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# Classify the following functions as injection, surjection or bijection : (i) f : N → N given by f(x) = x2 (ii) f : Z → Z given by f(x) = x2 (iii) f : N → N given by f(x) = x3 (iv) f : Z → Z given by f(x) = x3 (v) f : R → R, defined by f(x) = |x| (vi) f : Z → Z, defined by f(x) = x2 + x (vii) f : Z → Z, defined by f(x) = x − 5 (viii) f : R → R, defined by f(x) = sinx (ix) f : R → R, defined by f(x) = x3 + 1 (x) f : R → R, defined by f(x) = x3 − x (xi) f : R → R, defined by f(x) = sin2x + cos2x (xii) f : Q − {3} → Q, defined by $f\left(x\right)=\frac{2x+3}{x-3}$ (xiii) f : Q → Q, defined by f(x) = x3 + 1 (xiv) f : R → R, defined by f(x) = 5x3 + 4 (xv) f : R → R, defined by f(x) = 3 − 4x (xvi) f : R → R, defined by f(x) = 1 + x2 (xvii) f : R → R, defined by f(x) = $\frac{x}{{x}^{2}+1}$ [NCERT EXEMPLAR]

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## (i) f : N → N, given by f(x) = x2 Injection test: Let x and y be any two elements in the domain (N), such that f(x) = f(y). f(x)=f(y) ${x}^{2}={y}^{2}\phantom{\rule{0ex}{0ex}}x=y\text{(We do not get ± because}x\text{and}y\text{are in}\mathbf{N}\text{)}$ So, f is an injection . Surjection test: Let y be any element in the co-domain (N), such that f(x) = y for some element x in N (domain). f(x) = y ${x}^{2}=y\phantom{\rule{0ex}{0ex}}x=\sqrt{y},\text{which may not be in}\mathbf{N}\text{.}\phantom{\rule{0ex}{0ex}}\text{For example, if}y\text{=3,}\phantom{\rule{0ex}{0ex}}x=\sqrt{3}\text{is not in}\mathbf{N}\text{.}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ So, f is not a surjection. So, f is not a bijection. (ii) f : Z → Z, given by f(x) = x2 Injection test: Let x and y be any two elements in the domain (Z), such that f(x) = f(y). f(x) = f(y) ${x}^{2}={y}^{2}\phantom{\rule{0ex}{0ex}}x=±y$ So, f is not an injection . Surjection test: Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain). f(x) = y ${x}^{2}=y\phantom{\rule{0ex}{0ex}}x=±\sqrt{y}\text{which may not be in}\mathbf{Z}.\phantom{\rule{0ex}{0ex}}\text{For example, if}y\text{=3,}\phantom{\rule{0ex}{0ex}}x=±\sqrt{3}\text{is not in}\mathbf{Z}\text{.}$ So, f is not a surjection. So, f is not a bijection. (iii) f : N → N, given by f(x) = x3 Injection test: Let x and y be any two elements in the domain (N), such that f(x) = f(y). f(x) = f(y) ${x}^{3}={y}^{3}\phantom{\rule{0ex}{0ex}}x=y$ So, f is an injection . Surjection test: Let y be any element in the co-domain (N), such that f(x) = y for some element x in N (domain). f(x) = y ${x}^{3}=y\phantom{\rule{0ex}{0ex}}x=\sqrt[3]{y}\text{which may not be in}\mathbf{N}\text{.}\phantom{\rule{0ex}{0ex}}\text{For example, if}y\text{=3,}\phantom{\rule{0ex}{0ex}}x=\sqrt[3]{3}\text{is not in}\mathbf{N}\text{.}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ So, f is not a surjection and f is not a bijection. (iv) f : Z → Z, given by f(x) = x3 Injection test: Let x and y be any two elements in the domain (Z), such that f(x) = f(y) f(x) = f(y) ${x}^{3}={y}^{3}\phantom{\rule{0ex}{0ex}}x=y$ So, f is an injection. Surjection test: Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain). f(x) = y ${x}^{3}=y\phantom{\rule{0ex}{0ex}}x=\sqrt[3]{y}\text{which may not be in}\mathbf{Z}\text{.