Question 12 Marks
Let $\text{f(x)=}\begin{cases}1+\text{x,}&0\leq\text{x}\leq2\\3-\text{x,}&2<\text{x}\leq3\end{cases}.$ Find fof.
Answer$\text{f(x)=}\begin{cases}1+\text{x,}&0\leq\text{x}\leq2\\3-\text{x,}&2<\text{x}\leq3\end{cases}$
$\therefore$ Range of $\text{f}=[0,3]\subseteq$ Domain of f.
$\therefore$ fof(x) = f(f(x))
$=\text{f}\begin{cases}1+\text{x,}&0\leq\text{x}\leq2\\3-\text{x,}&2<\text{x}\leq3\end{cases}$
$\text{fof(x)}=\begin{cases}2+\text{x},&0\leq\text{x}\leq1\\2-\text{x},&1<\text{x}\leq2\\4-\text{x},&2<\text{x}\leq3\end{cases}$
View full question & answer→Question 22 Marks
If $f : R → R$ is defined by $f(x) = x^2,$ find $f^{-1}(-25).$
Answer$f : R → R$ defined by $f(x) = x^2 $
$\therefore f^{-1}(x^2) = x$
$\Rightarrow\ \text{f}^{-1}(-25)=\phi$
$[\because\ \sqrt{-25}\notin\text{R}]$
View full question & answer→Question 32 Marks
Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. State whether f is one-one or not.
Answerf = {(1, 4), (2, 5), (3, 6)}
Here, different elements of the domain have different images in the co-domain.
So, f is one-one.
View full question & answer→Question 42 Marks
If $A = \{1, 2, 3, 4\}$ and $B =\{a, b, c, d\}$ define any four bijections from $A$ to $B.$ Also give their inverse functions.
Answer$f_1 = \{(1, a), (2, b), (3, c), (4, d)\} \Rightarrow\ \text{f}_1^{-1}=\{(\text{a},1), (\text{b},2), (\text{c},3),(\text{d},4)\}$
$f_2 = \{(1, b), (2, a), (3, c), (4, d)\} \Rightarrow\ \text{f}_2^{-1}=\{(\text{b},1), (\text{a},2), (\text{c},3),(\text{d},4)\}$
$f_3 = \{(1, a), (2, b), (4, c), (3, d)\} \Rightarrow\ \text{f}_3^{-1}=\{(\text{a},1), (\text{b},2), (\text{c},4),(\text{d},3)\}$
$f_4 = \{(1, b), (2, a), (4, c), (3, d)\} \Rightarrow\ \text{f}_3^{-1}=\{(\text{b},1), (\text{a},2), (\text{c},4),(\text{d},3)\}$
Clearly, all these are bijections because they are one-one and onto.
View full question & answer→Question 52 Marks
If $f(x) = 2x + 5$ and $g(x) = x^2 + 1$ be two real functions, then describe the following functions:
fog
Also, show that fof ≠ $f^2$
Answerf(x) and g(x) are polynomials.
$\Rightarrow f : R → R$ and $g : R → R$.
So, $fog : R → R$ and $gof : R → R$.
$(fog)(x) = f(g(x))$
$= f(x^2 + 1)$
$= 2(x^2 + 1) + 5$
$= 2x^2 + 2 + 5$
$= 2x^2 + 7$
View full question & answer→Question 62 Marks
Let $f : R → R, g : R → R$ be two functions defined by $f(x) = x^2 + x + 1$ and $g(x) = 1 - x^2$. Write fog $(-2)$.
Answer$(f \circ g)(-2)=f(g(-2))$
$=f\left(1-(-2)^2\right)$
$=f(-3)$
$=(-3)^2+(-3)+1$
$=9-3+1$
$=7$
View full question & answer→Question 72 Marks
If f : R → R is defined by f(x) = 3x + 2, find f(f(x)).
Answerf(f(x)) = f(3x + 2)
= 3(3x + 2) + 2
= 9x + 6 + 2
= 9x + 8
View full question & answer→Question 82 Marks
If $f : R → R, g : R → R$ are given by $f(x) = (x + 1)^2$ and $g(x) = x^2 + 1,$ then write the value of $fog(-3).$
Answer$(fog)(-3) = f(g(-3))$
$= f((-3)^2 + 1)$
$= f(10)$
$= (10 + 1)^2$
$= 121$
View full question & answer→Question 92 Marks
Let A and B be two sets, each with a finite number of elements. Assume that there is an injective map from A to B and that there is an injective map from B to A. Prove that there is a bijection from A to B.
