Question 12 Marks
Let A = {12, 13, 14, 15, 16, 17} and f : A → Z be a function given by f(x) = highest prime factor of x. Find range of f.
AnswerWe have,
f(x) = highest prime factor of X.
$\therefore$ 12 = 3 × 4,
13 = 13 × 1,
14 = 7 × 2,
15 = 5 × 3,
16 = 2 × 8,
17 = 17 × 1
$\therefore$ f = {(12, 3), (13, 13), (14, 7), (15, 5), (16, 2), (17, 17)}
$\therefore$ Range(f) = {3, 13, 7, 5,2, 17}
View full question & answer→Question 22 Marks
If $\text{f(x)}=\frac{2\text{x}}{1+\text{x}^2},$ show that $\text{f}(\tan\theta)=\sin2\theta$
AnswerWe have,
$\text{f(x)}=\frac{2\text{x}}{1+\text{x}^2}$
Now,
$\text{f}(\tan\theta)=\frac{2(\tan\theta)}{1+\tan^2\theta}$
$=\sin2\theta$ $\Big[\because\ \sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}\Big]$
$\therefore\text{ f}(\tan\theta)=\sin2\theta$ Hence, proved.
View full question & answer→Question 32 Marks
If $f : R \rightarrow R$ be defined by $f(x) = x^2 + 1,$ then find $f^{-1}\{17\}$ and $f^{-1}\{-3\}.$
AnswerWe know that,
if $f : A \rightarrow 13$
such that $\text{y}\in 3.$ Then,
$\text{f}^{-1}\text{(y)}=\{\text{x}\in \text{A}:\text{f(x)}=\text{y}\}$ In other words, $f^{-1}(y)$ is the set of pre-images of $y.$
Let $f^{-1}\{17\} = x.$ Then, $f(x) = 17$
$\Rightarrow x^2 + 1 = 17$
$\Rightarrow x^2 = 17 - 1 = 16$
$\Rightarrow\text{ x}=\pm4$
Let $f^{-1}{-3} = x.$ Then, $f(x) = -3$
$\Rightarrow x^2 + 1 = -3$
$\Rightarrow x^2 = -3 - 1 = -4$
$\Rightarrow\text{x}=\sqrt{-4}$
$\therefore\ \text{f}^{-1}\{-3\}=\theta$
View full question & answer→Question 42 Marks
Let A = {9, 10, 11, 12, 13} and let f : A → N be defined by f(n) = the highest prime factor of n. Find the range of f.
AnswerWe have,
f(n) = The highest prime factor of n.
Now,
9 = 3 × 3,
10 = 5 × 2,
11 = 11 × 1,
12 = 3 × 4,
13 = 13 × 1,
$\therefore$ f = {(9, 3), (10, 5), (11, 11), (12, 3), (13, 13)}
Clearly, Range(f) = {3, 5, 11, 13}
View full question & answer→Question 52 Marks
Find the domain and range of the following real valued functions:
$\text{f(x)}=\frac{\text{ax}+\text{b}}{\text{bx}-\text{a}}$
AnswerGiven,
$\text{f(x)}=\frac{\text{ax}+\text{b}}{\text{bx}-\text{a}}$
Domain of f: Clearly, f(x) is a rational function of x as $\frac{\text{ax}+\text{b}}{\text{bx}-\text{a}}$ is a rational expression.
Clearly, f(x) assumes real values for all x except for all those values of x for which (bx -a) = 0, i.e., bx = a
$\Rightarrow\ \text{x}=\frac{\text{a}}{\text{b}}$
Hence, domain $(\text{f})=\text{R}-\Big\{\frac{\text{a}}{\text{b}}\Big\}$
Range of f,
Let f(x) = y
$\Rightarrow\ \frac{\text{ax}+\text{b}}{\text{bx}-\text{a}}=\text{y}$
$\Rightarrow\ (\text{ax}+\text{b})=\text{y}(\text{bx}-\text{a})$
$\Rightarrow\ (\text{ax}+\text{b})=(\text{bxy}-\text{ay})$
$\Rightarrow\ \text{b}+\text{ay}=\text{bxy}-\text{ax}$
$\Rightarrow\ \text{b}+\text{ay}=\text{x}(\text{by}-\text{a})$
$\Rightarrow\ \text{x}=\frac{\text{b}+\text{ay}}{\text{by}-\text{a}}$
Clearly, f(x) assumes real values for all x except for all those values of x for which (by - a) = 0, i.e. by = a.
