Question 13 Marks
If $\text{f(x)}=\frac{\text{x}-1}{\text{x}+1},$ then show that:
- $\text{f}\Big(\frac{1}{\text{x}}\Big)=-\text{f(x)}$
- $\text{f}\Big(-\frac{1}{\text{x}}\Big)=-\frac{1}{\text{f(x)}}$
Answer
- Given, $\text{f(x)}=\frac{\text{x}-1}{\text{x}+1}$
$\text{f}\Big(\frac{1}{\text{x}}\Big)=\frac{\frac{1}{\text{x}}-1}{\frac{1}{\text{x}}+1}=\frac{\frac{1-\text{x}}{\text{x}}}{\frac{1+\text{x}}{\text{x}}}$
$=\frac{1-\text{x}}{1+\text{x}}=-\text{f(x)}$
- $\text{f(x)}=\frac{\text{x}-1}{\text{x}+1}$
$\text{f}\Big(-\frac{1}{\text{x}}\Big)=\frac{-\frac{1}{\text{x}}-1}{-\frac{1}{\text{x}}+1}=\frac{\frac{-1-\text{x}}{\text{x}}}{\frac{-1+\text{x}}{\text{x}}}$
$=\frac{-1-\text{x}}{-1+\text{x}}=-\frac{1}{\frac{1+\text{x}}{\text{x}-1}}=-\frac{1}{\text{f(x)}}$ View full question & answer→Question 23 Marks
If $\text{f(x)}=\begin{cases}\text{x}^2,&\text{when }\text{ x}<0\\\text{x},&\text{when }\ 0\leq\text{x}<1\\\frac{1}{\text{x}},&\text{when }\text{ x}>0\end{cases}$
Find:
- $\text{f}\Big(\frac{1}{2}\Big)$
- $\text{f}(-2)$
- $\text{f}(1)$
- $\text{f}(\sqrt{3})$
- $\text{f}(\sqrt{-3})$
AnswerWe have,
$\text{f(x)}=\begin{cases}\text{x}^2,&\text{when }\text{ x}<0\\\text{x},&\text{when }\ 0\leq\text{x}<1\\\frac{1}{\text{x}},&\text{when }\text{ x}>0\end{cases}$
- $\text{f}\Big(\frac{1}{2}\Big)=\frac{1}{2}$
- $\text{f}(-2)=(-2)^2=4$
- $\text{f}(1)=\frac{1}{1}=1$
- $\text{f}(\sqrt{3})=\frac{1}{\sqrt{3}}$
- $\text{f}(\sqrt{-3})=$ does not exist because $\sqrt{3}\notin$ domain $(f).$
View full question & answer→Question 33 Marks
If $\text{y}=\text{f(x)}=\frac{\text{ax}-\text{b}}{\text{bx}-\text{a}},$ show that x = f(y).
AnswerWe have,
$\text{y}=\text{f(x)}=\frac{\text{ax}-\text{b}}{\text{bx}-\text{a}}$
$\Rightarrow\ \text{y}=\frac{\text{ax}-\text{b}}{\text{bx}-\text{a}}$
$\Rightarrow\ \text{y}(\text{bx}-\text{a})=\text{ax}-\text{b}$
$\Rightarrow\ \text{xyb}-\text{ay}=\text{ax}-\text{b}$
$\Rightarrow\ \text{xyb}-\text{ax}=\text{ay}-\text{b}$
$\Rightarrow\ \text{x}(\text{by}-\text{a})=\text{ay}-\text{b}$
$\Rightarrow\ \text{x}=\frac{\text{ay}-\text{b}}{\text{by}-\text{a}}$
$\Rightarrow\ \text{x}=\text{f(y)}$
Hence, proved.
