Question 15 Marks
Let $\text{f}:[-1,\infty)\rightarrow[-1,\infty)$ be given by $f(x) = (x + 1)^2 - 1, \text{x}\geq-1.$ Show that f is invertible. Also, find the set $S = \{x : f(x) = f^{-1}(x)\}.$
AnswerInjectivity: Let $x$ and $\text{y}\in[-1,\infty),$ such that
$f(x) = f(y)$
$\Rightarrow (x + 1)^2 - 1 = (y + 1)^2 - 1$
$\Rightarrow (x + 1)^2 = (y + 1)^2$
$\Rightarrow (x + 1) = (y + 1)$
$\Rightarrow x = y$
So, $f$ is a injection.
Surjectivity: Let $\text{y}\in[-1,\infty)$
Then, $f(x) = y$
$\Rightarrow (x + 1)^2 - 1 = y$
$\Rightarrow\ \text{x}+1=\sqrt{\text{y}+1}$
$\Rightarrow\ \text{x}=\sqrt{\text{y}+1}-1$
Clearly, $\text{x}=\sqrt{\text{y}+1}-1$ is real for all $\text{y}\geq-1.$
Thus, every element $\text{y}\in[-1,\infty)$ has its pre-image $\text{x}\in[-1,\infty)$ given by $\text{x}=\sqrt{\text{y}+1}-1$
$\Rightarrow f$ is a surjection.
So, $f$ is a bijection.
Hence, $f$ is invertible.
Let $f^{-1}(x) = y ......(1)$
$\Rightarrow f(y) = x$
$\Rightarrow (y + 1)^2 - 1 = x$
$\Rightarrow (y + 1)^2 = x + 1$
$\Rightarrow\ \text{y}+1=\sqrt{\text{x}+1}$
$\Rightarrow\ \text{y}=\pm\sqrt{\text{x}+1}-1$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\pm\sqrt{\text{x}+1}-1 [$from $(1)]$
$\text{f}(\text{x})=\text{f}^{-1}(\text{x})$
$\Rightarrow\ (\text{x}+1)^2-1=\pm\sqrt{\text{x}+1}-1$
$\Rightarrow\ (\text{x}+1)^2=\pm\sqrt{\text{x}+1}$
$\Rightarrow (x + 1)^4 = x + 1$
$\Rightarrow (x + 1)[(x + 1)^3 - 1] = 0$
$\Rightarrow x + 1 = 0$ or $(x + 1)^3 - 1 = 0$
$\Rightarrow x = -1$ or $(x + 1)^3 = 1$
$\Rightarrow x = -1$ or$ x + 1 = 1$
$\Rightarrow x = -1$ or $x = 0$
$\Rightarrow S = \{0, -1\}$
View full question & answer→Question 25 Marks
Consider $\text{f}:\text{R}_+\rightarrow[-5,\infty)$ given by $f(x) = 9x^2 + 6x - 5.$ Show that f is invertible with $\text{f}^{-1}(\text{x})=\frac{\sqrt{\text{x}+6}-1}{3}.$
AnswerInjectivity of f: Let $x$ and $y$ be two elements of domain $(R_+),$ such that $f(x) = f(y)$
$\Rightarrow 9x^2 + 6x - 5 = 9y^2 + 6y - 5$
$\Rightarrow 9x^2 + 6x = 9y^2 + 6y$
$\Rightarrow x = y ($As, $\text{x, y}\in\text{R}_+$)
So, $f$ is one-one.
Surjectivity of f: Let $y$ is in the co domain $(Q)$ such that $f(x) = y$
$\Rightarrow 9x^2 + 6x - 5 = y$
$\Rightarrow 9x^2 + 6x = y + 5$
$\Rightarrow 9x^2 + 6x + 1 = y + 6 ($Adding $1$ on both sides$)$
$\Rightarrow (3x + 1)^2 = y + 6$
$\Rightarrow\ \text{3x}+1=\sqrt{\text{y}+6}$
$\Rightarrow\ 3\text{x}=\sqrt{\text{y}+6}-1$
$\Rightarrow\ \text{x}=\frac{\sqrt{\text{y}+6}-1}{3}\in\text{R}^+$ (domain)
$\Rightarrow f$ is onto.
So, $f$ is a bijection and hence, it is invertible.
Finding $f^{-1}:$ Let $f^{-1}(x) = y .....(1)$
$\Rightarrow x = f(y)$
$\Rightarrow x = 9y^2 + 6y - 5$
$\Rightarrow x + 5 = 9y^2 + 6y$
$\Rightarrow x + 6 = 9y^2 + 6y + 1 ($adding $1$ on both sides$)$
$\Rightarrow x + 6 = (3y + 1)^2$
$\Rightarrow\ 3\text{y}+1=\sqrt{\text{x}+6}$
$\Rightarrow\ 3\text{y}=\sqrt{\text{x}+6}-1$
$\Rightarrow\ \text{y}=\frac{\sqrt{\text{x}+6}-1}{3}$
So, $\text{f}^{-1}(\text{x})=\frac{\sqrt{\text{x}+6}-1}{3} [$from $(1)]$
View full question & answer→Question 35 Marks
If $f : Q \rightarrow Q, g : Q \rightarrow Q$ are two functions defined by $f(x) = 2x$ and $g(x) = x + 2$, show that f and g are bijective maps. Verify that $(gof)^{-1} = f^{-1}og^{-1}$.
AnswerInjectivity of f : Let $x$ and $y$ be two elements of domain (Q),
such that $f(x)=f(y) \Rightarrow 2 x=2 y \Rightarrow x=y$ So, $f$ is one-one.
Surjectivity of $f$ : Let $y$ be in the co-domain (Q), such that $f(x)=y . \Rightarrow 2 x=y$
$\Rightarrow x =\frac{ y }{2} \in Q$ (domain) $\Rightarrow f$ is onto.
So, f is a bijection and hence, it is invertible.
Finding $f ^{-1}$ : Let $f ^{-1}( x )= y$
$\Rightarrow x=f(y) \Rightarrow x=2 y$
$\Rightarrow y=\frac{x}{2}$
So, $f ^{-1}( x )=\frac{ x }{2}[$ from (1)]
Injectivity of g : Let x and y be two elements of domain ( Q ),
such that $g(x)=g(y) \Rightarrow x+2=y+2 \Rightarrow x=y$ So, $g$ is one-one.
Surjectivity of g : Let y be in the co domain ( Q ),
such that $g(x)=y . \Rightarrow x+2=y$
$\Rightarrow x=y-2 \in Q \text { (domain) }$
$\Rightarrow g$ is onto. $So , g$ is a bijection and hence, it is invertible.
Finding $g ^{-1}$ : Let $g ^{-1}( x )= y$
$\qquad$
$\Rightarrow x=g(y)$
$\Rightarrow x=y+2$
$\Rightarrow y=x-2$
So, $g^{-1}(x)=x-2[$ from (2)]
Verification of (gof)$^{-1}=f^{-1}$ og$^{-1}: f(x)=2 x ; g(x)=x+2$ and
$f^{-1}(x)=\frac{x}{2}$
$g^{-1}(x)=x-2$
Now, $\left(f^{-1} g^{-1}\right)(x)=f^{-1}\left(g^{-1}(x)\right)$
$\Rightarrow\left(f^{-1} og^{-1}\right)(x)=f^{-1}(x-2)$
$\Rightarrow\left(f^{-1} og^{-1}\right)(x)=\frac{x-2}{2} \ldots \ldots .$
$(g \circ f)(x)=g(f(x))=g(2 x)=2 x+2$
Let $(g \circ f)^{-1}(x)=y$......(4)
$x=(g \circ f)(y) \Rightarrow x=2 y+2 \Rightarrow 2 y=x-2$
$\Rightarrow y=\frac{x-2}{2}$
$\Rightarrow( gof )^{-1}( x )=\frac{ x -2}{2} \ldots \ldots$ (5) [from (4)] From (3) and (5),
$( g \circ f )^{-1}= f ^{-1} og ^{-1}$^
View full question & answer→Question 45 Marks
If $\text{f(x)}=\sqrt{1-\text{x}}$ and $\text{g(x)}=\log_\text{e}\text{x}$ are two real functions, then describe, functions fog and gof.
Answer$\text{f(x)}={1-\text{x}}$For domain,
$1-\text{x}\geq0$
$\text{x}\leq1$
⇒ domain of g
$\text{f}:(-\infty,1]\rightarrow0,\infty=\log_\text{e}\text{x}$
Clearly, $\text{g}:0,\infty\rightarrow\text{R}$
Computation of fog: Clearly, the range of g is not a subset of the domain of f.
Therefore,
We need to compute the domain of fog.