$\Rightarrow\ \text{y}=\frac{\text{a}}{\text{b}}$
Hence, range $(\text{f})=\text{R}-\Big\{\frac{\text{a}}{\text{b}}\Big\}$
View full question & answer→Question 62 Marks
Find the domain of the following real valued functions of real variable:
$\text{f(x)}=\frac{\text{x}^2+2\text{x}+1}{\text{x}^2-8\text{x}+12}$
AnswerGiven,
$\text{f(x)}=\frac{\text{x}^2+2\text{x}+1}{\text{x}^2-8\text{x}+12}$
$=\frac{\text{x}^2+2\text{x}+1}{\text{x}^2-6\text{x}-2\text{x}+12}$
$=\frac{\text{x}^2+2\text{x}+1}{\text{x}(\text{x}-6)-2(\text{x}-6)}$
$=\frac{\text{x}^2+2\text{x}+1}{(\text{x}-6)(\text{x}-2)}$
Domain of f, Clearly, $f (x)$ is a rational function of $x$ as $\frac{\text{x}^2+2\text{x}+1}{\text{x}^2-8\text{x}+12}$ is a rational expression.
Clearly, $f(x)$ assumes real values for all x except for all those values of x for which $x^2 - 8x + 12 = 0,$ i.e. $x = 2, 6.$
Hence, domain $(f) = R - \{2, 6\}$
View full question & answer→Question 72 Marks
Find the domain and range of the following real valued functions:
$\text{f(x)}=\frac{\text{ax}-\text{b}}{\text{cx}-\text{d}}$
AnswerGiven,
$\text{f(x)}=\frac{\text{ax}-\text{b}}{\text{cx}-\text{d}}$
Domain of f: Clearly, f(x) is a rational function of x as $\frac{\text{ax}-\text{b}}{\text{cx}-\text{d}}$ is a rational expression.
Clearly, f(x) assumes real values for all x except for all those values of x for which (cx - d) = 0, i.e., cx = d
$\Rightarrow\ \text{x}=\frac{\text{d}}{\text{c}}$
Hence, domain $(\text{f})=\text{R}-\Big\{\frac{\text{d}}{\text{c}}\Big\}$
Range of f,
Let f(x) = y
$\Rightarrow\ \frac{\text{ax}-\text{b}}{\text{cx}-\text{d}}=\text{y}$
$\Rightarrow\ (\text{ax}-\text{b})=\text{y}(\text{cx}-\text{d})$
$\Rightarrow\ (\text{ax}-\text{b})=(\text{cxy}-\text{dy})$
$\Rightarrow\ \text{dy}-\text{b}=\text{cxy}-\text{ax}$
$\Rightarrow\ \text{dy}-\text{b}=\text{x}(\text{cy}-\text{a})$
$\Rightarrow\ \text{x}=\frac{\text{dy}-\text{b}}{\text{cy}-\text{a}}$
Clearly, f(x) assumes real values for all x except for all those values of x for which (by - a) = 0, i.e. by = a.
$\Rightarrow\ \text{y}=\frac{\text{a}}{\text{c}}$
Hence, range $(\text{f})=\text{R}-\Big\{\frac{\text{a}}{\text{c}}\Big\}$
View full question & answer→Question 82 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:
Range of f i.e., $f(A)$
AnswerWe have,
$f(x) = x^2 - 2x - 3$
Now, $f(-2) = (2)^2 - 2(-2) - 3$
$= 4 + 4 - 3$
$= 5$
$f(-1) = (-1)^2 - 2(-1) - 3$
$= 1 + 2 - 3$
$= 0$
$f(-0) = (-0)^2 - 2 \times 0 - 3$
$= -3$
$f(1) = (1)^2 - 2 \times 1 - 3$
$= 1 - 2 - 3$
$= -4$
$f(2) = (2)^2 - 2 \times 2 - 3$
$= 4 - 4 - 3$
$= -3$
Rang $(f) =\{-4, -3, 0, 5\}$
View full question & answer→Question 92 Marks
A function $f : R \rightarrow R$ is defined by $f(x) = x^2.$ Determine:
- Range of $f$
- $\{x : f(x) = 4\}$
- $\{y : f(y) = -1\}$
AnswerWe have,
$f(x) = x^2 ....(i)$
- Clearly range of $f = R^+ ($Set of all real numbers greater than or equal to zero$)$
- We have,
$\{x : f(x) = 4\}$
$\Rightarrow f(x) = 4 ...(4) ...(ii)$
Using equation $(i)$ and equation $(ii),$ we get
$\text{x}^2=4$
$\Rightarrow\text{ x}=\pm2$
$\therefore\ \{\text{x}:\text{f(x)}=4\}=\{-2,2\}$
- $\{\text{y}:\text{f(y)}=-1\}$
$\Rightarrow\text{f(y)}=-1\ ....(\text{iii})$
Clearly, $\text{x}^2\neq-1$ or $\text{x}^2\geq0$
$\Rightarrow\text{f(y)}\neq-1$
$\therefore\ \{\text{y}:\text{f(y)}=-1\}=\phi$
View full question & answer→Question 102 Marks
Find the domain of the following real valued functions of real variable:
$\text{f(x)}={\sqrt{9-\text{x}^2}}$
AnswerGiven,
$\text{f(x)}={\sqrt{9-\text{x}^2}}$
We observe that f(x) is defined for all satisfying,
$9-\text{x}^2\geq0$
$\Rightarrow\ \text{x}^2-9\leq0$
$\Rightarrow\ (\text{x}+3)(\text{x}-3)\leq0$
$\Rightarrow-3\leq\text{x}\leq3$
$\Rightarrow\ \text{x}\in[-3,3]$
Hence, domain (f) = [-3, 3]
View full question & answer→Question 112 Marks
Define a function as a correspondence between two sets.
