NCERT Solutions · Class 11 Mathematics Relations and Functions

Check the overlap point of the two pieces. In a piecewise rule the only place a second image can sneak in is where the two intervals meet; there the two formulas must agree.For \(\displaystyle f \). The pieces are \(\displaystyle x^{2} \) on \(\displaystyle 0\le x\le 3 \) and \(\displaystyle 3x \) on \(\displaystyle 3\le x\le 10 \); they overlap only at \(\displaystyle x=3 \).At \(\displaystyle x=3 \): \[x^{2}=3^{2}=9, \qquad 3x=3(3)=9 . \] Both formulas give the same value $\displaystyle 9$, so \(\displaystyle x=3 \) has the single image 9. Every other \(\displaystyle x \) in \(\displaystyle [0,10] \) lies in exactly one piece, so it has exactly one image.Hence each element of the domain \(\displaystyle [0,10] \) has a unique image, and \(\displaystyle f \) is a function.For \(\displaystyle g \). The pieces are \(\displaystyle x^{2} \) on \(\displaystyle 0\le x\le 2 \) and \(\displaystyle 3x \) on \(\displaystyle 2\le x\le 10 \); they overlap at \(\displaystyle x=2 \).At \(\displaystyle x=2 \): \[x^{2}=2^{2}=4, \qquad 3x=3(2)=6 . \] The two formulas disagree, so \(\displaystyle 2 \) is paired with both $\displaystyle 4$ and $\displaystyle 6$, i.e. \(\displaystyle (2,4)\in g \) and \(\displaystyle (2,6)\in g \).An element of the domain with two different images violates the definition of a function.\(\displaystyle f \) is a function (its pieces agree at \(\displaystyle x=3 \), both giving $\displaystyle 9$), while \(\displaystyle g \) is not a function (at \(\displaystyle x=2 \) it gives both $\displaystyle 4$ and $\displaystyle 6$).

Evaluate the two function values, then divide. With \(\displaystyle f(x)=x^{2} \):\[f(1.1)=(1.1)^{2}=1.21, \qquad f(1)=1^{2}=1 . \]Therefore \[\frac{f(1.1)-f(1)}{1.1-1}=\frac{1.21-1}{0.1}=\frac{0.21}{0.1}=2.1 . \]\(\displaystyle \dfrac{f(1.1)-f(1)}{(1.1-1)}=2.1 \).

A rational function is defined wherever its denominator is non-zero. The numerator \(\displaystyle x^{2}+2x+1 \) is a polynomial, so it causes no trouble; only the denominator can fail.Set the denominator to zero and factorise: \[x^{2}-8x+12=0 \;\Longrightarrow\; (x-2)(x-6)=0 \;\Longrightarrow\; x=2 \ \text{ or } \ x=6 . \]These two values must be excluded; every other real number is allowed.Domain \(\displaystyle =\mathbf{R}-\{2,\,6\} \).

A real square root needs a non-negative radicand.Domain. We need \[x-1\ge 0 \;\Longrightarrow\; x\ge 1 , \] so the domain is \(\displaystyle [1,\infty) \).Range. As \(\displaystyle x \) runs over \(\displaystyle [1,\infty) \), the expression \(\displaystyle x-1 \) runs over \(\displaystyle [0,\infty) \), and the (principal) square root of a non-negative number is non-negative. Every value \(\displaystyle y\ge 0 \) is attained, since \(\displaystyle x=y^{2}+1 \) lies in the domain and gives \(\displaystyle f(x)=\sqrt{y^{2}}=y \).So the range is \(\displaystyle [0,\infty) \).Domain \(\displaystyle =[1,\infty) \), range \(\displaystyle =[0,\infty) \).

The modulus is defined everywhere and never negative.Domain. \(\displaystyle |x-1| \) can be computed for every real \(\displaystyle x \), so the domain is \(\displaystyle \mathbf{R} \).Range. By definition \(\displaystyle |x-1|\ge 0 \) for all \(\displaystyle x \). Every non-negative value occurs: for \(\displaystyle y\ge 0 \), taking \(\displaystyle x=y+1 \) gives \(\displaystyle f(x)=|y|=y \). The value $\displaystyle 0$ occurs at \(\displaystyle x=1 \).So the range is the set of all non-negative reals, \(\displaystyle [0,\infty) \).Domain \(\displaystyle =\mathbf{R} \), range \(\displaystyle =[0,\infty) \).

