Question 14 Marks
Show that the binary operation $\ast \text{ on A = R - {-1}}$ defined as a $\text{a} \ast \text{b} = \text{a + b + ab}$ for all $\text{a, b}\in \text{A}$ is communicative and associative on A. Also find the identity element of $\ast$ in A and prove that every element of a is invertible.
AnswerCommutative: For any elements $\text{a, b}\in \text{A}$
$\text{a}\ast\text{b} = \text{a + b + ab = b + a + ba = b}\ast \text{a}.$ Hence $\ast$ is commutative
Associative: For any three elements $\text{a, b, c,}\in \text{A}$
$\text{a}\ast \text{(b}\ast\text{c}) = \text{a}\ast\text{(b + c + bc) = a + b + c + bc + ab + ac + abc}$
$\text{(a} \ast\text{b}) \ast\text{c} = \text{(a + b + ab)}\ast \text{c} = \text{a + b + ab + c + ac + bc + abc}$
$\therefore \text{a}{\ast} \text{(b}{\ast}\text{c}) = \text{(a}{\ast} \text{b}) \ast\text{c}, \text{Hence } \ast \text{ is Associative.}$
Identity element: let e $\in$ A be the identity element them $\text{a}\ast \text{e = e}\ast \text{a = a}$
$\Rightarrow\text{a + e + ae = e + a + ea = a}\Rightarrow\text{e(1 + a) = 0, as a} \neq -1 $
$\text{e = 0}$ is the identity element
Invertible: let $\text{a, b}\in \text{A}$ so that ‘b’ is inverse of a
$\therefore \text{a}\ast \text{b = b}\ast\text{a = e}$
$\Rightarrow\text{a + b + ab = b + a + ba = 0}$
$\text{As a}\neq -1, \text{b} = \frac{\text{-a}}{1 + \text{a}} \in \text{A}.$ Hence every element of A is invertible.
View full question & answer→Question 24 Marks
Determine whether the relation R defined on the set $\Re$ of all real numbers as R =$(\text{a,b) : a, b} \in \Re$ and $\text{a - b} + \sqrt{3} \in \text{S},$where S is the set of all irrational numbers, is reflexive, symmetric and transitive.
AnswerHere R ={ $\text{(a, b) : a, b} \in \Re$ and $\text{a - b + }\sqrt{3} \in \text{S}, $ where S is the set of all irrational numbers.}
$\text{(i)} \forall \text{a}\in \Re, \text{(a, a)} \in \text{R as a - a +} \sqrt{3}$ is irrational
$\therefore \text{R is reflexive}$
$\text{(ii)} \text{Let for a, b} \in \Re, \text{(a, b)} \in \text{R i. e. a - b +}\sqrt{3}$ is irrational
$\text{a - b +}\sqrt{3}$ is irrational $\Rightarrow \text{b - a +}\sqrt{3} \in \text{S} \therefore \text{(b, a)} \in \text{R}$
Hence R is symmetric
$\text{(iii)} \text{Let (a, b)} \in \text{R and (b, c)}\in \text{R for a, b, c}\in \Re$
$\therefore \text{a - b +} \sqrt{3} \in \text{S and b - c +} \sqrt{3} \in \text{S}$
$\text{adding to get a - c +} 2\sqrt{3} \in \text{S Hence (a, c)}\in \text{R}$
$\therefore\text{R is Transitive}$
View full question & answer→Question 34 Marks
$\text{Let A = Q} \times \text{Q}$ and let * be a binary operation on A defined by$\text{(a, b)} * \text{(c, d) = (ac, b + ad)} \text{ for (a, b), (c, d)} \in \text{A}.$ Determine, whether * is commutative and associative. Then, with respect to * on A.
- Find the identity element in A.
- Find the invertible elements of A.
Answer$\text{(a, b)}{*} \text{(c, d) = (ac, b + ad); (a, b), (c, d)} \in \text{A}$
$\text{(c, d)}{*} \text{(a, b) = (ca, d + bc)}$
Since $\text{b + ad} \neq \text{d + bc} \Rightarrow \text{ }{*}$is NOT comutative
for associativity, we have,
$\text{(a, b)}{*} \text{[(c, d)}{*} \text{(e, f)] = (a, b)}{*} \text{(ce, d + cf) = (ace, b + ad + acf)}$
$\Rightarrow \text{ }{*} \text{ is associative}$
Let (e, f) be the identity element in A
$\text{Then (a, b)}{*} \text{(e, f) = (a, b) = (e, f)}{*} \text{(a, b)}$
$\Rightarrow\text{(ae, b + af) = (a, b) = (ae, f + be)}$
$\Rightarrow \text{e = 1, f = 0} \Rightarrow \text{(1, 0) is the identity element}$
Let (c, d) be the inverse element for (a, b)
$\Rightarrow \text{(a, b)}{*} \text{(c, d) = (1, 0) = (c, d)}{*} \text{(a, b)}$
$\Rightarrow \text{(ac, b + ad) = (1, 0) = (ac, d + bc)}$
$\Rightarrow \text{ac} = 1 \Rightarrow \text{c} = \frac{1}{\text{a}} \text{ and b + ad} = 0 \Rightarrow \text{d} = - \frac{\text{b}}{\text{a}} \text{ and d + bc =0 } \Rightarrow \text{d = -bc = -b} \bigg(\frac{1}{\text{a}}\bigg)$
$\Rightarrow \bigg(\frac{1}{\text{a}}, - \frac{\text{b}}{\text{a}}\bigg), \text{a} \neq 0 \text{ is the inverse of (a, b)} \in \text{A}$
View full question & answer→Question 44 Marks
Consider $\text{f : R} - \left\{-\frac{4}{3}\right\} \rightarrow \text{R} - \left\{\frac{4}{3}\right\} \text{given by f(x)} = \frac{\text{4x + 3}}{\text{3x + 4}}.$ Show that f is bijective. Find the inverse of f and hence find f –1(0) and x such that f –1(x) = 2.
Answer$\text{Let } \text{x}_{1}, \text{x}_{2} \in \text{R} - \left\{-\frac{4}{3}\right\} \text{and } \text{f(x}_{1}) = \text{f(x}_{2})$
$\Rightarrow \frac{\text{4x}_{1} + 3}{\text{3x}_{1} + 4} = \frac{\text{4x}_{2} + 3}{\text{3x}_{2} + 4} \Rightarrow \text{(4x}_{1} + 3) \text{(3x}_{2} + 4) = \text{(3x}_{1} + 4) \text{(4x}_{2} + 3)$
$\Rightarrow \text{12x}_{1} \text{x}_{2} + \text{16x}_{1} + \text{9x}_{2} + 12 = 12_{1} \text{x}_{2} + 16\text{x}_{2} + 9\text{x}_{1} + 12$
$\Rightarrow 16(\text{x}_{1} - \text{x}_{2}) - 9 \text{(x}_{1} - \text{x}_{2}) = 0 \Rightarrow \text{x}_{1} - \text{x}_{2} = 0 \Rightarrow \text{x}_{1} = \text{x}_{2}$
Hence f is a 1–1 function
$\text{Let} \text{ y} = \frac{\text{4x + 3}}{\text{3x + 4}}, \text{for y} \in \text{R} - \left\{\frac{4}{3}\right\}$
$\text{3xy + 4y = 4x + 3} \Rightarrow \text{4x – 3xy = 4y – 3}$
$\Rightarrow \text{x} = \frac{\text{4y - 3}}{\text{4 - 3y}} \therefore \forall \text{ y} \in \text{R} - \left\{\frac{4}{3}\right\}, \text{x} \in \text{R} - \left\{-\frac{4}{3}\right\}$
Hence f is ONTO and so bijective
$\text{and } \text{f}^{-1} \text{(y)} = \frac{\text{4y - 3}}{\text{4 - 3y}} ; \text{y} \in \text{R} - \left\{\frac{4}{3}\right\}$
$\text{f}^{-1} (0) = - \frac{3}{4}$
$\text{and } \text{f}^{-1} \text{(x)} = 2 \Rightarrow \frac{\text{4x - 3}}{\text{4 - 3x}} = 2$
$\Rightarrow \text{4x - 3 = 8 - 6x}$
$\Rightarrow \text{10x} = 11 \Rightarrow \text{x} = \frac{11}{10}$
View full question & answer→Question 54 Marks
Show that the binary operation $\ast \text{ on A = R - {-1}}$ defined as a $\text{a} \ast \text{b} = \text{a + b + ab}$ for all $\text{a, b}\in \text{A}$ is communicative and associative on A. Also find the identity element of $\ast$ in A and prove that every element of a is invertible.
AnswerCommutative: For any elements $\text{a, b}\in \text{A}$
$\text{a}\ast\text{b} = \text{a + b + ab = b + a + ba = b}\ast \text{a}.$ Hence $\ast$ is commutative
Associative: For any three elements $\text{a, b, c,}\in \text{A}$
$\text{a}\ast \text{(b}\ast\text{c}) = \text{a}\ast\text{(b + c + bc) = a + b + c + bc + ab + ac + abc}$
$\text{(a} \ast\text{b}) \ast\text{c} = \text{(a + b + ab)}\ast \text{c} = \text{a + b + ab + c + ac + bc + abc}$
$\therefore \text{a}{\ast} \text{(b}{\ast}\text{c}) = \text{(a}{\ast} \text{b}) \ast\text{c}, \text{Hence } \ast \text{ is Associative.}$
Identity element: let e $\in$ A be the identity element them $\text{a}\ast \text{e = e}\ast \text{a = a}$
$\Rightarrow\text{a + e + ae = e + a + ea = a}\Rightarrow\text{e(1 + a) = 0, as a} \neq -1 $
$\text{e = 0}$ is the identity element
Invertible: let $\text{a, b}\in \text{A}$ so that ‘b’ is inverse of a
$\therefore \text{a}\ast \text{b = b}\ast\text{a = e}$
$\Rightarrow\text{a + b + ab = b + a + ba = 0}$
$\text{As a}\neq -1, \text{b} = \frac{\text{-a}}{1 + \text{a}} \in \text{A}.$ Hence every element of A is invertible.