View full question & answer→Question 43 Marks
If $\text{f(x)}=\log_\text{e}(1-\text{x})$ and $\text{g(x)}=[\text{x}],$ then determine the following functions:
(fg)(0)
AnswerWe have,
$\text{f(x)}=\log_\text{e}(1-\text{x})$
and $\text{g(x)}=[\text{x}]$
$\text{f(x)}=\log_\text{e}(1-\text{x})$ is defined, if 1 - x > 0
$\Rightarrow1>\text{x}$
$\Rightarrow\text{x}<1$
$\Rightarrow\text{x}\in(-\infty,1)$
$\therefore\text{ Domain(f)}=(-\infty,1)$
$\text{g(x)}=[\text{x}]$ is defined for all $\text{x}\in\text{R}$
$\therefore\ \text{Domain(g)}=\text{R}$
$\therefore\ \text{Domain(f)}\cap\text{R}\text{ Domain(g)}=(-\infty,1)\cap\text{R}$
$=(-\infty,1)$
$\text{fg}(0)=\log_\text{e}(1-0)\times[0]=0$
View full question & answer→Question 53 Marks
Find $\text{f}+\text{g},\text{ f}-\text{g},\text{ cf}(\text{c}\in\text{ R},\text{c}\neq0),\text{ fg},\frac{1}{\text{f}}$ and $\frac{\text{f}}{\text{g}}$ in the following:
If $f(x) = x^3 + 1$ and $g(x) = x + 1$
AnswerWe have,
$f(x) = x^3 + 1$ and $g(x) = x + 1$
Now,
$f+g: R \rightarrow R$ is given by $(f+g)(x)=x^3+x+2$
$f-g: R \rightarrow R$ is given by $(f-g)(x)=x^3+1-(x+1)=x^3-x$.
$c f: R \rightarrow R$ is given by (cf) $(x)=c\left(x^3+1\right)$.
$(f g)(x): R \rightarrow R$ is given by $(f g)(x)=\left(x^3+1\right)(x+1)=x^4+x^3+x+1$
$\frac{1}{\text{f}}:\text{R}-\{-1\}\rightarrow\text{R}$ is given by $\Big(\frac{1}{\text{f}}\Big)(\text{x})=\frac{1}{\text{x}^3+1}$
$\frac{\text{f}}{\text{g}}:\text{R}-\{-1\}\rightarrow\text{R}$ is given by $\Big(\frac{\text{f}}{\text{g}}\Big)\text{(x)}=\frac{(\text{x}+1)(\text{x}^2-\text{x}+1)}{(\text{x}+1)}=\text{x}^2-\text{x}+1$
View full question & answer→MCQ 63 Marks
Let $f(x) = 2x + 5$ and $g(x) = x^2 + x$. Describe: Find the domain in each case.
AnswerCorrect option: D. $\frac{\text{f}}{\text{g}}$
We have,
$f(x) = 2x + 5$ and $g(x) = x^2 + x$
We observe that $f(x) = 2x + 5$ is defined for all $\text{x}\in\text{R}$
So, domain $(f) = R$
- Clearly $g(x) = x^2 + x$ is defined for all $\text{x}\in\text{R}$
So, domain(g)$ = R$
$\therefore\text{ Domain(f)}\cap\text{Domain(g)}=\text{R}$
Clearly, $(f + g) : R \rightarrow R$ is given by
$(f + g)(x) = f(x) + g(x)$
$= 2x + 5 + x^2 + x$
$= x^2 + 3x + 5$
Domain $(f + g) = R$
- We find that $f - g : R \rightarrow R$ is defined as
$(f - g)(x) = f(x) - g(x)$
$= 2x + 5 - (x^2+ x)$
$= 2x + 5 - x^2 - x$
$= -x^2 + x + 5$
Domain $(f - g) = R$
- We find that $fg : R \rightarrow R$ is given by
$(fg)(x) = f(x) \times g(x)$
$= (2x + 5) \times (x^2 + x)$
$= 2x^3 + 2x^2 + 5x^2 + 5x$
$= 2x^3 + 7x^2 + 5x$
Domain $(fg) = R$
- We have,
$g(x) = x^2 + x$
$\therefore$ $f(x) = 0$
$\Rightarrow x^2 + x = 0$
$\Rightarrow x(x + 1) = 0$
$\Rightarrow x = 0$ or $x = -1$
So, domain $\Big(\frac{\text{f}}{\text{g}}\Big)$ = domain $\text{(f)}\cap$ domain $(g) - {x : g(x) = 0}$
$=\text{R}-\{-\phi,0\}$
We find that, $\frac{\text{f}}{\text{g}}:\text{R}-\{-1,0\}\rightarrow\text{R}$ is given by $\Big(\frac{\text{f}}{\text{g}}\Big)\text{(x)}=\frac{\text{f(x)}}{\text{g(x)}}=\frac{2\text{x}+5}{\text{x}^2+\text{x}}$
Domain $\Big(\frac{\text{f}}{\text{g}}\Big)=\text{R}-\{-1,0\}$ View full question & answer→Question 73 Marks
The function $f: X \rightarrow R$ is defined by $f(x)=x^3+1$, where $X=\{-1,0,3,9,7\}$.
