NCERT Solutions: Class 11 Maths Chapter 2 - Relations and Functions

Equal ordered pairs have equal corresponding components. Thus $\dfrac{x}{3}+1=\dfrac{5}{3}$, so $\dfrac{x}{3}=\dfrac{2}{3}$ and $x=2$. Also $y-\dfrac{2}{3}=\dfrac{1}{3}$, so $y=1$.

$B$ has 3 elements. Therefore $n(A\times B)=n(A)n(B)=3\times3=9$.

In $G\times H$, the first component is chosen from $G$ and the second from $H$. In $H\times G$, the order is reversed.

(i) Since $P=\{m,n\}$ and $Q=\{n,m\}$, each element of $P$ pairs with each element of $Q$, giving four ordered pairs. (ii) This is the definition of Cartesian product for non-empty sets. (iii) Since $B\cap\phi=\phi$, $A\times(B\cap\phi)=A\times\phi=\phi$.

Each coordinate can be either $-1$ or $1$, so there are $2^3=8$ ordered triples.

The first components of the ordered pairs form $A$, and the second components form $B$.

Here $B\cap C=\phi$, so $A\times(B\cap C)=A\times\phi=\phi$. Also $A\times B$ has second coordinates in $B$, while $A\times C$ has second coordinates in $C$; since $B\cap C=\phi$, their intersection is $\phi$. For (ii), every first coordinate in $A=\{1,2\}$ lies in $B$, and every second coordinate in $C=\{5,6\}$ lies in $D$. Hence every ordered pair of $A\times C$ is in $B\times D$.

$A\times B$ has 4 elements because $n(A)n(B)=2\times2=4$. A set with 4 elements has $2^4=16$ subsets, listed by choosing no elements, one element, two elements, three elements and all four elements.

The first coordinates $x,y,z$ must belong to $A$, and they are distinct. Since $n(A)=3$, these are all elements of $A$. The second coordinates 1 and 2 must belong to $B$, and since $n(B)=2$, these are all elements of $B$.

Since $n(A\times A)=9$, $n(A)^2=9$ and $n(A)=3$. The pairs $(-1,0)$ and $(0,1)$ show that $-1,0,1\in A$. Hence $A=\{-1,0,1\}$. Listing all $3\times3$ ordered pairs and removing the two given pairs gives the remaining seven pairs.

The condition $3x-y=0$ gives $y=3x$. With $x,y\in A=\{1,2,\ldots,14\}$, the possible values are $x=1,2,3,4$, giving $y=3,6,9,12$.

The natural numbers less than 4 are $1,2,3$. For these, $y=x+5$ gives $6,7,8$ respectively.

The difference between two integers is odd exactly when one is odd and the other is even. In $A$, $1,3,5$ are odd and pair with $4,6$ from $B$; $2$ is even and pairs with $9$ from $B$.

From Fig. 2.7, the arrows are $5\to3$, $6\to4$ and $7\to5$. These pairs satisfy $y=x-2$.

The condition says that $a$ divides $b$. For $a=1$, every element of $A$ is possible. For $a=2$, $b=2,4,6$. For $a=3$, $b=3,6$. For $a=4$, $b=4$. For $a=6$, $b=6$.

The first coordinates are the given values of $x$. The second coordinates are $x+5$, giving $5,6,7,8,9,10$.

The prime numbers less than 10 are $2,3,5,7$. Cubing them gives $8,27,125,343$.

$n(A\times B)=3\times2=6$. A relation from $A$ to $B$ is any subset of $A\times B$, so the number of relations is $2^6=64$.

For any integers $a$ and $b$, $a-b$ is an integer. Hence $R=Z\times Z$, so every integer appears as a first coordinate and every integer appears as a second coordinate.

A relation is a function when no first element has more than one image and every domain element has exactly one image. In (i) and (ii), each first coordinate occurs only once. In (iii), the first coordinate 1 is associated with both 3 and 5, so it is not a function.