AnswerA and B are two non empty sets.
Let f be a function from A to B. It is given that there is injective map from A to B. That means f is one-one function. It is also given that there is injective map from B to A. That means every element of set B has its image in set A.
f is onto function or surjective.
$\therefore$ f is bijective.
If a function is both injective and surjective, then the function is bijective.
View full question & answer→Question 102 Marks
Write the domain of the real function $\text{f(x)}=\sqrt{[\text{x}]-\text{x}}.$
Answer[x] is the greatest integral function.
Therefore, $0\leq\text{x}-[\text{x}]<1$
$\Rightarrow\ \sqrt{\text{x}-[\text{x}]}$ exists for every $\text{x}\in\text{R}$
⇒ Domain = R
View full question & answer→Question 112 Marks
Let f be a real function given by $\text{f(x)}=\sqrt{\text{x}-2}.$ Find the following:
$f^2$
Also, show that fof ≠ $f^2$.
AnswerWe have, $\text{fof}=\sqrt{\sqrt{\text{x}-2}-2}$
$\Rightarrow f^2(x) = f(x) \times f(x)$
$=\sqrt{\text{x}-2}\times\sqrt{\text{x}-2}=\text{x}-2$
So, $\text{fof}\neq\text{f}^2$
View full question & answer→Question 122 Marks
Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective:
{(a, b): a is a person, b is an ancestor of a}
Answerg = {(a, b): a is a person, h is an ancestor of a}
Since,the ordered map (a, b) does not map 'a' - a person to a living person.
Therefore, g is not a function.
View full question & answer→Question 132 Marks
Classify the following functions as injection, surjection or bijection:
$f : N → N$ given by $f(x) = x^2$
Answer$f : N → N$ given by $f(x) = x^2$ Let $x_1 = x_2$ for $\text{x}_1,\text{ x}_2\in\text{N}$$\text{x}_1^2=\text{x}_2^2\Rightarrow\ \text{f}(\text{x}_1)=\text{f}(\text{x}_2)$
$\therefore f$ is one-one.
Surjectivity: Since $f$ takes only square value like $1, 4, 9, 16 .....$
So, non-perfect square values in $N (\infty-\text{domain})$ do not have pre image in domain $N.$
Thus, $f$ is not onto.
View full question & answer→Question 142 Marks
Write the domain of the real function $\text{f(x)}=\sqrt{\text{x}-[\text{x}]}.$
AnswerWe have, $\text{f(x)}=\sqrt{\text{x}-[\text{x}]}$
The domain of will be real only if
$\text{x}-[\text{x}]\geq0$
⇒ Domain of f = x for all $\text{x}\in\text{R}$
$\therefore$ Domain of f = R
$[\because\ \text{f(x)}=\text{x}-[\text{x}]=\text{x}\ \forall\ \text{x}\in\text{R}]$
View full question & answer→Question 152 Marks
If $f : R → R$ is defined by $f(x) = x^2$, write $f^{-1}(25)$.
Answer$\text { Let } f^{-1}(25)=x \ldots .(1)$
$\Rightarrow f(x)=25 $
$\Rightarrow x^2=25 $
$\Rightarrow x^2-25=0 $
$\Rightarrow(x-5)(x+5)=0 $
$\Rightarrow x= \pm 5 $
$\Rightarrow f^{-1}(25)=\{-5,5\}[\text { from (1)] }$
View full question & answer→Question 162 Marks
Find fog $(2)$ and gof $(1)$ when $: f : R \rightarrow R; f(x) = x^2 + 8$ and $g : R \rightarrow R; g(x) = 3x^3 + 1.$
Answer$(fog)(2) = f(g(2)) = f(3 \times 2^3 + 1) = f(25) = 25^2 + 8 = 633$
$(gof)(1) = g(f(1)) = g(1^2 + 8) = g(9) = 3 \times 9^3 + 1 = 2188$
View full question & answer→Question 172 Marks
If $f : R → R$ is defined by $f(x) = 10x - 7,$ then write $f^{-1}(x).$
AnswerLet $f^{-1}(x) = y ......(1)$
$\Rightarrow f(y) = x$
$\Rightarrow 10y - 7 = x$
$\Rightarrow 10y = x + 7$
$\Rightarrow\ \text{y}=\frac{\text{x}+7}{10}$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{\text{x}+7}{10}$ [from (1)]
View full question & answer→Question 182 Marks
Let $\text{f}:\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\rightarrow\ \text{A}$ be defined by f(x)= sinx. If f is a bijection, write set A.
