
$\Rightarrow \log x>1 \Rightarrow \log (\log x)>\log 1$
$\Rightarrow \log (\log x)>0$
$\Rightarrow \log (\log (\log (x))>\log 0$
$\Rightarrow \log (\log (\log x)) \in(-\infty, \infty)$
$\Rightarrow$ codomain of $f(x)=$ Range of $f(x) \Rightarrow f$ is onto
Again logarithmic functions are always one-one.
$\therefore f(x)$ is both one - one and onto.
View full question & answer→Question 61 Mark
Assertion $(A)$ : Relation $R$ defined in the set $A$ as $R\{(x, y): y-x$ is an integer, $x, y \in R\}$ is an equivalence relation.
Reason $(R)$ : Relation $R$ defined in the set $B$ as $R\{(x, y): x=\alpha y$ for some rational number $\alpha$, $x, y \in R\}$ is an equivalence relation.
AnswerFor every $(x, x) \in R$
$ \Rightarrow x-x=0$, which is an integer.
$\therefore A$ is reflexive
Let, $(x, y) \in R \Rightarrow y-x$ is an integer.
$\Rightarrow x-y$ is an integer.
$\Rightarrow (y, x) \in R \therefore R$ is symmetric.
Let $(x, y),(y, z) \in R$. Let $x, y, z \in R$
$\left.\begin{array}{l}(x, y) \in R \Rightarrow y-x \text { is an integer } \\ (y, z) \in R \Rightarrow z-y \text { is an integer }\end{array}\right\}$
$\Rightarrow z-x$ is an integer.
$\Rightarrow (x, z) \in R \therefore R$ is transitive
Hence, $R$ is an equivalence relation.
Now, $R=\{(x, y): x=\alpha y, \alpha \in Q\}$
Since, $(0, x) \in R \Rightarrow 0=\alpha x$ for $\alpha=0$
But $(x, 0) \notin R \Rightarrow R$ is not symmetric
$\therefore R$ is not an equivalence relation.
$\therefore $ Assertion is true, Reason is false.
View full question & answer→Question 71 Mark
Assertion $(A) :$ The function $f: R \rightarrow[0,1)$ defined by $f(x)=\frac{x^2}{x^2+1}$ is surjective.
Reason $(R) :$ For surjection, Range of $f(x)=$ codomain of $f(x)$
Answer$(a) :$ For onto function, codomain of $f=$ range of $f$.
We have, $f(x)=\frac{x^2}{1+x^2}$. Then $y \geq 0$.
$\Rightarrow y\left(1+x^2\right)=x^2 \Rightarrow x^2(y-1)=-y$
$\Rightarrow x^2=\frac{y}{1-y} \geq 0$
$\Rightarrow \frac{y-0}{y-1} \leq 0 \Rightarrow 0 \leq y<1 (\because y \neq 1)$
$\Rightarrow$ codomain of $f=$ Range of $f$, as $f: R \rightarrow[0,1)$.
View full question & answer→Question 81 Mark
Assertion (A) : The relation $R$ in a set
$A=\{1,2,3,4\}$ defined by $R=\{(x, y): 3 x-y=0\}$
have the Domain $=\{1,2,3,4\}$ and Range $=$ $\{3,6,9,12\}$.
Reason (R) : Domain & Range of the relation $(R)$ is respectively the set of all first & second entries of the distinct ordered pair of the relation.
Answer(d) $: R=\{(x, y): y=3 x, x \in A\} \therefore R=\{(1,3)\}$
$\therefore \quad$ Domain of the relation $=\{1\}$ and Range of the relation $=\{3\}$.
View full question & answer→Question 91 Mark
Assertion (A) : Let $A=\{-1,1,2,3\}$ and $B=\{1,4,9\}$, where $f: A \rightarrow B$ given by $f(x)=x^2$, then $f$ is not one-one function.
Reason (R): If $x_1 \neq x_2 \Rightarrow f\left(x_1\right) \neq f\left(x_2\right)$, for every $x_1, x_2 \in$ domain, then $f$ is one-one.
Answer(a) : Here $f(-1)=1, f(1)=1$, $f(2)=4, f(3)=9$
Two elements 1 and -1 have the same image $1 \in B$.
So, $f$ is not one-one function. View full question & answer→