⇒ Domain fog = x : x $\in$ Domain g and $\text{g}(\text{x})\in$ Domain of f
⇒ Domain fog = x : x : x $\in0,\ \infty$ and $\log_\text{e}\text{x}\in(-\infty,1]$
⇒ Domain fog = x : x : x $\in0,\ \infty$ and $\text{x}\in(0,\text{e}]$
⇒ Domain fog = x : x $\in(0,\text{e}]$
⇒ Domain fog = (0, e]
⇒ fog : 0, e → R
Therefore,
(fog)(x) = f(g(x))
$=\text{f}(\log_\text{e}\text{x})$
$=\sqrt{1-\log_\text{e}\text{x}}$
Computation of gof: Clearly, the range of f is a subset of the domain of g.
$\Rightarrow\ \text{gof}:(-\infty,1]\rightarrow\text{R}$
$(\text{gof})(\text{x})=\text{g(f(x))}$
$=\text{g}(\sqrt{1-\text{x}})$
$\Rightarrow\ \log_\text{e}\sqrt{1-\text{x}}$
$=\log_\text{e}(1-\text{x})^\frac{1}{2}$
$=\frac{1}{2}\log_\text{e}(1-\text{x})$
View full question & answer→Question 55 Marks
Let $f : N \rightarrow N$ be a function as$ f(x) = 9x^2 + 6x - 5$. Show that $f : N \rightarrow S$, where S is the range of f, is invertible. Find the inverse of f and hence find $f^{-1}(43$) and $f^{-1}(163).$
AnswerWe have, $f : N \rightarrow N$ is a function defined as $f(x) = 9x^2 + 6x - 5.$
Let $y = f(x) = 9x^2 + 6x - 5$
$\Rightarrow y = 9x^2 + 6x - 5$
$\Rightarrow y = 9x^2 + 6x + 1 - 1 - 5$
$\Rightarrow y = (9x^2 + 6x + 1) - 6$
$\Rightarrow y = (3x + 1)^2 - 6$
$\Rightarrow y + 6 = (3x + 1)^2$
$\Rightarrow\ \sqrt{\text{y}+6}=3\text{x}+1\ \ (\because\ \text{y}\in\text{N})$
$\Rightarrow\ \sqrt{\text{y}+6}-1=3\text{x}$
$\Rightarrow\ \text{x}=\frac{\sqrt{\text{y}+6}-1}{3}$
$\Rightarrow\ \text{g(y)}=\frac{\sqrt{\text{y}+6}-1}{3}$ [Let x = g(y)]
Now, fog(y) = f[g(y)] $=\text{f}\bigg(\frac{\sqrt{\text{y}+6}-1}{3}\bigg)$
$=9\bigg(\frac{\sqrt{\text{y}+6}-1}{3}\bigg)^2+6\bigg(\frac{\sqrt{\text{y}+6}-1}{3}\bigg)-5$
$=9\bigg(\frac{\text{y}+6-2\sqrt{\text{y}+6}+1}{9}\bigg)+2\Big(\sqrt{\text{y}+6}-1\Big)-5$
$=\text{y}+6-2\sqrt{\text{y}+6}+1+2\sqrt{\text{y}+6}-2-5$
$=\text{y}$
$=\text{I}_\text{y}$ (Identity function)$\text{gof}(\text{x})=\text{g[f(x)]}$
$=\text{g}(9\text{x}^2+6\text{x}-5)$
$=\frac{\sqrt{(9\text{x}^2+6\text{x}-5)+6}-1}{3}$
$=\frac{(3\text{x}+1)-1}{3}$
$=\frac{3\text{x}}{3}$
$=\text{x}$
$=\text{I}_\text{X}$ (Identity function)
Since, fog(y) and gof(x) are identity function. Thus, f is invertible.
So, $\text{f}^{-1}(\text{x})=\text{g(x)}=\frac{\sqrt{\text{x}+6}-1}{3}$
Now, $\text{f}^{-1}(43)=\frac{\sqrt{43+6}-1}{3}=\frac{\sqrt{49}-1}{3}$
$=\frac{7-1}{3}=\frac{6}{3}=2$ And, $\text{f}^{-1}(163)=\frac{\sqrt{163+6}-1}{3}=\frac{\sqrt{169}-1}{3}$
$=\frac{13-1}{3}=\frac{12}{3}=4$
View full question & answer→Question 65 Marks
Let A = {a, b, c}, B = {u, v, w} and let f and g be two functions from A to B and from B to A, respectively, defined as:
f = {(a, v), (b, u), (c, w)}, g = {(u, b), (v, a), (w, c)}.
Show that f and g both are bijections and find fog and gof.
AnswerProving f is a bijection:f = {(a, v), (b, u), (c, w)} and f : A → B
Injectivity of f: No two elements of A have the same image in B.
So, f is one-one.
Surjectivity of f: Co-domain of f = {u v, w}
Range of f = {u v, w}
Both are same.
So, f is onto.
Hence, f is a bijection.
Proving g is a bijection:
g = {(u, b), (v, a), (w, c)} and g : B → A
Injectivity of g: No two elements of B have the same image in A.
So, g is one-one.
Surjectivity of g: Co-domain of g = {a, b, c}
Range of g = {a, b, c}
Both are the same.
So, g is onto.
Hence, g is a bijection.
Finding fog: Co-domain of g is same as the domain of f.
So, fog exists and fog: {u v, w} → {u v, w}
(fog)(u) = f(g(u)) = f(b) = u
(fog)(v) = f(g(v)) = f(a) = v
(fog)(w) = f(g(w)) = f(c) = w
So, fog = {(u, u), (v, v), (w, w)}
Finding gof.
Co-domain of f is same as the domain of g.
So, fog exists and gof: {a, b, c} → {a, b, c}
(gof)(a) = g(f(a)) = g(v) = a
(gof)(b) = g(f(b)) = g(u) = b
(gof)(c) = g(f(c)) = g(w) = c
So, gof = {(a, a), (b, b), (c, c)}
View full question & answer→Question 75 Marks
Classify the following functions as injection, surjection or bijection: $f : R \rightarrow R,$ defined by $\text{f(x)}=\frac{\text{x}}{\text{x}^2+1}$
Answer$f : R \rightarrow R,$ defined by $\text{f(x)}=\frac{\text{x}}{\text{x}^2+1}$
Injection test: Let x and y be any two elements in the domain $(R),$
such that $f(x) = f(y).$
$f(x) = f(y)$
$\frac{\text{x}}{\text{x}^2+1}=\frac{\text{y}}{\text{y}^2+1}$
$xy^2 + x = x^2y + y$
$xy^2 - x^2y + x - y = 0$
$-xy(-y + x) + 1(x - y) = 0$
$(x - y)(1 - xy) = 0$
$x = y$ or $\text{x}=\frac{1}{\text{y}}$
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$).$
$f(x) = y$
$\frac{\text{x}}{\text{x}^2+1}=\text{y}$
$yx^2 - x + y = 0$
$\text{x}=\frac{-(-1)\pm\sqrt{1-4\text{y}^2}}{2\text{y}},$ if $\text{y}\neq0$
$=\frac{1\pm\sqrt{1-4\text{y}^2}}{2\text{y}},$ which may not be in $R$
For example, if $y = 1$, then
$\text{x}=\frac{1\pm\sqrt{1-4}}{2}=\frac{1\pm\text{i}\sqrt{3}}{2},$ which is not in $R$
So, $f$ is not surjection and $f$ is not bijection.
View full question & answer→Question 85 Marks
Show that the exponential function $f : R → R$, given by $f(x) = e^x$, is one-one but not onto. What happens if the co-domain is replaced by R + 0R0 + (set of all positive real numbers)?
AnswerThen the co-domain and range become the same and in that case,
f is onto and hence, it is a bijection.
View full question & answer→Question 95 Marks
If $f : R \rightarrow (-1, 1)$ defined by $\text{f(x)}=\frac{10^{\text{x}}-10^{-\text{x}}}{10^{\text{x}}+10^{-\text{x}}}$ is invertible, find $f^{-1}$.
AnswerInjectivity of f: Let x and y be two elements of domain (R), such that
$f(x) = f(y)$$\Rightarrow\ \frac{10^{\text{x}}-10^{-\text{x}}}{10^{\text{x}}-10^{-\text{x}}}=\frac{10^{\text{y}}-10^{-\text{y}}}{10^{\text{y}}-10^{-\text{y}}}$
$\Rightarrow\ \frac{10^{-\text{x}}(10^{2\text{x}}-1)}{10^{-\text{x}}(10^{2\text{x}}+1)}=\frac{10^{-\text{y}}(10^{2\text{y}}-1)}{10^{-\text{y}}(10^{2\text{y}}+1)}$
$\Rightarrow\ \frac{(10^{2\text{x}}-1)}{(10^{2\text{x}}+1)}=\frac{(10^{2\text{y}}-1)}{(10^{2\text{y}}+1)}$
$\Rightarrow (10^{2x} - 1)(10^{2y} + 1) = (10^{2x} + 1)(10^{2y} - 1) $
$\Rightarrow 10^{2x+2y} + 10^{2x} - 10^{2y} - 1 = 10^{2x+2y} - 10^{2x} + 10^{2y} - 1$
$ \Rightarrow 2 \times 10^{2x} = 2 \times 10^{2y} $
$\Rightarrow 10^{2x} = 10^{2y} $
$\Rightarrow 2x = 2y$
$ \Rightarrow x = y$So, f is one-one.