View full question & answer→Question 64 Marks
Let $\text{A} = \Re\times\Re$ and$\ast$ be the binary operation on A defined by $\text(a, b) \ast \text{(c, d)} = \text{(a + c, b + d)}. $ Prove that $\ast$ is commutative and associative. Find the identity element for $\ast$ on A. Also write the inverse element of the element (3, – 5) in A.
Answer$\forall \text{a, b, c, d, e, f}\in \Re$
$\text{((a, b)}\ast \text{(c,d))}\ast(\text{e, f})= \text{a + c, b + d)}\ast\text{(e, f)}$
$= \text{(a + c + e, b + d + f)} \rightarrow\text{(3)}$
$\text{((a, b)}\ast \text{(c,d))}\ast(\text{e, f}) = \text{(a + b)}\ast\text{(c + e, d + f)}$
$ = \text{(a + c = e, b + d + f)} \rightarrow \text{(4)}$
$\therefore \ast \text { is Associative}$
Let (x, y) be on identity element in $\Re\times\Re$
$\Rightarrow \text{(a, b)}\ast\text{(x, y)} = \text{(a, b)} = \text{(x, y)}\ast \text{(a,b)}$
$\Rightarrow \text{a + x = a, b + y = b}$
$\text{x = 0, y = 0}$
$\therefore \text{(0, 0) is identity element} $
Let the inverse element of $\text{(3, -5) be } \text{(x}_{1},\text{y}_{1},) $
$\Rightarrow (3, - 5)\ast \text{(x}_{1},\text{y}_{1},) = (0, 0) = \text{(x}_{1},\text{y}_{1},) \ast(3, -5)$
$\text{3 + x}_{1} = 0,\text{-5 + y}_{1} = 0 $
$\text{x}_{1} = -3, \text{y}_{1} = 5 $
$\Rightarrow (-3, 5) \text{is an inverse of (3, -5)}$
View full question & answer→Question 74 Marks
Let $\text{A} = \text{R} - \left\{3\right\}, \text{B} = \text{R} - \left\{1\right\}. \text{Let f : A} \rightarrow \text{B}$ be defined by $\text{f(x)}\frac{\text{x - 2}}{\text{x - 3}}, \forall \text{ x} \in \text{A}.$ Show that f is bijective. Also, find
- $\text{x, if f}^{-1} \text{(x) = 4}$
- $\text{f}^{-1} (7)$
Answer$\text{x}_{1}, \text{x}_{2} \in \text{A and } \text{f(x}_{1}) = \text{f(x}_{2})$
$\Rightarrow \frac{\text{x}_{1} - 2}{\text{x}_{1} - 3} = \frac{\text{x}_{2} - 2}{\text{x}_{2} - 3} \Rightarrow \text{(x}_{1} - 2) \text{(x}_{2} - 3) = \text{(x}_{1} - 3) \text{(x}_{2} - 2)$
$\Rightarrow \text{x}_{1} \text{x}_{2} - \text{3x}_{1} - 2\text{x}_{2} + 6 = \text{x}_{1} \text{x}_{2} - \text{2x}_{1} - \text{3x}_{2} + 6$
$\Rightarrow \text{x}_{1} = \text{x}_{2}$
Hence f is a one-one function
Let $\text{y} = \frac{\text{x - 2}}{\text{x - 3}} \text{ for y } \in \text{R} - \left\{1\right\}$
$\Rightarrow \text{x} = \frac{\text{3y - 2}}{\text{y - 1}} ; \text{y} \neq 1$
$\therefore \forall \text{ y} \in \text{R} - \left\{1\right\}, \text{x} \in \text{R} - \left\{3\right\}$
i.e. Range of f = co-domain of f.
Hence f is onto and so bijective.
Also, $\text{f}^{-1} \text{(x)} = \frac{\text{3x - 2}}{\text{x - 1}} ; \text{x} \neq 1$
Now, $\text{f}^{-1} \text{(x)} = 4 \Rightarrow \frac{\text{3x - 2}}{\text{x - 1}} = 4 \Rightarrow \text{x = 2}$
and $ \text{f}^{-1}(7) = \frac{19}{6}$
View full question & answer→Question 84 Marks
If f, $\text{f, g : R}\rightarrow \text{R}$ be two functions defined as $\text{f}(x) = |x| + x \text{ and } \text{g} (x) = |x| - x, \forall \text{ }x \in \text{R}.$Then find fog and gof. Hence find fog(–3), fog(5) and gof (–2).
Answer$\text{Given f}(x) = |\text{x}| + \text{x}$
$\text{and g} (x) = |{\text{x}}| - x , \forall x \in \text{R}$
$\text{fog} = \text{f(g(x))} = |\text{x}| + \text{g(x)}$
$ = \text{||x| - x| + (|x| - x)}$
Therefore,
$\text{f(g(x))} = \begin{cases} 0 & \text{x} \geq 0\\ 4x & \text{x} < 0\\ \end{cases}$
$\text{fog} = \begin{cases} 4\text{x} & \text{x} < 0\\ 0 & \text{x} \geq 0\\ \end{cases}$
$\text{gof = g(f(x)) = |f(x)| - f(x)}$
$= ||\text{x| + x| - (|x| + x)}$
Therefore, $\text{g(f(x)) = gof = 0}$
| Now, fog(−3) =(4)(−3) = −12 | (since, fog = 4x for x < 0) |
| fog(5) = 0 | (since, fog = 0 for x ≥ 0) |
| gof(−2) = 0 | (since, gof = 0 for x < 0) |
View full question & answer→Question 94 Marks
Let $\text{A} = \text{R} \times \text{R}$ and let * be a binary operation on A defined by $\text{(a, b)} {*} \text{(c, d)} = \text{(ad + bc, bd)}$ for all $\text{(a, b), (c, d)} \in \text{R} \times \text{R}.$
- Show that * is commutative on A.
- Show that * is associative on A.
- Find the identity element of * in A.
Answer - $\text{(a, b)}{*} \text{(c, d) = (ad + bc, bd)}$
$\text{Now}, \text{(c, d)}{*} \text{(a, b) = (cb + da, db) = (ad + bc, bd) = (a, b)}{*} \text{(c, d)}$
$\Rightarrow \text{ }{*} \text{ is Communicative}$
- $\text{[(a, b) }{*} \text{(c, d)]}{*} \text{(e, f) = (ad + bc, bd)}{*} \text{(e, f) = (adf + bcf + bde, bdf)]}$
$\text{(a,b)}{*} \text{[(c, d)}{*} \text{(e, f)] = (a, b)} {*} \text{(cf + de, df) = (adf + bcf + bde, bdf)}$
$\Rightarrow \text{ } {*} \text{ is associative.}$
- Let (e1, e2) be the identity element of A.
$\Rightarrow \text{(a, b}) {*} \text{(e}_{1}, {\text{e}_{2}) = \text{(a, b)} = \text{(e}_{1}, \text{e}_{2}) {*} \text{(a, b)}}$
$\Rightarrow \text{(ae}_{2} + \text{be}_{1} , \text{be}_{2}) = \text{(a, b)} = \text{(e}_{1}\text{b} + \text{e}_{2}\text{a}, \text{e}_{2} \text{b})$
$\Rightarrow \text{a}\text{e}_{2} + \text{b}\text{e}_{1} = \text{a and b}\text{e}_{2} = \text{b} \Rightarrow \text{e}_{1} = 0, \text{e}_{2} = 1$
$\Rightarrow$ (0, 1) is the identity on A.
View full question & answer→Question 104 Marks
Let N denote the set of all natural numbers and R be the relation on$\text{N} \times \text{N}$defined by $\text{(a, b) R (c, d)}$ if ad$\text{(b + c) = bc(a + d)}$. Show that R is an equivalence relation.
Answer$\forall \text{a}, \text{b}\in \text{N},\text{(a, b)R (a, b) as ab ( b + a) = ba(a + b)} $
$\therefore \text{R is reflexive} \dots\dots\dots\dots\dots\dots \text{(i)}$
$ \text{Let (a, b) R (c, d) for (a, b), (c, d)} \in \text{N} \times \text{N}$
$\therefore \text{ad (b + c) = bc (a + d)} \dots\dots\dots\dots\dots\dots\text{(ii)}$
$ \text{Also (c, d) R (a, b)} \because\text{cb (d + a) = da (c + b) (using ii)}$
$\therefore \text{R is symmetric} \dots\dots\dots\dots\text{(iii)}$
$\text{Let (a, b) R (c,d) and (c, d) R(e,f), for a, b, c, d, e, f,} \in\text{N}$
$\therefore \text{ad (b + c) = bc (a + d) and cf (d + e) = de (c + f)}$
$\therefore \frac{\text{b + c}}{\text{bc}} = \frac{\text{a + d}}{\text{ad}} \text{and} \frac{\text{d + e}}{\text{de}} = \frac{\text{c + f}}{\text{cf}}$
$\text{i.e} \frac{1}{\text{c}} + \frac{1}{\text{b}} =\frac{1}{\text{d}} +\frac{1}{\text{a}} \text{and} \frac{1}{\text{e}} + \frac{1}{\text{d}} = \frac{1}{\text{f}}+ \frac{1}{\text{c}}$
$\text{adding we get} \frac{1}{\text{c}}+ \frac{1}{\text{b}} + \frac{1}{\text{e}} + \frac{1}{\text{d}} = \frac{1}{\text{d}} + \frac{1}{\text{a}} + \frac{1}{\text{f}} + \frac{1}{\text{c}}$
$\Rightarrow \text{af (b + e) = be (a + f)}$
$\text{Hence (a, b) R (e,f)} \therefore \text{R is transitive}\dots\dots\dots\dots\text{(iv)}$
$\text{From (i),(iii) and (iv) R is an equivalence relation}$
View full question & answer→Question 114 Marks
Discuss the commutativity and associativity of binary operation $^{‘*’}$ defined on A = Q – {1} by the rule $\text{a} ^{*} \text{b = a – b + ab}$ for all a, b $\in$ A. Also find the identity element of $^{*}$ in A and hence find the invertible elements of A.
