NCERT Solutions · Class 12 Mathematics Relations and Functions

Use the definitions: \(\displaystyle f\) is one-one if \(\displaystyle f(x_1)=f(x_2)\Rightarrow x_1=x_2\), and onto if every \(\displaystyle y\) in the co-domain is \(\displaystyle f(x)\) for some \(\displaystyle x\) in the domain.\(\displaystyle f:\mathbf{R}_{*}\to\mathbf{R}_{*},\ f(x)=\dfrac{1}{x}\) is one-one. Let \(\displaystyle x_1,x_2\in\mathbf{R}_{*}\) with \(\displaystyle f(x_1)=f(x_2)\). Then \[\frac{1}{x_1}=\frac{1}{x_2}\ \Rightarrow\ x_2=x_1 ,\] the cross-multiplication being legitimate because \(\displaystyle x_1,x_2\neq 0\). Hence \(\displaystyle f\) is one-one.It is onto. Take any \(\displaystyle y\in\mathbf{R}_{*}\). Since \(\displaystyle y\neq 0\), the number \(\displaystyle x=\dfrac{1}{y}\) is a well-defined non-zero real, so \(\displaystyle x\in\mathbf{R}_{*}\), and \[f\!\left(\frac{1}{y}\right)=\frac{1}{\,1/y\,}=y .\] So every element of the co-domain has a pre-image, and \(\displaystyle f\) is onto. Therefore \(\displaystyle f\) is a bijection on \(\displaystyle \mathbf{R}_{*}\).Domain replaced by \(\displaystyle \mathbf{N}\). Let \(\displaystyle g:\mathbf{N}\to\mathbf{R}_{*}\), \(\displaystyle g(n)=\dfrac{1}{n}\) (this is a genuine function, since \(\displaystyle 0\notin\mathbf{N}\) and \(\displaystyle 1/n\neq 0\)). Hence the result is not true when the domain is replaced by \(\displaystyle \mathbf{N}\) with the same co-domain: \(\displaystyle g\) is one-one but not onto, so it is not a bijection.

Throughout, injective (one-one) means \(\displaystyle f(x_1)=f(x_2)\Rightarrow x_1=x_2\); surjective (onto) means every element of the co-domain is an image. A single counter-example kills either property.(i) \(\displaystyle f:\mathbf{N}\to\mathbf{N},\ f(x)=x^{2}\). Injective: if \(\displaystyle x_1^{2}=x_2^{2}\) then \(\displaystyle (x_1-x_2)(x_1+x_2)=0\); since \(\displaystyle x_1,x_2\in\mathbf{N}\) are positive, \(\displaystyle x_1+x_2\neq 0\), so \(\displaystyle x_1=x_2\). Injective. Surjective: \(\displaystyle 2\in\mathbf{N}\) is not a perfect square — \(\displaystyle x^{2}=2\) gives \(\displaystyle x=\sqrt2\notin\mathbf{N}\). Not surjective.(ii) \(\displaystyle f:\mathbf{Z}\to\mathbf{Z},\ f(x)=x^{2}\). Injective: \(\displaystyle f(-1)=1=f(1)\) but \(\displaystyle -1\neq 1\). Not injective — this is exactly where the sign matters, the negative integers are now available. Surjective: \(\displaystyle -1\in\mathbf{Z}\) has no pre-image, since \(\displaystyle x^{2}\ge 0\) for all \(\displaystyle x\). Not surjective.(iii) \(\displaystyle f:\mathbf{R}\to\mathbf{R},\ f(x)=x^{2}\). Injective: \(\displaystyle f(-1)=f(1)=1\), so not injective. Surjective: \(\displaystyle -2\in\mathbf{R}\) has no pre-image, as \(\displaystyle x^{2}\ge 0\). Not surjective.(iv) \(\displaystyle f:\mathbf{N}\to\mathbf{N},\ f(x)=x^{3}\). Injective: if \(\displaystyle x_1^{3}=x_2^{3}\) then \(\displaystyle (x_1-x_2)\left(x_1^{2}+x_1x_2+x_2^{2}\right)=0\), and the second factor is positive for \(\displaystyle x_1,x_2\in\mathbf{N}\); hence \(\displaystyle x_1=x_2\). Injective. Surjective: \(\displaystyle 2\in\mathbf{N}\) is not a cube of a natural number \(\displaystyle \left(1^{3}=1<2<8=2^{3}\right)\). Not surjective.(v) \(\displaystyle f:\mathbf{Z}\to\mathbf{Z},\ f(x)=x^{3}\). Injective: the identity \(\displaystyle x_1^{3}-x_2^{3}=(x_1-x_2)\left(x_1^{2}+x_1x_2+x_2^{2}\right)\) again applies; the quadratic factor \(\displaystyle \left(x_1+\tfrac{x_2}{2}\right)^{2}+\tfrac{3x_2^{2}}{4}\) vanishes only when \(\displaystyle x_1=x_2=0\), which also gives \(\displaystyle x_1=x_2\). So \(\displaystyle x_1^{3}=x_2^{3}\Rightarrow x_1=x_2\). Injective (unlike \(\displaystyle x^{2}\), cubing preserves sign, so \(\displaystyle -1\) and \(\displaystyle 1\) are no longer glued together). Surjective: \(\displaystyle 2\in\mathbf{Z}\) has no pre-image, since \(\displaystyle 1^{3}=1\) and \(\displaystyle 2^{3}=8\) and no integer lies strictly between \(\displaystyle 1\) and \(\displaystyle 2\). Not surjective.Summary: (i) injective, not surjective; (ii) neither; (iii) neither; (iv) injective, not surjective; (v) injective, not surjective. None of the five is a bijection.