Surjectivity of f: Let y is in the co domain (R), such that f(x) = y
$\Rightarrow\ \frac{10^{\text{x}}-10^{-\text{x}}}{10^{\text{x}}+10^{-\text{x}}}=\text{y}$
$\Rightarrow\ \frac{10^{-\text{x}}(10^{2\text{x}}-1)}{10^{-\text{x}}(10^{2\text{x}}+1)}=\text{y}$
$\Rightarrow\ 10^{2\text{x}}-1=\text{y}\times10^{2\text{x}}+\text{y}$
$\Rightarrow\ 10^{2\text{x}}(1-\text{y})=1+\text{y}$ $\Rightarrow\ 10^{2\text{x}}=\frac{1+\text{y}}{1-\text{y}}$
$\Rightarrow\ 2\text{x}=\log\Big(\frac{1+\text{y}}{1-\text{y}}\Big)$
$\Rightarrow\ \text{x}=\frac{1}{2}\log\Big(\frac{1+\text{y}}{1-\text{y}}\Big)\in\text{R}$ (domain) $\Rightarrow f$ is onto.
So, f is a bijection and hence, it is invertible.
Finding $f^{-1}$: Let $f^{-1}(x) = y .......(1)$
$\Rightarrow f(y) = x$
$\Rightarrow\ \frac{10^{\text{y}}-10^{-\text{y}}}{10^{\text{y}}+10^{-\text{y}}}=\text{x}$
$\Rightarrow\ \frac{10^{-\text{y}}(10^{2\text{y}}-1)}{10^{-\text{y}}(10^{2\text{y}}+1)}=\text{x}$
$\Rightarrow\ 10^{2\text{y}}-1=\text{x}\times10^{2\text{y}}+\text{x}$
$\Rightarrow\ 10^{2\text{y}}(1-\text{x})=1+\text{x}$
$\Rightarrow\ 10^{2\text{y}}=\frac{1+\text{x}}{1-\text{x}}$
$\Rightarrow\ 2\text{y}=\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$
$\Rightarrow\ \text{y}=\frac{1}{2}\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$
So, $\text{f}^{-1}(\text{x})=\frac{1}{2}\log\Big(\frac{1+\text{x}}{1-\text{x}}\Big)$ [from (1)]
View full question & answer→Question 105 Marks
Consider $f : \{1, 2, 3\} \rightarrow \{a, b, c\}$ and $g : \{a, b, c\} \rightarrow \{$apple, ball, cat$\}$ defined as $f(1) = a, f(2) = b, f(3) = c, g(a) =$ apple, $g(b) =$ ball and $g(c) =$ cat. Show that $f, g $ and gof are invertible. Find $f^{-1}, g^{-1} $ and $gof^{-1} $ and show that $(gof)^{-1} = f^{-1}og^{-1}.$
Answer$f = \{(1, a), (2, b), (3, c)\}$ and $g = \{(a, apple), (b, ball), (c, cat)\}$
Clearly, $f$ and $g$ are bijections.
So, f and g are invertible.
Now,
$f^{-1} = \{(a, 1), (b, 2), (c, 3)\}$ and $g^{-1} = \{(apple, a), (ball, b), (cat, c)\}$
So, $f^{-1}og^{-1} = \{(apple, 1), (ball, 2), (cat, 3)\} .....(1)$
$f : \{1, 2, 3\} \rightarrow \{a, b, c\}$ and $g : \{a, b, c\} \rightarrow \{$apple, ball, cat$\}$
So, $gof :\{1, 2, 3\} \rightarrow \{$apple, ball, cat$\}$
$\Rightarrow (gof)(1) = g(f(1)) = g(a) =$ apple
$(gof)(2) = g(f(2)) = g(b) =$ ball
and $(gof)(3) = g(f(3)) = g(c) = cat$
$\therefore gof = \{(1, apple), (2, ball), (3, cat)\}$
Clearly, $gof$ is a bijection.
So, gof is invertible.
$(gof)^{-1} = \{(apple, 1), (ball, 2), (cat, 3)\} ......(2)$
From $(1)$ and $(2)$, we get
$(gof)^{-1} = f^{-1}og^{-1}$
View full question & answer→Question 115 Marks
Consider the function $\text{f}:\text{R}^{+}\rightarrow[-9,\infty]$ given by $f(x) = 5x2 + 6x - 9.$ Prove that $f$ is invertible with $\text{f}^{-1}\text{(y)}=\frac{\sqrt{54+5\text{y}}-3}{5}.$
Answer$\text{f}:\text{R}^{+}\rightarrow\ [-9,\infty)$ given by $f(x) = 5x^2 + 6x - 9$ For any $\text{x, y}\in\text{R}^{+}$
$f(x) = f(y)$
$\Rightarrow 5x^2 + 6x - 9 = 5y^2 + 6y - 9$
$\Rightarrow 5(x^2 - y^2) + 6(x - y) = 0$
$\Rightarrow (x - y)[5(x + y) + 6] = 0$
$\Rightarrow (x - y) = 0 [\because5(\text{x}+\text{y})+6\neq0\text{ as x, y}\in\text{R}^{+}]$
$\Rightarrow x = y$
So, $f$ is an injection.
Let y be an arbitrary element of $[-9,\infty).$
$f(x) = y$
$\Rightarrow 5x^2 + 6x - 9 = y$
$\Rightarrow 25x^2 + 30x - 45 = 5y$
$\Rightarrow 25x^2 + 30x + 9 - 54 = 5y$
$\Rightarrow (5x + 3)^2 = 5y + 54$
$\Rightarrow(5\text{x}+3)=\sqrt{5\text{y}+54}$
$\Rightarrow\ \text{x}=\frac{\sqrt{5\text{y}+54}-3}{5}$
Now, $\text{y}\in[-9,\infty)$
$\Rightarrow\ \text{y}\geq-9$
$\Rightarrow\ 5\text{y}+54\geq9$
$\Rightarrow\ \sqrt{5\text{y}+54}\geq3$
$\Rightarrow\ \sqrt{5\text{y}+54}-3\geq0$
$\Rightarrow\ \frac{\sqrt{5\text{y}+54}-3}{5}\geq0$
$\Rightarrow\ \text{x}\geq0\Rightarrow\ \text{x}\in\text{R}^{+}$
Thus, for every $\text{y}\in[-9,\infty)$ there exist $\text{x}=\frac{\sqrt{5\text{y}+54}-3}{5}\in\text{R}^{+}$ such that $f(x) = y$.
So, $\text{f}:\text{R}^{+}\rightarrow[-9,\infty)$ is onto.
Thus, $\text{f}:\text{R}^{+}\rightarrow[-9,\infty)$ is a bijection and hence invertible.
Let $f^{-1} $ denotes the inverse of $f.$
Then,
$(fof^{-1})(y) = y$ for all $\text{y}\in[-9,\infty)$
$f(f^{-1}(y)) = y$ for all$ \text{y}\in[-9,\infty)$
$\Rightarrow 5(f^{-1}(y))^2 + 6(f^{-1}(y)) - 9 = y$ for all $\text{y}\in[-9,\infty)$
$\Rightarrow 25(f^{-1}(y))^2 + 30(f^{-1}(y)) - 45 = 5y$ for all $\text{y}\in[-9,\infty)$
$\Rightarrow 25(f^{-1}(y))^2 + 30(f^{-1}(y)) + 9 = 5y + 54$ for all $\text{y}\in[-9,\infty)$
$\Rightarrow {5f^{-1}(y) + 3}^2 = 5y + 54 $ for all $\text{y}\in[-9,\infty)$
$\Rightarrow 5f^{-1}(y) + 3 =\sqrt{5\text{y}+54} $ for all $ \text{y}\in[-9,\infty)$
$\Rightarrow\ \text{f}^{-1}(\text{y})=\frac{\sqrt{5\text{y}+54}-3}{5}$
View full question & answer→Question 125 Marks
Give an example of a function:
Which is one-one but not onto.
AnswerWhich is one-one but not onto.
f : Z → Z given by f(x) = 3x + 2
Infectivity: Let x andy be any two elements in the domain (Z), such that f(x) = f(y).
f(x) = f(y)
Implies that 3x + 2 = 3y + 2
Implies that 3x = 3y
Implies that x = y
Implies that f(x) = f(y)
Implies that x = y
Therefore,
f is one-one.
Subjectivity: 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
Implies that 3x + 2 = y
Implies that 3x = y - 2
Implies that $\text{x}=\frac{\text{y}-2}{3}.$
It may not be in the domain (Z) because if we take y = 3, $\text{x}=\frac{\text{y}-2}{3}=\frac{3-2}{3}=\frac{1}{3}\notin$ domain Z.
Therefore, for every element in the domain there need not be any element in the domain such that f(x) = y.
Thus, f is not onto.