Answer$f : X \rightarrow R $given by $f(x) = x^3 + 1$
$f(-1) = (-1)^3 + 1 = -1 + 1 = 0$
$f(0) = (0)^3 + 1 = 0 + 1 = 1$
$f(3) = (3)^3 + 1 = 27 + 1 = 28$
$f(9) = (9)^3 + 1 = 81 + 1 = 82$
$f(7) = (7)^3 + 1 = 343 + 1 = 344$
Set of ordered pairs are {$(-1, 0), (0, 1), (3, 28), (9, 82), (7, 344)}$
View full question & answer→Question 83 Marks
Let $A = \{p, q, r, s\}$ and $B = \{1, 2, 3\}$. Which of the following relations from $A$ to $B$ is not a function?
AnswerWe have,
$A = \{p, q, r, s\}$ and $B = \{1, 2, 3\}$
- Now,
$R_1 = \{(p, 1), (q, 2), (r, 1), (s, 2)\}$
$R_1$ is a function
- Now,
$R_2 = \{(p, 1), (q, 1), (r, 1), (s, 1)\}$
$R_2$ is a function.
- Now,
$R_3 = \{(p, 1), (q, 2), (p, 2), (s, 3)\}$
$R_3$ is not a function because an element $\text{p}\in\text{A}$ is associated to two elements $1$ and $2$ in $B.$
- Now,
$R_4 = \{(p, 2), (q, 3), (r, 2), (s, 2)\}$
$R_4$ is a function. View full question & answer→Question 93 Marks
Find the domain of the following real valued functions of real variable:
$\text{f(x)}=\frac{1}{\text{x}}$
AnswerGiven, $\text{f(x)}=\frac{1}{\text{x}}$
Domain of f,
We observe that f(x) is defined for all x except at x = 0
At x = 0, f(x) takes the intermediate form $\frac{1}{0}$
Hence, domain (f) = R - {0}
View full question & answer→Question 103 Marks
Let $A = {-2, -1, 0, 1, 2}$ and f : $A \rightarrow Z$ be a function defined by $f(x) = x^2 - 2x - 3.$ Find:
Pre-images of $6, −3$ and $5.$
AnswerWe have,
$f(x) = x^2 - 2x - 3$
Now, $f(-2) = (2)^2 - 2(-2) - 3$
$= 4 + 4 - 3$
$= 5$
$f(-1) = (-1)^2 - 2(-1) - 3$
$= 1 + 2 - 3$
$= 0$
$f(-0) = (-0)^2 - 2 \times 0 - 3$
$= -3$
$f(1) = (1)^2 - 2 \times 1 - 3$
$= 1 - 2 - 3$
$= -4$
$f(2) = (2)^2 - 2 \times 2 - 3$
$= 4 - 4 - 3= -3$
Clearly, pre-images of 6,-3 and is $\phi,$ {0, 2}, -2 respectively.
View full question & answer→Question 113 Marks
Write the following relations as sets of ordered pairs and find which of them are functions:
$\big\{(\text{x},\text{y})=\text{x}+\text{y}=3,\text{x},\text{y}\in\{0,1,2,3\}\big\}$
AnswerWe have,
$\big\{(\text{x},\text{y})=\text{x}+\text{y}=3,\text{x},\text{y}\in\{0,1,2,3\}\big\}$
Now, y = 3 - x
Putting x = 0, 1, 2, 3, we get
y = 3, 2, 1, 0 respectively
$\therefore$ R = {(0, 3), (1, 2), (2, 1), (3, 0)}
Yes, this relation is a function.
View full question & answer→Question 123 Marks
Find the domain and range of the following real valued functions:
f(x) = |x - 1|
AnswerThe given real function is f(x) = |x - 1|
It is clear that |x - 1| is defined for all real numbers.
