2 Marks Questions

Question 12 Marks

If f, g, h are three function defined from R to R as follows:
h(x) = x² + 1

Answer

We have,
h(x) = x² + 1
Range of $\text{h(x)}=\{\text{x}\in\text{R}:\text{x}\geq1\}$

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Question 22 Marks

If f(x) = (x - a)²(x - b)², find f(a + b).

Answer

We have,
f(x) = (x - a)²(x - b)²
Now, f(a + b) = (a + b - a)²(a + b - b)²
= b²a²
Hence, f(a + b) = a²b²

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Question 32 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.

Answer

We 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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Question 42 Marks

If $\text{f(x)}=\frac{2\text{x}}{1+\text{x}^2},$ show that $\text{f}(\tan\theta)=\sin2\theta$

Answer

We 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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Question 52 Marks

If f : R → R be defined by f(x) = x² + 1, then find f^-1{17} and f^-1{-3}.

Answer

We know that,
if f : A → 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
⇒ x² + 1 = 17
⇒ x² = 17 - 1 = 16
$\Rightarrow\text{ x}=\pm4$
Let f^-1{-3} = x. Then, f(x) = -3
⇒ x² + 1 = -3
⇒ x² = -3 - 1 = -4
$\Rightarrow\text{x}=\sqrt{-4}$
$\therefore\ \text{f}^{-1}\{-3\}=\theta$

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Question 62 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.

Answer

We 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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Question 72 Marks

Find the domain and range of the following real valued functions:
$\text{f(x)}=\frac{\text{ax}+\text{b}}{\text{bx}-\text{a}}$

Answer

Given,
$\text{f(x)}=\frac{\text{ax}+\text{b}}{\text{bx}-\text{a}}$
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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Question 82 Marks

Find the domain of the following real valued functions of real variable:
$\text{f(x)}=\frac{\text{x}^2+2\text{x}+1}{\text{x}^2-8\text{x}+12}$

Answer

Given,
$\text{f(x)}=\frac{\text{x}^2+2\text{x}+1}{\text{x}^2-8\text{x}+12}$
$=\frac{\text{x}^2+2\text{x}+1}{\text{x}^2-6\text{x}-2\text{x}+12}$
$=\frac{\text{x}^2+2\text{x}+1}{\text{x}(\text{x}-6)-2(\text{x}-6)}$
$=\frac{\text{x}^2+2\text{x}+1}{(\text{x}-6)(\text{x}-2)}$
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 x² - 8x + 12 = 0, i.e. x = 2, 6.
Hence, domain (f) = R - {2, 6}

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Question 92 Marks

Find the domain and range of the following real valued functions:
$\text{f(x)}=\frac{\text{ax}-\text{b}}{\text{cx}-\text{d}}$

Answer

Given,
$\text{f(x)}=\frac{\text{ax}-\text{b}}{\text{cx}-\text{d}}$
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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Question 102 Marks

Let A = {-2, -1, 0, 1, 2} and f : A → Z be a function defined by f(x) = x² - 2x - 3. Find:
Range of f i.e., f(A)

Answer

We have,
f(x) = x² - 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 × 0 - 3
= -3
f(1) = (1)² - 2 × 1 - 3
= 1 - 2 - 3
= -4
f(2) = (2)² - 2 × 2 - 3
= 4 - 4 - 3
= -3
Rang (f) = {-4, -3, 0, 5}

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Question 112 Marks

A function f : R → R is defined by f(x) = x². Determine:

Range of f
{x : f(x) = 4}
{y : f(y) = -1}

Answer

We have,

f(x) = x² ....(i)

Clearly range of f = R⁺ (Set of all real numbers greater than or equal to zero)
We have,

{x : f(x) = 4}

⇒ 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 $\text{x}^2\geq0$

$\Rightarrow\text{f(y)}\neq-1$

$\therefore\ \{\text{y}:\text{f(y)}=-1\}=\phi$

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Question 122 Marks

Find the domain of the following real valued functions of real variable:
$\text{f(x)}={\sqrt{9-\text{x}^2}}$

Answer

Given,
$\text{f(x)}={\sqrt{9-\text{x}^2}}$
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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Question 132 Marks

Define a function as a correspondence between two sets.

Answer

Function: 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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Question 142 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:
f₁ = {(1, 1), (2, 11), (3, 1), (4, 15)}

Answer

We have,
f₁ = {(1, 1), (2, 11), (3, 1), (4, 15)}
f₁is a function from X to Y.

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Question 152 Marks

Find the domain and range of the following real valued functions:
$\text{f(x)}=\frac{1}{\sqrt{16-\text{x}^2}}$

Answer

Given,
$\text{f(x)}=\frac{1}{\sqrt{16-\text{x}^2}}$
$(16-\text{x}^2)>0$
$\Rightarrow16>\text{x}^2$
$\Rightarrow\text{x}\in(-4,4)$
$\frac{1}{\sqrt{16-\text{x}^2}}$ 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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Question 162 Marks

Write the following relations as sets of ordered pairs and find which of them are functions:
$\{(\text{x},\text{y}):\text{y}=3\text{x},\text{x}\in\{1,2,3\},\text{y}\in\{3,6,9,12\}\}$

Answer

We 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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Question 172 Marks

Find the domain and range of the following real valued functions:
$\text{f(x)}=\sqrt{\text{x}-3}$

Answer

Given, $\text{f(x)}=\sqrt{\text{x}-3}$
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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