View full question & answer→Question 135 Marks
Classify the following functions as injection, surjection or bijection:
$f : R \rightarrow R,$ defined by $f(x) = x^3 - x$
Answer$f : R \rightarrow R,$ defined by $f(x) = x^3 - x$
Injective: Let $\text{x, y}\in\text{R}$
such that, $f(x) = f(y)$
$\Rightarrow x^3 - x = y^3 - y $
$\Rightarrow x^3 - y^3 - (x - y) = 0 $
$\Rightarrow (x - y)(x^2 + xy + y^2 - 1) = 0$
$\because\ \text{x}^2+\text{xy}+\text{y}^2\geq0$
$\Rightarrow\ \text{x}^2+\text{xy}+\text{y}^2-1\geq-1$
$\therefore\ \text{x}^2+\text{xy}+\text{y}^2-1\neq0$
$\Rightarrow \text{x}-\text{y}=0\Rightarrow \text{x}=\text{y}$
$\therefore$ f is one-one.
Surjective: Let $\text{y}\in\text{R},$
then$ f(x) = y $
$\Rightarrow x^3 - x - y = 0$
We know that a degree 3 equation has atleast one real solution.
Let $\text{x}=\alpha$ be that real solution
$\therefore\ \alpha^2-\alpha=\text{y}$
$\Rightarrow\ \text{f}(\alpha)=\text{y}$
$\therefore$ For each $\text{y}\in\text{R,}$
there exist $\text{x}=\alpha\in\text{R}$
such that $\text{f}(\alpha)=\text{y}$
$\therefore f$ is onto.
View full question & answer→Question 145 Marks
Let f be a real function given by $\text{f(x)}=\sqrt{\text{x}-2}.$ Find the following:
(fofof)(38)
Also, show that fof ≠ $f^2.$
AnswerWe have, $\text{f(x)}=\sqrt{\text{x}-2}$ Clearly, Domain $(\text{f})=[2,\infty)$ and Range $(\text{f})=[0,\infty).$ We observe that range (f) is not a subset of domain of f. $\therefore$ Domain of (fof) = {x : x $\in$ Domain (f) and f(x) $\in$ Domain (f)} = {x : x $\in[2,\infty)$ and $\sqrt{\text{x}-2}\in[2,\infty)$} = {x : x $\in[2,\infty)$ and $\sqrt{\text{x}-2}\geq2$} = {x : x $\in[2,\infty)$ and $\text{x}-2\geq4$} = {x : x $\in[2,\infty)$ and $\text{x}\geq6$} $=[6,\infty)$ Clearly, range of $\text{f}=[0,\infty)⊄$ domain of fof.$\therefore$ Domain of ((fof)of) = {x : x $\in$ domain (f) and f(x) $\in$ Domain (fof)}
= {x : x $\in[2,\infty)$ and $\sqrt{\text{x}-2}\in[6,\infty)$}
= {x : x $\in[2,\infty)$ and $\sqrt{\text{x}-2}\geq6$}
= {x : x $\in[2,\infty)$ and $\text{x}-2\geq36$}
= {x : x $\in[2,\infty)$ and $\text{x}\geq38$}
$=[38,\infty)$ Now, (fof)(x) = f(f(x)) $=\text{f}(\sqrt{\text{x}-2})=\sqrt{\sqrt{\text{x}-2}-2}$ (fofof)(x) = (fof)(f(x)) $=(\text{fof})(\sqrt{\text{x}-2})=\sqrt{\sqrt{\sqrt{\text{x}-2}-2}-2}$ $\therefore\ \text{fofof}:[38,\infty)\rightarrow\text{R}$ defined as $(\text{fofof})(\text{x})=\sqrt{\sqrt{\sqrt{\text{x}-2}-2}-2}$ $(\text{fofof})(38)=\sqrt{\sqrt{\sqrt{38-2}-2}-2}$ $=\sqrt{\sqrt{\sqrt{36}-2}-2}=\sqrt{\sqrt{6-2}-2}$ $=\sqrt{\sqrt{4}-2}=\sqrt{2-2}=0$
View full question & answer→Question 155 Marks
A function $f : R \rightarrow R$ is defined as $f(x) = x^3 + 4.$ Is it a bijection or not? In case it is a bijection, find $f^{-1}(3).$
AnswerWe have,
$f : R \rightarrow R$ in a function defined by
$f(x) = x^3 + 4$
Injectivity: Let f(x_1) = f(x_2) for $\text{x}_1,\text{x}_2\in\text{R}$
$\Rightarrow\ \text{x}_1^3+4=\text{x}_2^3+4$
$\Rightarrow\ \text{x}_1^3=\text{x}_2^3$
$\Rightarrow\ \text{x}_1=\text{x}_2$
$\Rightarrow f$ is one-one.
Surjectivity: Let $\text{y}\in\text{R}$ be artritrary such that
$f(x) = y$
$\Rightarrow x^3 + 4 = y$
$\Rightarrow x^3 + 4 - y = 0$
We know that an odd degree equation must have a real root.
$\Rightarrow\ \alpha^3+4=\text{y}$
$\Rightarrow\ \text{f}(\alpha)=\text{y}$
$\Rightarrow f$ is onto.
Since f is one-one and onto.
$\Rightarrow f$ is bijective.
finally, $f(x) = y$
$\Rightarrow x^3 + 4 = y$
$\Rightarrow x^3 = y - 4$
$\Rightarrow\ \text{x}=(\text{y}-4)^\frac{1}{3}$
$\therefore\ \text{f}^{-1}(\text{x})=(\text{x}-4)^\frac{1}{3}$
$\therefore\ \text{f}^{-1}(3)=(3-4)^\frac{1}{3}=-1$
View full question & answer→Question 165 Marks
Give an example of a function:
Which is neither one-one nor onto.
AnswerWhich is neither one-one nor onto.
$f : Z \rightarrow Z$ given by $f(x) = 2x^2 + 1$
Infectivity: Let x andy be any two elements in the domain $(Z),$ such that $f(x) = f(y).$
$f(x) = f(y)$
Implies that $2x^2 + 1 = 2y^2 + 1$
Implies that $2x^2 = 2y^2$
Implies that $x^2 = y^2$
Implies that $\text{x}=\pm\text{y}$
Therefore, different elements of domain/ may give the same image.
Thus, $f$ is not one-one.
Subjectivity: 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$
Implies that $2x^2 + 1 = y$
Implies that $2x^2 = y - 1$
Implies that $\text{x}^2=\frac{\text{y}-1}{2}$
Implies that $\text{x}=\pm\sqrt{\frac{\text{y}-1}{2}},\notin\text{Z}$ always.
For example, if we take, $y = 4,$
$\text{x}=\pm\sqrt{\frac{\text{y}-1}{2}}=\pm\sqrt{\frac{4-1}{2}}=\pm\sqrt{\frac{3}{2}},\notin\text{Z}$
Therefore, $x$ may not be in $Z ($domain$).$
Thus, $f$ is not onto.
View full question & answer→Question 175 Marks
Let f = {(1, -1), (4, -2), (9, -3), (16, 4)} and g = {(-1, -2), (-2, -4), (-3, -6), (4, 8)}. Show that gof is defined while fog is not defined. Also, find gof.
AnswerWe have, f = {(1, -1), (4, -2), (9, -3), (16, 4)} and
g = {(-1, -2), (-2, -4), (-3, -6), (4, 8)}
Now,
Domain of f = {1, 4, 9, 16}
Range of f = {-1, -2, -3, 4}
Domain of g = {-1, -2, -3, 4}
Range of g = {-2, -4, -6, 8}
Clearly range of f = domain of g
$\therefore$ gof is defined.
but, range of g $\neq$ domain of f
$\therefore$ fog in not defined.
Now,
gof(1) = g(-1) = -2
gof(4) = g(-2) = -4
gof(9) = g(-3) = -6
gof(16) = g(4) = 8
$\therefore$ gof = {(1, -2), (4, -4), (9, -6), (16, 8)}
View full question & answer→Question 185 Marks
Let f = {(3, 1), (9, 3), (12, 4)} and g = {(1, 3), (3, 3) (4, 9) (5, 9)}. Show that gof and fog are both defined. Also, find fog and gof.
Answerf = {(3, 1), (9, 3), (12, 4)} and g = {(1, 3), (3, 3) (4, 9) (5, 9)} f : {3, 9, 12} → {1, 3, 4} and g : {1, 3, 4, 5} → {3, 9}Co-domain of f is a subset of the domain of g.
Therefore, exists and (gof) : {3, 9, 12} → {3, 9}
(gof)(3) = g(f(3)) = g(1) = 3 (gof)(9) = g(f(9)) = g(3) = 3 (gof)(12) = g(f(12)) = g(4) = 9 (gof) = {(3, 3), (9, 3), (12, 9)} Co-domain of g is a subset of the domain of f. Therefore, (fog) exists and : {1, 3, 4, 5} → {3, 9, 12} (fog)(1) = f(g(1)) = f(3) = 1 (fog)(3) = f(g(3)) = f(3) = 1 (fog)(4) = f(g(4)) = f(9) = 3 (fog)(5) = f(g(5)) = f(9) = 3 fog = {(1, 1), (3, 1), (4, 3), (5, 3)}
View full question & answer→Question 195 Marks
Give examples of two surjective functions $f_1$ and $f_2$ from $Z$ to $Z$ such that $f_1 + f_2$ is not surjective.