}\phantom{\rule{0ex}{0ex}}\text{For example, if}y\text{=3,}\phantom{\rule{0ex}{0ex}}x=\sqrt[3]{3}\text{is not in}\mathbf{Z}\text{.}$ So, f is not a surjection and f is not a bijection. (v) f : R → R, defined by f(x) = |x| Injection test: Let x and y be any two elements in the domain (R), such that f(x) = f(y) f(x) = f(y) $\left|x\right|=\left|y\right|\phantom{\rule{0ex}{0ex}}x=±y$ So, f is not an injection . Surjection test: Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain). f(x) = y $\left|x\right|=y\phantom{\rule{0ex}{0ex}}x=±y\in \mathbf{Z}$ So, f is a surjection and f is not a bijection. (vi) f : Z → Z, defined by f(x) = x2 + x Injection test: Let x and y be any two elements in the domain (Z), such that f(x) = f(y). f(x) = f(y) ${x}^{2}+x={y}^{2}+y\phantom{\rule{0ex}{0ex}}\text{Here, we cannot say that}x\text{=}y\text{.}\phantom{\rule{0ex}{0ex}}\text{For example, x = 2 and y = - 3}\phantom{\rule{0ex}{0ex}}\text{Then},\phantom{\rule{0ex}{0ex}}{x}^{2}+x={2}^{2}+2=6\phantom{\rule{0ex}{0ex}}{y}^{2}+y={\left(-3\right)}^{2}-3=6\phantom{\rule{0ex}{0ex}}\text{So, we have two numbers 2 and -3 in the domain}\mathbf{Z}\text{whose image is same as 6.}$ So, f is not an injection . Surjection test: Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain). f(x) = y ${x}^{2}+x=y\phantom{\rule{0ex}{0ex}}\text{Here, we cannot say}x\in \mathbf{Z}.\phantom{\rule{0ex}{0ex}}\text{For example,}y\text{=-4.}\phantom{\rule{0ex}{0ex}}{x}^{2}+x=-4\phantom{\rule{0ex}{0ex}}{x}^{2}+x+4=0\phantom{\rule{0ex}{0ex}}x=\frac{-1±\sqrt{-15}}{2}=\frac{-1±i\sqrt{15}}{2}\text{which is not in}\mathbf{Z}.$ So, f is not a surjection and f is not a bijection. (vii) f : Z → Z, defined by f(x) = x − 5 Injection test: Let x and y be any two elements in the domain (Z), such that f(x) = f(y). f(x) = f(y) x $-$ 5 = y $-$ 5 x = y So, f is an injection . Surjection test: Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain). f(x) = y x $-$ 5 = y x = y + 5, which is in Z. So, f is a surjection and f is a bijection. (viii) f : R → R, defined by f(x) = sinx Injection test: Let x and y be any two elements in the domain (R), such that f(x) = f(y). f(x) = f(y) $\mathrm{sin}x=\mathrm{sin}y\phantom{\rule{0ex}{0ex}}\text{Here,}\mathit{\text{x}}\text{may not be equal to}\mathit{\text{y}}\text{because}\mathrm{sin}0=\mathrm{sin\pi }.\phantom{\rule{0ex}{0ex}}\text{So, 0 and π have the same image 0.