Recall the definition of the greatest integer function: \(\displaystyle [x]\) is the unique integer \(\displaystyle n\) with \(\displaystyle n\le x<n+1\).Not one-one. It is enough to exhibit two distinct arguments with the same image. Take \(\displaystyle x_1=1.2\) and \(\displaystyle x_2=1.7\). Both lie in \(\displaystyle [1,2)\), so \[f(1.2)=[1.2]=1=[1.7]=f(1.7),\qquad\text{but } 1.2\neq 1.7 .\] In fact every non-empty interval \(\displaystyle [n,n+1)\) is collapsed to the single value \(\displaystyle n\), so \(\displaystyle f\) is many-one.Not onto. By definition \(\displaystyle [x]\) is always an integer, so the range of \(\displaystyle f\) is \(\displaystyle \mathbf{Z}\), a proper subset of the co-domain \(\displaystyle \mathbf{R}\). Concretely, take \(\displaystyle y=0.5\in\mathbf{R}\); if \(\displaystyle [x]=0.5\) for some real \(\displaystyle x\), then \(\displaystyle 0.5\) would be an integer, which is false. So \(\displaystyle 0.5\) has no pre-image.Hence the greatest integer function \(\displaystyle f:\mathbf{R}\to\mathbf{R}\), \(\displaystyle f(x)=[x]\), is neither one-one nor onto.

The modulus function is \(\displaystyle f(x)=|x|=\begin{cases}x, & x\ge 0\\ -x, & x<0\end{cases}\)Not one-one. Take \(\displaystyle x_1=-1\) and \(\displaystyle x_2=1\). Then \[f(-1)=|-1|=1\quad\text{and}\quad f(1)=|1|=1,\] so \(\displaystyle f(-1)=f(1)\) although \(\displaystyle -1\neq 1\). Every non-zero \(\displaystyle x\) shares its image with \(\displaystyle -x\), so \(\displaystyle f\) is many-one.Not onto. For every real \(\displaystyle x\), \(\displaystyle |x|\ge 0\), so the range of \(\displaystyle f\) is \(\displaystyle [0,\infty)\), which is a proper subset of the co-domain \(\displaystyle \mathbf{R}\). Concretely \(\displaystyle y=-1\in\mathbf{R}\) has no pre-image, because \(\displaystyle |x|=-1\) is impossible.Hence the modulus function \(\displaystyle f:\mathbf{R}\to\mathbf{R}\), \(\displaystyle f(x)=|x|\), is neither one-one nor onto.