Hence, domain of f = R
Also, for $\text{x}\in\text{R},(\text{x}-1)$ assumes all real numbers.
Thus, the range of f is the set of all non-negative real numbers.
Hence, range of $\text{f}=[0,\infty)$
View full question & answer→Question 133 Marks
If a function f : R → R be defined by:
$\text{f(x)}=\begin{cases}3\text{x}-2,&\text{ x}<0\\1,&\text{x}=0\\4\text{x}+1,&\text{ x}>0\end{cases}$
Find: f(1), f(-1), f(0) and f(2)
AnswerWe have,
$\text{f(x)}=\begin{cases}3\text{x}-2,&\text{ x}<0\\1,&\text{x}=0\\4\text{x}+1,&\text{ x}>0\end{cases}$
Now, f(1) = 4 × 1 + 1 = 5
f(-1) = 3 × (-1) - 2
= -3 - 2 = -5
f(0) = 1
and, f(2) = 4 × 2 + 1 = 9
$\therefore$ f(1) = 5, f(-1) = -5
f(0) = 1, f(2) = 9
View full question & answer→Question 143 Marks
Find the domain and range of the following real valued functions:
f(x) = -|x|
Answer$\text{f(x)}=-|\text{x}|,\text{ x}\in\text{R}$
We know that,
$|\text{x}|=\begin{cases}\text{x},&\text{x}\geq0\\-\text{x},&\text{x}<0\end{cases}$
$\therefore\ \text{f(x)}=-|\text{x}|=\begin{cases}\text{x},&\text{x}\geq0\\-\text{x},&\text{x}<0\end{cases}$
Since f(x) is defined for $\text{x}\in\text{R},$ domain of f = R
It can be observed that the range of f (x) = -|x| is all real numbers except positive real numbers.
$\therefore$ The range of f is $(-\infty,0)$
View full question & answer→Question 153 Marks
The function f is defined by $\text{f(x)}=\begin{cases}\text{x}^2,& 0\leq\text{x}\leq3\\3\text{x},&3\leq\text{x}\leq10\end{cases}$
The relation g is defined by $\text{g(x)}=\begin{cases}\text{x}^2,& 0\leq\text{x}\leq2\\3\text{x},&2\leq\text{x}\leq10\end{cases}$
Show that f is a function and g is not a function.
AnswerWe have,
$\text{f(x)}=\begin{cases}\text{x}^2,& 0\leq\text{x}\leq3\\3\text{x},&3\leq\text{x}\leq10\end{cases}$
and $\text{g(x)}=\begin{cases}\text{x}^2,& 0\leq\text{x}\leq2\\3\text{x},&2\leq\text{x}\leq10\end{cases}$
Now,$ f(3) = (3)^2 = 9$ and $f(3) = 3 × 3 = 9$
and $g(2) = (2)^2 = 4$ and $g(2) = 3 × 2 = 6$
We observe f(x) takes unique value at each point in its domain [0, 10]. However g(x) does not takes unique value at each in its domain [0, 10]
Hence, g(x) is not a function.
View full question & answer→Question 163 Marks
If $\text{f(x)}=(\text{a}-\text{x}^\text{n})^{\frac{1}{\text{n}}},\text{a}>0$ and $\text{n}\in\text{N},$ then prove that f(f(x)) = x for all x.
AnswerGiven,
$\text{f(x)}=(\text{a}-\text{x}^\text{n})^{\frac{1}{\text{n}}},\text{a}>0$
Now,
$\text{f}(\text{f(x)})=\text{f}(\text{a}-\text{x}^\text{n})^{\frac{1}{\text{n}}}$
$=\bigg[\text{a}-\Big\{\big(\text{a}-\text{x}^{\text{n}}\big)^{\frac{1}{\text{n}}}\Big\}^{\text{n}}\bigg]^{\frac{1}{\text{n}}}$
$=\big[\text{a}-\big(\text{a}-\text{x}^{\text{n}}\big)\big]^{\frac{1}{\text{n}}}$
$=\big[\text{a}-\text{a}+\text{x}^\text{n}\big]^{\frac{1}{\text{n}}}$
$=(\text{x}^\text{n})^\frac{1}{\text{n}}$
$=(\text{x})^{\text{n}\times\frac{1}{\text{n}}}$
$=\text{x}$
$\therefore\ \text{f}(\text{f(x)})=\text{x}$ Hence, proved.