Answer$\text{a }{*} \text{ b} = \text{a - b + ab } \forall \text{ a}, \text{b} \in \text{A} = \text{Q} - [1]$
$\text{b }{*} \text{ a} = \text{b - a + ba}$
$\text{(a }{*} \text{ b}) \neq \text{b }{*} \text{ a} \Rightarrow$ is not commutative.
$\text{(a }{*} \text{ b) }{*} \text{ c} = \text{a - b + ab) }{*} \text{ c}$
$\text{= a – b – c + ab + ac – bc + abc}$
$\text{a }{*} \text{ (b }{*} \text{ c)} = \text{a }{*} \text{ (b - c + bc)}$
$\text{= a – b + c + ab – ac – bc + abc}$
$\text{(a }{*} \text{ b) }{*} \text{ c} \neq \text{a }{*} \ \text{(b } {*}\text{ c)}$
$\Rightarrow \text{ }{*}$ is not associative.
Existence of identity
| $\text{a }{*} \text{ e = a – e + ae = a}$ | $\text{e }{*} \text{ a = e – a + ea = a }$ |
| $\Rightarrow \text{e (a - 1) = 0}$ | $\Rightarrow \text{e (1 + a) = 2a}$ |
| $\Rightarrow \text{e = 0}$ | $\Rightarrow \text{e} = \frac{\text{2a}}{\text{1 + a}}$ |
$\because$ e is not unique
$\therefore$ No idendity element exists.
$\text{a }{*} \text{ b} = \text{e = b }{*} \text{ a}$
$\therefore$ No identity element exists.
$\Rightarrow$ Inverse element does not exist.
View full question & answer→Question 124 Marks
Consider $\text{f} : \text{R}_{+} \rightarrow [ - 5, \infty)$ given by $\text{f}(x) = 9x^{2} + 6x - 5.$ Show that f is invertible with $\text{f}^{-1}\text{(y)} = \bigg(\frac{\sqrt{\text{y} + 6} - 1}{3}\bigg).$
Hence Find:
- $\text{f}^{-1} (10)$
- $\text{y if }\text{f}^{-1} \text{(y)} = \frac{4}{3},$
where R+ is the set of all non-negative real numbers.
AnswerClearly $\text{f}^{-1} \text{(y) = g (y):} [ -5, \infty) \rightarrow \text{R}_{+} \text{ and},$
$\text{fog (y) = y} \bigg(\frac{\sqrt{\text{y}\text{ + } 6} - 1}{3}\bigg) = 9 \bigg(\frac{\sqrt{\text{y + 6}} - 1}{3} \bigg)^{2} + 6 \bigg(\frac{\sqrt{\text{y + 6}} - 1}{3}\bigg) - 5 = \text{y}$
$\text{and (gof) (x) = g} (\text{9x}^{2} + \text{6x - 5)} = \frac{\sqrt{\text{9x}^{2} + \text{6x + 1}} - 1}{3} = \text{x}$
$\therefore \text{g = f}^{-1}$
- $\text{f}^{-1} (10) = \frac{\sqrt{16} - 1}{3} = 1$
- $\text{f}^{-1} (\text{y)} = \frac{4}{3} \Rightarrow \text{y = 19}$
View full question & answer→Question 134 Marks
$\text{Let f : N}\rightarrow\text{N}$ be a function defined as $\text{f(x)} = 9x^{2} + 6x -5.$ Show that $\text{f : N}\rightarrow\text{S},$ where S is the range of f, is invertible. Find the inverse of f and hence find $\text{f}^{-1}(43) \text{and f}^{-1}(163). $
Answer$\text{Let x}_{1},\text{x}_{2}\in \text{N and f (x}_{1}) =\text{f(x}_{2}) $
$\Rightarrow\text{9x}^{2}_{1} + 6\text{x}_{1} - 5 = \text{9x}^{2}_{2} + 6\text{x}_{2} - 5$
$^2_1 - \text {x}^{2}_{2}) + 6(\text{x}_{1} - \text{x}_{2}) = 0 \Rightarrow\text{(x}_{1} - \text{x}_{2}) \text{(9x}_{1} + \text{9x}_{2} + 6) = 0$
$\Rightarrow\text{x}_{1} - \text{x}_{2} = \text{0 or x}_{1} = \text{x}_{2}\text{ as}\text{(9x}_{1} + \text{9x}_{2} + 6\neq 0,\text{x}_{1}, \text{x}_{2} \in \text{N}$
$\therefore$ f is one one function
$\text{f : N}\rightarrow\text{S is ONTO as co-domain = Range}$
Hence f is invertible
$\text{y} = \text{9x}^{2} + \text{6x} - 5 = (\text{3x + 1)}^{2} - 6 \Rightarrow\text{x} = \frac{\sqrt{\text{y} + 6} -1}{3}$
$\therefore\text{f}^{-1}\text{(y)} = \frac{\sqrt{\text{y} + 6} - 1}{3}, \text{y}\in\text{S}$
$\text{f}^{-1}(43) = \frac{\sqrt{49}-1}{3} = 2$
$\text{f}^{-1}(163) = \frac{\sqrt{169} - 1}{3} = 4$
View full question & answer→Question 144 Marks
Find the values of x for which f(x) = [x (x - 2)]2 is an increasing function. Also, find the points on the curve, where the tangent is parallel to x-axis.
Answerf (x) = [x (x – 2)]2
f′ (x) = 4x (x – 2) (x – 1)
f′ (x) = 0 gives x = 0, x = 1 or x = 2
∴ Intervals are (- $\propto$,0), (0, 1), (1, 2), (2, $\propto$)
Increasing in [0, 1] and [2, $\propto$]
$\Rightarrow\text{OR}$ 0 < x < 1 and x > 2
The point where tangents are parallel to x axis
are (0, 0), (1, 1), (2, 0).
View full question & answer→Question 154 Marks
Show that the function $\text{f} : \text{R}\rightarrow\text{R}$ defined by $\text{f(x})=\frac{\text{x}}{\text{x}^2+1},\forall\text{ x}\in\text{R}$ is neither one-one nor onto. Also, if $\text{g} : \text{R}\rightarrow\text{R}$ is defined as g(x) = 2x – 1, find fog(x).
Answer$\text{f(x})=\frac{\text{x}}{\text{x}^2+1}$
For one-one f(x) = f(y)
$\frac{\text{x}}{\text{x}^2+1}=\frac{\text{y}}{\text{y}^2+1}$
$\text{xy}^2+\text{x}=\text{yx}^2+\text{y}$
$\text{xy}(\text{y}-\text{x})=\text{y}-\text{x}$
$\text{xy}=1$
$\text{x}=\frac{1}{\text{y}}$
$\text{x}\neq\text{y}$
So, not one-one
For onto f(x) = y
$\frac{\text{x}}{\text{x}^2+1}=\text{y}$
$\text{x}=\text{yx}^2+\text{y}$
$\text{x}^2\text{y}+\text{y}-\text{x}=0$
x cannot be express in y so not onto
As g(x) = 2x – 1
$\text{fog(x})=\text{f[g(x})]=\text{f}(2\text{x}-1)=\frac{2\text{x}-1}{(2\text{x}-1)^2+1}$
$=\frac{2\text{x}-1}{4\text{x}^2-4\text{x}+2}$
View full question & answer→Question 164 Marks
Let $\text{A}=\{\text{x}\in\text{Z}:0\leq\text{x}\leq12\}.$ Show that, $\text{R}=\{(\text{a, b}):\text{a, b}\in\text{A},\ |\text{a}-\text{b}|$ is divisible by 4} is an equivalence relation. Find the set of all elements related to 1. Also write the equivalence class [2].
AnswerR = {(a, b) : a, b ∈ A, |a – b| is divisible by 4}
Reflexivity: for any a ∈ A
|a – a| = 0, which is divisible by 4
(a, a) ∈ R.
So, R is reflexive.
Symmetry: Let (a, b) ∈ R
|a – b| is divisible by 4
⇒ |b – a| is also divisible by 4
So, R is symmetry
Transitive: Let (a, b) ∈ R & (b, c) ∈ R
|a – b| is divisible by 4
$|\text{a}-\text{b}|=4\lambda$
$\text{a}-\text{b}=\pm4\lambda\ \ \ .....(1)$
$|\text{b}-\text{c}|$ is divisible by 4
$|\text{b}-\text{c}|=4\mu$
$\text{b}-\text{c}=\pm4\mu\ \ \ .....(2)$
Add (1) & (2)
$\text{a}-\text{b}+\text{b}-\text{c}=\pm4(\lambda+\mu)$
$\text{a}-\text{c}=\pm4(\lambda+\mu)$
$\Rightarrow(\text{a, c})\in\text{R}$
So, Transitive
Hence, R is reflexive, Symmetry & Transitive so, it is an equivalence relation
Let x be an element of A such that (x, 1) ∈ R, then
|x – 1| is divisible by 4
x – 1 = 0, 4, 8, 12
⇒ x = 1, 5, 9
Hence, the set of all element of A which are related to 1 in {1, 5, 9}.