Answer$\because$ f is a bijection,Co-domain of f = range of f
As $-1\leq\sin\text{x}\leq1,$
$-1\leq\text{y}\leq1$
Therefore, A = [-1, 1]
View full question & answer→Question 192 Marks
If $f(x) = 2x + 5$ and $g(x) = x^2 + 1$ be two real functions, then describe the following functions:$f^2$
Also, show that $fof \neq f^2$
Answer$f(x)$ and $g(x)$ are polynomials.
$\Rightarrow f : R \rightarrow R$ and $g : R \rightarrow R.$
So, $fog : R \rightarrow R$ and $gof : R \rightarrow R.$
$f^2(x) = f(x) \times f(x)$
$= (2x + 5)(2x + 5)$
$= (2x + 5)^2$
$= 4x^2 + 20x + 25$
View full question & answer→Question 202 Marks
If f(x) = x + 7 and g(x) = x - 7, x ∈ R, write fog (7).
Answer(fog)(7) = f(g(7))
= f(7 - 7)
= f(0)
= 0 + 7
= 7
View full question & answer→Question 212 Marks
If $f : R → R$ is given by $f(x) = x^3$, write $f^{-1}(1)$.
AnswerLet $f^{-1}(1) = x$ .....(1)
$\Rightarrow f(x) = 1$
$\Rightarrow x^3 = 1$
$\Rightarrow x^3 - 1 = 0$
$\Rightarrow (x - 1)(x^2 + x + 1) = 0$ [Using the identity: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$]
$\Rightarrow x = 1 (\text{as x}\in\text{R})$
$\Rightarrow f^{-1}(1) = {1}$ [from (1)]
View full question & answer→Question 222 Marks
If $f(x) = 2x + 5$ and $g(x) = x^2 + 1$ be two real functions, then describe the following functions:
fof
Also, show that $fof \neq f^2$
Answer$f(x)$ and $g(x)$ are polynomials.
$\Rightarrow f : R \rightarrow R$ and $g : R \rightarrow R.$
So, $fog : R \rightarrow R$ and $gof : R \rightarrow R.$
$(fof)(x) = f(f(x))$
$= f(2x + 5)$
$= 2(2x + 5) + 5$
$= 4x + 10 + 5$
$= 4x + 15$
View full question & answer→Question 232 Marks
Which of the following functions from $A$ to $B$ are one-one and onto$?$
$f_2 = \{(2, a), (3, b), (4, c)\}; A = \{2, 3, 4\}, B = \{a, b, c\}$
Answer$f_2 = \{(2, a), (3, b), (4, c)\} A =\{2, 3, 4\}, B = \{a, b, c\}$ It in not clear that different elements of $A$ have different images in $B.$
$\therefore f_2$ in not one-one.
Again, each element of $B$ is the image of some element of $A.$
$\therefore f_2$ in not on to.
View full question & answer→Question 242 Marks
Which of the following graphs represents a one-one function?
-
-
AnswerIn the graph of (b), different elements on the x-axis have different images on the y-axis. But in (a), the graph cuts the x-axis at 3 points, which means that 3 points on the x-axis have the same image as O and hence, it is not one-one.
View full question & answer→Question 252 Marks
Which one of the following graphs represents a function?
-
-
AnswerFigure (a) represents a function f : R → R
Whereas fig (b) does not represent a function.
View full question & answer→Question 262 Marks
Find the number of all onto functions from the set A = {1, 2, 3, ..., n} to itself.
AnswerWe know that every onto function from A to itself is one-one.