View full question & answer→Question 173 Marks
Find the domain and range of the following real valued functions:
$\text{f(x)}=\sqrt{9-\text{x}^2}$
AnswerGiven,
$\text{f(x)}=\sqrt{9-\text{x}^2}$
$\big(9-\text{x}^2\big)\geq\text{x}^2$
$\Rightarrow9\geq\text{x}^2$
$\Rightarrow\text{x}\in[-3,3]$
$\sqrt{9-\text{x}^2}$ is defined for all real numbers that are greater than or equal to -3 and less than or equal to 3.
Thus, domain of f(x) is $\{\text{x}:-3\leq\text{x}\leq3\}$ or [-3, 3]
For any value of x such that $-3\leq\text{x}\leq3,$ the value of f(x) will lie between 0 and 3.
Hence, the range of f(x) is $\{\text{x }0\leq\text{x}\leq3\}$ or [0, 3]
View full question & answer→Question 183 Marks
Let $X = {1, 2, 3, 4}$ and $Y = {1, 5, 9, 11, 15, 16}.$ Determine which of the following sets are functions from $X$ to $Y:$
$f_3 = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}$
AnswerWe have,
$f_3 = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}$
$f_3$ is not a function from $X$ to $Y$. because there is an element $2\in\text{x}$ is associated to two element 9 and 11 in Y.
View full question & answer→Question 193 Marks
Find the domain of the following real valued functions of real variable:
$\text{f(x)}=\frac{1}{\text{x}-7}$
AnswerGiven, $\text{f(x)}=\frac{1}{(\text{x}-7)}$
Domain of f,
Clearly, f(x) is not defined for all (x - 7) = 0 i.e., x = 7
At x = 7, f(x) takes the intermediate form $\frac{1}{0}$
Hence, domain (f) = R - {7}
View full question & answer→Question 203 Marks
If f(x) be defined on [-2, 2] and is given by $\text{f(x)}=\begin{cases}-1,&-2\leq\text{x}\leq0\\\text{x}-1,&0<\text{x}\leq2\end{cases}$ and g(x) = f(|x|) + |f(x)|. Find g(x)
AnswerWe have,
$\text{f(x)}=\begin{cases}-1,&-2\leq\text{x}\leq0\\\text{x}-1,&0<\text{x}\leq2\end{cases}$
Now, $\text{f}\big(|\text{x}|\big)=|\text{x}|-1, $ where $-2\leq\text{x}\leq2$
and $|\text{f(x)}|=\begin{cases}1,&-2\leq\text{x}\leq0\\-(\text{x}-1),&0\leq\text{x}\leq1\$\text{x}-1),&1\leq\text{x}\leq2\end{cases}$
$\therefore\ \text{g(x)}=\text{f}(|\text{x}|)+|\text{f}(\text{x})|$
$|\text{f(x)}|=\begin{cases}-\text{x},&-2\leq\text{x}\leq0\\0,&0<\text{x}<1\\2(\text{x}-1),&1\leq\text{x}\leq2\end{cases}$
View full question & answer→Question 213 Marks
Find the domain and range of the following real valued functions:
$\text{f(x)}=\sqrt{\text{x}^2-16}$
AnswerGiven,
$\text{f(x)}=\sqrt{\text{x}^2-16}$
$(\text{x}^2-16)\geq0$
$\Rightarrow\text{x}^2\geq16$
$\Rightarrow\text{x}\in(-\infty,-4)\cup[4,\infty)$
$\sqrt{\text{x}^2-16}$ is defined for all real numbers that are greater than or equal to 4 and less than or equal to -4.
Thus, domain of f(x) is $\{\text{x}:\text{x}\leq-4\text{ or x}\geq4\}$ or $(-\infty,-4]\cup[4,\infty)$
Range of f,
For $\text{x}\geq4,$ we have,
$\text{x}^2-16\geq0$
$\Rightarrow\sqrt{\text{x}^2-16}\geq0$
$\Rightarrow\text{f(x)}\geq0$
For $\text{x}\geq-4,$ we have
$\text{x}^2-16\geq0$
$\Rightarrow\sqrt{\text{x}^2-16}\geq0$
$\Rightarrow\text{f(x)}\geq0$
Thus, f(x) takes all real values greater than zero.