View full question & answer→Question 174 Marks
Prove that the relation R on Z defined by $(\text{a, b})\in\text{R}\Leftrightarrow\ \text{a}-\text{b}$ is divisible by 5 is an equivalence relation on Z.
AnswerWe observe the following properties of relation R.
Reflexivity: Let a be an arbitrary element of R. Then,
⇒ a - a = 0 = 0 × 5
⇒ a - a is divisible by 5
$\Rightarrow\ (\text{a, a})\in\text{R}$ for all $\text{a}\in\text{Z}$
So, R is reflexive on Z.
Symmetry: Let $(\text{a, b})\in\text{R}$
⇒ a - b is divisible by 5
⇒ a - b = 5p for some $\text{p}\in\text{Z}$
⇒ b - a = 5(-p)
Here, $-\text{p}\in\text{Z}$ [Since $\text{p}\in\text{Z}$]
⇒ b - a is divisible by 5
$\Rightarrow\ (\text{b, a})\in\text{R}$ for all $\text{a, b}\in\text{Z}$
So, R is symmetric on Z.
Transitivity: Let a, b and $\text{b},\text{c}\in\text{R}$
⇒ a - b is divisible by 5
⇒ a - b = 5p for some Z
Also, b - c is divisible by 5
⇒ b - c = 5q for some Z
Adding the above two, we get
a - b + b - c = 5p + 5q
⇒ a - c = 5(p + q)
⇒ a - c is divisible by 5
Here, $\text{p}+\text{q}\in\text{Z}$
$\Rightarrow\ \text{a, c}\in\text{R}$ for all $\text{a, c}\in\text{Z}$
So, R is transitive on Z.
Hence, R is an equivalence relation on Z.
View full question & answer→Question 184 Marks
Let A =R×R and * be a binary operation on A defined by,
(a, b) * (c, d) = (a + c, b + d).
Show that * is commutative and associative. Find the binary element for * on A, if any.
AnswerWe have,
A = R × R and * is a binary operation on A defined by (a, b) * (c, d) = (a + c, b + d).
Now,
(a, b) * (c, d) = (a + c, b + d) = (c + a, d + b)
⇒ (a, b) * (c, d) = (c, d) * (a, b)
So, * is commutative.
Also,
(a, b) * [(c, d) * (e, f)] = (a, b) * (c + e, d + f)
= (a, b) * (c + e, d + f)
= (a + c + e, b + d + f)
= (a + c, b + d) * (e, f)
= [(a, b) * (c, d)] * (e, f)
⇒ (a, b) * [(c, d) * (e, f)] = [(a, b) * (c, d)] * (e, f)
So, * is associative.
Let (x, y) be the binary element for * on A.
(a, b) * (x, y) = (a, b) = (x, y) * (a, b)
⇒ (a + x, b + y) = (a, b)
⇒ a + x = a and b + y = b
⇒ x = 0 and y = 0
Hence, (0, 0) is the binary element for * on A.
View full question & answer→Question 194 Marks
Classify the following functions as injection, surjection or bijection:
f : R → R, defined by f(x) = x3 - x
Answerf : R → R, defined by f(x) = x3 - x Injective: Let $\text{x, y}\in\text{R}$ such that, f(x) = f(y) ⇒ x3 - x = y3 - y ⇒ x3 - y3 - (x - y) = 0 ⇒ (x - y)(x2 + xy + y2 - 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 ⇒ x3 - 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 204 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 ≠ f2.
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 214 Marks
A function f : R → R is defined as f(x) = x3 + 4. Is it a bijection or not? In case it is a bijection, find f-1(3).
AnswerWe have,
f : R → R in a function defined by
f(x) = x3 + 4
Injectivity: Let f(x1) = f(x2) for $\text{x}_1,\text{x}_2\in\text{R}$
$\Rightarrow\ \text{x}_1^3+4=\text{x}_2^3+4$
$\Rightarrow\ \text{x}_1^3=\text{x}_2^3$
$\Rightarrow\ \text{x}_1=\text{x}_2$
⇒ f is one-one.
Surjectivity: Let $\text{y}\in\text{R}$ be artritrary such that
f(x) = y
⇒ x3 + 4 = y
⇒ x3 + 4 - y = 0
We know that an odd degree equation must have a real root.
$\Rightarrow\ \alpha^3+4=\text{y}$
$\Rightarrow\ \text{f}(\alpha)=\text{y}$
⇒ f is onto.
Since f is one-one and onto.
⇒ f is bijective.
finally, f(x) = y
⇒ x3 + 4 = y
⇒ x3 = 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 224 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 234 Marks
Give examples of two surjective functions f1 and f2 from Z to Z such that f1 + f2 is not surjective.
AnswerWe know that f1 : R → R, given by f1(x) = x, and f2(x) = -x are surjective functions.
Proving f1 is surjective: Let y be an element in the co-domain (R), such that f1(x) = y.
f1(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, f1 is surjective.
Proving f2 is surjective: Let f2(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, f2 is surjective.
Proving (f1 + f2) is not surjective:
Given: (f1 + f2)(x) = f1(x) + f2(x) = x + (-x) = 0
Therefore, for every real number x, (f1 + f2)(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, f1 + f2 is not surjective.
View full question & answer→Question 244 Marks
Define a binary operation * on the set {0, 1, 2, 3, 4, 5} as:
$\text{a}\times\text{b}=\begin{cases}\text{a + b},&\text{if }\text{a + b}<6\\\text{a + b}-6,&\text{if }\text{a + b}\geq6\end{cases}$
Show that 0 is the identity for this operation and each element a ≠ 0 of the set is invertible with 6 − a being the inverse of a.
AnswerLet X = {0, 1, 2, 3, 4, 5}.
The operation * on X is defined as:
$\text{a}\times\text{b}=\begin{cases}\text{a + b},&\text{if }\text{a + b}<6\\\text{a + b}-6,&\text{if }\text{a + b}\geq6\end{cases}$
An element $\text{e}\in\text{X}$ is the identity element for the operation *, if
a * e = a = e * a $\forall\text{ a}\in\text{X}$.
For $\text{a}\in\text{X}$, we observed that:
a * 0 = a + 0 = a $[\text{a}\in\text{X}\Rightarrow\text{a}+0<6]$
0 * a = a = 0 * a $[\text{a}\in\text{X}\Rightarrow0+\text{a}<6]$
$\therefore$ a * 0 = a = 0 * a $\forall\text{ a}\in\text{X}$
Thus, 0 is the identity element for the given operation *.
An element $\text{a}\in\text{X}$ is invertible if there exists be X such that a * b = 0 = b * a.
i.e.,$\begin{cases}\text{a + b}=0=\text{b + a},&\text{if }\text{a + b}<6\\\text{a + b}-6=0=\text{b + a}-6,&\text{if }\text{a + b}\geq6\end{cases}$
i.e.,
a = -b or b = 6 - a
But, X = {0, 1, 2, 3, 4, 5} and $\text{a, b}\in\text{X}$. Then, $\text{a}\neq-\text{b}$.
Therefore, b = 6 - a is the inverse of $\text{a}\in\text{X}$.
Hence, the inverse of an element $\text{a}\in\text{X},\text{a}\neq0$ is 6 - a i.e., a-1 = 6 - a.
View full question & answer→Question 254 Marks
Show that the function f : Q → Q, defined by f(x) = 3x + 5, is invertible. Also, find f-1.
AnswerGiven that f : Q → 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) ⇒ 3x + 5 = 3y + 5 ⇒ x = y So, f is an injection. Let y be an arbitrary element of Q such that f(x) = y. ⇒ 3x + 5 = y ⇒ 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 → Q is a bijection and hence invertible. Let f-1 denotes the inverse of f. Thus, fof-1(x) = x for all $\text{x}\in\text{Q}$ ⇒ f[f-1(x)] = x for all $\text{x}\in\text{Q}$ ⇒ 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 264 Marks
Show that the function $\text{f}:\text{R}_\ast\rightarrow\text{R}_\ast$ defined by $\text{f(x)}=\frac{1}{\text{x}}$ is one-one and onto, where $\text{R}_\ast$ is the set of all non-zero real numbers. Is the result true, if the domain $\text{R}_\ast$ is replaced by N with co-domain being same as $\text{R}_\ast?$
AnswerIt is given that $\text{f}:\text{R}_\ast\rightarrow\text{R}_\ast$ is defined by $\text{f(x)}=\frac{1}{\text{x}}$
One-one:
f(x) = f(y)
$\Rightarrow\frac{1}{\text{x}}=\frac{1}{\text{y}}$
⇒ x = y
$\therefore$ f is one-one.
Onto:
It is clear that for $\text{y}\in\text{R}_\ast,$ there exists$\text{x}=\frac{1}{\text{y}}\in\text{R}_\ast\ (\text{Exists as y}\neq0)$ such that $\text{f(x)}=\frac{1}{\frac{1}{\text{y}}}=\text{y}.$
$\therefore$ f is onto.
Thus, the given function (f) is one-one and onto.
Now, consider function $\text{g}:\text{N}\rightarrow\text{R}_\ast$ defined by
$\text{g(x)}=\frac{1}{\text{x}}.$
We have,
$\text{g}(\text{x}_1)=\text{g}(\text{x}_2)\Rightarrow\frac{1}{\text{x}_1}=\frac{1}{\text{x}_2}\Rightarrow\text{x}_1=\text{x}_2$
$\therefore$ g is one-one.
