3 Marks Question

Question 13 Marks

If $\text{f(x)}=\text{y}=\frac{\text{ax}-\text{b}}{\text{cx}-\text{a}}$ then prove that f(y) = x.

Answer

We have, $\text{f(x)}=\text{y}=\frac{\text{ax}-\text{b}}{\text{cx}-\text{a}}$
$\therefore\text{f(y)}=\frac{\text{ay}-\text{b}}{\text{cy}-\text{a}}=\frac{\text{a}\Big(\frac{\text{ax}-\text{b}}{\text{cx}-\text{a}}\Big)-\text{b}}{\text{c}\Big(\frac{\text{ax}-\text{b}}{\text{cx}-\text{a}}\Big)-\text{a}}$
$=\frac{\text{a}(\text{ax}-\text{b})-\text{b}(\text{cx}-\text{a})}{\text{c}(\text{ax}-\text{b})-\text{a}(\text{cx}-\text{a})}$
$=\frac{\text{a}^2\text{x}-\text{ab}-\text{b}\text{cx}+\text{ab}}{\text{a}\text{cx}-\text{bc}-\text{a}\text{cx}+\text{a}^2}$
$=\frac{\text{a}^2\text{x}-\text{b}\text{cx}}{\text{a}^2-\text{bc}}$
$=\frac{\text{x}(\text{a}^2-\text{b}\text{c})}{(\text{a}^2-\text{bc})}$
$\therefore\text{f(y)}=\text{x}$

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Question 23 Marks

Find the values of $x$ for which the functions:
$f(x) = 3x^2 – 1$ and $g(x) = 3 + x$ are equal.

Answer

Given that$: (x) = 3x^2 – 1$ and $g(x) = 3 + x$
Since $f(x) = g(x) ($given$)$
$\Rightarrow 3x^2 - 1 = 3 + x$
$\Rightarrow 3x^2 - x -4 = 0$
$\Rightarrow 3x^2 - 4x + 3x - 4 = 0$
$\Rightarrow x(3x - 4) + 1(3x - 4) = 0$
$\Rightarrow (3x - 4)(X + 1) = 3x - 4 = 0$ or $x + 1 = 0$
$\Rightarrow 3x = 4$ or $x = -1$
$\therefore\text{x}=\frac{4}{3}$
Hence$,$ the value of $x$ are $-1$ and $\frac{4}{3}.$

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Question 33 Marks

Find the domain and Range of the function $\text{f(x)}=\frac{1}{\sqrt{\text{x}-5}}.$

Answer

Given that: $\text{f(x)}=\frac{1}{\sqrt{\text{x}-5}}.$
Here, it is clear that f(x) is real when $\text{x} - 5 > 0 \Rightarrow\text{x} > 5$
Hence, the domain $=(5, \infty)$
Now to find the range put
$\text{f(x)}=\text{y}=\frac{1}{\sqrt{\text{x}-5}}$
$\Rightarrow\sqrt{\text{x}-5}=\frac{1}{\text{y}}$
$\Rightarrow\text{x}-5=\frac{1}{\text{y}^2}$
$\Rightarrow\text{x}=\frac{1}{\text{y}^2}+5$
For $\text{x}\in(5, \infty), \text{y}\in\text{R}^+.$
Hence, the range of $\text{f}=\text{R}^+.$

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Question 43 Marks

Express the following functions as set of ordered pairs and determine their range. $f : X \rightarrow R, f(x) = x^3 + 1,$ where $X = {–1, 0, 3, 9, 7}$

Answer

We have$, f : X \rightarrow R, f(x) = x^3 + 1.$
Where $x = {–1, 0, 3, 9, 7}$
Now $f(-1) = (-1)^3 + 1 = -1 + 1 + 0$
$f(0) = (0)^3 + 1 = 0 + 1 = 1$
$f(3) = (3)^3 + 1 = 27 + 1 = 28$
$f(9) = (9)^3 + 1 = 729 + 1 = 730$
$f(7) = (7)^3 + 1 = 343 + 1 = 344$
$f = {(-1, 0), (0, 1), (3, 28), (9, 730), (7, 344)}$
Range of $f = {0, 1, 28, 730, 344}$

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Question 53 Marks

If$\text{A}= \big\{\text{x} : \text{x}\in\text{W}, \text{x}\in2\big\},$$\text{B}= \big\{\text{x} : \text{x}\in\text{N},1< \text{x} < 5\big\},$$\text{C}=\big\{3, 5\big\}$ find.

$\text{A}\times(\text{B} \cap \text{C}) $
$\text{A}\times(\text{B} \cup \text{C}) $

Answer

Given that:
$\text{A}= \big\{\text{x} : \text{x}\in\text{W}, \text{x}\in2\big\}\Rightarrow \text{A}=\big\{0, 1\big\}$
$\text{B}= \big\{\text{x} : \text{x}\in\text{N},1< \text{x} < 5\big\}\Rightarrow\text{B}=\big\{2, 3, 4\big\}$
$\text{C}=\big\{3, 5\big\}$
Now $(\text{B} \cap \text{C}) =\big\{3\big\}$and $(\text{B} \cup \text{C}) =\big\{2, 3, 4, 5\big\}$

$\text{A}\times(\text{B} \cap \text{C}) =\big\{0, 1\big\}\times\big\{3\big\}=\big\{(0, 3), (1, 3)\big\}$

$\text{A}\times(\text{B} \cup \text{C}) =\big\{0, 1\big\}\times\big\{2, 3, 4, 5\big\}$

$=\big\{(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)\big\}$

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Question 63 Marks

Find the range of the following function given by:
$\text{f(x)}=\frac{3}{2-\text{x}^2}$

Answer

We have, $\text{f(x)}=\frac{3}{2-\text{x}^2}=\text{y}$ (let)
$\Rightarrow2-\text{x}^2=\frac{3}{\text{y}}$
$\Rightarrow\text{x}^2=2-\frac{3}{\text{y}}$
Since $\text{x}^2\geq0,2-\frac{3}{\text{y}}\geq0$
$\Rightarrow\frac{2\text{y}-3}{\text{y}}\geq0$
$\Rightarrow2\text{y}-3\geq0\text{y}>0\Rightarrow2\text{y}-3\leq0\text{y}<0$
$\Rightarrow\text{y}\geq\frac{3}{2}\text{y}<0$
$\Rightarrow\text{y}\in(-\infty, 0)\cup[\frac{3}{2},\infty)$
$\therefore$ Range of f is $(-\infty,0)\cup[\frac{3}{2}, \infty)$

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