NCERT Exemplar Solutions Class 11 Maths Relations and Functions — Objective, Fill in the Blanks and True or False Q24–42

NCERT Exemplar Problems, Class 11 Mathematics — Relations and Functions, Objective, Fill in the Blanks and True or False questions 24 to 42. All 19 questions solved.

Twelve multiple-choice questions (24 to 35) on counting relations, solving an equation with $[x]$, and domain and range of a formula; two fill-in-the-blank questions (36, 37) on combining functions given as sets of pairs; and five True or False statements (38 to 42) on ordered pairs and Cartesian products. For domain, check denominators and square roots; for range, solve for $x$ or bound the expression; for pair-sets, work only on the common first coordinates.

One printed answer is wrong, Question 26. It is explained where it occurs. Questions 32 and 35 also have two identical options in the book, flagged where they occur.

Key insight. A function formed from $f$ and $g$ lives only where both are defined (and, for $f/g$, where $g\ne0$). Once the domain is fixed, the question reduces to arithmetic on a few points or to sign analysis on an interval.

Question 24

Let $n(A)=m$ and $n(B)=n$. Then the total number of non-empty relations that can be defined from $A$ to $B$ is

Solution. A relation from $A$ to $B$ is any subset of $A\times B$, which has $mn$ elements. The number of subsets is $2^{mn}$, and removing the empty subset leaves $2^{mn}-1$.

Question 25

If $[x]^2-5[x]+6=0$, where $[\,\cdot\,]$ denotes the greatest integer function, then

Solution. Treat $[x]$ as the unknown: $[x]^2-5[x]+6=([x]-2)([x]-3)=0$, so $[x]=2$ or $[x]=3$. $[x]=2$ means $2\le x<3$ and $[x]=3$ means $3\le x<4$. The union is $[2,4)$. Option (A) wrongly includes $x=4$ (where $[x]=4$) and misses $[2,3)$; options (B) and (C) each miss one of the two unit intervals.

Question 26

Range of $f(x)=\dfrac1{1-2\cos x}$ is

Solution. Since $-1\le\cos x\le1$, we get $-1\le1-2\cos x\le3$, and $1-2\cos x$ takes every value in $[-1,3]$. The function is undefined where it is $0$. So $t=1-2\cos x\in[-1,0)\cup(0,3]$.

• For $t\in(0,3]$: $\frac1t\in\left[\frac13,\infty\right)$.
• For $t\in[-1,0)$: $\frac1t\in(-\infty,-1]$.

The range is therefore $(-\infty,-1]\cup\left[\frac13,\infty\right)$, option (C). The printed key says (B), which is wrong: $f(0)=\frac1{1-2}=-1$ and $f\left(\frac\pi2\right)=1$, but $1$ is not in $\left[-1,\frac13\right]$. In fact the function never takes any value strictly between $-1$ and $\frac13$, which is most of what (B) claims.

Question 27

Let $f(x)=\sqrt{1+x^2}$, then

Solution. Following the hint, $f(xy)=\sqrt{1+x^2y^2}$ and $f(x)f(y)=\sqrt{(1+x^2)(1+y^2)}=\sqrt{1+x^2+y^2+x^2y^2}$. The second has the extra non-negative terms $x^2+y^2$ under the root, so it is at least as big: $f(xy)\le f(x)f(y)$, with equality only when $x=y=0$.

Question 28

Domain of $\sqrt{a^2-x^2}$ $(a>0)$ is

Solution. We need $a^2-x^2\ge0$, i.e. $x^2\le a^2$, i.e. $|x|\le a$. A square root of zero is fine, so the end points are included.

Question 29

If $f(x)=ax+b$, where $a$ and $b$ are integers, $f(-1)=-5$ and $f(3)=3$, then $a$ and $b$ are equal to

Solution. $f(-1)=-a+b=-5$ and $f(3)=3a+b=3$. Subtracting the first from the second, $4a=8$, so $a=2$, and then $b=-5+a=-3$. Check: $f(3)=6-3=3$.

Question 30

The domain of the function $f(x)=\sqrt{4-x}+\dfrac1{\sqrt{x^2-1}}$ is equal to

Solution. Two conditions. From $\sqrt{4-x}$: $4-x\ge0$, so $x\le4$ (and $x=4$ is allowed). From $\frac1{\sqrt{x^2-1}}$: $x^2-1>0$ strictly, so $x<-1$ or $x>1$ (both $\pm1$ excluded, since they would make the denominator zero). Intersecting: $(-\infty,-1)\cup(1,4]$.

Question 31

The domain and range of the real function $f$ defined by $f(x)=\dfrac{4-x}{x-4}$ is given by

Solution. The denominator is zero at $x=4$, so the domain is $\mathbf R-\{4\}$. For every other $x$, $\frac{4-x}{x-4}=\frac{-(x-4)}{x-4}=-1$. So the only value taken is $-1$.

Question 32

The domain and range of the real function $f$ defined by $f(x)=\sqrt{x-1}$ is given by

Solution. $x-1\ge0$ gives the domain $[1,\infty)$. At $x=1$ the value is $0$, and as $x$ grows the root takes every non-negative value, so the range is $[0,\infty)$. Options (A) and (B) wrongly exclude $0$ from the range. Options (C) and (D) are printed with identical text in the book; the key marks (C), which is right.

Question 33

The domain of the function $f$ given by $f(x)=\dfrac{x^2+2x+1}{x^2-x-6}$ is

Solution. Factor the denominator: $x^2-x-6=(x-3)(x+2)$, zero at $x=3$ and $x=-2$. The numerator $(x+1)^2$ does not cancel either factor, so both points stay excluded. Options (C) and (D) remove a whole interval instead of two points.