Set the output equal to \(\displaystyle y \) and solve for \(\displaystyle x \); the range is the set of \(\displaystyle y \) for which a real \(\displaystyle x \) exists.Here \(\displaystyle f(x)=\dfrac{x^{2}}{1+x^{2}} \) (the denominator \(\displaystyle 1+x^{2} \) is never $\displaystyle 0$, so \(\displaystyle f \) is defined on all of \(\displaystyle \mathbf{R} \)). Put \[y=\frac{x^{2}}{1+x^{2}} . \]Cross-multiplying, \[y(1+x^{2})=x^{2} \;\Longrightarrow\; y = x^{2}-yx^{2} = x^{2}(1-y) \;\Longrightarrow\; x^{2}=\frac{y}{1-y}\quad (y\neq 1). \]A real \(\displaystyle x \) exists exactly when \(\displaystyle x^{2}\ge 0 \), i.e. when \[\frac{y}{1-y}\ge 0 . \] The quotient is non-negative precisely when \(\displaystyle y\ge 0 \) and \(\displaystyle 1-y>0 \), that is \(\displaystyle 0\le y<1 \). (If \(\displaystyle y\ge 1 \) the numerator and denominator have opposite signs or the denominator vanishes; if \(\displaystyle y<0 \) the quotient is negative.)A quick sanity check agrees: \(\displaystyle x=0 \) gives \(\displaystyle f=0 \); \(\displaystyle x=1 \) gives \(\displaystyle \tfrac12 \); and as \(\displaystyle |x| \) grows, \(\displaystyle \dfrac{x^{2}}{1+x^{2}}=1-\dfrac{1}{1+x^{2}} \) rises towards $\displaystyle 1$ without ever reaching it.Range of \(\displaystyle f = [0,\,1) \).

Algebra of real functions. For functions with the same domain, \[(f+g)(x)=f(x)+g(x), \qquad (f-g)(x)=f(x)-g(x), \qquad \left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)},\ g(x)\neq 0 . \]Here \(\displaystyle f(x)=x+1 \) and \(\displaystyle g(x)=2x-3 \), both defined on all of \(\displaystyle \mathbf{R} \).Sum. \[(f+g)(x)=(x+1)+(2x-3)=3x-2, \qquad x\in\mathbf{R}. \]Difference. \[(f-g)(x)=(x+1)-(2x-3)=x+1-2x+3=-x+4, \qquad x\in\mathbf{R}. \]Quotient. We must exclude the zero of \(\displaystyle g \): \(\displaystyle 2x-3=0 \Rightarrow x=\dfrac{3}{2} \). So \[\left(\frac{f}{g}\right)(x)=\frac{x+1}{2x-3}, \qquad x\in\mathbf{R}-\left\{\tfrac{3}{2}\right\}. \]\(\displaystyle (f+g)(x)=3x-2 \) and \(\displaystyle (f-g)(x)=4-x \), both on \(\displaystyle \mathbf{R} \); \(\displaystyle \left(\dfrac{f}{g}\right)(x)=\dfrac{x+1}{2x-3} \) on \(\displaystyle \mathbf{R}-\left\{\dfrac{3}{2}\right\} \).

Turn two of the ordered pairs into simultaneous equations, then check the rest.Each pair \(\displaystyle (x,y) \in f \) means \(\displaystyle y=f(x)=ax+b \).From \(\displaystyle (1,1) \): \[a(1)+b=1 \quad \Longrightarrow \quad a+b=1 . \]From \(\displaystyle (2,3) \): \[a(2)+b=3 \quad \Longrightarrow \quad 2a+b=3 . \]Subtracting the first equation from the second gives \(\displaystyle a=2 \), and then ($\displaystyle 1$) gives \(\displaystyle b=1-2=-1 \).Check the remaining pairs with \(\displaystyle f(x)=2x-1 \): \[f(0)=2(0)-1=-1 \ \checkmark \ \big[(0,-1)\big], \qquad f(-1)=2(-1)-1=-3 \ \checkmark \ \big[(-1,-3)\big]. \]Both are consistent, and \(\displaystyle a=2,\ b=-1 \) are integers as required.\(\displaystyle a=2 \) and \(\displaystyle b=-1 \), so \(\displaystyle f(x)=2x-1 \).