Hence, range $(\text{f})=[0,\infty)$
View full question & answer→Question 223 Marks
Find the domain and range of the following real valued functions:
$\text{f(x)}=\sqrt{\text{x}-1}$
AnswerGiven, $\text{f(x)}=\sqrt{\text{x}-1}$
Domain (f): Clearly, f(x) assumes real values if $\text{x}-1\geq0$
$\Rightarrow\ \text{x}\geq1$
$\Rightarrow\ \text{x}\in[1,\infty)$
Hence, domain $(\text{f})=[1,\infty)$
Range of f: For $\text{x}\geq1,$ we have,
$\text{x}-1\geq0$
$\Rightarrow\ \sqrt{\text{x}-1}\geq0$
$\Rightarrow\ \text{f(x)}\geq0$
Thus, f(x) takes all real values greater than zero.
Hence, range $(\text{f})=[0,\infty)$
View full question & answer→Question 233 Marks
If f, g, h are three function defined from $R$ to $R$ as follows:
$f(x) = x^2$
AnswerWe have,
$f(x) = x^2$
Range of $f(x) = R^+$ (set of all real numbers greater than or equal to zero)
$=\{\text{x}\in\text{R}|\text{x}\geq0\}$
View full question & answer→Question 243 Marks
Let $f : R^+ \rightarrow R$, where $R^+$ is the set of all positive real numbers, such that $\text{f(x)}=\log_\text{e}\text{x}.$ Determine:
- The image set of the domain of $f$
- ${x : f(x) = -2}$
- Whether $f(xy) = f(x) + f(y)$ holds.
AnswerWe have,
$\text{f}=\text{R}^+\rightarrow\text{R}$
and $\text{f(x)}=\log_\text{e}\text{x}\ ...(\text{i})$
- Now, $\text{f}=\text{R}^+\rightarrow\text{R}$
$\therefore$ The image set of the domain of $f = R$
- Now, $\{\text{x}:\text{f(x)}=-2\}$
$\Rightarrow\text{ f(x)}=-2\ ...(\text{ii})$
Using equation $(i)$ and equation $(ii)$, we get
$\log_\text{e}\text{x}=-2$
$\Rightarrow\ \text{x}=\text{e}^{-2}$ $\big[\because\ \log_\text{a}\text{b}=\text{c}\Rightarrow\text{b}=\text{a}^{\text{c}}\big]$
$\therefore\ \{\text{x}:\text{f(x)}=-2\}=\big\{\text{e}^{-2}\big\}$
- Now, $\text{f(xy)}=\log_\text{e}(\text{xy})$
$=\log_\text{e}\text{x}+\log_\text{e}\text{y}$
$\text{f(x)}+\text{f(y)}$
$\therefore\ \text{f(xy)}=\text{f(x)}+\text{f(y)}$
Yes, $\text{f(xy)}=\text{f(x)}+\text{y}$ View full question & answer→Question 253 Marks
If $f(x) = x^2,$ find $\frac{\text{f}(1.1)-\text{f}(1)}{(1.1)-1}$
Answer$\text{f(x)}=\text{x}^2$
$\text{f}(1.1)=1.21$
$\text{f}(1)=1$
$\frac{\text{f}(1.1)-\text{f}(1)}{(1.1)-1}=\frac{1.21-1}{1.1-1}$
$=\frac{0.21}{0.1}=2.1$
View full question & answer→Question 263 Marks
Find the domain of the following real valued functions of real variable:
$\text{f(x)}=\sqrt{\text{x}-2}$
AnswerGiven,
$\text{f(x)}=\sqrt{\text{x}-2}$
Clearly, f(x) assumes real values if $\text{x}\geq2=0$
$\Rightarrow\ \text{x}\geq2$
$\Rightarrow\text{x}\in\big[2,\infty)$
Hence, domain $(\text{f})=[2,\infty)$
View full question & answer→Question 273 Marks
Find the domain of the following real valued functions of real variable:
$\text{f(x)}={\sqrt{\frac{\text{x}-2}{3-\text{x}}}}$
AnswerGiven,
$\text{f(x)}={\sqrt{\frac{\text{x}-2}{3-\text{x}}}}$
Clearly, f(x) assumes real values if
$\text{x}-2\geq0$ and $3-\text{x}>0$
$\Rightarrow\ \text{x}\leq2$ and $3>\text{x}$
