5 Marks Questions - Page 2 - MATHS STD 12 Science Questions - Vidyadip

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5 Marks Questions

Question 515 Marks

On R − {1}, a binary operation * is defined by a * b = a + b − ab. Prove that * is commutative and associative. Find the identity element for * on R − {1}. Also, prove that every element of R − {1} is invertible.

Answer

Commutativity:Let, $\text{a},\text{b}\in\text{R}-\{1\}.$ Then,
a * b = a + b - ab
= b + a - ba
= b * a
Therefore,
a * b = b * a, $\forall\ \text{a},\text{b}\in\text{R}-\{1\}$
Thus, * is commutative on R - {1}.
Associativity:
Let, $\text{a},\text{b}\in\text{R}-\{1\}.$ 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, $\forall\text{ a},\text{b},\text{c}\in\text{R}-\{1\}$
Thus, * is associative on R - {1}.
Finding identity element:
Let e be the element in R - {1} with respect to * such that
a * e = a = e * a, $\forall\text{ a}\in\text{R}-\{1\}$
a * e = a and e * a = a, $\forall\text{ a}\in\text{R}-\{1\}$
⇒ a + e - ae = a and e + a - ea = a, $\forall\text{ a}\in\text{R}-\{1\}$
e(1 - a) = 0, $\forall\text{ a}\in\text{R}-\{1\}$
$\text{e}=0\in\forall\text{ a}\in\text{R}-\{1\},\forall\text{ a}\in\text{R}-\{1\}$ $[\because\ \text{a}\neq1]$
Thus, 0 is the identity element in R - {1} with respect to *.
Finding inverse:
Let $\text{a}\in\text{R}-\{1\}$ and $\text{b}\in\text{R}-\{1\}$ be the inverse of a. Then,
a * b = e = b * a
a * b = e and b * a = e
⇒ a + b - ab = 0 and b + a - ba = 0
⇒ a = ab - b
⇒ a = b(a - 1)
$\Rightarrow\text{b}=\frac{\text{a}}{\text{a}-1}$
Thus, $\frac{\text{a}}{\text{a}-1}$ is inverse of $\text{a}\in\text{R}-\{1\}.$

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Question 525 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 $f^{-1}(163).$

Answer

We 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, $\text{f}^{-1}(163)=\frac{\sqrt{163+6}-1}{3}=\frac{\sqrt{169}-1}{3}$
$=\frac{13-1}{3}=\frac{12}{3}=4$

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Question 535 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.

Answer

Proving 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)}

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Question 545 Marks

Let A = {1, 2, 3, ... 9} and R be the relation in A × A defined by (a, b)R(c, d) if a + d = b + c for (a, b), (c, d) in A × A. Prove that R is an equivalence relation and also obtain the equivalent class [(2, 5)].

Answer

Given that, A = {1, 2, 3, ... 9} and (a, b)R(c, d) if a + d = b + c for (a, b) ∈ A × A and (c, d) ∈ A × A.Let (a, b)R(a, b)
⇒ a + b = b + a, ∀ a, b ∈ A which is true for any a, b ∈ A.
Since, the sum of two numbers doesn’t depend on the order.
Hence, R is reflexive.
Let (a, b)R(c, d)
Now, a + d = b + c
Also, c + b = d + a
⇒ (c, d)R(a, b)
Hence, R is symmetric.
Let (a, b)R(c, d) and (c, d)R(e, f)
$\therefore$ a + d = b + c and c + f = d + e
⇒ a + d = b + c and d + e = c + f
⇒ (a + d) - (d + e) = (b + c) - (c + f)
⇒ a - e = b - f
⇒ a + f = b + e
⇒ (a, b)R(e, f)
Hence, R is transitive.
Since, R is symmetric, reflexive and transitive.
Hence, R is an equivalence relation.
The equivalence class containing [(2, 5)] is given by {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}.

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Question 555 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.

Answer

A = 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.

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Question 565 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.

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Question 575 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.

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Question 585 Marks

Let $f: W \rightarrow W$ 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.