Disprove a "for all" claim with a single counter-example. Recall \(\displaystyle \mathrm{R}=\{(a,b): a,b\in\mathbf{N},\ a=b^{2}\} \), so \(\displaystyle (a,b)\in\mathrm{R} \) means the first entry is the square of the second.(i) \(\displaystyle (a,a)\in\mathrm{R} \) for all \(\displaystyle a\in\mathbf{N} \) — false.This would require \(\displaystyle a=a^{2} \) for every natural number, which holds only for \(\displaystyle a=1 \).Counter-example: take \(\displaystyle a=2 \). Then \(\displaystyle a^{2}=4\neq 2 \), so \(\displaystyle (2,2)\notin\mathrm{R} \).(ii) \(\displaystyle (a,b)\in\mathrm{R} \Rightarrow (b,a)\in\mathrm{R} \) — false.Counter-example: \(\displaystyle (4,2)\in\mathrm{R} \) because \(\displaystyle 4=2^{2} \). But for \(\displaystyle (2,4) \) we would need \(\displaystyle 2=4^{2}=16 \), which is false. So \(\displaystyle (2,4)\notin\mathrm{R} \).(iii) \(\displaystyle (a,b)\in\mathrm{R} \) and \(\displaystyle (b,c)\in\mathrm{R} \Rightarrow (a,c)\in\mathrm{R} \) — false.Counter-example: \(\displaystyle (16,4)\in\mathrm{R} \) because \(\displaystyle 16=4^{2} \), and \(\displaystyle (4,2)\in\mathrm{R} \) because \(\displaystyle 4=2^{2} \). But for \(\displaystyle (16,2) \) we would need \(\displaystyle 16=2^{2}=4 \), which is false. So \(\displaystyle (16,2)\notin\mathrm{R} \).All three statements are false; counter-examples: (i) \(\displaystyle (2,2)\notin\mathrm{R} \); (ii) \(\displaystyle (4,2)\in\mathrm{R} \) but \(\displaystyle (2,4)\notin\mathrm{R} \); (iii) \(\displaystyle (16,4),(4,2)\in\mathrm{R} \) but \(\displaystyle (16,2)\notin\mathrm{R} \).

Two different tests. A relation from A to B is any subset of \(\displaystyle \mathrm{A}\times\mathrm{B} \); a function additionally needs every element of A to have exactly one image.(i) Is \(\displaystyle f \) a relation from A to B? Yes.Check each ordered pair has its first entry in \(\displaystyle \mathrm{A}=\{1,2,3,4\} \) and its second in \(\displaystyle \mathrm{B}=\{1,5,9,11,15,16\} \): So \(\displaystyle f\subseteq \mathrm{A}\times\mathrm{B} \), which makes \(\displaystyle f \) a relation from A to B.(ii) Is \(\displaystyle f \) a function from A to B? No.The element \(\displaystyle 2\in\mathrm{A} \) occurs as a first entry twice: \(\displaystyle (2,9) \) and \(\displaystyle (2,11) \). So $\displaystyle 2$ has two distinct images, $\displaystyle 9$ and $\displaystyle 11$, and the uniqueness requirement fails.(i) True — \(\displaystyle f \) is a relation from A to B, since every pair lies in \(\displaystyle \mathrm{A}\times\mathrm{B} \). (ii) False — \(\displaystyle f \) is not a function, because $\displaystyle 2$ has two images, $\displaystyle 9$ and 11.

Look for one first-coordinate carrying two different second coordinates. In \(\displaystyle f=\{(ab,\,a+b): a,b\in\mathbf{Z}\} \) the input is the product \(\displaystyle ab \) and the output is the sum \(\displaystyle a+b \). Different factor pairs can give the same product but different sums.Counter-example: take the product 6. Thus the single input $\displaystyle 6$ is assigned two different outputs, $\displaystyle 5$ and $\displaystyle 7$, which contradicts the requirement that a function give each element of the domain exactly one image.(As a second flaw, \(\displaystyle f \) is not even defined on all of \(\displaystyle \mathbf{Z} \) in the way a function \(\displaystyle \mathbf{Z}\to\mathbf{Z} \) must be — but the counter-example above already settles it.)No, \(\displaystyle f \) is not a function from \(\displaystyle \mathbf{Z} \) to \(\displaystyle \mathbf{Z} \), since \(\displaystyle (6,5) \) and \(\displaystyle (6,7) \) both belong to \(\displaystyle f \).

Factorise each element of A into primes and take the largest prime that appears. The range is the set of all images \(\displaystyle f(n) \) for \(\displaystyle n\in\mathrm{A}=\{9,10,11,12,13\} \).\[\begin{aligned} 9 &= 3\times 3 &&\Rightarrow\ f(9)=3\\ 10 &= 2\times 5 &&\Rightarrow\ f(10)=5\\ 11 &= 11 \ \text{(prime)} &&\Rightarrow\ f(11)=11\\ 12 &= 2\times 2\times 3 &&\Rightarrow\ f(12)=3\\ 13 &= 13 \ \text{(prime)} &&\Rightarrow\ f(13)=13 \end{aligned} \]Collecting the images and listing each only once (a set has no repetitions, so the two $\displaystyle 3$'s count once):Range of \(\displaystyle f = \{3,\,5,\,11,\,13\} \).