The signum function is \[f(x)=\begin{cases}\ \ 1, & \text{if } x>0\\ \ \ 0, & \text{if } x=0\\ -1, & \text{if } x<0\end{cases}\]Not one-one. Take \(\displaystyle x_1=1\) and \(\displaystyle x_2=2\). Both are positive, so \[f(1)=1=f(2),\qquad\text{but } 1\neq 2 .\] Indeed the whole of \(\displaystyle (0,\infty)\) is sent to \(\displaystyle 1\) and the whole of \(\displaystyle (-\infty,0)\) to \(\displaystyle -1\), so \(\displaystyle f\) is many-one.Not onto. The range of \(\displaystyle f\) is the three-element set \(\displaystyle \{-1,0,1\}\), a proper subset of the co-domain \(\displaystyle \mathbf{R}\). Concretely, take \(\displaystyle y=2\in\mathbf{R}\): there is no \(\displaystyle x\in\mathbf{R}\) with \(\displaystyle f(x)=2\), since \(\displaystyle f(x)\) only ever takes the values \(\displaystyle -1,0,1\).Hence the signum function \(\displaystyle f:\mathbf{R}\to\mathbf{R}\) is neither one-one nor onto.

Here \(\displaystyle f=\{(1,4),(2,5),(3,6)\}\) is given as a set of ordered pairs, so its values are read straight off: \[f(1)=4,\qquad f(2)=5,\qquad f(3)=6 .\] (Each element of \(\displaystyle \mathrm{A}=\{1,2,3\}\) occurs exactly once as a first coordinate, so \(\displaystyle f\) is indeed a function from A to B.)One-one. For a finite listed function it suffices to check that no two distinct elements of A share an image. The three images \(\displaystyle 4,5,6\) are distinct, i.e. \[f(1)=4\neq 5=f(2),\quad f(2)=5\neq 6=f(3),\quad f(1)=4\neq 6=f(3).\] Equivalently, if \(\displaystyle x_1,x_2\in\mathrm{A}\) and \(\displaystyle f(x_1)=f(x_2)\), then since distinct elements of A have distinct images we must have \(\displaystyle x_1=x_2\).Hence \(\displaystyle f\) is one-one (injective).Note: \(\displaystyle f\) is not onto, since \(\displaystyle 7\in\mathrm{B}\) is not an image — but only injectivity was asked for.

(i) \(\displaystyle f:\mathbf{R}\to\mathbf{R},\ f(x)=3-4x\).One-one: let \(\displaystyle f(x_1)=f(x_2)\). Then \[3-4x_1=3-4x_2\ \Rightarrow\ -4x_1=-4x_2\ \Rightarrow\ x_1=x_2 .\] So \(\displaystyle f\) is one-one.Onto: let \(\displaystyle y\in\mathbf{R}\) be arbitrary and solve \(\displaystyle y=3-4x\) for \(\displaystyle x\): \[4x=3-y\ \Rightarrow\ x=\frac{3-y}{4}.\] This \(\displaystyle x\) is a real number for every real \(\displaystyle y\), so it lies in the domain, and \[f\!\left(\frac{3-y}{4}\right)=3-4\cdot\frac{3-y}{4}=3-(3-y)=y .\] So \(\displaystyle f\) is onto.Being both one-one and onto, \(\displaystyle f\) is bijective.(ii) \(\displaystyle f:\mathbf{R}\to\mathbf{R},\ f(x)=1+x^{2}\).One-one: take \(\displaystyle x_1=-1,\ x_2=1\). Then \(\displaystyle f(-1)=1+1=2\) and \(\displaystyle f(1)=1+1=2\), so \(\displaystyle f(-1)=f(1)\) with \(\displaystyle -1\neq 1\). Hence \(\displaystyle f\) is not one-one (it is many-one).Onto: since \(\displaystyle x^{2}\ge 0\) for all real \(\displaystyle x\), we have \(\displaystyle f(x)=1+x^{2}\ge 1\); the range is \(\displaystyle [1,\infty)\). So \(\displaystyle y=0\in\mathbf{R}\) has no pre-image, because \(\displaystyle 1+x^{2}=0\) gives \(\displaystyle x^{2}=-1\), impossible for real \(\displaystyle x\). Hence \(\displaystyle f\) is not onto.So \(\displaystyle f\) is neither one-one nor onto, and therefore not bijective.