}$ So, f is not an injection . Surjection test: Range of f = [$-$1, 1] Co-domain of f = R Both are not same. So, f is not a surjection and f is not a bijection. (ix) f : R → R, defined by f(x) = x3 + 1 Injection test: Let x and y be any two elements in the domain (R), such that f(x) = f(y). f(x) = f(y) ${x}^{3}+1={y}^{3}+1\phantom{\rule{0ex}{0ex}}{x}^{3}={y}^{3}\phantom{\rule{0ex}{0ex}}x=y$ So, f is an injection. Surjection test: Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain). f(x) = y ${x}^{3}+1=y\phantom{\rule{0ex}{0ex}}x=\sqrt[3]{y-1}\in \mathbf{R}$ So, f is a surjection. So, f is a bijection. (x) f : R → R, defined by f(x) = x3 − x Injection test: Let x and y be any two elements in the domain (R), such that f(x) = f(y). f(x) = f(y) ${x}^{3}-x={y}^{3}-y\phantom{\rule{0ex}{0ex}}\text{Here, we cannot say}x=y.\phantom{\rule{0ex}{0ex}}\text{For example,}x\text{=1 and}\mathit{\text{y}}\text{=-1}\phantom{\rule{0ex}{0ex}}{x}^{3}-x=1-1=0\phantom{\rule{0ex}{0ex}}{y}^{3}-y={\left(-1\right)}^{3}-\left(-1\right)-1+1=0\phantom{\rule{0ex}{0ex}}\text{So, 1 and -1 have the same image 0.}$ So, f is not an injection. Surjection test: Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain). f(x) = y ${x}^{3}-x=y\phantom{\rule{0ex}{0ex}}\text{By observation we can say that there exist some}x\text{in}\mathbf{R},\text{such that}{x}^{3}\text{-x=y.}$ So, f is a surjection and f is not a bijection. (xi) f : R → R, defined by f(x) = sin2x + cos2x f(x) = sin2x + cos2x = 1 So, f(x) = 1 for every x in R. So, for all elements in the domain, the image is 1. So, f is not an injection. Range of f = {1} Co-domain of f = R Both are not same. So, f is not a surjection and f is not a bijection. (xii) f : Q − {3} → Q, defined by $f\left(x\right)=\frac{2x+3}{x-3}$ Injection test: Let x and y be any two elements in the domain (Q − {3}), such that f(x) = f(y). f(x) = f(y) $\frac{2x+3}{x-3}=\frac{2y+3}{y-3}\phantom{\rule{0ex}{0ex}}\left(2x+3\right)\left(y-3\right)=\left(2y+3\right)\left(x-3\right)\phantom{\rule{0ex}{0ex}}2xy-6x+3y-9=2xy-6y+3x-9\phantom{\rule{0ex}{0ex}}9x=9y\phantom{\rule{0ex}{0ex}}x=y$ So, f is an injection. Surjection test: Let y be any element in the co-domain (Q − {3}), such that f(x) = y for some element x in Q (domain). f(x) = y $\frac{2x+3}{x-3}=y\phantom{\rule{0ex}{0ex}}2x+3=xy-3y\phantom{\rule{0ex}{0ex}}2x-xy=-3y-3\phantom{\rule{0ex}{0ex}}x\left(2-y\right)=-3\left(y+1\right)\phantom{\rule{0ex}{0ex}}x=\frac{3\left(y+1\right)}{y-2},\text{which is not defined at}y\text{=2.}\phantom{\rule{0ex}{0ex}}$ So, f is not a surjection and f is not a bijection. (xiii) f : Q → Q, defined by f(x) = x3 + 1 Injection test: Let x and y be any two elements in the domain (Q), such that f(x) = f(y). f(x) = f(y) ${x}^{3}+1={y}^{3}+1\phantom{\rule{0ex}{0ex}}{x}^{3}={y}^{3}\phantom{\rule{0ex}{0ex}}x=y$ So, f is an injection . Surjection test: Let y be any element in the co-domain (Q), such that f(x) = y for some element x in Q (domain). f(x) = y ${x}^{3}+1=y\phantom{\rule{0ex}{0ex}}x=\sqrt[3]{y-1,}\text{which may not be in}\mathbf{Q}.\phantom{\rule{0ex}{0ex}}\text{For example, if}y\text{= 8,}\phantom{\rule{0ex}{0ex}}{x}^{3}+1=8\phantom{\rule{0ex}{0ex}}{x}^{3}=7\phantom{\rule{0ex}{0ex}}x=\sqrt[3]{7},\text{which is not in}\mathbf{\text{Q}}\text{.