(i) $|x|$ is defined for every real $x$, and $-|x|\le0$ with maximum 0. (ii) For the square root to be real, $9-x^2\ge0$, so $-3\le x\le3$. The value of $\sqrt{9-x^2}$ is least 0 and greatest 3.

Substitute into $f(x)=2x-5$: $f(0)=0-5=-5$, $f(7)=14-5=9$, and $f(-3)=-6-5=-11$.

$t(0)=32$. $t(28)=\dfrac{9\times28}{5}+32=50.4+32=82.4$. $t(-10)=\dfrac{-90}{5}+32=-18+32=14$. If $t(C)=212$, then $\dfrac{9C}{5}+32=212$, so $\dfrac{9C}{5}=180$ and $C=100$.

(i) If $x\gt0$, then $3x\gt0$, so $2-3x\lt2$, and values can decrease without bound. (ii) Since $x^2\ge0$, $x^2+2\ge2$, with equality at $x=0$. (iii) The identity function takes every real value.

For $f$, the only overlap of the intervals is at $x=3$. Both rules give the same value there: $3^2=9$ and $3\times3=9$. Hence each input has exactly one image. For $g$, the overlap is at $x=2$. The first rule gives $2^2=4$, while the second gives $3\times2=6$. The same input has two images, so $g$ is not a function.

$f(1.1)=(1.1)^2=1.21$ and $f(1)=1$. Therefore $\dfrac{f(1.1)-f(1)}{1.1-1}=\dfrac{1.21-1}{0.1}=\dfrac{0.21}{0.1}=2.1$.

The denominator must not be zero. Since $x^2-8x+12=(x-2)(x-6)$, the function is undefined at $x=2$ and $x=6$. Hence the domain is all real numbers except 2 and 6.

For the square root to be real, $x-1\ge0$, so $x\ge1$. Also $\sqrt{x-1}$ is always non-negative and can take every non-negative real value.

The modulus expression $|x-1|$ is defined for every real $x$. Its value is never negative, and it equals 0 at $x=1$, so the range is $[0,\infty)$.

For every real $x$, $x^2\ge0$, so $\dfrac{x^2}{1+x^2}\ge0$. Also $1+x^2>x^2$, so $\dfrac{x^2}{1+x^2}\lt1$. The value 0 occurs at $x=0$. Values get arbitrarily close to 1 as $|x|$ becomes large, but never equal 1. Hence the range is $[0,1)$.

Add and subtract the function values: $(f+g)(x)=(x+1)+(2x-3)=3x-2$ and $(f-g)(x)=(x+1)-(2x-3)=4-x$. For the quotient, divide $f(x)$ by $g(x)$ and exclude values where $g(x)=0$, i.e. $2x-3=0$.

Since $(0,-1)\in f$, $f(0)=b=-1$. Since $(1,1)\in f$, $f(1)=a+b=1$. Substituting $b=-1$ gives $a-1=1$, so $a=2$.

(i) $(a,a)\in R$ would require $a=a^2$, which is not true for all natural numbers; for example $a=2$ fails. (ii) $(4,2)\in R$ because $4=2^2$, but $(2,4)\notin R$ because $2\ne4^2$. (iii) $(16,4)\in R$ and $(4,2)\in R$, but $(16,2)\notin R$ would require $16=2^2$, which is false.

(i) Every ordered pair in $f$ has first coordinate in $A$ and second coordinate in $B$, so $f\subset A\times B$ and is a relation from $A$ to $B$. (ii) It is not a function because the element 2 has two images, 9 and 11.

The same first coordinate can have different second coordinates. For example, with $a=0,b=0$, we get $(ab,a+b)=(0,0)$. With $a=0,b=1$, we get $(0,1)$. Thus 0 has two images, 0 and 1, so $f$ is not a function.

The highest prime factors are: $9\mapsto3$, $10\mapsto5$, $11\mapsto11$, $12\mapsto3$ and $13\mapsto13$. Hence the set of images is $\{3,5,11,13\}$.