Therefore, the number of one-one functions = number of bijections =n!
View full question & answer→Question 272 Marks
If f : R → R be defined by $\text{f(x)} = (3 - \text{x}^3)^\frac{1}{3},$ then find fof(x).
Answerf : R → R defined by $\text{f(x)}=(3-\text{x}^3)^\frac{1}{3}$
$\therefore$ fof(x) = f(f(x))
$=\text{f}(3-\text{x}^3)^\frac{1}{3}$
$=\bigg\{3-\Big[(3-\text{x}^3)^\frac{1}{3}\Big]^3\bigg\}^\frac{1}{3}$
$=\{3-3+\text{x}^3\}^\frac{1}{3}$
$\therefore$ fof(x) = x
View full question & answer→Question 282 Marks
Let $f : R → R$ and $g : R → R$ be defined by $f(x) = x^2$ and $g(x) = x + 1$. Show that $fog \neq gof.$
AnswerGiven, $f : R \rightarrow R$ and $g : R \rightarrow R.$
So, the domains of f and g are the same.
$(fog)(x) = f(g(x)) = f(x + 1) = (x + 1)^2 = x^2 + 1 + 2x$
$(gof)(x) = g(f(x)) = g(x^2) = x^2 + 1$
So, $fog \neq gof.$
View full question & answer→Question 292 Marks
If f, g : R → R be two functions defined as f(x) = |x| + x and g(x) = |x| – x, $\forall\ \text{x}\in\text{R}.$ Then find fog and gof. Hence find fog(-3), fog(5) and gof(-2).
Answer$\text{fog(x)}=\begin{cases}0,\ \text{x}\geq0\\-4\text{x},\ \text{x}<0\end{cases}$gof(x) = 0, for all x fog(-3) = 12
fog(5) = 0
gof(-2) = 0
View full question & answer→Question 302 Marks
Let $\text{f}:\text{R}-\Big\{-\frac{3}{5}\Big\}\rightarrow\ \text{R}$ be a function defined as $\text{f(x)}=\frac{2\text{x}}{5\text{x}+3}.$ Write $f^{-1}$: Range of $\text{f}\rightarrow\ \text{R}-\Big\{-\frac{3}{5}\Big\}.$
AnswerLet $f^{-1}(x) = y$ ......(1) ⇒ f(y) = x$\Rightarrow\ \frac{2\text{y}}{5\text{y}+3}=3\text{x}$
⇒ 2y = 5xy + 3x ⇒ 2y - 5xy = 3x ⇒ y(2 - 5x) = 3x $\Rightarrow\ \text{y}=\frac{3\text{x}}{2-5\text{x}}$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{3\text{x}}{2-5\text{x}}$ [from 1]
View full question & answer→Question 312 Marks
Show that the function $f : R → {3} → R - {2}$ given by $\text{f(x)}=\frac{\text{x}-2}{\text{x}-3}$ is a bijection.
AnswerWe have, $f : R^+ → R^+$ given by
$f(x) = x^2$
$g : R^+ → R^+$ given by
$\text{g(x)}=\sqrt{\text{x}}$
$\therefore$ fog(x) = f(g(x))
$=\text{f}(\sqrt{\text{x}})=(\sqrt{\text{x}})^2=\text{x}$
Also, gof(x) = g(f(x))
$=\text{g}(\text{x}^2)=\sqrt{\text{x}^2}=\text{x}$
Thus,
fog(x) = gof(x)
View full question & answer→Question 322 Marks
If $f : C → C$ is defined by $f(x) = (x - 2)^3$, write $f^{-1}(-1)$.
AnswerLet $f^{-1}(1) = x ......(1)$
$\Rightarrow f(x) = -1$
$\Rightarrow (x - 2)^3 = -1$
$\Rightarrow\ \text{x}-2=-1\ \text{or }-\omega\text{ or }-\omega^2$
as the roots of $(-1)^\frac{1}{3}$ are $-1,-\omega\text{ and }-\omega^2,$ where $\omega=\frac{1\pm\text{i}\sqrt{3}}{2}$
$\Rightarrow\ \text{x}=-1+2\text{ or }2-\omega\text{ or }2-\omega^2=1,2-\omega,2-\omega$
$\Rightarrow\ \text{f}^{-1}(-1)=1,2-\omega,2-\omega^2$ [from 1]
View full question & answer→Question 332 Marks
If A = {a, b, c} and B = {-2, -1, 0, 1, 2}, write the total number of one-one functions from A to B.