Recall that elements of \(\displaystyle \mathrm{A}\times\mathrm{B}\) are ordered pairs \(\displaystyle (a,b)\) with \(\displaystyle a\in\mathrm{A},\ b\in\mathrm{B}\), and that two ordered pairs are equal exactly when their corresponding coordinates are equal. The map is \(\displaystyle f(a,b)=(b,a)\), which does land in \(\displaystyle \mathrm{B}\times\mathrm{A}\).One-one. Let \(\displaystyle (a_1,b_1),(a_2,b_2)\in\mathrm{A}\times\mathrm{B}\) with \[f(a_1,b_1)=f(a_2,b_2)\ \Rightarrow\ (b_1,a_1)=(b_2,a_2).\] Comparing coordinates gives \(\displaystyle b_1=b_2\) and \(\displaystyle a_1=a_2\), hence \(\displaystyle (a_1,b_1)=(a_2,b_2)\). So \(\displaystyle f\) is one-one.Onto. Let \(\displaystyle (b,a)\) be an arbitrary element of \(\displaystyle \mathrm{B}\times\mathrm{A}\), so \(\displaystyle b\in\mathrm{B}\) and \(\displaystyle a\in\mathrm{A}\). Then \(\displaystyle (a,b)\in\mathrm{A}\times\mathrm{B}\) — this is the step to get right: the pre-image is the same two entries written in the other order — and \[f(a,b)=(b,a).\] So every element of \(\displaystyle \mathrm{B}\times\mathrm{A}\) has a pre-image, and \(\displaystyle f\) is onto.Hence \(\displaystyle f:\mathrm{A}\times\mathrm{B}\to\mathrm{B}\times\mathrm{A}\), \(\displaystyle f(a,b)=(b,a)\), is bijective. (Its inverse is the corresponding swap map \(\displaystyle g:\mathrm{B}\times\mathrm{A}\to\mathrm{A}\times\mathrm{B}\), \(\displaystyle g(b,a)=(a,b)\).)

Here \[f(n)=\begin{cases}\dfrac{n+1}{2}, & n \text{ odd}\\[6pt] \dfrac{n}{2}, & n \text{ even}\end{cases}\qquad n\in\mathbf{N}.\]Test one-one. Compute on the smallest values, taking care to use the correct branch for each: \[f(1)=\frac{1+1}{2}=1\quad(1\text{ is odd}),\qquad f(2)=\frac{2}{2}=1\quad(2\text{ is even}).\] So \(\displaystyle f(1)=f(2)=1\) while \(\displaystyle 1\neq 2\). Hence \(\displaystyle f\) is not one-one. (Generally \(\displaystyle f(2m-1)=f(2m)=m\), so every value is taken twice.)Test onto. Let \(\displaystyle m\in\mathbf{N}\) be arbitrary. Then \(\displaystyle 2m\in\mathbf{N}\) is even, and \[f(2m)=\frac{2m}{2}=m .\] So every natural number has a pre-image, and \(\displaystyle f\) is onto.Conclusion. \(\displaystyle f\) is onto but not one-one, therefore \(\displaystyle f\) is not bijective.

