NCERT Exemplar Solutions Class 11 Maths Relations and Functions — Short Answer Type Q1–15

NCERT Exemplar Problems, Class 11 Mathematics — Relations and Functions, Short Answer Type questions 1 to 15. All 15 questions solved.

This band is about the vocabulary of the chapter: Cartesian products (Q1 to Q3), equal ordered pairs (Q4), listing a relation from a rule (Q5 to Q9), deciding whether a relation is a function (Q10), and evaluating or combining functions (Q11 to Q15). Two facts do most of the work: $(a,b)=(c,d)$ exactly when $a=c$ and $b=d$, and a relation is a function when every first coordinate appears with exactly one second coordinate.

One printed answer is wrong: Question 11(b) has $1364/4$ in the key; the value is $1363/4$. It is explained where it occurs.

Key insight. Everything here is bookkeeping done carefully. Write the sets out in full, test each candidate pair against the rule, and for functions substitute first and simplify afterwards.

Question 1

Let $A=\{-1,2,3\}$ and $B=\{1,3\}$. Determine (i) $A\times B$, (ii) $B\times A$, (iii) $B\times B$, (iv) $A\times A$.

Solution. $X\times Y$ is the set of all ordered pairs with first entry from $X$ and second from $Y$. Take each element of the first set in turn and pair it with every element of the second. Note that $A\times B\neq B\times A$, because order matters inside a pair. The sizes are $3\times2=6$, $6$, $4$ and $9$.

(i) $A\times B=\{(-1,1),(-1,3),(2,1),(2,3),(3,1),(3,3)\}$

(ii) $B\times A=\{(1,-1),(1,2),(1,3),(3,-1),(3,2),(3,3)\}$

(iii) $B\times B=\{(1,1),(1,3),(3,1),(3,3)\}$

(iv) $A\times A=\{(-1,-1),(-1,2),(-1,3),(2,-1),(2,2),(2,3),(3,-1),(3,2),(3,3)\}$

Question 2

If $P=\{x: x<3,\ x\in\mathbf N\}$, $Q=\{x: x\le 2,\ x\in\mathbf W\}$, find $(P\cup Q)\times(P\cap Q)$, where $\mathbf W$ is the set of whole numbers.

Solution. The catch is that $\mathbf N$ starts at $1$ and $\mathbf W$ starts at $0$. So $P=\{1,2\}$ and $Q=\{0,1,2\}$. Then $P\cup Q=\{0,1,2\}$ and $P\cap Q=\{1,2\}$, and the product has $3\times2=6$ pairs.

Question 3

If $A=\{x: x\in\mathbf W,\ x<2\}$, $B=\{x: x\in\mathbf N,\ 1<x<5\}$, $C=\{3,5\}$, find (i) $A\times(B\cap C)$ (ii) $A\times(B\cup C)$.

Solution. $A=\{0,1\}$, $B=\{2,3,4\}$, $C=\{3,5\}$. Work out the set inside the bracket first, then form the product.

(i) $B\cap C=\{3\}$, so $A\times(B\cap C)=\{(0,3),(1,3)\}$.

(ii) $B\cup C=\{2,3,4,5\}$, so $A\times(B\cup C)$ has $2\times4=8$ pairs.

Question 4

In each case find $a$ and $b$: (i) $(2a+b,\ a-b)=(8,3)$ (ii) $\left(\dfrac a4,\ a-2b\right)=(0,\ 6+b)$.

Solution. Equal ordered pairs have equal coordinates, which gives two equations.

(i) $2a+b=8$ and $a-b=3$. Adding, $3a=11$, so $a=\frac{11}{3}$, and then $b=a-3=\frac23$.

(ii) $\frac a4=0$ gives $a=0$. Then $a-2b=6+b$ becomes $-2b=6+b$, so $b=-2$.

Question 5

Given $A=\{1,2,3,4,5\}$ and $S=\{(x,y): x\in A,\ y\in A\}$, find the ordered pairs satisfying (i) $x+y=5$ (ii) $x+y<5$ (iii) $x+y>8$.