}$ So, f is not a surjection and f is not a bijection. So, f is a surjection and f is a bijection. (xiv) f : R → R, defined by f(x) = 5x3 + 4 Injection test: Let x and y be any two elements in the domain (R), such that f(x) = f(y). f(x) = f(y) $5{x}^{3}+4=5{y}^{3}+4\phantom{\rule{0ex}{0ex}}5{x}^{3}=5{y}^{3}\phantom{\rule{0ex}{0ex}}{x}^{3}={y}^{3}\phantom{\rule{0ex}{0ex}}x=y$ So, f is an injection . Surjection test: Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain). f(x) = y $5{x}^{3}+4=y\phantom{\rule{0ex}{0ex}}5{x}^{3}=y-4\phantom{\rule{0ex}{0ex}}{x}^{3}=\frac{y-4}{5}\phantom{\rule{0ex}{0ex}}x=\sqrt[3]{\frac{y-4}{5}}\in \mathbf{R}$ So, f is a surjection and f is a bijection. (xv) f : R → R, defined by f(x) = 3 − 4x Injection test: Let x and y be any two elements in the domain (R), such that f(x) = f(y). f(x) = f(y) $3-4x=3-4y\phantom{\rule{0ex}{0ex}}-4x=-4y\phantom{\rule{0ex}{0ex}}x=y$ So, f is an injection . Surjection test: Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain). f(x) = y $3-4x=y\phantom{\rule{0ex}{0ex}}4x=3-y\phantom{\rule{0ex}{0ex}}x=\frac{3-y}{4}\in \mathbf{R}$ So, f is a surjection and f is a bijection. (xvi) f : R → R, defined by f(x) = 1 + x2 Injection test: Let x and y be any two elements in the domain (R), such that f(x) = f(y). f(x) = f(y) $1+{x}^{2}=1+{y}^{2}\phantom{\rule{0ex}{0ex}}{x}^{2}={y}^{2}\phantom{\rule{0ex}{0ex}}x=±y$ So, f is not an injection. Surjection test: Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain). f(x) = y $1+{x}^{2}=y\phantom{\rule{0ex}{0ex}}{x}^{2}=y-1\phantom{\rule{0ex}{0ex}}x=±\sqrt{y-1}\text{which may not be in}\mathbf{R}\phantom{\rule{0ex}{0ex}}\text{For example, if}y\text{=0,}\phantom{\rule{0ex}{0ex}}x=±\sqrt{-1}=±i\text{is not in}\mathbf{R}.$ So, f is not a surjection and f is not a bijection. (xvii) f : R → R, defined by f(x) = $\frac{x}{{x}^{2}+1}$ Injection test: Let x and y be any two elements in the domain (R), such that f(x) = f(y). f(x) = f(y) $\frac{x}{{x}^{2}+1}=\frac{y}{{y}^{2}+1}\phantom{\rule{0ex}{0ex}}x{y}^{2}+x={x}^{2}y+y\phantom{\rule{0ex}{0ex}}x{y}^{2}-{x}^{2}y+x-y=0\phantom{\rule{0ex}{0ex}}-xy\left(-y+x\right)+1\left(x-y\right)=0\phantom{\rule{0ex}{0ex}}\left(x-y\right)\left(1-xy\right)=0\phantom{\rule{0ex}{0ex}}x=y\mathrm{or}x=\frac{1}{y}$ So, f is not an injection. Surjection test: Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain). f(x) = y $\frac{x}{{x}^{2}+1}=y\phantom{\rule{0ex}{0ex}}y{x}^{2}-x+y=0\phantom{\rule{0ex}{0ex}}x=\frac{-\left(-1\right)±\sqrt{1-4{y}^{2}}}{2y},\mathrm{if}y\ne 0\phantom{\rule{0ex}{0ex}}=\frac{1±\sqrt{1-4{y}^{2}}}{2y},\mathrm{which}\mathrm{may}\mathrm{not}\mathrm{be}\mathrm{in}\mathbf{R}\phantom{\rule{0ex}{0ex}}\mathrm{For}\mathrm{example},\mathrm{if}y=1,\mathrm{then}\phantom{\rule{0ex}{0ex}}x=\frac{1±\sqrt{1-4}}{2}=\frac{1±\mathrm{i}\sqrt{3}}{2},\mathrm{which}\mathrm{is}\mathrm{not}\mathrm{in}\mathbf{R}\phantom{\rule{0ex}{0ex}}\mathrm{So},f\mathrm{is}\mathrm{not}\mathrm{surjection}\mathrm{and}f\mathrm{is}\mathrm{not}\mathrm{bijection}.$ So, f is not a surjection and f is not a bijection.

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