AnswerA = {a, b, c}, B = {-2, -1, 0, 1, 2}
The total number of one-one function is
$^{\text{n}}\text{C}_\text{m}\times\text{m}!$ (where n = number of elements of B = 5, m = number of elements of A = 3.)
$=\ ^5\text{C}_3\times3!$
$=10\times6=60$
View full question & answer→Question 342 Marks
Let C denote the set of all complex numbers. A function $f : C → C$ is defined by $f(x) = x^3$. Write $f^{-1}(1)$.
Answer$f : R → R$ defined by $f(x) = x^3$
$\therefore f^{-1}(x^3) = x$
$\Rightarrow\ \text{f}^{-1}(1)=\{1,\omega,\omega^2\}$ $[\because\ \sqrt[3]{1}=\{1,\omega,\omega^2\}]$
View full question & answer→Question 352 Marks
Find gof and fog when $f : R \rightarrow R$ and $g : R \rightarrow R$ are defined by:
$f(x) = 2x + x^2 $ and $g(x) = x^3$
AnswerGiven: $f : R \rightarrow R$ and $g : R \rightarrow R$
Therefore,$ gof : R \rightarrow R$ and $fog : R \rightarrow R$
$f(x) = 2x + x^2$ and $g(x) = x^3$
$gof(x) = g(f(x)) = g(2x + x^2)$
$gof(x) = g(2x + x^2)^3$
$fog(x) = f(g(x)) = f(x^3)$
$\therefore fog(x) = 2x^3 + x^6$
View full question & answer→Question 362 Marks
Let $f : R - {-1} → R - {1}$ be given by $\text{f(x)}=\frac{\text{x}}{\text{x}+1}.$ Write $f^{-1}(x)$.
Answerf : R - [-1] → R - [1] given by $\text{f(x)}=\frac{\text{x}}{\text{x}+1}$
$\Rightarrow\ \text{f}^{-1}\Big(\frac{\text{x}}{\text{x}+1}\Big)=\text{x}$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{\text{x}}{1-\text{x}}$
$\because$ Let $\frac{\text{x}}{\text{x}+1}=\text{y}$
$\Rightarrow\ \text{x}=\text{xy}+\text{y}$
$\Rightarrow\ \text{x}(1-\text{y})=\text{y}$
$\Rightarrow\ \text{x}=\frac{\text{y}}{1-\text{y}}$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{\text{x}+7}{10}$
View full question & answer→Question 372 Marks
Let $f : R → R^+$ be defined by $f(x) = ax, a > 0$ and $\text{a}\neq1.$ Write $f^{-1}(x)$.
AnswerLet $f^{-1}(x) = y$ .......(1)
$\Rightarrow f(y) = x$
$\Rightarrow a^y = x$
$\Rightarrow\ \text{y}=\log_\text{a}\text{x}$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\log_\text{a}\text{x}$ [from (1)]
View full question & answer→Question 382 Marks
If $f : C \rightarrow C$ is defined by $f(x) = x^4,$ write $f^{-1}(1).$
AnswerLet $f^{-1}(1) = x ......(1)$
$\Rightarrow f(x) = 1$
$\Rightarrow x^4 = 1$
$\Rightarrow x^4 - 1 = 0$
$\Rightarrow (x^2 - 1)(x^2 + 1) = 0 [$Using identity: $a^2 - b^2 = (a - b)(a + b)]$
$\Rightarrow (x - 1)(x + 1)(x - i)(x + i) = 0$ where $\text{i}=\sqrt{-1} [$Using identity$: a^2 - b^2 = (a - b)(a + b)]$
$\Rightarrow\ \text{x}=\pm1,\pm\text{i}$
$\Rightarrow f^{-1}(1) = {-1, 1, i, -i} [$from $(1)]$
View full question & answer→Question 392 Marks
If f : A → A, g : A → A are two bijections, then prove that:
fog is an injection.