Solution. Fix $x$ and see which $y$ in $A$ work. $y$ must stay between $1$ and $5$.

(i) $x+y=5$: $(1,4),(2,3),(3,2),(4,1)$. The pair $(5,0)$ is out because $0\notin A$.

(ii) $x+y<5$ means $x+y\le4$: $(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)$.

(iii) $x+y>8$ means $x+y\ge9$: $(4,5),(5,4),(5,5)$.

Question 6

Given $R=\{(x,y): x,y\in\mathbf W,\ x^2+y^2=25\}$, find the domain and range of $R$.

Solution. Whole-number solutions of $x^2+y^2=25$: the squares available are $0,1,4,9,16,25$, and the pairs summing to $25$ are $0+25$, $9+16$, $16+9$, $25+0$. So $R=\{(0,5),(3,4),(4,3),(5,0)\}$. The domain is the set of first entries and the range the set of second entries; here they coincide.

Question 7

If $R_1=\{(x,y)\mid y=2x+7,\ x\in\mathbf R,\ -5\le x\le5\}$, find the domain and range of $R_1$.

Solution. The domain is given: $[-5,5]$. The line $y=2x+7$ is increasing, so the range runs from the value at the left end, $2(-5)+7=-3$, to the value at the right end, $2(5)+7=17$, and every value in between is reached.

Question 8

If $R_2=\{(x,y)\mid x,y \text{ are integers and } x^2+y^2=64\}$, find $R_2$.

Solution. Squares not exceeding $64$ are $0,1,4,9,16,25,36,49,64$. Only $0+64$ and $64+0$ add to $64$: subtracting each of the others from $64$ leaves $63,60,55,48,39,28,15$, none of which is a square. So one coordinate is $0$ and the other is $\pm8$. Since $x,y$ may be negative, there are four pairs.

Question 9

If $R_3=\{(x,|x|)\mid x \text{ is a real number}\}$, find the domain and range of $R_3$.

Solution. $|x|$ is defined for every real $x$, so the domain is $\mathbf R$. The second entry $|x|$ is never negative and takes every non-negative value ($|y|=y$ for $y\ge0$), so the range is $\mathbf R^+\cup\{0\}=[0,\infty)$.

Question 10

Is the given relation a function? Give reasons. (i) $h=\{(4,6),(3,9),(-11,6),(3,11)\}$ (ii) $f=\{(x,x)\mid x\in\mathbf R\}$ (iii) $g=\left\{\left(n,\frac1n\right)\mid n \text{ a positive integer}\right\}$ (iv) $s=\{(n,n^2)\mid n \text{ a positive integer}\}$ (v) $t=\{(x,3)\mid x\in\mathbf R\}$.

Solution. The test: no first coordinate may be paired with two different second coordinates.

(i) $3$ is paired with both $9$ and $11$. Not a function. (Two inputs sharing the output $6$ is allowed; one input with two outputs is not.)

(ii) Each $x$ goes to $x$ only. A function (the identity).

(iii) Each positive integer $n$ has the single value $1/n$. A function.

(iv) Each $n$ has the single square $n^2$. A function.

(v) Every $x$ goes to $3$. A function, a constant function.

Question 11

If $f(x)=x^2+7$ and $g(x)=3x+5$, find (a) $f(3)+g(-5)$ (b) $f\left(\frac12\right)\times g(14)$ (c) $f(-2)+g(-1)$ (d) $f(t)-f(-2)$ (e) $\dfrac{f(t)-f(5)}{t-5}$, $t\ne5$.

Solution. Substitute the number into whichever function is named.

(a) $f(3)=9+7=16$ and $g(-5)=-15+5=-10$, so the sum is $6$.