Further, it is clear that g is not onto as for $1.2\in\text{R}_\ast$ there does not exit any x in N such that $\text{g(x)}=\frac{1}{1.2}.$
Hence, function g is one-one but not onto.
View full question & answer→Question 274 Marks
Relation R in the set A of human beings in a town at a particular time given by
- R = {(x, y) : x and y work at the same place}
- R = {(x, y) : x and y live in the same locality}
- R = {(x, y) : x is exactly 7 cm taller than y}
- R = {(x, y) : x is wife of y}
- R = {(x, y) : x is father of y}
AnswerRelation R in the set A of human being in a town at a particular time. - R = {( x, y) : x and y work at the same place}
| Since $ (\text{x},\text{x})\in\text{R},$ because x and x work at the same place. | $\therefore$ R is reflexive. |
| Now, if $ (\text{x},\text{y})\in\text{R}\ \text{and}\ (\text{y},\text{x})\in\text{R},$ since x and y work at | |
| the same place and y and x work at the same place. | $\therefore$ R is symmetric. |
| Now, if $ (\text{x},\text{y})\in\text{R}\ \text{and}\ (\text{y},\text{x})\in\text{R}\Rightarrow(\text{x},\text{z})\in\text{R}.$ | $\therefore$ R is transitive. |
Therefore, R is reflexive, symmetric and transitive.
- R = {( x, y) : x and y live in the same locality}
| Since $ (\text{x},\text{x})\in\text{R},$ because x and x live in the same locality. | $\therefore$ R is reflexive. | |
| Also $ (\text{x},\text{y})\in\text{R}\ \Rightarrow (\text{y},\text{x})\in\text{R},$ because x and y live | | |
| in same locality and y and x also live in same locality. | $\therefore$ R is symmetric. | |
| $\text{Again }(\text{x},\text{y})\in\text{R}\ \ \Rightarrow\ \ (\text{y},\text{x})\in\text{R}\ \Rightarrow\ \ (\text{x},\text{z})\in\text{R}$ | $\therefore$ R is transitive. | |
Therefore, R is reflexive, symmetric and transitive.
- R = {( x, y) : x is exactly 7 cm taller than y}
| x is not exactly 7 cm taller than x , so $ (\text{x},\text{x})\notin\text{R},$ | $\therefore$ | R is not reflexive. |
| Also x is exactly 7 cm taller than y but y is not | | |
| 7 cm taller than x , so $ (\text{x},\text{y})\in\text{R}\ \text{but}\ (\text{y},\text{x})\notin\text{R}$ | $\therefore$ | R is not symmetric. |
Now x is exactly 7 cm taller than y and y is
exactly 7 cm taller than z then it does not | | |
| imply that x is exactly 7 cm taller than z. | $\therefore$ | R is not transitive. |
Therefore, R is neither reflexive, nor symmetric and nor transitive.
- R = {( x, y) : x is wife of y}
| x is not wife of x , so $ (\text{x},\text{x})\notin\text{R},$ | $\therefore$ | R is not reflexive. |
| Also x is wife of y but y is not wife of x , | | |
| so $ (\text{x},\text{y})\in\text{R}\ \text{but}\ (\text{y},\text{x})\notin\text{R}$ | $\therefore$ | R is not symmetric. |
| $\text{Also }(\text{x},\text{y})\in\text{R}\text{ and }(\text{y},\text{z})\in\text{R}\text{ then }(\text{x},\text{z})\notin\text{R}$ | $\therefore$ | R is not transitive. |
Therefore, R is neither reflexive, nor symmetric and nor transitive.
- R = {( x, y) : x is father of y}
| x is not father of x , so $ (\text{x},\text{x})\notin\text{R},$ | $\therefore$ | R is not reflexive. |
| Also x is father of y but y is not father of x , | | |
| so $ (\text{x},\text{y})\in\text{R}\ \text{but}\ (\text{y},\text{x})\notin\text{R}$ | $\therefore$ | R is not symmetric. |
| $\text{Also }(\text{x},\text{y})\in\text{R}\text{ and }(\text{y},\text{z})\in\text{R}\text{ then }(\text{x},\text{z})\notin\text{R}$ | $\therefore$ | R is not transitive. |
Therefore, R is neither reflexive, nor symmetric and nor transitive.
View full question & answer→Question 284 Marks
Let S be the set of all real numbers except -1 and let '*' be an operation defined by a * b = a + b + ab for all a, b ∈ S. Determine whether '*' is a binary operation on S. If yes, check its commutativity and associativity. Also, solve the equation (2 * x) * 3 = 7.
AnswerChecking for binary operation: Let $\text{a, b}\in\text{S.}$ Then, $\text{a, b}\in\text{R}$ and $\text{a}\neq-1,\text{b}\neq-1$ a * b = a + b + ab We need to prove that $\text{a}+\text{b}+\text{ab}\in\text{S.}$ $\big[$For this we have to prove that $\text{a}+\text{b}+\text{ab}\in\text{R}$ and $\text{a}+\text{b}+\text{ab}\neq-1\big]$ Since, $\text{a, b}\in\text{R},\ \text{a}+\text{b}+\text{ab}\in\text{R},$ let us assume that a + b + ab = -1. a + b + ab + 1 = 0 a + ab + b + 1 = 0 a(1 + b) + 1(1 + b) = 0 (a + 1)(b+ 1) = 0 a = -1, b = -1 [which is false] Hence, $\text{a}+\text{b}+\text{ab}\neq-1$ Therefore, $\text{a}+\text{b}+\text{ab}\in\text{S}$ Thus, * is a binary operation on S. Commutativity: Let $\text{a, b}\in\text{S.}$ Then, a * b = a + b + ab = b + a + ba = b * a Therefore,
a * b = b * a, $\forall\ \text{a, b}\in\text{S}$ Thus, * is commutative on N. Associativity: Let $\text{a, b, c}\in\text{S.}$ a * (b * c) = a * (b + c + bc) = a + b + c + bc + a(b + ac + bc) = a + b + c + bc + ab + ac + abc (a * b) * c = (a + b + ab) * c = a + b + ab + c + (a + b + ab)c = a + b + ab + c + ac + bc + abc Therefore, a * (b * c) = (a * b) * c, $\forall\ \text{a, b, c}\in\text{S}$ Thus, * is associative on S. Now, Given: (2 * x) * 3 = 7 Implies that (2 + x + 2x) * 3 = 7 Implies that (2 + 3x) * 3 = 7 View full question & answer→Question 294 Marks
If f : Q → Q, g : Q → 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-1og-1.
AnswerInjectivity of f: Let x and y be two elements of domain (Q), such that f(x) = f(y) ⇒ 2x = 2y ⇒ x = y So, f is one-one. Surjectivity of f: Let y be in the co-domain (Q), such that f(x) = y. ⇒ 2x = y $\Rightarrow\ \text{x}=\frac{\text{y}}{2}\in\text{Q}$ (domain) ⇒ f is onto. So, f is a bijection and hence, it is invertible. Finding f-1: Let f-1(x) = y .....(1) ⇒ x = f(y) ⇒ x = 2y $\Rightarrow\ \text{y}=\frac{\text{x}}{2}$ So, $\text{f}^{-1}(\text{x})=\frac{\text{x}}{2}$ [from (1)] Injectivity of g: Let x and y be two elements of domain (Q), such that g(x) = g(y) ⇒ x + 2 = y + 2 ⇒ x = y So, g is one-one. Surjectivity of g: Let y be in the co domain (Q), such that g(x) = y. ⇒ x + 2 = y $\Rightarrow\ \text{x}=\text{y}-2\in\text{Q}$ (domain)
⇒ g is onto. So, g is a bijection and hence, it is invertible. Finding g-1: Let g-1(x) = y .....(2) ⇒ x = g(y) ⇒ x = y + 2 ⇒ y = x - 2 So, g-1(x) = x - 2 [from (2)] Verification of (gof)-1 = f-1og-1: f(x) = 2x; g(x) = x + 2 and $\text{f}^{-1}(\text{x})=\frac{\text{x}}{2};$ g-1(x) = x - 2 Now, (f-1og-1)(x) = f-1(g-1(x)) ⇒ (f-1og-1)(x) = f-1(x - 2) $\Rightarrow\ (\text{f}^{-1}\text{og}^{-1})(\text{x})=\frac{\text{x}-2}{2}\ ......(3)$ (gof)(x) = g(f(x)) = g(2x) = 2x + 2 Let (gof)-1(x) = y ......(4) x = (gof)(y) ⇒ x = 2y + 2 ⇒ 2y = x - 2 $\Rightarrow\ \text{y}=\frac{\text{x}-2}{2}$ $\Rightarrow\ (\text{gof})^{-1}(\text{x})=\frac{\text{x}-2}{2}\ .....(5)$ [from (4)] From (3) and (5), (gof)-1 = f-1og-1 View full question & answer→Question 304 Marks
Let S be the set of all rational numbers except 1 and * be defined on S by a * b = a + b - ab, for all a, b ∈ S.
Prove that:
- * is a binary operation on S.
- * is commutative as well as associative.
AnswerWe have, S = R - {1} and * is defined on S as a * b = a + b - ab, for all a, b ∈ S. - It is seen that for each a, b ∈ S, there is a unique element a + b - ab in S.
This means that * carries each pair (a, b) to a unique element a * b = a + b - ab in S.
Therefore, * is a binary operation on S.
