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}
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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.
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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 $\mathrm{f}: \mathrm{A} \rightarrow 13$ such that $\mathrm{y} \in 3$. Then, $\mathrm{f}^{-1}(\mathrm{y})=\{\mathrm{x} \in \mathrm{A}: \mathrm{f}(\mathrm{x})=\mathrm{y}\}$ In other words, $\mathrm{f}^{-1}(\mathrm{y})$ is the set of pre-images of $y$. Let $f^{-1}\{17\}=x$.
Then, $f(x\}=17 \Rightarrow x^2+1=17 \Rightarrow x^2=17-1=16 \Rightarrow x= \pm 4$ Let $f^{-1}\{-3\}=x$. Then, $f(x)$
$=-3 \Rightarrow x^2+1=-3 \Rightarrow x^2=-3-1=-4 \Rightarrow x=\sqrt{-4} \therefore f^{-1}\{-3\}=\theta$
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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}
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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 $(\text{f})=\text{R}-\Big\{\frac{\text{a}}{\text{b}}\Big\}$
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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 $(f)=R-\{2,6\}$
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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 $(\text{f})=\text{R}-\Big\{\frac{\text{a}}{\text{c}}\Big\}$
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Let $A=\{-2,-1,0,1,2\}$ and $f: A \rightarrow Z$ be a function defined by $f(x)=x^2-2 x-3$. Find: Range of $f$ i.e., $f(A)$
AnswerWe have, $f(x)=x^2-2 x-3$ Now, $f(-2)=(2)^2-2(-2)-3=4+4-3=5 f(-1)=(-1)^2-2(-1)-3=1+2-3=0 f(-0)=$ $(-0)^2-2 \times 0-3=-3 f(1)=(1)^2-2 \times 1-3=1-2-3=-4 f(2)=(2)^2-2 \times 2-3=4-4-3=-3$ Rang $(f)=\{-4,-3,0,5\}$
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A function f : R → R is defined by $f(x) = x^2$. Determine:
- Range of f
- {x : f(x) = 4}
- {y : f(y) = -1}
AnswerWe have, $f(x)=x^2 \ldots$ (i)
i. Clearly range of $f=R^{+}$(Set of all real numbers greater than or equal to zero)
ii. We have,
$\{x: f(x)=4\}$
$\Rightarrow f(x)=4 \ldots(4) \ldots(i i)$
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]
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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.
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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 $Y: f_1=$ $\{(1,1),(2,11),(3,1),(4,15)\}$
AnswerWe have, $f_1=\{(1,1),(2,11),(3,1),(4,15)\} f_1$ is a function from $X$ to $Y$.
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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 $(\text{f})=\Big[\frac{1}{4},\infty\Big]$
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If $f(x)=(x-a)^2(x-b)^2$, find $f(a+b)$
AnswerWe have, $f(x)=(x-a)^2(x-b)^2$ Now, $f(a+b)=(a+b-a)^2(a+b-b)^2=b^2 a^2$ Hence, $f(a+b)=a^2 b^2$
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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 $(\text{f})=[0,\infty)$
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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 $\text{h(x)}=\{\text{x}\in\text{R}:\text{x}\geq1\}$
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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.
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