AnswerGiven: A → A, g : A → A are two bijections.
Then, fog : A → A
Injectivity of fog: Let x and y be two elements of the domain (A), such that
(fog)(x) = (fog)(y)
⇒ f(g(x)) = f(g(y))
⇒ g(x) = g(y) (As, f is one-one)
⇒ x = y (As, g is one-one)
So, fog is an injection.
View full question & answer→Question 402 Marks
Which of the following functions from A to B are one-one and onto?
$f_1 = \{(1, 3), (2, 5), (3, 7)\}; A = \{1, 2, 3\}, B =\{3, 5, 7\}$
Answer$f_1 = \{(1, 3), (2, 5), (3, 7)\}$
$A = \{1, 2, 3\}, B = \{3, 5, 7\}$
We can earily observe that in $f_1$ every element of $A$ has different image from $B.$
$\therefore f_1$ in not one-one.
Also, each element of $B$ is the image of some element of $A.$
$\therefore f_1$ in not on to.
View full question & answer→Question 412 Marks
If $f : R → $R be defined by $f(x) = x^4$, write $f^{-1}(1).$
Answer$ \text { Let } f^{-1}(1)=x \ldots . . .(1) $
$ \Rightarrow f(x)=1 $
$ \Rightarrow x^4=1 $
$ \Rightarrow x^4-1=0 $
$ \Rightarrow\left(x^2-1\right)\left(x^2+1\right)=0\left[\text { Using identity: } a^2-b^2=(a-b)(a+b)\right] $
$ \Rightarrow(x-1)(x+1)\left(x^2+1\right)=0\left[\text { Using identity: } a^2-b^2=(a-b)(a+b)\right] $
$ \Rightarrow x= \pm 1[\text { as } x \in R] $
$ \Rightarrow f^{-1}(1)=\{-1,1\}[\text { from (1)] }$
View full question & answer→Question 422 Marks
If f : A → B is an injection, such that range of f = {a}, determine the number of elements in A.
AnswerRange of f = {a}
Therefore, the number of images of f = 1
Since, is an injection, there will be exactly one image for each element of f.
Therefore, number of element in A = 1.
View full question & answer→Question 432 Marks
Write the domain of the real function $\text{f(x)}=\frac{1}{\sqrt{|\text{x}|-\text{x}}}.$
AnswerCase-1: When x > 0
|x| = x
$\Rightarrow\ \frac{1}{\sqrt{|\text{x}|-\text{x}}}=\frac{1}{\sqrt{\text{x}-\text{x}}}=\frac{1}{0}=\infty$
Case-2: When x < 0
|x| = -x
$\Rightarrow\ \frac{1}{\sqrt{|\text{x}|-\text{x}}}=\frac{1}{\sqrt{-\text{x}-\text{x}}}=\frac{1}{\sqrt{-2\text{x}}}$ (exists because when x < 0, -2x > 0)
⇒ f(x) is defined when x < 0
So, domain $=(-\infty,0)$
View full question & answer→Question 442 Marks
Write the total number of one-one functions from set A = {1, 2, 3, 4} to set B = {a, b, c}.
AnswerA has 4 elements and B has 3 elements.
Also, one-one function is only possible from A to B if nA ≤ nB.
But, here nA > nB
So, the number of one-one functions from A to B is 0.
View full question & answer→Question 452 Marks
Let A = {1, 2, 3, 4} and B = {a, b} be two sets. Write the total number of onto functions from A to B.
Question 462 Marks
Let $A = \{a, b, c, d\}$ and f$ : A \rightarrow A$ be given by $f = \{(a, b), (b, d), (c, a), (d, c)\}$. Write $f ^{-1}.$
AnswerWe have,
$A = \{a, b, c, d\}$ and $f : A \rightarrow A$ be given by
$f = \{(a, b), (b, d), (c, a), (d, c)\}$
$($Since, the elements of a function when interchanged gives inverse function. Therefore, $f^{-1} = \{(b, a), (d, b), (a, c), (c, d)\})$
View full question & answer→Question 472 Marks
If $f : R → R$ defined by $f(x) = 3x - 4$ is invertible, then write $f^{-1}(x)$.