- Commutativity: Let a, b ∈ S. Then,
a * b = a + b - ab
= b + a - ba
= b * a
Therefore, a * b = b * a, $\forall\ \text{a, b}\in\text{S}$
Thus, * is commutative on S.
Associativity: Let a, b, c ∈ S. Then,
a * (b * c) = a * (b + c - bc)
= a + b + c - bc - a(b + c - bc)
= a + b + c - bc - ab - ac + abc
(a * b) * c = (a + b - ab) * c
= a + b - ab + c = (a + b - ab)c
= a + b + c - ab - ac - bc + abc
Therefore,
a * (b * c) = (a * b) * c, $\forall\ \text{a, b, c}\in\text{S}$
Thus, * is associative on S.
Therefore, * is commutative as well as associative.
View full question & answer→Question 314 Marks
Show that the function f: R → {x ∈ R: –1 < x < 1} defined by $f(\text{x})=\frac{\text{x}}{1+|\text{x}|},\text{x}\in\text{R}$ is one-one and onto function.
AnswerIt is given that $f:\text{R}\rightarrow\{\text{x}\in\text{R}:-1<\text{x}<1\}$ is defined as $f(\text{x})=\frac{\text{x}}{1+|\text{x}|},\text{x}\in\text{R}.$
Suppose f(x) = f(y), where $\text{x},\text{y}\in\text{R}.$
$\Rightarrow\frac{\text{x}}{1+|\text{x}|}=\frac{\text{y}}{1+|\text{y}|}$
It can be observed that if x is positive and y is negative, then we have: $\frac{\text{x}}{1+\text{x}}=\frac{\text{y}}{1+\text{y}}\Rightarrow2\text{xy}=\text{x}-\text{y}$
Since x is positive and y is negative:
x > y ⇒ x - y > 0
But, 2xy is negative.
Then, $2\text{xy}\neq\text{x}-\text{y}$
Thus, the case of x being positive and y being negative can be ruled out.
Under a similar argument, x being negative and y being positive can also be ruled out
$\therefore$ x and y have to be either positive or negative.
When x and y are both positive, we have:
$f(\text{x})=f(\text{y})\Rightarrow\frac{\text{x}}{1+\text{x}}=\frac{\text{y}}{1+\text{y}}\Rightarrow\text{x}+\text{xy}=\text{y}+\text{xy}\Rightarrow\text{x}=\text{y}$
When x and y both are negative, we have:
$f(\text{x})=f(\text{y})\Rightarrow\frac{\text{x}}{1-\text{x}}=\frac{\text{y}}{1-\text{y}}\Rightarrow\text{x}-\text{xy}=\text{y}-\text{xy}\Rightarrow\text{x}=\text{y}$
$\therefore$ f is one-one.
Now, let $\text{y}\in\text{R}$ such that -1 < y < 1.
If y is negative, then there exists $\text{x}=\frac{\text{y}}{1+\text{y}}\in\text{R}$ such that
$f(\text{x})=f\Big(\frac{\text{y}}{1+\text{y}}\Big) =\frac{\Big(\frac{\text{y}}{1+\text{y}}\Big)}{1+\Big|\frac{\text{y}}{1+\text{y}}\Big|}=\frac{\frac{\text{y}}{1+\text{y}}}{1+\Big(\frac{\text{-y}}{1+\text{y}}\Big)}=\frac{\text{y}}{1+\text{y}-\text{y}}=\text{y}.$
If y is positive, then there exists $\text{x}=\frac{\text{y}}{1+\text{y}}\in\text{R}$ such that
$f(\text{x})=f\Big(\frac{\text{y}}{1-\text{y}}\Big) =\frac{\Big(\frac{\text{y}}{1-\text{y}}\Big)}{1+\Big|\frac{\text{y}}{1-\text{y}}\Big|}=\frac{\frac{\text{y}}{1-\text{y}}}{1+\frac{\text{y}}{1+\text{y}}}=\frac{\text{y}}{1-\text{y}+\text{y}}=\text{y}.$
$\therefore$ f is onto.
Hence, f is one-one and onto.
View full question & answer→Question 324 Marks
Let f: R → R be defined as f(x) = 10x + 7. Find the function g: R → R such that gof = fog = 1R.
AnswerIt is given that f: R → R is defined as f(x) = 10x + 7.
One-one:
Let f(x) = f(y), where $\text{x},\text{y}\in\text{R}.$
⇒ 10x + 7 = 10y + 7
⇒ x = y
$\therefore$ f is a one-one function.
Onto:
For $\text{y}\in\text{R},$ let y = 10x + 7.
$\Rightarrow\text{x}=\frac{\text{y}-7}{10}\in\text{R}$
Therefore, for any $\text{y}\in\text{R},$ there exists $\text{x}=\frac{\text{y}-7}{10}\in\text{R}$ such that
$f(\text{x})=f\Big(\frac{\text{y}-7}{10}\Big)=10\Big(\frac{\text{y}-7}{10}\Big)+7=\text{y}-7+7=\text{y}.$
$\therefore$ f is onto.
Therefore, f is one-one and onto.
Thus, f is an invertible function.
Let us define g: R → R as $\text{g(y)}=\frac{\text{y}-7}{10}.$
Now, we have:
gof(x) = g(f(x)) = g(10x + 7) $=\frac{(10\text{x}+7)}{10}=\frac{10\text{x}}{10}=\text{x}$
And,
$fo\text{g(y)}=f(\text{g(y)})=f\Big(\frac{\text{y}-7}{10}\Big)=10\Big(\frac{\text{y}-7}{10}\Big)+7=\text{y}-7+7=\text{y}$
$\therefore$ gof = IR and fog = IR
Hence, the required function g: R → R is defined as $\text{g(y)}=\frac{\text{y}-7}{10}.$
View full question & answer→Question 334 Marks
Let Z be the set of integers. Show that the relation R = {(a, b): a, b ∈ Z and a + b is even} is an equivalence relation on Z.
AnswerWe have, Z be set of integers and R = {(a, b): a, b ∈ Z and a + b is even} be a relation on Z. We want to prove that R is an equivalence relation on Z. Now, Reflexivity: Let $\text{a}\in\text{Z}$ ⇒ a + a is even [if a is even ⇒ a + a is even, if a is odd ⇒ a + a is even] $\Rightarrow\ (\text{a, a})\in\text{R}$ ⇒ R is reflexive. Symmetric: Let $\text{a, b}\in\text{Z}$ and $(\text{a, b})\in\text{R}$ ⇒ a + b is even ⇒ b + a is even $\Rightarrow\ (\text{b, a})\in\text{R}$ ⇒ R is symmetric. Transitivity: Let $(\text{a, b})\in\text{R}$ and $(\text{b, c})\in\text{R}$ For some $\text{a, b, c}\in\text{Z}$
⇒ a + b is even and b + c is even [if b is odd, then a and c must be odd ⇒ a + c is even, if b is even, then a and c must be even ⇒ a + c is even] ⇒ a + c is even $\Rightarrow\ (\text{a, c})\in\text{R}$ ⇒ R is transitive. Hence, R is an equivalence relation on Z. View full question & answer→Question 344 Marks
Consider the binary operations *: R × R → R and o: R × R → R defined as $\text{a}∗\text{b} = |\text{a } – \text{ b}|\text{ and}\text{ a} o \text{b} = \text{a},\forall\text{a},\text{b}\in\text{R}.$ Show that * is commutative but not associative, o is associative but not commutative. Further, show that $\forall\text{a},\text{b},\text{c}\in\text{R},\text{a} *(\text{b}o\text{c}) = (\text{a} *\text{b})o(\text{a}*\text{b}).$ [If it is so, we say that the operation * distributes over the operation o]. Does o distribute over *? Justify your answer.
AnswerIt is given that *: R × R → and o: R × R → R is defined as
a * b = |a - b| and aob = a, $\Box\text{a},\text{b}\in\text{R}.$
For $\text{a},\text{b}\in\text{R},$ we have:
a * b = |a - b|
b * a = |b - a| = |-(a - b)| = |a - b|
$\therefore$ The operation * is commutative.
It can be observed that,
(1 * 2) * 3 = (|1 - 2|) * 3 = 1 * 3 = |1 - 3| = 2
1 * (2 * 3) = 1 * (|2 - 3|) = 1 * 1 = |1 - 1| = 0
$\therefore(1*2) *3\neq1*(2*3)(\text{where }1, 2,3\in\text{R})$
$\therefore$ The operation * is not associative.
Now, consider the operation o:
It can be observed that 1o2 = 1 and 2o1 = 2.
$\therefore1o2\neq2o1\ (\text{where }1,2\in\text{R})$
$\therefore$ The operation o is not commutative.
Let $\text{a},\text{b},\text{c}\in\text{R}.$ Then, we have:
(aob)oc = aoc = a
ao(boc) = aob = a
⇒ (aob)oc = ao(boc)
$\therefore$ The operation o is associative.
Now, let $\text{a},\text{b},\text{c}\in\text{R},$ then we have:
a * (boc) = a * b = |a - b|
(a * b)o(a * c) = (|a - b|)o(|a - c|) = |a - b|
Hence, a * (boc) = (a * b)o(a * c).
Now,
1o(2 * 3) = 1o(|2 - 3|) = 1o1 = 1
(1o2) * (1o3) = 1 * 1 = |1 - 1| = 0
$\therefore1o(2 * 3)\neq(1o2)*(1o3)\ (\text{where }1,2,3\in\text{R})$
$\therefore$ The operation o does not distribute over *.
View full question & answer→Question 354 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 364 Marks
Show that the relation R on the set Z of integers, given by R = {(a, b): 2 divides a - b}, is an equivalence relation.