AnswerWe know that $f_1 : R \rightarrow R,$ given by $f_1(x) = x,$ and $f_2(x) = -x $ are surjective functions.
Proving $f_1$ is surjective: Let $y$ be an element in the co-domain $(R),$ such that $f_1(x) = y.$
$f_1(x) = y$
Implies that $x = y$, which is in $R.$
Therefore, for every element in the co-domain, there exists some pre-image in the domain.
Therefore, $f_1$ is surjective.
Proving $f_2$ is surjective: Let $f_2(x) = y$
$x = y,$ which is in $R$.
Therefore, for every element in the co-domain, there exists some pre-image in the domain.
Therefore, $f_2$ is surjective.
Proving $(f_1 + f_2)$ is not surjective:
Given: $(f_1 + f_2)(x) = f_1(x) + f_2(x) = x + (-x) = 0$
Therefore, for every real number $x, (f_1 + f_2)(x) = 0$
Therefore, the image of every number in the domain is same as $0.$
Implies that Range $= \{0\}$
Co-domain $= R$
Therefore, both are not same.
Therefore, $f_1 + f_2$ is not surjective.
View full question & answer→Question 205 Marks
Let f : N → N be defined by $\text{f(n)}=\begin{cases}\text{n}+1,&\text{if n is odd}\\\text{n}-1,&\text{if n is even}\end{cases}$ Show that f is a bijection.
AnswerWe have, $\text{f(n)}=\begin{cases}\text{n}+1,&\text{if n is odd}\\\text{n}-1,&\text{if n is even}\end{cases}$Injection test:
Case I: If n is odd,
Let $\text{x, y}\in\text{N}$ such that f(x) = f(y) As, f(x) = f(y) ⇒ x + 1 = y + 1 ⇒ x = yCase II: If n is even,
Let $\text{x, y}\in\text{N}$ such that f(x) = f(y) As, f(x) = f(y) ⇒ x - 1 = y - 1 ⇒ x = y So, f is injective. Surjection test: Case I: If n is odd, As, for every $\text{n}\in\text{N},$ there exists y = n - 1 in N such that f(y) = f(n - 1) = n - 1 + 1 = nCase II: If n is even,
As, for every $\text{n}\in\text{N},$ there exists y = n + 1 in N such that f(y) = f(n + 1) = n + 1 - 1 = n So, f is surjective. So, f is bijection.
View full question & answer→Question 215 Marks
Show that the function $f : Q \rightarrow Q$, defined by $f(x) = 3x + 5$, is invertible. Also, find $f^{-1}$.
AnswerGiven that $f : Q \rightarrow Q$ defined by $f(x) = 3x + 5$.
To prove that $f$ is invertible, we need to prove that $f$ is one-one and onto.
Let $\text{x, y}\in\text{Q}$ be such that, $f(x) = f(y)$
$\Rightarrow 3x + 5 = 3y + 5 $
$\Rightarrow x = y$
So, $f$ is an injection. Let $y$ be an arbitrary element of $Q$ such that $f(x) = y$.
$\Rightarrow 3x + 5 = y $
$\Rightarrow 3x = y - 5$
$\Rightarrow\ \text{x}=\frac{\text{y}-5}{3}$
Thus, for any $\text{y}\in\text{Q}$ there exists $\text{x}=\frac{\text{y}-5}{3}\in\text{Q}$
such that $\text{f(x)}=\text{f}\Big(\frac{\text{y}-5}{3}\Big)=3\frac{\text{y}-5}{3}+5=\text{y}$
Thus, $f : Q \rightarrow Q$ is a bijection and hence invertible.
Let $f^{-1}$ denotes the inverse of $f$.
Thus, $fof^{-1}(x) = x$ for all $\text{x}\in\text{Q}$
$\Rightarrow f[f^{-1}(x)] = x$ for all $\text{x}\in\text{Q}$
$\Rightarrow 3f^{-1}(x) + 5 = x$ for all $\text{x}\in\text{Q}$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{\text{x}-5}{3}$ for all $\text{x}\in\text{Q}$
View full question & answer→Question 225 Marks
Consider $f : R \rightarrow R$ given by $f(x) = 4x + 3$. Show that f is invertible. Find the inverse of f.
AnswerInjectivity of f: Let x and y be two elements of domain (R), such that
$f(x) = f(y) \Rightarrow 4x + 3 = 4y + 3 $
$\Rightarrow 4x = 4y $
$\Rightarrow x = y$ So, f is one-one.
Surjectivity of f: Let y be in the co-domain (R), such that
$f(x) = y. \Rightarrow 4x + 3 = y \Rightarrow 4x = y - 3$
$\Rightarrow\ \text{x}=\frac{\text{y}-3}{4}\in\text{R}$ (Domain) $\Rightarrow f$ is onto.
So, $f$ is a bijection and hence is invertible.
Finding $f^{-1}$: Let $f^{-1}(x) = y ....(1) $
$\Rightarrow x = f(y) $
$\Rightarrow x = 4y + 3$
$ \Rightarrow x - 3 = 4y$
$\Rightarrow\ \text{y}=\frac{\text{x}-3}{4}$
So, $\text{f}^{-1}(\text{x})=\frac{\text{x}-3}{4}$ [from (1)]
View full question & answer→Question 235 Marks
Give an example of a function:
Which is not one-one but onto.
AnswerWhich is not one-one but onto.
$\text{f}:\text{Z}\rightarrow\text{N}\cup\{0\}$ given by f(x) = |x|
Infectivity: Let x and y be any two elements in the domain (Z), such that f(x) = f(y).
Implies that |x| = |y|
Implies that $\text{x}=\pm\text{y}$
Therefore, different elements of domain f may give the same image.
Therefore, f is not one-one.
Subjectivity: 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
Implies that |x| = y
Implies that $\text{x}=\pm\text{y},$
Which is an element in Z (domain).
Therefore, for every element in the co-domain, there exists a pre-image in the domain.
Thus, f is onto.
View full question & answer→Question 245 Marks
Let $\text{f}:\text{R}-\Big\{-\frac{4}{3}\Big\}\rightarrow\text{R}$ be a function defined as $\text{f(x)}=\frac{4\text{x}}{3\text{x}+4}.$ Show that $\text{f}:\text{R}-\Big\{-\frac{4}{3}\Big\}\rightarrow\text{Range (f)}$ is one-one and onto. Hence find $f^{-1}$.
AnswerWe have given that
$f : R → (0, 2)$ defined by
$\text{f(x)}=\frac{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+1$ is invertible.
let $f(x) = y$
$\Rightarrow\ \frac{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+1=\text{y}$
$\Rightarrow\ \frac{2\text{e}^{\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}=\text{y}$
$\Rightarrow\ \frac{2\text{e}^{2\text{x}}}{\text{e}^{2\text{x}}+1}=\text{y}$
$\Rightarrow\ 2\text{e}^{2\text{x}}=\text{y}(\text{e}^{2\text{x}}+1)$
$\Rightarrow\ \text{e}^{2\text{x}}(2-\text{y})=\text{y}$
$\Rightarrow\ \text{e}^{2\text{x}}=\frac{\text{y}}{2-\text{y}}\Rightarrow\ \text{x}=\frac{1}{2}\log_\text{e}\Big(\frac{\text{y}}{2-\text{y}}\Big)$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{1}{2}\log_\text{e}\Big(\frac{\text{x}}{2-\text{x}}\Big)$
View full question & answer→Question 255 Marks
If $f : R \rightarrow R$ be defined by $f(x) = x^3 - 3$, then prove that $f^{-1}$ exists and find a formula for $f^{-1}$. Hence, find $f^{-1}(24)$ and $f^{-1}(5)$.
AnswerInjectivity of f: Let x and y be two elements in domain (R),
Such that, $x^3 - 3 = y^3 - 3 \Rightarrow x^3 = y^3 \Rightarrow x = y$
So, $f$ is one-one.
Surjectivity of f: Let y be in the co-domain (R) such that $f(x) = y$
$\Rightarrow x^3 - 3 = y $
$\Rightarrow x^3 = y + 3$
$\Rightarrow\ \text{x}=(\text{y}+3)^3\in\text{R}$
$\Rightarrow f$ is onto. So, f is a bijection and hence, it is invertible.
Finding $f^{-1}$:
Let $f^{-1}(x) = y ....(1) $
$\Rightarrow x = y^3 - 3 $
$\Rightarrow x + 3 = y^3 $
$\Rightarrow y = (x + 3)^3 $
$= f^{-1}(x)$ [from $1$]
So, $f^1(x) = (x + 3)^3$
Now, $f^1(24) = (24 + 3)^3 = 27^3 = 19683$ and, $f^{-1}(5) = (5 + 3)^3 = 8^3 = 512$
View full question & answer→Question 265 Marks
Find fog and gof if:$f(x) = \sin^{-1}x, g(x) = x^2$
Answer$f(x) = \sin^{-1}x, g(x) = x^2$
$\text{f}:[-1,1]\rightarrow\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big];\ \text{g}:\text{R}\rightarrow[0,\infty)$
Computing fog: Clearly, the range of g is not a subset of the domain of f.