AnswerLet $f^{-1}(x) = y$ .....(1)
$\Rightarrow f(y) = x$
$\Rightarrow 3y - 4 = x$
$\Rightarrow 3y = x + 4$
$\Rightarrow\ \text{y}=\frac{\text{x}+4}{3}$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{\text{x}+4}{3}$ [from (1)]
View full question & answer→Question 482 Marks
If $A = \{1, 2, 3\}$ and $B = \{a, b\},$ write the total number of functions from $A$ to $B.$
AnswerIf set $A$ has $m$ elements and set $B$ has $n$ elements, then the number of functions from $A$ to $B$ is $n m$.
Given: $A=\{1,2,3\}$ and $B=\{a, b\}$
$\Rightarrow \mathrm{n}(\mathrm{A})=3$ and $\mathrm{n}(\mathrm{B})=2$
$\therefore$ Number of functions from $A$ to $B=2^3=8$
View full question & answer→Question 492 Marks
If $f(x) = 2x + 5$ and $g(x) = x^2 + 1$ be two real functions, then describe the following functions:
gof
Also, show that fof ≠ $f^2$
Answerf(x) and g(x) are polynomials.
$\Rightarrow f : R → R$ and $g : R → R$.
So, $fog : R → R$ and $gof : R → R$.
$(gof)(x) = g(f(x))$
$= g(2x + 5)$
$= (2x + 5)^2 + 1$
$= 4x^2 + 20x + 26$
View full question & answer→Question 502 Marks
If $f: C \rightarrow C$ is defined by $f(x)=x^2$, write $f^{-1}(-4)$. Here, $C$ denotes the set of all complex numbers.
Answer$ f: C \rightarrow C \text { defined by } f(x)=x^2 \Rightarrow f^{-1}\left(x^2\right)=x$
$ \Rightarrow f^{-1}(-4)=f^{-1}\left[(2 i,-2 i)^2\right]=(2 i,-2 i)$
$ \therefore f^{-1}(-4)=(2 i,-2 i)$
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Write whether f : R → R, given by $\text{f(x)}=\text{x}+\sqrt{\text{x}^2},$ is one-one, many-one, onto or into.
Answerf : R → R given by $\text{f(x)}=\text{x}+\sqrt{\text{x}^2}$
$\because\ \text{f(x)=}\begin{cases}2\text{x};&\text{for x}>0\\0;&\text{for x}<0\end{cases}$
$\therefore$ f is many-one function.
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Which one the following relations on A = {1, 2, 3} is a function?
f = {(1, 3), (2, 3), (3, 2)}, g = {(1, 2), (1, 3), (3, 1)}
AnswerAs, each element of the domain set has unique image in the relation f = {(1, 3), (2, 3), (3, 2)}
So, f is a function.
Also, the element 1 of the domain set has two images 2 and 3 of the range set in the relation g = {(1, 2), (1, 3), (3, 1)}
So, g is not a function.
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Let f be a function from C (set of all complex numbers) to itself given by $f(x) = x^3$. Write $f^{-1}(-1)$.
AnswerLet $f^{-1}(-1) = x$ .....(1)
$\Rightarrow f(x) = -1$
$\Rightarrow x^3 = -1$
$\Rightarrow x^3 + 1 = 0$
$\Rightarrow (x + 1)(x^2 - x + 1) = 0$
[Using the identity: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$]
$\Rightarrow\ (\text{x}+1)(\text{x}+\omega)(\text{x}+\omega^2)=0,$ where $\omega=\frac{1\pm\text{i}\sqrt{3}}{2}$
$\Rightarrow\ \text{x}=-1,-\omega,-\omega^2$ $(\text{as x}\in\text{C})$
$\Rightarrow\ \text{f}^{-1}(-1)=\{-1,-\omega,-\omega^2\}$ [from 1]
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If $f(x) = 4 - (x - 7)^3$, then write $f^{-1}(x)$.