AnswerWe have, R = {(a, b): a - b is divisible by 2; a, b $\in\text{Z}$} To prove: R is an equivalence relation. Proof:
Reflexivity: Let $\text{a}\in\text{Z}$ ⇒ a - a = 0 ⇒ a - a is divisible by 2 $\Rightarrow\ (\text{a, a})\in\text{R}$ ⇒ R is reflexive. Symmetric: Let $\text{a, b}\in\text{Z}$ and $(\text{a, b})\in\text{R}$ ⇒ a - b is divisible by 2 ⇒ a - b = 2p For some $\text{p}\in\text{Z}$ ⇒ b - a = 2 × (-p) $\Rightarrow\ \text{b}-\text{a}\in\text{R}$
⇒ R is symmetric. Transitive: Let $\text{a, b, c}\in\text{Z}$ and such that $(\text{a, b})\in\text{R}$ and $(\text{b, c})\in\text{R}$ ⇒ a - b = 2p and b - c = q For some $\text{p, q}\in\text{Z}$ ⇒ a - c = 2(p + q) ⇒ a - c is divisible by 2 $\Rightarrow\ (\text{a, c})\in\text{R}$ ⇒ R is transitive. View full question & answer→Question 374 Marks
Consider f : R → 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) ⇒ 4x + 3 = 4y + 3 ⇒ 4x = 4y ⇒ x = y So, f is one-one. Surjectivity of f: Let y be in the co-domain (R), such that f(x) = y. ⇒ 4x + 3 = y ⇒ 4x = y - 3 $\Rightarrow\ \text{x}=\frac{\text{y}-3}{4}\in\text{R}$ (Domain) ⇒ f is onto. So, f is a bijection and hence is invertible. Finding f-1: Let f-1(x) = y ....(1) ⇒ x = f(y) ⇒ x = 4y + 3 ⇒ 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 384 Marks
Let f : N → N be a function as f(x) = 9x2 + 6x - 5. Show that f : N → 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 → N is a function defined as f(x) = 9x2 + 6x - 5. Let y = f(x) = 9x2 + 6x - 5 ⇒ y = 9x2 + 6x - 5 ⇒ y = 9x2 + 6x + 1 - 1 - 5 ⇒ y = (9x2 + 6x + 1) - 6 ⇒ y = (3x + 1)2 - 6 ⇒ 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 394 Marks
Let A = R – {3} and B = R – {1}. Consider the function f: A → B defined by $f(\text{x})=\Big(\frac{\text{x}-2}{\text{x}-3}\Big).$ Is f one-one and onto? Justify your answer.
AnswerA = R - {3}, B = R - {1}
f: A → B is defined as $f(\text{x})=\Big(\frac{\text{x}-2}{\text{x}-3}\Big)$
Let $\text{x},\text{y}\in\text{A}$ such that f(x) = f(y)
$\Rightarrow\frac{\text{x}-2}{\text{x}-3}=\frac{\text{y}-2}{\text{y}-3}$
⇒ (x - 2)(y - 3) = (y - 2)(x - 3)
⇒ xy - 3x - 2y + 6 = xy - 3y - 2x + 6
⇒ -3x - 2y = -3y - 2x
⇒ 3x - 2x = 3y - 2y
⇒ x = y
$\therefore$ f is one-one.
Let $\text{y}\in\text{B}=\text{R}-\{1\}.$ Then, $\text{y}\neq1.$
The function is onto if there exists $\text{x}\in\text{A}$ such that f(x) = y.
Now,
f(x) = y
$\Rightarrow\frac{\text{x}-2}{\text{x}-3}=\text{y}$
⇒ x - 2 = xy - 3y
⇒ x(1 - y) = -3y + 2
$\Rightarrow\text{x}=\frac{2-3\text{y}}{1-\text{y}}\in\text{A}\ \ \ \ \ [\text{y}\neq1]$
Thus, for any $\text{y}\in\text{B},$ there exists $\frac{2-3\text{y}}{1-\text{y}}\in\text{A}$ such nthat
$f\Big(\frac{2-3\text{y}}{1-\text{y}}\Big)=\frac{\Big(\frac{2-3\text{y}}{1-\text{y}}\Big)-2}{\Big(\frac{2-3\text{y}}{1-\text{y}}\Big)-3}=\frac{2-3\text{y}-2+2\text{y}}{2-3\text{y}-3+3\text{y}}=\frac{-\text{y}}{-1}=\text{y}$
$\therefore$ f is onto.
Hence, function f is one-one and onto.
View full question & answer→Question 404 Marks
Let f: N → N be defined by ${f(n)}=\begin{cases}\frac{\text{n}+1}{2},\text{ if n is odd}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{for all}\text{ n }\in\text{ N}.\\\frac{\text{n}}{2}\ \ \ \ ,\text{if n is even}\end{cases}$ State whether the function f is bijective. Justify your answer.
Answer${f(n)}=\begin{cases}\frac{\text{n}+1}{2},\text{ if n is odd}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{for all}\text{ n }\in\text{ N}.\\\frac{\text{n}}{2}\ \ \ \ ,\text{if n is even}\end{cases}$
f: N → N is defined as
It can be observed that:
$\therefore{f}(1)={f}(2),\ \text{where }1\neq2.$
$\therefore$ f is not one-one.
Consider a natural number (n) in co-domain N.
Case I: n is odd
$\therefore$ n = 2r + 1 for some $\text{r}\in\text{N}.$ Then, there exists $4\text{r}+1\in\text{N}$ such that
$f(4\text{r}+1)=\frac{4\text{r}+1+1}{2}=2\text{r}+1$
Case II: n is even
$\therefore$ n = 2r for some $\text{r}\in\text{N}.$ Then, there exists $4\text{r}\in\text{N}$ such that $f(4\text{r})=\frac{4\text{r}}{2}=2\text{r}$
$\therefore$ f is onto.
Hence, f is not a bijective function.
View full question & answer→Question 414 Marks
Let f: W → W be defined as f(n) = n – 1, if n is odd and f (n) = n + 1, if n is even. Show that f is invertible. Find the inverse of f. Here, W is the set of all whole numbers.
AnswerIt is given that: f: W → W is defined as $f(\text{n})=\begin{cases}\text{n}-1,\ \text{if n is odd}\\\text{n}+1,\ \text{if n is even}\end{cases}$ One-one: Let f(n) = f(m). It can be observed that if n is odd and m is even, then we will have n - 1 = m + 1. ⇒ n - m = 2 However, this is impossible. Similarly, the possibility of n being even and m being odd can also be ignored under a similar argument. $\therefore$ Both n and m must be either odd or even. Now, if both n and m are odd, then we have: f(n) = f(m) ⇒ n - 1 = m - 1 ⇒ n = m Again, if both n and m are even, then we have: f(n) = f(m) ⇒ n + 1 = m + 1 ⇒ n = m $\therefore$ f is one-one. It is clear that any odd number 2r + 1 in co-domain N is the image of 2r in domain N and any even number 2r in co-domain N is the image of 2r + 1 in domain N. $\therefore$ f is onto. Hence, f is an invertible function. Let us define g: W → W as: $\text{g(m)}=\begin{cases}\text{m}+1,\ \text{if m is odd}\\\text{m}-1,\ \text{if m is even}\end{cases}$ Now, when n is odd: gof(n) = g(f(n)) = g(n - 1) = n - 1 + 1 = n And, when n is even: gof(n) = g(f(n)) = g(n + 1) = n + 1 - 1 = n Similarly, when m is odd: fog(m)=f(g(m)) = f(m - 1) = m - 1 + 1 = m When m is even: fog(m) = f(g(m)) = f(m + 1) = m + 1 - 1 = m $\therefore$ gof = IW and fog = IW Thus, f is an invertible and the inverse of f is given by f-1 = g, which is the same as f. Hence, the inverse of f is fitself.
View full question & answer→Question 424 Marks
Consider f: R+ → [– 5, ∞) given by f(x) = 9x2 + 6x – 5. Show that f is invertible with $f^{-1}(\text{y})=\left(\frac{\Big(\sqrt{\text{y}+6}\Big)-1}{3}\right).$
AnswerConsider $f:\text{R}_{+}\rightarrow[-5,\infty]$ and f(x) = 9x2 + 6x - 5. Let $\text{x}_1,\text{x}_2\in\text{R}\rightarrow[-5,\infty],\text{ then }f(\text{x}_1)=9\text{x}_{1}^{2}+6\text{x}_1-5\text{ and }f(\text{x}_2)=9\text{x}_{2}^{2}+6\text{x}_2-5$ Now, f(x1) = f(x2) then
$9\text{x}_{1}^{2}+6\text{x}_1-5 =9\text{x}_{2}^{2}+6\text{x}_2-5$ $\Rightarrow9\text{x}_{1}^{2}+6\text{x}_1 =9\text{x}_{2}^{2}+6\text{x}_2\ \ \Rightarrow\ 9(\text{x}_{1}^{2}-\text{x}_{2}^{2})+6(\text{x}_1-\text{x}_2)=0$ ⇒ (x1 - x2)[9(x1 + x2) +6] = 0 ⇒ x1 - x2 = 0 ⇒ x1 = x2 $\therefore$ f is one-one. Now, again y = 9x2 + 6x - 5 ⇒ 9x2 + 6x - (5 + y) = 0 $\Rightarrow\ \text{x}=\frac{-6\pm\sqrt{(6)^2+4\times9(5+\text{y})}}{18}=\frac{-6\pm6\sqrt{1+5+\text{y}}}{18}=\frac{-6\pm6\sqrt{\text{y}+6}}{18}=\frac{\sqrt{\text{y}+6}-1}{3}$ $\therefore f(\text{x})=f\left(\frac{\sqrt{\text{y}+6}-1}{3}\right)=9\left(\frac{\sqrt{\text{y}+6}-1}{3}\right)^2+6\left(\frac{\sqrt{\text{y}+6}-1}{3}\right)-5$ $=9\left(\frac{\text{y}+6+1-2\sqrt{\text{y}+6}-1}{9}\right)+2(\sqrt{\text{y}+6}-1)-5$ $=\text{y}+7-2\sqrt{\text{y}+6}+2\sqrt{\text{y}+6}-2-5=\text{y}$ Therefore, f(x) is invertible and $f^{-1}(\text{x})=\frac{\sqrt{\text{y}+6}-1}{3}.$ View full question & answer→Question 434 Marks
Let A = {– 1, 0, 1, 2}, B = {– 4, – 2, 0, 2} and f, g: A → B be functions defined by $f(\text{x})=\text{x}^2-\text{x},\ \text{x}\in\text{A}\ \text{and }\text{g(x)}=2\Big|\text{x}-\frac{1}{2}\Big|,\ \text{x}\in\text{A}.$ Are f and g equal? Justify your answer. (Hint: One may note that two functions f: A → B and g: A → B such that $f(\text{a}) = \text{g(a)}\ \forall \text{a} \in \text{A},$ are called equal functions).