Question 34

The domain and range of the function $f$ given by $f(x)=2-|x-5|$ is

Solution. $|x-5|$ is defined for all real $x$, so the domain is $\mathbf R$. $|x-5|$ takes all values in $[0,\infty)$, so $2-|x-5|$ takes all values in $(-\infty,2]$. The maximum $2$ is reached at $x=5$, so $2$ is included, which rules out (C).

Question 35

The domain for which the functions $f(x)=3x^2-1$ and $g(x)=3+x$ are equal is

Solution. Equal functions means $3x^2-1=3+x$, i.e. $3x^2-x-4=(3x-4)(x+1)=0$, so $x=-1$ or $x=\frac43$. The answer is the two-element set $\left\{-1,\frac43\right\}$, not an interval. Options (A) and (B) are printed identically in the book; the key marks (A).

Question 36

Fill in the blank. Let $f=\{(0,1),(2,0),(3,-4),(4,2),(5,1)\}$ and $g=\{(1,0),(2,2),(3,-1),(4,4),(5,3)\}$ be two real functions. Then the domain of $f\cdot g$ is given by ________.

Solution. The product is defined only where both functions are defined. The domain of $f$ is $\{0,2,3,4,5\}$ and the domain of $g$ is $\{1,2,3,4,5\}$. Their intersection is $\{2,3,4,5\}$: $0$ is missing from $g$ and $1$ from $f$.

Question 37

Match the following. Let $f=\{(2,4),(5,6),(8,-1),(10,-3)\}$ and $g=\{(2,5),(7,1),(8,4),(10,13),(11,5)\}$ be two real functions. Match (a) $f-g$, (b) $f+g$, (c) $f\cdot g$, (d) $\frac fg$ with (i) $\left\{\left(2,\frac45\right),\left(8,\frac{-1}4\right),\left(10,\frac{-3}{13}\right)\right\}$, (ii) $\{(2,20),(8,-4),(10,-39)\}$, (iii) $\{(2,-1),(8,-5),(10,-16)\}$, (iv) $\{(2,9),(8,3),(10,10)\}$.

Solution. The common domain is $\{2,8,10\}$ (the first coordinates found in both $f$ and $g$). At these points $f$ takes $4,-1,-3$ and $g$ takes $5,4,13$. Then:

• (a) $f-g$: $4-5=-1$, $-1-4=-5$, $-3-13=-16$, giving (iii).
• (b) $f+g$: $9,\ 3,\ 10$, giving (iv).
• (c) $f\cdot g$: $20,\ -4,\ -39$, giving (ii).
• (d) $f/g$: $\frac45,\ \frac{-1}4,\ \frac{-3}{13}$, giving (i).

Question 38

True or False. The ordered pair $(5,2)$ belongs to the relation $R=\{(x,y): y=x-5,\ x,y\in\mathbf Z\}$.

Solution. Put $x=5$: $y=5-5=0\ne2$. So $(5,2)$ does not satisfy the rule. False.

Question 39

True or False. If $P=\{1,2\}$, then $P\times P\times P=\{(1,1,1),(2,2,2),(1,2,2),(2,1,1)\}$.

Solution. $P\times P\times P$ has $2^3=8$ ordered triples, including $(1,1,2)$ and $(2,2,1)$, which are not listed. The given set has only four. False.

Question 40

True or False. If $A=\{1,2,3\}$, $B=\{3,4\}$ and $C=\{4,5,6\}$, then $(A\times B)\cup(A\times C)=\{(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,3),(3,4),(3,5),(3,6)\}$.

Solution. $(A\times B)\cup(A\times C)=A\times(B\cup C)$ and $B\cup C=\{3,4,5,6\}$. So the union is $A\times\{3,4,5,6\}$, which has $3\times4=12$ pairs: every first entry $1,2,3$ with every second entry $3,4,5,6$. That is exactly the set given. True.

Question 41

True or False. If $(x-2,\ y+5)=\left(-2,\ \frac13\right)$ are two equal ordered pairs, then $x=4$, $y=\dfrac{-14}3$.

Solution. Equal pairs give $x-2=-2$, so $x=0$ (not $4$), and $y+5=\frac13$, so $y=\frac13-5=-\frac{14}3$. The value of $y$ is right but $x=4$ is wrong. False.

Question 42

True or False. If $A\times B=\{(a,x),(a,y),(b,x),(b,y)\}$, then $A=\{a,b\}$ and $B=\{x,y\}$.

Solution. The first coordinates that appear are $a,b$, so $A=\{a,b\}$, and the second coordinates are $x,y$, so $B=\{x,y\}$. Also $|A|\,|B|=2\times2=4$ matches the four pairs. True.

Common mistakes

• Q26: picking the interval between $-1$ and $\frac13$ because those are the numbers that appear. The reciprocal of a number in $[-1,3]$ jumps over the gap around $0$, so the range is two outside pieces, not the middle.
• Q25: forgetting that $[x]=3$ also allows $x$ up to but not including $4$, and writing $[2,3]$. Each integer value of $[x]$ contributes a half-open unit interval.
• Q30: keeping $x=1$ or $x=-1$ because the square root of zero is defined. Here the root sits in a denominator, so zero must go.
• Q31: cancelling to $-1$ and then claiming the domain is all of $\mathbf R$. The function is still undefined at $x=4$.
• Q35, Q15 (same idea): answering with an interval. Two functions are equal only at isolated points, so the answer is a set of two numbers.
• Q36, Q37: adding or dividing the pairs at points like $5$ and $7$. A first coordinate that appears in only one function is dropped.

Practise next

• This unit’s other pages: Short Answer Type, Q1-15 and Long Answer Type, Q16-23.
• NCERT textbook: Exercise 2.3 for domain and range, Exercise 2.1 for Cartesian products.
• Miscellaneous Exercise, Chapter 2.