AnswerFunction: Let $A$ and $B$ be two non$-$empty sets. Then a function $'f\ '$ from set $A$ to swt $B$ is a rule or method or correspondence which associates element of set $A$ to elements of set $b$ such that,
- All elements of set $A$ are associated to element in set $B.$
- An element of se $A$ is associated to a unique element in set $B.$
In other words, a function $'f\ '$ from a set a to set $B$ associates each element of set $A$ to a unique element of set $b.$ View full question & answer→Question 122 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 $Y:$
$f_1 = \{(1, 1), (2, 11), (3, 1), (4, 15)\}$
AnswerWe have,
$f_1 = \{(1, 1), (2, 11), (3, 1), (4, 15)\}$
$f_1$ is a function from $X$ to $Y.$
View full question & answer→Question 132 Marks
Find the domain and range of the following real valued functions:
$\text{f(x)}=\frac{1}{\sqrt{16-\text{x}^2}}$
AnswerGiven,
$\text{f(x)}=\frac{1}{\sqrt{16-\text{x}^2}}$
$(16-\text{x}^2)>0$
$\Rightarrow16>\text{x}^2$
$\Rightarrow\text{x}\in(-4,4)$
$\frac{1}{\sqrt{16-\text{x}^2}}$ is defined for all real numbers that are greater than -4 and less than 4.
Thus, domain of f(x) is {x : - 4 < x < 4} or (-4, 4).
Range of f,
Let f(x) = y
$\Rightarrow\frac{1}{\sqrt{16-\text{x}^2}}=\text{y}$
$\Rightarrow\frac{1}{\sqrt{16-\text{x}^2}}=\text{y}^2$
$\Rightarrow\frac{1}{\text{y}^2}=16-\text{x}^2$
$\Rightarrow\text{x}^2=16-\frac{1}{\text{y}^2}$
Since, $-4<\text{x}<4$
$\Rightarrow0\leq\text{x}^2<16$
$\Rightarrow0\leq16-\frac{1}{\text{y}^2}<16$
$\Rightarrow-16\leq-\frac{1}{\text{y}^2}<0$
$\Rightarrow16\geq\frac{1}{\text{y}^2}>0$
$\Rightarrow\frac{1}{16}\leq\text{y}^2<\infty$
$\Rightarrow\frac{1}{4}\leq\text{y}<\infty$ $(\because\text{y}\geq0)$
Hence, range $(\text{f})=\Big[\frac{1}{4},\infty\Big]$
View full question & answer→Question 142 Marks
If $f(x) = (x - a)^2(x - b)^2,$ find $f(a + b).$
AnswerWe have,
$f(x) = (x - a)^2(x - b)^2$
Now, $f(a + b) = (a + b - a)^2(a + b - b)^2$
$= b^2a^2$
Hence, $f(a + b) = a^2b^2$
View full question & answer→Question 152 Marks
Find the domain and range of the following real valued functions:
$\text{f(x)}=\sqrt{\text{x}-3}$
AnswerGiven, $\text{f(x)}=\sqrt{\text{x}-3}$
Domain(f): Clearly, f(x) assumes real values if $\text{x}-3\geq0$
$\Rightarrow\ \text{x}\geq3$
$\Rightarrow\text{x}\in[3,\infty)$
Hence, domain $(\text{f})=[3,\infty)$
Range of f : For $\text{x}\geq3,$ we have
$\text{x}-3\geq0$
$\Rightarrow\ \sqrt{\text{x}-3}\geq0$
$\Rightarrow\ \text{f(x)}\geq0$
Thus, f(x) takes all real values greater than zero.
Hence, range $(\text{f})=[0,\infty)$
View full question & answer→Question 162 Marks
If $f, g, h$ are three function defined from $R$ to $R$ as follows:
$h(x) = x^2 + 1$
AnswerWe have,
$h(x) = x^2 + 1$
Range of $\text{h(x)}=\{\text{x}\in\text{R}:\text{x}\geq1\}$
View full question & answer→Question 172 Marks
Write the following relations as sets of ordered pairs and find which of them are functions:
$\{(\text{x},\text{y}):\text{y}=3\text{x},\text{x}\in\{1,2,3\},\text{y}\in\{3,6,9,12\}\}$
AnswerWe have,
$\{(\text{x},\text{y}):\text{y}=3\text{x},\text{x}\in\{1,2,3\},\text{y}\in\{3,6,9,12\}\}$
Putting x = 1, 2, 3 in y = 3x, we get
y = 3, 6, 9 respectively
$\therefore$ R = {(1, 3), (2, 6), (3, 9)}
Yes, it is a function.
View full question & answer→