(b) $f\left(\frac12\right)=\frac14+7=\frac{29}{4}$ and $g(14)=42+5=47$. The product is $\frac{29\times47}{4}=\frac{1363}{4}$. The printed key gives $\frac{1364}{4}$, which is a slip: $29\times47=1363$, not $1364$, and $\frac{1364}{4}=341$ is a whole number, which the product is not.

(c) $f(-2)=4+7=11$ and $g(-1)=-3+5=2$, so $13$.

(d) $f(t)-f(-2)=(t^2+7)-11=t^2-4$.

(e) $f(t)-f(5)=t^2-25=(t-5)(t+5)$, so dividing by $t-5$ gives $t+5$.

Question 12

Let $f(x)=2x+1$ and $g(x)=4x-7$. (a) For what real $x$ is $f(x)=g(x)$? (b) For what real $x$ is $f(x)<g(x)$?

Solution. (a) $2x+1=4x-7$ gives $8=2x$, so $x=4$. (b) $2x+1<4x-7$ gives $8<2x$, i.e. $x>4$. Dividing by the positive number $2$ keeps the inequality direction.

Question 13

If $f(x)=2x+1$ and $g(x)=x^2+1$, find (i) $f+g$ (ii) $f-g$ (iii) $fg$ (iv) $\dfrac fg$.

Solution. Combine the expressions pointwise.

(i) $(f+g)(x)=2x+1+x^2+1=x^2+2x+2$.

(ii) $(f-g)(x)=2x+1-x^2-1=2x-x^2$.

(iii) $(fg)(x)=(2x+1)(x^2+1)=2x^3+2x+x^2+1=2x^3+x^2+2x+1$. (The key prints this as $2x^3+x^3+2x+1$, a misprint of the middle term.)

(iv) $\left(\frac fg\right)(x)=\dfrac{2x+1}{x^2+1}$, defined for every real $x$ because $x^2+1\ge1$ is never zero.

Question 14

Express $f:X\to\mathbf R$, $f(x)=x^3+1$, where $X=\{-1,0,3,9,7\}$, as a set of ordered pairs and find its range.

Solution. Compute $x^3+1$ for each member of $X$: $(-1)^3+1=0$, $0+1=1$, $27+1=28$, $729+1=730$, $343+1=344$. The range is the set of second entries.

Question 15

Find the values of $x$ for which $f(x)=3x^2-1$ and $g(x)=3+x$ are equal.

Solution. Set $3x^2-1=3+x$, so $3x^2-x-4=0$. Factor: $3x^2-x-4=(3x-4)(x+1)$, giving $x=\frac43$ or $x=-1$. Check: $f(-1)=2=g(-1)$ and $f\left(\frac43\right)=\frac{16}3-1=\frac{13}3=g\left(\frac43\right)$.

Common mistakes

• Q2, Q3: treating $\mathbf N$ as starting at $0$. Natural numbers begin at $1$ and whole numbers at $0$, so $P$ and $Q$ differ by exactly one element, and that changes the union.
• Q1: writing $A\times B$ and $B\times A$ as the same set. The pairs are ordered, so $(-1,1)\in A\times B$ but $(1,-1)\in B\times A$.
• Q5(ii): reading “$x+y<5$” as “$\le5$” and including the pairs that sum to exactly $5$. A strict inequality means a maximum sum of $4$.
• Q8: listing only $(0,8)$ and $(8,0)$. Integers include negatives, so $(0,-8)$ and $(-8,0)$ belong too.
• Q10(i): saying $h$ is not a function because $6$ appears twice as a second entry. Repeated outputs are fine; it fails because the input $3$ has two outputs.
• Q11(b): multiplying $\frac{29}{4}$ by $47$ carelessly. Work it as $29\times47=29\times50-29\times3=1450-87=1363$.

Practise next

• The rest of this unit: Long Answer Type, Q16-23 and Objective, Fill in the Blanks and True or False, Q24-42.
• NCERT textbook: Exercise 2.1 for Cartesian products and relations, Exercise 2.2 for functions and their values.
• Miscellaneous Exercise, Chapter 2.