Answer

It is given that: $f: W \rightarrow W$ 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 $n - 1 = m + 1.$
$\Rightarrow n - m = 2$
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: $f$
$(n) = f(m) \Rightarrow n - 1 = m - 1$
$\Rightarrow n = m$
Again, if both n and m are even, then we have:
$f(n) = f(m)$
$\Rightarrow n + 1 = m + 1 $
$\Rightarrow n = m$
$\therefore$ 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 \rightarrow 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: $fog(m) = f(g(m)) = f(m + 1) = m + 1 - 1 = m$
$\therefore gof = I_W and fog = I_W$
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.

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Question 595 Marks

Show that the relation R defined by R = {(a, b): a - b is divisible by 3; a, b ∈ Z} is an equivalence relation.

Answer

We 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.

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Question 605 Marks

Consider $f: R_+\rightarrow [– 5, \infty )$ given by $f(x) = 9x^2 + 6x – 5.$ Show that f is invertible with $f^{-1}(\text{y})=\left(\frac{\Big(\sqrt{\text{y}+6}\Big)-1}{3}\right).$

Answer

Consider $f:\text{R}_{+}\rightarrow[-5,\infty]$ and $f(x) = 9x^2 + 6x - 5$. 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, $f(x_1) = f(x_2)$ then $9\text{x}_{1}^{2}+6\text{x}_1-5 =9\text{x}_{2}^{2}+6\text{x}_2-5$
$\Rightarrow9\text{x}_{1}^{2}+6\text{x}_1 =9\text{x}_{2}^{2}+6\text{x}_2\ \ \Rightarrow\ 9(\text{x}_{1}^{2}-\text{x}_{2}^{2})+6(\text{x}_1-\text{x}_2)=0$
$\Rightarrow (x_1 - x_2)[9(x_1 + x_2) +6] = 0$
$\Rightarrow x_1 - x_2 = 0$
$\Rightarrow x_1 = x_2 $
$\therefore$ f is one-one. Now, again $y = 9x^2 + 6x - 5$
$\Rightarrow 9x^2 + 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 $f^{-1}(\text{x})=\frac{\sqrt{\text{y}+6}-1}{3}.$

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Question 615 Marks

Let C be the set of all complex numbers and $C_0$ be the set of all no-zero complex numbers. Let a relation R on $C_0$ be defined as $\text{z}_1\text{R z}_2\Leftrightarrow\frac{\text{z}_1-\text{z}_2}{\text{z}_1+\text{z}_2}$ is real for all $\text{z}_1,\ \text{z}_2\in\text{C}_0.$ Show that R is an equivalence relation.

Answer

Test for reflexivity: Since, $\frac{\text{z}_1-\text{z}_1}{\text{z}_1+\text{z}_1}=0,$ which is a real number.

So, $(\text{z}_1,\text{ z}_1)\in\text{R}$
Hence, R is relexive relation.

Test for symmetric: Let $(\text{z}_1,\text{ z}_2)\in\text{R.}$

Then, $\frac{\text{z}_1-\text{z}_2}{\text{z}_1+\text{z}_2}=\text{x,}$ where x is real
$\Rightarrow\ -\Big(\frac{\text{z}_1-\text{ z}_2}{\text{z}_1+\text{ z}_2}\Big)=-\text{x}$
$\Rightarrow\ \Big(\frac{\text{z}_2-\text{ z}_1}{\text{z}_2+\text{ z}_1}\Big)=-\text{x},$ is also a real number
So, $(\text{z}_2,\text{ z}_1)\in\text{R}$
Hence, R is symmetric relation.

Test for transivity: Let $(\text{z}_1,\text{ z}_2)\in\text{R}$ and $(\text{z}_2,\text{ z}_3)\in\text{R}.$