AnswerIt is given that A = {-1, 0, 1, 2}, B ={-4, -2, 0, 2}.
Also, it is given that f, g: A → B are defined by
$f(\text{x})=\text{x}^2-\text{x},\ \text{x}\in\text{A}\ \text{and }\text{g(x)}=2\Big|\text{x}-\frac{1}{2}\Big|,\ \text{x}\in\text{A}.$
It is observed that:
f(-1) = (-1)2 - (-1) = 1 + 1 = 2
$\text{g}(-1)=2\Big|(-1)-\frac{1}{2}\Big|-1=2\Big(\frac{3}{2}\Big)-1=3-1=2$
⇒ f(-1) = g(-1)
f(0) = (0)2 - 0 = 0
$\text{g}(0)=2\Big|0-\frac{1}{2}\Big|-1=2\Big(\frac{1}{2}\Big)-1=1-1=0$
⇒ f(0) = g(0)
f(1) = (1)2 - 1 = 1 - 1 = 0
$\text{g}(1)=2\Big|1-\frac{1}{2}\Big|-1=2\Big(\frac{1}{2}\Big)-1=1-1=0$
⇒ f(1) = g(1)
f(2) = (2)2 - 2 = 4 - 2 = 2
$\text{g}(2)=2\Big|2-\frac{1}{2}\Big|-1=2\Big(\frac{3}{2}\Big)-1=3-1=2$
$\therefore f(\text{a})=\text{g(a)}\ \forall\text{a}\in\text{A}$
Hence, the functions f and g are equal.
View full question & answer→Question 444 Marks
Show that the relation R defined by R = {(a, b): a - b is divisible by 3; a, b ∈ Z} is an equivalence relation.
AnswerWe observe the following relations of relation R.
Reflexivity: Let a be an arbitrary element of R. Then,
a - a = 0 = 0 × 3
⇒ a - a is divisible by 3
$\Rightarrow\ (\text{a, a})\in\text{R}$ for all $\text{a}\in\text{Z}$
So, R is reflexive on Z.
Symmetry: Let $(\text{a, b})\in\text{R}$
⇒ a - b is divisible by 3
⇒ a - b = 3p for some $\text{p}\in\text{Z}$
⇒ b - a = 3(-p)
Here, $-\text{p}\in\text{Z}$
⇒ b - a is divisible by 3
⇒ (b, a)∈R for all a, $\text{b}\in\text{Z}$
So, R is symmetric on Z.
Transitivity: Let a, b and b, c $\in\text{R}$
⇒ a - b and b - c are divisible by 3
⇒ a - b = 3p for some $\text{p}\in\text{Z}$
and b - c = 3q for some $\text{q}\in\text{Z}$
Adding the above two, we get
a - b + b - c = 3p + 3q
⇒ a - c = 3(p + q)
Here, p + q $\in\text{Z}$
⇒ a - c is divisible by 3
$ \Rightarrow \text{a, c} \in \text{R}$ for all a, c $\in\text{Z}$
So, R is transitive on Z.
Hence, R is an equivalence relation on Z.
View full question & answer→Question 454 Marks
Show that the exponential function f : R → R, given by f(x) = ex, 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 464 Marks
Consider f : {1, 2, 3} → {a, b, c} and g : {a, b, c} → {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-1og-1.
Answerf = {(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-1og-1 = {(apple, 1), (ball, 2), (cat, 3)} .....(1)
f : {1, 2, 3} → {a, b, c} and g : {a, b, c} → {apple, ball, cat}
So, gof : {1, 2, 3} → {apple, ball, cat}
⇒ (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-1og-1
View full question & answer→Question 474 Marks
Let A = R - {3}, B = R - {1}. Let f : A → B be defined by $\text{f}(\text{x})=\frac{\text{x}-2}{\text{x}-3}\ \forall\ \text{x}\in\text{A}.$ Then show that f is bijective.
AnswerWe are given that, A = R - {3} and B = R - {1}. Consider, $\text{f}(\text{x})=\frac{\text{x}-2}{\text{x}-3}\ \forall\ \text{x}\in\text{A}$
Injectivity:
Let f (x1) = f (x2) $\Rightarrow\ \frac{\text{x}_1-2}{\text{x}_1-3}=\frac{\text{x}_2-2}{\text{x}_2-3}$ ⇒ (x1 - 2)(x2 - 3) = (x2 - 2)(x1 - 3) ⇒ x1x2 - 3x1 - 2x2 + 6 = x1x2 - 3x2 - 2x1 + 6 ⇒ -3x1 - 2x2 = -3x2 - 2x1 ⇒ -x1 = -x2 ⇒ x1 = x2 Hence, f(x) is an injective function. Surjectivity:
Let f(x) = y
$\Rightarrow\ \frac{\text{x}-2}{\text{x}-3}=\text{y}$
⇒ x - 2 = xy - 3y ⇒ x - xy = -3y + 2 ⇒ x(1 - y) = 2 - 3y $\Rightarrow\ \text{x}=\frac{2-3\text{y}}{1-\text{y}}$ $\Rightarrow\ \text{x}=\frac{3\text{y}-2}{\text{x}-3}\in\text{A},\ \forall\ \text{y}\in\text{B}$ Hence, f(x) is surjective function. Since, f(x) is both injective and surjective function. Hence, f(x)is a bijective function. View full question & answer→Question 484 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 494 Marks
Show that the relation R in the set A of points in a plane given by R = {(P, Q) : distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.
AnswerIt is given that
R = {(P, Q): distance of the point P from the origin is same as the distance of the point Q from the origin},
Now, it is clear that
$(\text{P},\text{P})\in\text{R}$ since the distance of point P from origin is always the same as the distance of the same point P from the origin.
Therefore, R is reflexive.
Now, Let us take $(\text{P},\text{Q})\in\text{R},$
⇒ The distance of point P from origin is always the same as the distance of the same point Q from the origin.
⇒ The distance of point Q from origin is always the same as the distance of the same point P from the origin.
$\Rightarrow(\text{Q},\text{P})\in\text{R}$
Therefore, R is symmetric.
Now, Let $(\text{P},\text{Q}),(\text{Q},\text{S})\in\text{R}$
⇒ The distance of point P and Q from the origin is always the same as the distance of the same point Q and S
from the origin.
⇒ The distance of points P and S from the origin is the same.
$\Rightarrow(\text{P},\text{S})\in\text{R}$
Therefore, R is transitive.
Therefore, R is equivalence relation.
The set of all points related to $\text{P}\neq(0,0)$ will be those the set of all points related to $\text{P}\neq(0,0)$ will be those points whose distance from the origin is the same as the distance of point P from the origin.
So, if O(0,0) is the origin and OP = k, then the set of all points related to P is at a distance of k from the origin.
Therefore, this set of points forms a circle with the centre as the origin and this circle passes through point P.
View full question & answer→Question 504 Marks
Show that the relation R, defined on the set A of all polygons as R = {(P1, P2): P1 and P2 have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?
AnswerWe observe the following properties on R.
Reflexivity: Consider P1 be an arbitrary element of A.
Then, polygon P1 and P1 have the same number of sides.
Since they are one and the same.
Implies that$\text{P}_1, \text{P}_1\in\text{R}$ $$ for all $\text{P}_1\in\text{A.}$
So,R is reflexive on A.
Symmetry: Consider $\text{P}_1,\text{ P}_2\in\text{R}$
Implies that P1 and P2 have the same number of sides.
Implies that P2 and P1 have the same number of sides.
Implies that $\text{P}_2,\text{ P}_1\in\text{R}$ for all $\text{P}_1,\text{ P}_2\in\text{A}$
So, R is symmetric on A.
Transitivity: Consider $\text{P}_1, \text{P}_2, \text{P}_3\in\text{R}$ $$
Implies that P1 and P2 have the same number of sides and P2 and P3 have the same number of sides
Implies that P1, P2 and P3 have the same number of sides.
Implies that P1 and P3 have the same number of sides.
Implies that $\text{P}_1,\text{ P}_3\in\text{R}$ for all $\text{P}_1,\text{ P}_3\in\text{A.}$
So, R is transitive on A.
Hence, R is an equivalence relation on the set A. Also, the set of all the triangles $\in\text{A}$ is related to the right angle triangle T with the sides 3, 4, 5.
View full question & answer→