$\Rightarrow\ \text{x}\in[2,3)$
Hence, domain (f) = [2, 3)
View full question & answer→Question 283 Marks
Find the domain of the following real valued functions of real variable:
$\text{f(x)}=\frac{3\text{x}-2}{\text{x}+1}$
AnswerGiven, $\text{f(x)}=\frac{3\text{x}-2}{\text{x}+1}$
Domain of f,
Clearly, f(x) is not defined for all (x + 1) = 0 i.e., x = -1
At x = -1, f(x) takes the intermediate form $\frac{1}{0}$
Hence, domain (f) = R - {-1}
View full question & answer→Question 293 Marks
Find the domain and range of the following real valued functions:
$\text{f(x)}=\frac{\text{x}-2}{2-\text{x}}$
AnswerGiven,
$\text{f(x)}=\frac{\text{x}-2}{2-\text{x}}$
Domain(f):
Clearly, f(x) is defined for all x satisfying, if $2-\text{x}\neq0$
$\Rightarrow\ \text{x}\neq2$
Hence, domain (f ) = R - {2}.
Range of f,
Let f(x) = y
$\Rightarrow\ \frac{\text{x}-2}{2-\text{x}}=\text{y}$
⇒ x - 2 = y(2 - x)
⇒ x - 2 = - y(x - 2)
⇒ y = - 1
Hence, range (f) = {-1}.
View full question & answer→Question 303 Marks
If $\text{f(x)}=\log_\text{e}(1-\text{x})$ and $\text{g(x)}=[\text{x}],$ then determine the following functions:
f + g
AnswerWe have,
$\text{f(x)}=\log_\text{e}(1-\text{x})$
and $\text{g(x)}=[\text{x}]$
$\text{f(x)}=\log_\text{e}(1-\text{x})$ is defined, if 1 - x > 0
$\Rightarrow1>\text{x}$
$\Rightarrow\text{x}<1$
$\Rightarrow\text{x}\in(-\infty,1)$
$\therefore\text{ Domain(f)}=(-\infty,1)$
$\text{g(x)}=[\text{x}]$ is defined for all $\text{x}\in\text{R}$
$\therefore\ \text{Domain(g)}=\text{R}$
$\therefore\ \text{Domain(f)}\cap\text{R}\text{ Domain(g)}=(-\infty,1)\cap\text{R}$
$=(-\infty,1)$
$\text{f}+\text{g}:(-\infty,1)\rightarrow\text{R}$ defined by $(\text{fg)(x)}=\text{f(x)}\times\text{g(x)}$
$=\log_\text{e}(1-\text{x})+[\text{x}]$
View full question & answer→Question 313 Marks
Let $f : R \rightarrow R$ and $g : C \rightarrow C$ be two functions defined as $f(x) = x^2$ and $g(x) = x^2$. Are they equal functions?
AnswerWe have,
$f : R \rightarrow R$ and $g : C \rightarrow C$
$\therefore$ Domain (f) = R and Domain (g) = c
$\therefore\ \text{Domain(f)}\neq\text{g}=\text{c}$
$\therefore$ f(x) and g(x) are not equal functions.
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Find the domain of the following real valued functions of real variable:
$\text{f(x)}=\frac{2\text{x}+1}{\text{x}^2-9}$
AnswerGiven, $\text{f(x)}=\frac{2\text{x}+1}{\text{x}^2-9}$
Domain of f,
Clearly, f(x) is defined for all $\text{x}\in\text{R}$ except for $\text{x}^2-9\neq0,\text{i.e.,}\text{ x}=\pm3$
At x = -3, 3, f(x) takes the intermediate form $\frac{1}{0}$
Hence, domain (f) = R - {-3, 3}
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Write the following relations as sets of ordered pairs and find which of them are functions:
{(x, y) : y > x + 1, x = 1, 2 and y = 2, 4, 6}
AnswerWe have,
{(x, y) : y > x + 1, x = 1, 2 and y = 2, 4, 6}
Putting x = 1, 2 in y > x + 1, we get
y > 2, y > 3 respectively.