Then,
$\frac{\text{z}_1-\text{z}_2}{\text{z}_1+\text{z}_2}=\text{x},$ where x is a real number.
$\Rightarrow\ \text{z}_1-\text{z}_2=\text{xz}_1+\text{xz}_2$
$\Rightarrow\ \text{z}_1-\text{xz}_1=\text{z}_2+\text{xz}_2$
$\Rightarrow\ \text{z}_1(1-\text{x})=\text{z}_2(1+\text{x})$
$\Rightarrow\ \frac{\text{z}_1}{\text{z}_2}=\frac{(1+\text{x})}{(1-\text{x})}\ .....(1)$
Also,
$\frac{\text{z}_2-\text{z}_3}{\text{z}_2+\text{z}_3}=\text{y},$ where y is a real number.
$\Rightarrow\ \text{z}_2-\text{z}_3=\text{yz}_2+\text{yz}_3$
$\Rightarrow\ \text{z}_2-\text{yz}_2=\text{z}_3+\text{yz}_3$
$\Rightarrow\ \text{z}_2(1-\text{y})=\text{z}_3(1+\text{y})$
$\Rightarrow\ \frac{\text{z}_2}{\text{z}_3}=\frac{(1+\text{y})}{(1-\text{y})}\ .....(2)$
Dividing (1) and (2), we get
$\frac{\text{z}_1}{\text{z}_3}=\Big(\frac{1+\text{x}}{1-\text{x}}\Big)\times\Big(\frac{1-\text{y}}{1+\text{y}}\Big)=\text{z,}$ where z is a real number.
$\Rightarrow\ \frac{\text{z}_1-\text{z}_3}{\text{z}_1+\text{z}_3}=\frac{\text{z}-1}{\text{z}+1},$ which is real
$\Rightarrow\ (\text{z}_1,\text{ z}_3)\in\text{R}$
Hence, R is transitive relation.
From (i), (ii) and (iii), R is an equivalenve relation.

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Question 625 Marks

Let $A = \{– 1, 0, 1, 2\},$
$B = \{– 4, – 2, 0, 2\}$ and $f, g: A \rightarrow B$ 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 $f: A \rightarrow B$ and $g: A \rightarrow B$ such that $f(\text{a}) = \text{g(a)}\ \forall \text{a} \in \text{A},$ are called equal functions).

Answer

It is given that $A =\{-1, 0, 1, 2\}, B =\{-4, -2, 0, 2\}.$
Also, it is given that $f, g: A \rightarrow 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:
$f(-1) = (-1)^2 - (-1) = 1 + 1 = 2$
$\text{g}(-1)=2\Big|(-1)-\frac{1}{2}\Big|-1=2\Big(\frac{3}{2}\Big)-1=3-1=2$
$\Rightarrow f(-1) = g(-1)$
$f(0) = (0)^2 - 0 = 0$
$\text{g}(0)=2\Big|0-\frac{1}{2}\Big|-1=2\Big(\frac{1}{2}\Big)-1=1-1=0$
$\Rightarrow f(0) = g(0)$
$f(1) = (1)^2 - 1 = 1 - 1 = 0$
$\text{g}(1)=2\Big|1-\frac{1}{2}\Big|-1=2\Big(\frac{1}{2}\Big)-1=1-1=0$
$\Rightarrow f(1) = g(1)$
$f(2) = (2)^2 - 2 = 4 - 2 = 2$
$\text{g}(2)=2\Big|2-\frac{1}{2}\Big|-1=2\Big(\frac{3}{2}\Big)-1=3-1=2$
$\therefore f(\text{a})=\text{g(a)}\ \forall\text{a}\in\text{A}$
Hence, the functions f and g are equal.

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Question 635 Marks

Show that the exponential function $f: R \rightarrow R$, given by $f(x)=e^x$, is one-one but not onto. What happens if the codomain is replaced by $\mathrm{R}+0 \mathrm{R} 0+$ (set of all positive real numbers)?

Answer

Then the co-domain and range become the same and in that case,
f is onto and hence, it is a bijection.

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Question 645 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 $f^{-1}.$

Answer

Injectivity 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)]

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Question 655 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
$(gof)^{-1} = f^{-1}og^{-1}$

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Question 665 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 $f(x) = 5x^2 + 6x - 9F$ or 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 $\text{y}\in[-9,\infty)$
$\Rightarrow 5f^{-1}(y) + 3 =\sqrt{5\text{y}+54} for all \text{y}\in[-9,\infty)$
$\Rightarrow\ \text{f}^{-1}(\text{y})=\frac{\sqrt{5\text{y}+54}-3}{5}$

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