Domain(fog) = {x: $\text{x}\in$ domain of g and $\text{g(x)}\in$ domain of f}
Domain(fog) = {x: $\text{x}\in\text{R}$ and $\text{x}^2\in[-1,1]$}
Domain(fog) = {x: $\text{x}\in\text{R}$ and $\text{x}\in[-1,1]$}
Domain of (fog) $= [-1, 1]$
fog :$ [-1, 1] \rightarrow R$
$(fog)(x) = f(g(x))$
$= f(x^2)$
$= \sin^{-1}(x^2)$
Computing gof: Clearly, the range of f is a subset of the domain of g.
fog :$ [-1, 1] \rightarrow R$
$(gof)(x) = g(f(x))$
$= g(\sin^{-1}x)$
$= (\sin^{-1}x)^2$
View full question & answer→Question 275 Marks
Show that if $f_1$ and $f_2$ are one-one maps from R to R, then the product $f_1 \times f_2 : R \rightarrow R$ defined by $(f_1 \times f_2)(x) = f_1(x)f_2(x)$ need not be one-one.
AnswerWe know that $f_1: R \rightarrow R$,
given by $f_1(x)=x$, and $f_2(x)=x$ are one-one.
Proving $f_1$ is one-one: Let $x$ and $y$ be two elements in the domain $R$, such that $f_1(x)=f_1(y)$
$\Rightarrow x=y$
So, $f _1$ is one-one.
Proving $f_2$ is one-one: Let $x$ and $y$ be two elements in the domain $R$, such that $f_2(x)=f_2(y)$ $\Rightarrow x=y$ So, $f_2$ is one-one. Proving $f_1 \times f_2$ is not one-one:
$\text { Given: }\left(f_1 \times f_2\right)(x)$
$=f_1(x) \times f_2(x)$
$=x \times x=x^2$
Let $x$ and $y$ be two elements in the domain $R$, such that $\left(f_1 \times f_2\right)(x)=\left(f_1 \times f_2\right)(y)$
$\Rightarrow x^2=y^2$
$\Rightarrow x= \pm y$
So, $\left(f_1 \times f_2\right)$ is not one-one.
View full question & answer→Question 285 Marks
Classify the following functions as injection, surjection or bijection:
f : Q - {3} → Q, defined by $\text{f(x)}=\frac{2\text{x}+3}{\text{x}-3}$
Answerf : Q - {3} → Q, defined by $\text{f(x)}=\frac{2\text{x}+3}{\text{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{2\text{x}+3}{\text{x}-3}=\frac{2\text{y}+3}{\text{y}-3}$
(2x + 3)(y - 3) = (2y + 3)(x - 3)
2xy - 6x + 3y - 9 = 2xy -6y + 3x - 9
9x = 9y
x = y
Therefore, 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{2\text{x}+3}{\text{x}-3}=\text{y}$
2x + 3 = xy - 3y
2x - xy = -3y - 3
x(2 - y) = -3(y + 1)
$\text{x}=\frac{3\text{y}+1}{\text{y}-2},$ which is not defined at y = 2.
Therefore, f is not a surjection and f is not a bijection.
View full question & answer→Question 295 Marks
Classify the following functions as injection, surjection or bijection:f: $Q \rightarrow Q$, defined by $f(x)=x^3+1$
Answer$f: Q \rightarrow Q$, defined by $f(x)=x^3+1$ Injective: Let $x, y \in Q$ such that
$f(x) = f(y)$
$\Rightarrow x^3 + 1 = y^3 + 1$
$\Rightarrow (x^3 - y^3) = 0$
$\Rightarrow (x - y)(x^2 + xy + y^2) = 0$
but $\text{x}^2+\text{xy}+\text{y}^2\geq0$
$\therefore$ $x - y = 0$
$\Rightarrow x = y$
$\therefore$ f is injective.
Surjective: Let $\text{y}\in\text{Q}$ be arbitrary, then
$f(x) = y$
$\Rightarrow x^3 + 1 - y = 0$
We know that a degree 3 equation has alteast one real solution.
Let $\text{x}=\alpha$ be that solution
$\therefore\ \alpha^3+1=\text{y}$
$\therefore\ \text{f}(\alpha)=\text{y}$
$\therefore$ f is onto.
View full question & answer→Question 305 Marks
If $\text{f(x)}=\frac{4\text{x}+3}{6\text{x}-4},\ \text{x}\neq\frac{2}{3},$ show that fof(x) = x for all $\text{x}\neq\frac{2}{3}.$ What is the inverse of f?
AnswerIt is given that, $\text{f(x)}=\frac{4\text{x}+3}{6\text{x}-4},\ \text{x}\neq\frac{2}{3}$
(fof)(x) = f(f(x))
$=\text{f}\Big(\frac{4\text{x}+3}{6\text{x}-4}\Big)=\frac{4\Big(\frac{4\text{x}+3}{6\text{x}-4}\Big)+3}{6\Big(\frac{4\text{x}+3}{6\text{x}-4}\Big)-4}$
$=\frac{16\text{x}+12+18\text{x}-12}{24\text{x}+18-24\text{x}+16}=\frac{34\text{x}}{34}=\text{x}$
Therefore, fof(x) = x, for all $\text{x}\neq\frac{2}{3}.$
⇒ fof = 1
Hence, the given function f is invertible and the inverse of f is f itself.
View full question & answer→Question 315 Marks
Given A = {2, 3, 4}, B = {2, 5, 6, 7}. Construct an example of each of the following:
- An injective map from A to B.
- A mapping from A to B which is not injective.
- A mapping from A to B.
AnswerGiven A = {2, 3, 4}, B = {2, 5, 6, 7}
Let f : A → B, f : A → B be a mapping from A to B f = {(2, 5), (3, 6), (4, 7)}
f is an injective mapping.
Since for every element $\text{a}\in\text{A}$ there is an unique element $\text{b}\in\text{B}$
Let us define a mapping: A → B given by g = {(2, 2), (2, 5), (3, 6), (4, 7)}
g is not an injective mapping.
Since the element $2\in\text{A}$ is not uniquely mapped
Since (2, 2) and (2, 5) both belong to the mapping g, g is not injective.
Let us define a mapping h : A → B
h : A → B given by h = {(2, 2), (5, 3), (7, 4)}
h is a mapping from A to B.
B to A since the every ordered puts {2, 5, 7} $\in\text{B}$ to elements in {2, 3, 4} $\in\text{A}$
View full question & answer→Question 325 Marks
Classify the following functions as injection, surjection or bijection:
$f : Z \rightarrow Z$ given by $f(x) = x^3$
Answer$f : Z → Z$ given by $f(x) = x^3$ Injectivity: Let $\text{x, y}\in\text{Z}$ such that $x = y \Rightarrow x^3 = y^3 \Rightarrow f(x) = f(y) \Rightarrow f$ is one-one.Surjective: Since f attains only cubic values like $\pm1,\pm8,\pm27, ....$
So, no non-cubic value of Z (co-domain) have pre image in Z (domain). $\therefore$ f is not onto.
View full question & answer→Question 335 Marks
Classify the following functions as injection, surjection or bijection:$f : R \rightarrow R$, defined by$ f(x) = 5x^3 + 4$
Answerf : R → R, defined by $f(x) = 5x^3 + 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)$
$5x^3 + 4 = 5y^3 + 4$
$5x^3 = 5y^3$
$x^3 = y^3$
$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$
$5x^3 + 4 = y$
$5x^3 = y - 4$
$\text{x}^3=\frac{\text{y}-4}{5}$
$\text{x}=\sqrt[3]{\frac{\text{y}-4}{5}}\in\text{R}$
So, f is a surjection and f is a bijection.
View full question & answer→Question 345 Marks
Find gof and fog when $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by: $f(x)=2 x+3$ and $g(x)=x^2+5$
AnswerGiven: $f : R \rightarrow R$ and $g : R \rightarrow R$
Therefore, gof $: R \rightarrow R$ and fog$ : R \rightarrow R$
$f(x) = 2x + 3$ and $g(x) = x^2 + 5$
Now, $(gof)(x) = g(f(x))$
$= g(2x + 3)$
$= (2x + 3)^2 + 5$
$= 4x^2 + 9 + 12x + 5$
$= 4x^2 + 12x + 14$
$(fog)(x) = f(g(x))$
$= f(x^2 + 5)$
$= 2(x^2 + 5) + 3$
$= 2x^2 + 10 + 3$
$= 2x^2 + 13$
View full question & answer→Question 355 Marks
Suppose $f_1$ and $f_2$ are non-zero one-one functions from R to R. Is $\frac{\text{f}_1}{\text{f}_2}$ necessarily one-one? Justify your answer. Here, $\frac{\text{f}_1}{\text{f}_2}:\text{R}\rightarrow\ \text{R}$ is given by $\Big(\frac{\text{f}_1}{\text{f}_2}\Big)(\text{x})=\frac{\text{f}_1(\text{x})}{\text{f}_2(\text{x})}$ for all $\text{x}\in\text{R}.$
AnswerWe know that $f_1 : R \rightarrow R$, given by $f_1(x) = x^3$ and $f_2(x) = x$ are one-one.