AnswerWe have, $f(x) = 4 - (x - 7)^3$ Let $y = 4 - (x - 7)^3$
$\Rightarrow\ (\text{x} - 7)^3 = 4 - \text{y}$
$\Rightarrow\ \text{x}-7=\sqrt[3]{4-\text{y}}$
$\Rightarrow\ \text{x}=7+\sqrt[3]{4-\text{y}}$
$\Rightarrow\ \text{f}^{-1}(\text{y})=7+\sqrt[3]{4-\text{y}}$
$\therefore\ \text{f}^{-1}(\text{x})=7+\sqrt[3]{4-\text{x}}$
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Which of the following functions from A to B are one-one and onto?
$f_3 = \{(a, x), (b, x), (c, z), (d, z)\}; A = \{a, b, c, d,\}, B = \{x, y, z\}$
Answer$ f_3=\{(a, x),(b, x),(c, z),(d, z)\}$
$ A=\{a, b, c, d\}, B=\{x, y, z\}$
Since, $f_3(a)=x=f_3(b)$ and $f_3(c)=z=f_3(d)$
$\therefore f_3$ in not one-one.
Again, $\mathrm{y} \in \mathrm{B}$ in not the image of any of the element of A .
$\therefore f_3$ in not on to.
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Write the domain of the real function f defined by $\text{f(x)}=\sqrt{25-\text{x}^2}.$
AnswerWe have, $\text{f(x)}=\sqrt{25-\text{x}^2}$ The function is defined only when $25-\text{x}^2\geq0$$\text{x}^2-25\leq0$
$(\text{x}+5)(\text{x}-5)\leq0$
$\text{x}\in[-5,5]$
Therefore, the domain of the given function is [-5, 5].
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Let A = {x ∈ R : -4 ≤ x ≤ 4 and x ≠ 0} and f : A → R be defined by $\text{f(x)}=\frac{|\text{x}|}{\text{x}}.$ Write the range of f.
AnswerWe have, $\text{A}=\{\text{x}\in\text{R}:-4\leq\text{x}\leq4\text{ and x }\neq0\}$
f : A → R defined by $\text{f(x)}=\frac{|\text{x}|}{\text{x}}$
Clearly, $\text{f(x)}=\begin{cases}1;&\text{x}>0\\-1;&\text{x}<0\end{cases}$
$\therefore$ Range of f = {-1, 1}
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What is the range of the function $\text{f(x)}=\frac{|\text{x}-1|}{\text{x}-1}?$
Answer$\text{f(x)}=\frac{|\text{x}-1|}{\text{x}-1}=\frac{\pm(\text{x}-1)}{\text{x}-1}=\pm1$Range of f = {-1, 1}
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If f : A → A, g : A → A are two bijections, then prove that:
fog is a surjection.
AnswerGiven: A → A, g : A → A are two bijections. Then, fog : A → A Surjectivity of fog: let z be an element in the co-domain of fog (A).Now, $\text{z}\in\text{A}$ (co-domain of f) and f is a surjection.
So, z = f(y), where $\text{y}\in\text{A}$ (domain of f) .....(1)
Now, $\text{y}\in\text{A}$ (co-domain of g) and g is a surjection.
So, y = g(x), where $\text{x}\in\text{A}$ (domain of g) .....(2)
From (1) and (2),
z = f(y) = f(g(x)) = (fog)(x), where $\text{x}\in\text{A}$ (domain of fog)
So, fog is a surjection.
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Let $\text{f}:\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\rightarrow\ \text{R}$ be a function defined by f(x) = cos[x]. write range (f).
Answer$\text{f}:\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\rightarrow\ \text{R}$ given by f(x) = cos[x]$\because\ \cos\text{x}$ in position in $\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)$
$\therefore$ cos[x] will be $\{1, \cos1, \cos2\}$
$\therefore$ Range of $\text{f}=\{1, \cos1, \cos2\}$
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Let $f, g : R \rightarrow R$ be defined by $f(x) = 2x + 1$ and $g(x) = x^2 - 2$ for all $x \in R$, respectively. Then, find gof.
AnswerWe have,
$f, g : R \rightarrow R$ are defined by $f(x) = 2x + 1 $ and$ g(x) = x^2 - 2$ for all $x \in R$, respectively
Now,
$gof(x) = g(f(x))$
$= g(2x + 1)$
$= (2x + 1)^2 - 2$
$= 4x^2 + 4x + 1 - 2$
$= 4x^2 + 4x - 1$
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