$\therefore$ R = {(1, 4), (1, 6), (2, 4), (2, 6)}
It is not a function from A to b because two ordered pairs in R have the same first element.
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If $\text{f(x)}=\text{x}^3-\frac{1}{\text{x}^3},$ show that $\text{f(x)}+\text{f}\Big(\frac{1}{\text{x}}\Big)=0$
AnswerWe have,
$\text{f(x)}=\text{x}^3-\frac{1}{\text{x}^3}\ ....(\text{i})$
Now, $\text{f}\Big(\frac{1}{\text{x}}\Big)=\Big(\frac{1}{\text{x}}\Big)^3-\frac{1}{\big(\frac{1}{\text{x}}\big)^3}$
$=\frac{1}{\text{x}^3}-\frac{1}{\frac{1}{\text{x}^3}}$
$\text{f}\Big(\frac{1}{\text{x}}\Big)=\frac{1}{\text{x}^3}-\text{x}^3\ ....(\text{ii})$
Adding equation (i) and equation (ii), we get
$\text{f(x)}+\text{f}\Big(\frac{1}{\text{x}}\Big)=\Big(\text{x}^3-\frac{1}{\text{x}^3}\Big)+\Big(\frac{1}{\text{x}^3}-\text{x}^3\Big)$
$=\text{x}^3-\frac{1}{\text{x}^3}+\frac{1}{\text{x}^3}-\text{x}^3$
$=0$
$\therefore\ \text{f(x)}+\text{f}\Big(\frac{1}{\text{x}}\Big)=0$ Hence, proved.
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Let f and g be two real functions defined by $\text{f(x)}=\sqrt{\text{x}+1}$ and $\text{g(x)}=\sqrt{9-\text{x}^2}$ Then describe the following functions:
f + g
AnswerWe have,
$\text{f(x)}=\sqrt{\text{x}+1}$ and $\text{g(x)}=\sqrt{9-\text{x}^2}$
We observe that $\text{f(x)}=\sqrt{\text{x}+1}$ is defined for all $\text{x}\geq-1$
So, domain $\text{f}=[-1,\infty]$
Clearly, $\text{g(x)}=\sqrt{9-\text{x}^2}$ is defined for
$9-\text{x}^2\geq0$
$\Rightarrow\text{x}^2-9\leq0$
$\Rightarrow\text{x}^2-3^2\leq0$
$\Rightarrow\text{x }\in[-3,3]$
$\therefore\ \text{domain(g)}=[-3,3]$
Now,
$\text{domain(f)}\cap\text{domain(g)}$
$=[-1,\infty]\cap[-3,3]$
$=[-1,3]$
f + g : [-1, 3] → R is given by (f + g)(x) = f(x) + g(x)
$=\sqrt{\text{x}+1}+\sqrt{9-\text{x}^2}$
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Let $X=\{1,2,3,4\}$ and $Y=\{1,5,9,11,15,16\}$. Determine which of the following sets are functions from $X$ to $Y$ : $f_2=\{(1,1),(2,7),(3,5)\}$
AnswerWe have,
$f_2 = {(1, 1), (2, 7), (3, 5)}$
$f_2$ is not a function from X to Y. because there is an element $4\in\text{x}$ which is not. associated to any element of Y.
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Find the domain of the following real valued functions of real variable:
$\text{f(x)}=\frac{1}{\sqrt{\text{x}^2-1}}$
AnswerGiven,
$\text{f(x)}=\frac{1}{\sqrt{\text{x}^2-1}}$
Clearly, f(x) is defined for $x^2 - 1 > 0$
$(\text{x}+1)(\text{x}-1)>0$ [Since $a^2 - b^2 = (a + b)(a - b)]$
$\text{x}<-1$ and $\text{x}>1$
$\text{x }\in(-\infty,-1)\cup(1,\infty)$
Hence, domain $(\text{f})=(-\infty,-1)\cup(1,\infty)$
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