Here \(\displaystyle \mathrm{A}=\mathbf{R}-\{3\}\), \(\displaystyle \mathrm{B}=\mathbf{R}-\{1\}\) and \(\displaystyle f(x)=\dfrac{x-2}{x-3}\); the point \(\displaystyle x=3\) is removed so the formula makes sense.One-one. Let \(\displaystyle x_1,x_2\in\mathrm{A}\) with \(\displaystyle f(x_1)=f(x_2)\): \[\frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3}.\] Since \(\displaystyle x_1\neq 3\) and \(\displaystyle x_2\neq 3\), the denominators are non-zero and we may cross-multiply: \[(x_1-2)(x_2-3)=(x_2-2)(x_1-3)\] \[x_1x_2-3x_1-2x_2+6=x_1x_2-3x_2-2x_1+6 .\] Cancelling \(\displaystyle x_1x_2\) and \(\displaystyle 6\), \[-3x_1-2x_2=-3x_2-2x_1\ \Rightarrow\ -x_1=-x_2\ \Rightarrow\ x_1=x_2 .\] So \(\displaystyle f\) is one-one.Onto. Let \(\displaystyle y\in\mathrm{B}\), i.e. \(\displaystyle y\in\mathbf{R},\ y\neq 1\), and solve \(\displaystyle y=\dfrac{x-2}{x-3}\): \[y(x-3)=x-2\ \Rightarrow\ xy-3y=x-2\ \Rightarrow\ x(y-1)=3y-2\ \Rightarrow\ x=\frac{3y-2}{y-1},\] which is legitimate precisely because \(\displaystyle y\neq 1\) — this is why \(\displaystyle 1\) had to be deleted from the co-domain.This \(\displaystyle x\) must also lie in the domain A, i.e. \(\displaystyle x\neq 3\). If \(\displaystyle \dfrac{3y-2}{y-1}=3\), then \(\displaystyle 3y-2=3y-3\), i.e. \(\displaystyle -2=-3\), which is false; so \(\displaystyle x\neq 3\) and \(\displaystyle x\in\mathrm{A}\).Check: \[f\!\left(\frac{3y-2}{y-1}\right)=\frac{\dfrac{3y-2}{y-1}-2}{\dfrac{3y-2}{y-1}-3}=\frac{3y-2-2(y-1)}{3y-2-3(y-1)}=\frac{y}{1}=y .\] So \(\displaystyle f\) is onto.Conclusion. \(\displaystyle f\) is both one-one and onto, hence a bijection from \(\displaystyle \mathrm{A}\) to \(\displaystyle \mathrm{B}\), with inverse \(\displaystyle f^{-1}(y)=\dfrac{3y-2}{y-1}\).

\(\displaystyle f:\mathbf{R}\to\mathbf{R},\ f(x)=x^{4}\). Test the two properties separately.One-one? Take \(\displaystyle x_1=-1\) and \(\displaystyle x_2=1\): \[f(-1)=(-1)^{4}=1,\qquad f(1)=1^{4}=1 .\] So \(\displaystyle f(-1)=f(1)\) although \(\displaystyle -1\neq 1\). An even power destroys the sign, so \(\displaystyle f\) is many-one, not one-one.Onto? For every real \(\displaystyle x\), \(\displaystyle x^{4}=\left(x^{2}\right)^{2}\ge 0\), so the range of \(\displaystyle f\) is \(\displaystyle [0,\infty)\). Take \(\displaystyle y=-1\in\mathbf{R}\): \(\displaystyle x^{4}=-1\) has no real solution, so \(\displaystyle -1\) has no pre-image. Hence \(\displaystyle f\) is not onto.Being neither one-one nor onto, the correct option is (D).

\(\displaystyle f:\mathbf{R}\to\mathbf{R},\ f(x)=3x\). Test both properties.One-one. Let \(\displaystyle f(x_1)=f(x_2)\). Then \[3x_1=3x_2\ \Rightarrow\ x_1=x_2\] (dividing by \(\displaystyle 3\neq 0\)). So \(\displaystyle f\) is one-one.Onto. Let \(\displaystyle y\in\mathbf{R}\) be arbitrary. Put \(\displaystyle x=\dfrac{y}{3}\); since \(\displaystyle y\) is real and \(\displaystyle 3\neq 0\), this \(\displaystyle x\) is a real number, i.e. it lies in the domain, and \[f\!\left(\frac{y}{3}\right)=3\cdot\frac{y}{3}=y .\] So every real number has a pre-image and \(\displaystyle f\) is onto. (Contrast with \(\displaystyle f:\mathbf{Z}\to\mathbf{Z}\), \(\displaystyle f(x)=3x\), which would fail here — the pre-image \(\displaystyle y/3\) need not be an integer. Over \(\displaystyle \mathbf{R}\) division by \(\displaystyle 3\) is always available.)Being one-one and onto, \(\displaystyle f\) is a bijection, so the correct option is (A) \(\displaystyle f\) is one-one onto.