Injectivity of $f_1$: Let x and y be two elements in the domain R, such that
$f_1(x) = f_2(y)$
$\Rightarrow x^3 = y$
$\Rightarrow\ \text{x}=\sqrt[3]{\text{y}}\in\text{R}$
So, $f_1$ is one-one.
Injectivity of $f_2$: Let x and y be two elements in the domain R, such that
$f_2(x) = f_2(y)$
$\Rightarrow x = y$ $\text{x}\in\text{R.}$
So, $f_2$ is one-one.
Proving $\frac{\text{f}_1}{\text{f}_2}$ is not one-one.
Given that $\frac{\text{f}_1}{\text{f}_2}(\text{x})=\frac{\text{f}_1(\text{x})}{\text{f}_2(\text{x})}=\frac{\text{x}^3}{\text{x}}=\text{x}^2$
Let x and y be two elements in the domain R, such that
$\frac{\text{f}_1}{\text{f}_2}(\text{x})=\frac{\text{f}_1}{\text{f}_2}(\text{y})$
$\Rightarrow\ \text{x}^2=\text{y}^2$
$\Rightarrow\ \text{x}=\pm\text{y}$
So, $\frac{\text{f}_1}{\text{f}_2}$ is not one-one.
View full question & answer→Question 365 Marks
Classify the following functions as injection, surjection or bijection:
$f : Z \rightarrow Z,$ defined by $f(x) = x^2 + x$
Answer$f : Z \rightarrow Z$, given by $f(x) = x^2 + 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$
Here, we cannot say that $x = y.$
For example, $x = 2$ and $y = -3$
Then, $x^2 + x = 2^2 + 2 = 6$
$y^2 + y = (-3)^2 - 3 = 6$
So, we have two numbers 2 and -3 in the domain Z whose image is same as 6.
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$
Here, we cannot say $\text{x}\in\text{Z}$
For example, $y = -4$
$x^2 + x = -4$
$x^2 + x + 4 = 0$
$\text{x}=\frac{-1\pm\sqrt{-15}}{2}=\frac{-1\pm\text{i}\sqrt{15}}{2}$ which is not in Z.
So, f is not a surjection and f is not a bijection.
View full question & answer→Question 375 Marks
Classify the following functions as injection, surjection or bijection:f : R → R, defined by f(x) = 3 - 4x
Answerf : R → R, defined by f(x) = 3 - 4xInjection 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
-4x = -4y
x = y
Therefore, 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
4x = 3 - y
$\text{x}=\frac{3-\text{y}}{4}\in\text{R}$
Therefore, f is a surjection and f is a bijection.
View full question & answer→Question 385 Marks
Let $A=\{1,2,3,4\} ; B=\{3,5,7,9\} ; C=\{7,23,47,79\}$ and $f: A \rightarrow B, g: B \rightarrow C$ be defined as $f(x)=2 x+1$ and $g(x)=x^2-$ 2. Express $(g \circ f)^{-1}$ and $f^{-1} og ^{-1}$ as the sets of ordered pairs and verify that $(g \circ f)^{-1}=f^{-1} \circ g^{-1}$.
Answer$= 2x + 1$
$\Rightarrow f = 1, 21 + 1, 2, 22 + 1, 3, 23 + 1, 4, 24 + 1 = 1, 3, 2, 5, 3, 7, 4, 9$
$g(x) = x^2 - 2$
$\Rightarrow g = 3, 32 - 2, 5, 52 - 2, 7, 72 - 2, 9, 92 - 2$
$= 3, 7, 5, 23, 7, 47, 9, 79$
Clearly f and g are bijections and,
hence, $f^{-1} : B \rightarrow A$ and $g^{-1} : C \rightarrow B$ exist.
So,
$f^{-1}=3,1,5,2,7,3,9,4 \text { and } g^{-1}=7,3,23,5,47,7,79,9$
Now, $f ^{-1} og ^{-1}: C \rightarrow A _{ r } f ^{-1} og ^{-1}=7,1,23,2,47,3,79,4 \ldots$
Also, $f: A \rightarrow B$ and $g: B \rightarrow C$,
$\Rightarrow \text { gof : } A \rightarrow C, \text { gof }^{-1}: C \rightarrow A$
So, $f ^{-1} og ^{-1}$ and gof $f ^{-1}$ have same domains.
$=g f x=g(2 x+1)=2 x+12-2$
$=4 \times 2+4 x+1-2$
$=4 \times 2+4 x-1$
Then,
gof(1) = g(f(1)) $= 4 + 4 - 1 = 7,$
gof(2) = g(f(2)) $= 4 + 4 - 1 = 23,$
gof(3) = g(f(3)) $= 4 + 4 - 1 = 47$ and
gof(4) = g(f(4)) $= 4 + 4 - 1 = 79$
So, $= 1, 7, 2, 23, 3, 47, 4, 79$
$\Rightarrow gof^{-1} = 7, 1, 23, 2, 47, 3, 79, 4 ....(2)$
We get: $gof^{-1} = f^{-1}og^{-1}$
View full question & answer→Question 395 Marks
If $f : R \rightarrow R$ be the function defined by $f(x) = 4x^3 + 7$, show that f is a bijection.
AnswerInjectivity: Let x and y be any two elements in the domain (R), such that $f(x) = f(y)$
$\Rightarrow 4x^3 + 7 = 4y^3 + 7$
$\Rightarrow 4x^3 = 4y^3$
$\Rightarrow x^3 = y^3$
$\Rightarrow x = y$
So, f is one-one.
Surjectivity: 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$
$\Rightarrow 4x^3 + 7 = y$
$\Rightarrow 4x^3 = y - 7$
$\Rightarrow\ \text{x}^3=\frac{\text{y}-7}{4}$
$\Rightarrow\ \text{x}=\sqrt[3]{\frac{\text{y}-7}{4}}\in\text{R}$
So, for every element in the co-domain, there exists some pre-image in the domain.
$\Rightarrow f$ is onto. Since, f is both one-to-one and onto, it is a bijection.
View full question & answer→Question 405 Marks
Let $A=\{x \in R \mid-1 \leq x \leq 1\}$ and let $f: A \rightarrow A, g: A \rightarrow A$ be two functions defined by $f(x)=x^2$ and $g(x)=\sin \left(\frac{\pi x}{2}\right)$. Show that $g ^{-1}$ exists but $f ^{-1}$ does not exist. Also, find $g ^{-1}$.
Answerf is not one-one because $f^{-1} = (-1)^2 = 1$ and $f^1 = 1^2 = 1$
$\Rightarrow -1$ and 1 have the same image under f.
$\Rightarrow f$ is not a bijection. So,$ f^{-1}$ does not exist.
Injectivity of g: Let x and y be any two elements in the domain (A), such that $\Rightarrow\ \sin\pi\text{x}^2=\sin\pi\text{y}^2$$\Rightarrow\ \pi\text{x}^2=\pi\text{y}^2$
$\Rightarrow x = y$ So, g is one-one.
Surjectivity of g: Range of $\text{g}=\sin\pi^{-1}2,\ \sin\pi^{1}2=\sin-\pi^2,$
$\sin\pi^2=-1,1=\text{A}$ (co-domain of g)
$\Rightarrow $ g is onto. ⇒ g is a bijection. So, $g^{-1} $ exists. Also, let $g^{-1}x = y ......(1)$
$\Rightarrow gy = x$ $\Rightarrow\ \sin\pi\text{y}^2=\text{x}\Rightarrow\ \pi\text{y}^2=\sin^{-1}\text{x}$
$\Rightarrow\ \text{y}=2\pi\sin^{-1}\text{x}\Rightarrow\ \text{g}^{-1}\text{x}=2\pi\sin^{-1}\text{x}$ [from (1)]
View full question & answer→Question 415 Marks
Let f be a real function given by $\text{f(x)}=\sqrt{\text{x}-2}.$ Find the following:
fof
Also, show that fof ≠ $f^2.$
AnswerWe have, $\text{f(x)}=\sqrt{\text{x}-2}$
Clearly, Domain $(\text{f})=[2,\infty)$ and Range $(\text{f})=[0,\infty).$
We observe that range (f) is not a subset of domain of f.
