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 $(f-g)(x)=x^3+1-(x+1)=x^3-x$. cf: $R \rightarrow R$ is given by $(c f)(x)=c\left(x^3+1\right) .(f g)(x): R \rightarrow R$ is given by $(f g)(x)=\left(x^3+1\right)(x+$ 1) $=x^4+x^3+x+1 \frac{1}{f}: R-\{-1\} \rightarrow R$ is given by $\left(\frac{1}{f}\right)(x)=\frac{1}{x^3+1} \frac{f}{g}: R-\{-1\} \rightarrow R$ 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→Question 63 Marks
Let $f(x) = 2x + 5$ and $g(x) = x^2 + x$. Describe:
- $f + g$
- $f - g$
- $fg$
- $\frac{\text{f}}{\text{g}}$
Find the domain in each case. AnswerWe 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 $B=\{1,2,3\}$
a. Now,
$R_1=\{(p, 1),(q, 2),(r, 1),(s, 2)\}$
$R_1$ is a function
b. Now,
$R_2=\{(p, 1),(q, 1),(r, 1),(s, 1)\}$
$R_2$ is a function.
c. Now,
$R_3=\{(p, 1),(q, 2),(p, 2),(s, 3)\}$
$\mathrm{R}_3$ is not a function because an element $\mathbf{p} \in \mathbf{A}$ is associated to two elements 1 and 2 in $B$.
d. 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 → Z be a function defined by $f(x) = x^2 - 2x - 3$. Find: Pre-images of $6, −3$ and $5$.
AnswerWe have, $f(x)=x^2-2 x-3$ 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.
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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 $2 \in x$ 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 : $R^+ → R,$ 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 $(\text{fg)(x)}=\text{f(x)}\times\text{g(x)}$ $=\log_\text{e}(1-\text{x})+[\text{x}]$
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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 $
$\therefore$ Domain $(f) \neq g=c $
$\therefore f(x)$ and $g(x)$ are not equal functions.
View full question & answer→Question 323 Marks
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}
View full question & answer→Question 333 Marks
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.
View full question & answer→Question 343 Marks
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.
View full question & answer→Question 353 Marks
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) $=\sqrt{\text{x}+1}+\sqrt{9-\text{x}^2}$
View full question & answer→Question 363 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 $Y: f_2=$ $\{(1,1),(2,7),(3,5)\}$
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 $4 \in \mathrm{x}$ which is not. associated to any element of Y .
View full question & answer→Question 373 Marks
Find the domain of the following real valued functions of real variable: $\text{f(x)}=\frac{1}{\sqrt{\text{x}^2-1}}$
AnswerGiven, $f(x)=\frac{1}{\sqrt{x^2-1}}$
Clearly, $f(x)$ is defined for $x^2-1>0(x+1)(x-1)>0\left[\right.$
Since $\left.a^2-b^2=(a+b)(a-b)\right]$
$\mathrm{x}<-1$ and $\mathrm{x}>1 \mathrm{x} \in(-\infty,-1) \cup(1, \infty)$
Hence, domain $(f)=(-\infty,-1) \cup(1, \infty)$
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