$\therefore$ Domain of (fof) = {x : x $\in$ Domain (f) and f(x) $\in$ Domain (f)}
= {x : x $\in [2,\infty)$ and $\sqrt{\text{x}-2}\in[2,\infty)$}
= {x : x $\in [2,\infty)$ and $\sqrt{\text{x}-2}\in2$}
= {x : x $\in [2,\infty)$ and $\text{x}-2\geq4$}
= {x : x $\in [2,\infty)$ and $\text{x}\geq6$}
$=[6,\infty)$
Now,
(fof)(x) = f(f(x))
$=\text{f}(\sqrt{\text{x}-2})=\sqrt{\sqrt{\text{x}-2}-2}$
$\therefore\ \text{fof}:[6,\infty)\rightarrow\text{R}$ defined as
$(\text{fof})(\text{x})=\sqrt{\sqrt{\text{x}-2}-2}$
View full question & answer→Question 425 Marks
Prove that the function $f : N \rightarrow N$, defined by $f(x) = x^2 + x + 1$, is one-one but not onto.
Answer$f : N \rightarrow N,$ defined by$ f(x) = x^2 + x + 1$
Injectivity: Let x and y be any two elements in the domain (N), such that f(x) = f(y).
$\Rightarrow x^2 + x + 1 = y^2 + y + 1$
$\Rightarrow (x^2 - y^2) + (x - y) = 0$
$\Rightarrow (x + y)(x - y) + (x - y) = 0$
$\Rightarrow (x - y)(x + y + 1) = 0$
$\Rightarrow x - y = 0 [(x + y + 1)$
cannot be zero because x and y are natural numbers]
$\Rightarrow x = y $So, f is one-one.
Surjectivity: The minimum number in N is 1. When $x = 1, x^2 + x + 1 = 1 + 1 + 1 = 3$
$\Rightarrow\ \text{x}^2 + \text{x} + 1\geq3,$
for every x in N. ⇒ f(x) will not assume the values 1 and 2. So, f is not onto.
View full question & answer→Question 435 Marks
Consider $\text{f}:\text{R}\rightarrow\text{R}_+\rightarrow[4,\infty)$ given by $f(x) = x^2 + 4.$ Show that f is invertible with inverse of f given by $\text{f}^{-1}(\text{x})=\sqrt{\text{x}-4,}$ where $R^+$ is the set of all non-negative real numbers.
AnswerInjectivity of f: Let x and y be two elements of the domain (Q), such that f(x) = f(y)
$\Rightarrow x^2 + 4 = y^2 + 4$
$\Rightarrow x^2 = y^2$
$\Rightarrow x = y$ as co-domain as $R^+ $
So, f is one-one.
Surjectivity of f: Let y be in the co-domain (Q), such that $f(x) = y$
$\Rightarrow x^2 + 4 = y$
$\Rightarrow x^2 = y - 4$
$\Rightarrow\ \text{x}=\text{y}-4\in\text{R}$
$\Rightarrow f$ is onto. So, f is a bijection and, hence, it is invertible.
Finding $f^{-1}:$ Let $f^{-1}(x) = y .......(1)$
$\Rightarrow x = y^2 + 4$
$\Rightarrow x - 4 = y^2$
$\Rightarrow y = x - 4$
So, $f^{-1}(x) = x - 4 [from (1)]$
View full question & answer→Question 445 Marks
Classify the following functions as injection, surjection or bijection:
$f : N \rightarrow N $given by $f(x) = x^3$
Answer$f : N \rightarrow N$ given by$ f(x) = x^3$
Injection test: Let x and y be any two elements in the domain (N), such that f(x) = f(y).
$f(x) = f(y)$
$x^3 = y^3$
$x = y$
Therefore, 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).
$f(x) = y$
$x^3 = y$
$\text{x}=\sqrt[3]{\text{y}}$ which may not be in N.
For example, if y = 3,
$\text{x}=\sqrt[3]{3}$ is not in N.
Therefore, f is not a surjection and f is not a bijection.
View full question & answer→Question 455 Marks
Give examples of two functions f : N → Z and g : Z → Z, such that gof is injective but g is not injective.
AnswerDefine f : N → Z as f(x) = x and g : Z → Z as g(x) = |x|. We first show that g is not injective. It can be observed that:g(-1) = |-1| = 1
g(1) = |1| = 1
Therefore, g(-1) = g(1), but $-1\neq1.$
Therefore, g is not injective.
Now, gof : N → Z is defined as gof(x) = g(f(x)) = g(x) = |x|.
Let $\text{x, y}\in\text{N}$ such that gof(x) = gof(y). ⇒ |x| = |y| Since x and y $\in\text{N,}$ both are positive. $\therefore$ |x| = |y| ⇒ x = y Hence, gof is injective.
View full question & answer→Question 465 Marks
If $\text{f}:\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\rightarrow\text{R}$ and g $: [-1, 1] \rightarrow R$ be defined as f(x) = tanx and $\text{g(x)}=\sqrt{1-\text{x}^2}$ respectively, describe fog and gof.
Answer$\text{f}:\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\rightarrow\text{R}$ and $g : [-1, 1] \rightarrow R$ defined as f(x) = tanx and $\text{g(x)}=\sqrt{1-\text{x}^2}$Range of f: let $y = f(x)$
$\Rightarrow y = tanx$
$\Rightarrow x = \tan^{-1}y$
Since, $\text{x}\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big),\ \text{y}\in(-\infty,\infty)$
$\therefore$ Range of f $\subset$ domain of $g = [-1, 1]$
$\therefore$ gof exists.
By similar argument fog exists.
Now,
fog(x) = f(g(x))
$=\text{f}(\sqrt{1-\text{x}^2})$
$=\tan(\sqrt{1-\text{x}^2})$
Again,
gof(x) = g(f(x))
= g(tanx)
$=\sqrt{1-\tan^2\text{x}}$
View full question & answer→Question 475 Marks
Let f be a real function given by $\text{f(x)}=\sqrt{\text{x}-2}.$ Find the following:
fofof
Also, show that fof ≠ $f^2.$
Answerfofof = (fof)of
We have, $\text{f}:[2,\infty)\rightarrow(0,\infty)$ and $\text{fof}:[6,\infty)\rightarrow\text{R}$
⇒ Range of f is not a subset of the domain of fof.
Then, domain ((fof)of) = {x : x $\in$ doamin of fand f(x) $\in$ domain of fof}
⇒ Domain ((fof)of) = {x : x $\in[2,\infty)$ and $\sqrt{\text{x}-2}\in[6,\infty)$}
⇒ Domain ((fof)of) = {x : x $\in[2,\infty)$ and $\sqrt{\text{x}-2}\geq6$}
⇒ Domain ((fof)of) = {x : x $\in[2,\infty)$ and $\text{x}-2\geq36$}
⇒ Domain ((fof)of) = {x : x $\in[2,\infty)$ and $\text{x}\geq38$}
⇒ Domain ((fof)of) = {x : x $\geq38$}
⇒ Domain ((fof)of) = $[38,\infty)$
$\text{fof}:[38,\infty)\rightarrow\text{R}$
So, ((fof)of)(x) = (fof)(f(x))
$=(\text{fof})(\sqrt{\text{x}-2})$
$=\sqrt{\sqrt{\sqrt{\text{x}-2}-2}-2}$
View full question & answer→Question 485 Marks
Let $A = {-1, 0, 1}$ and $f = {(x, x^2): x \in A}$. Show that $f : A \rightarrow A$ is neither one-one nor onto.
Answer$A = {-1, 0, 1}$ and $f = {(x, x^2): x \in A}$ Given, $f(x) = x^2$
Injectivity: $f(1) = 1^2 = 1$ and $f(-1) = (-1)^2 = 1$
Implies that 1 and -1 have the same images. Therefore, f is not one-one.
Surjectivity: Co-domain of $f = {-1, 0, 1}f(1) = 1^2 = 1,$
$f(-1) = (-1)^2 = 1$ and
$f(0) = 0$
Implies that Range of $f = {0, 1}$
Therefore, both are not same.
Hence, f is not onto.
View full question & answer→Question 495 Marks
Verify associativity for the following three mappings : $f : N \rightarrow Z_0$_ (the set of non-zero integers), $g : Z_0 \rightarrow Q$ and $h : Q \rightarrow R$ given by$ f(x) = 2x$, $\text{g(x)}=\frac{1}{\text{x}}$ and $h(x) = e^x.$
AnswerWe have, $f : N \rightarrow Z _0, g: Z _0 $
$\rightarrow Q$ and $h : Q \rightarrow R$
Also, $f ( x )=2 x , g ( x )=\frac{1}{ x }$ and $h ( x )= e ^{ x }$
Now, $f : N \rightarrow Z _0$ and hog: $Z _0 \rightarrow R$
$\therefore($ hog)of: $N \rightarrow R$
also, gof: $N \rightarrow Q$ and $h : Q \rightarrow R$
$\therefore$ ho(gof) : $N \rightarrow R$
Thus, (hog)of and ho(gof) exist and are function from N to set R.
Finally, (hog)of(x) = (hog)(f(x)) = (hog)(2x)
$=\text{h}\Big(\frac{1}{2\text{x}}\Big)$
$=\text{e}^\frac{1}{2\text{x}}$
Now, ho(gof)(x) = ho(g(2x))
$=\text{h}\Big(\frac{1}{2\text{x}}\Big)$
$=\text{e}^\frac{1}{2\text{x}}$
Hence, associativity verified.
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