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 $f : R \rightarrow R$ is defined by $f(x) = x^2$, find $f^{-1}(-25)$.
Answer$f : R \rightarrow R$ 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$f_1=\{(1, a),(2, b),(3, c),(4, d)\} \Rightarrow f_1^{-1}=\{(a, 1),(b, 2),(c, 3),(d, 4)\} f_2=\{(1, b),(2, a),(3, c),(4, d)\}$
$\Rightarrow f_2^{-1}=\{(b, 1),(a, 2),(c, 3),(d, 4)\}$
$f_3=\{(1, a),(2, b),(4, c),(3, d)\} \Rightarrow f_3^{-1}=\{(a, 1),(b, 2),(c, 4),(d, 3)\}$
$f_4=\{(1, b),(2, a),(4, c),(3, d)\} \Rightarrow f_3^{-1}=\{(b, 1),(a, 2),(c, 4),(d, 3)\}$
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 $\neq f^2$
Answer$f(x) \text { and } g(x) \text { are polynomials. }$
$\Rightarrow f: R \rightarrow R \text { and } g: R \rightarrow R \text {. }$
$\text { So, fog: } R \rightarrow R \text { and gof: } R \rightarrow R \text {. }$
$(f \circ g)(x)=f(g(x))$
$=f\left(x^2+1\right)$
$=2\left(x^2+1\right)+5$
$=2 x^2+2+5$
$=2 x^2+7$
View full question & answer→Question 62 Marks
Let $f: R \rightarrow R, g: R \rightarrow R$ 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 $f : R \rightarrow R, g : R \rightarrow R$ 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 $\neq f^2$.
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:
$f : N \rightarrow N$ given by $f(x) = x^2$
Answer$f : N \rightarrow N$ 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 \rightarrow R$ is defined by $f(x) = x^2$, write $f^{-1}(25).$
AnswerLet $f^{-1}(25) = x ....(1)$
$\Rightarrow f(x) = 25$
$\Rightarrow x^2 = 25$
$\Rightarrow x^2 - 25 = 0$
$\Rightarrow (x - 5)(x + 5) = 0$
$\Rightarrow\ \text{x}=\pm5$
$\Rightarrow f^{-1}(25) = {-5, 5}$ [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)=3 x^3+1$.
Answer$(f \circ g)(2)=f(g(2))=f\left(3 \times 2^3+1\right)=f(25)=25^2+8=633$
$(g \circ f)(1)=g(f(1))=g\left(1^2+8\right)=g(9)=3 \times 9^3+1=2188$
View full question & answer→Question 172 Marks
If $f: R \rightarrow R$ is defined by $f(x)=10 x-7$, then write $f^{-1}(x)$.
Answer$\text { Let } f^{-1}(x)=y$
$\Rightarrow f(y)=x$
$\Rightarrow 10 y-7=x$
$\Rightarrow 10 y=x+7$
$\Rightarrow y=\frac{x+7}{10}$
$\Rightarrow f^{-1}(x)=\frac{x+7}{10}[\text { 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)=2 x+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)$
$=(2 x+5)(2 x+5)$
$=(2 x+5)^2$
$=4 x^2+20 x+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 \rightarrow R$ is given by $f(x)=x^3$, write $f^{-1}(1)$.
AnswerLet $f ^{-1}(1)= x$
$\Rightarrow f ( x )=1$
$\Rightarrow x^3=1$
$\Rightarrow x ^3-1=0$
$\Rightarrow(x-1)\left(x^2+x+1\right)=0\left[\right.$ Using the identity: $\left.a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right]$
$\Rightarrow x=1($ as $x \in R )$
$\Rightarrow f ^{-1}(1)=\{1\}[$ from (1)]
View full question & answer→Question 222 Marks
If $f(x)=2 x+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$.
$\text { (fof }(x)=f(f(x))$
$=f(2 x+5)$
$=2(2 x+5)+5$
$=4 x+10+5$
$=4 x+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 \rightarrow R$ and $g: R \rightarrow R$ be defined by $f(x)=x^2$ and $g(x)=x+1$. Show that fog $\neq g$ gof.
AnswerGiven, $f: R \rightarrow R$ and $g: R \rightarrow R$.
So, the domains of $f$ and $g$ are the same.
$(f \circ g)(x)=f(g(x))=f(x+1)=(x+1)^2=x^2+1+2 x$
$(g \circ f)(x)=g(f(x))=g\left(x^2\right)=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)$
$\Rightarrow f(y) = x$
$\Rightarrow\ \frac{2\text{y}}{5\text{y}+3}=3\text{x}$
$\Rightarrow 2y = 5xy + 3x \Rightarrow 2y - 5xy = 3x \Rightarrow 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 \rightarrow\{3\} \rightarrow R-\{2\}$ given by $f(x)=\frac{x-2}{x-3}$ is a bijection.
AnswerWe have, $f: R^{+} \rightarrow R^{+}$given by
$f(x)=x^2$
$g: R^{+} \rightarrow R^{+} \text {given by }$
$g(x)=\sqrt{x}$
$\therefore f(x)=f(g(x))$
$=f(\sqrt{x})=(\sqrt{x})^2=x$
Also, $g \circ f(x)=g(f(x))$
$=g\left(x^2\right)=\sqrt{x^2}=x$
Thus,
$f \circ g(x)=\operatorname{gof}(x)$
View full question & answer→Question 322 Marks
If $f: C \rightarrow C$ is defined by $f(x)=(x-2)^3$, write $f^{-1}(-1)$.
Answer
Let $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 \rightarrow C$ is defined by $f(x)=x^3$. Write $f^{-1}(1)$.
Answer$f : R \rightarrow 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)=2 x+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)=2 x+x^2 \text { and } g(x)=x^3$
$g \circ f(x)=g(f(x))=g\left(2 x+x^2\right)$
$g \circ f(x)=g\left(2 x+x^2\right)^3$
$f \circ g(x)=f(g(x))=f\left(x^3\right)$
$\therefore f \circ g(x)=2 x^3+x^6$
View full question & answer→Question 362 Marks
Let $f: R-\{-1\} \rightarrow R-\{1\}$ be given by $f(x)=\frac{x}{x+1}$. Write $f^{-1}(x)$.
Answer$f: R-[-1] \rightarrow R-[1]$ given by $f(x)=\frac{x}{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 \rightarrow R^{+}$be defined by $f(x)=a x, a>0$ and $a \neq 1$. Write $f^{-1}(x)$.
AnswerLet $f ^{-1}( x )= y$
$\Rightarrow f ( y )= x$
$\Rightarrow a^y= x$
$\Rightarrow y =\log _{ a } x$
$\Rightarrow f ^{-1}( x )=\log _{ a } 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$ $\qquad$
$\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[\right.$ Using identity: $\left.a^2-b^2=(a-b)(a+b)\right]$
$\Rightarrow(x-1)(x+1)(x-i)(x+i)=0$ where $i=\sqrt{-1}$ [Using identity: $\left.a^2-b^2=(a-b)(a+b)\right]$
$\Rightarrow x = \pm 1, \pm 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 \rightarrow R$ be defined by $f(x)=x^4$, write $f^{-1}(1)$.
Answer$\text { Let } f^{-1}(1)=x \ldots \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$
$\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.
AnswerWhen two sets A and B have m and n elements respectively, then the number of onto functions from A to B is,$\begin{cases}\sum_{\text{r}=1}^\text{n}(-1)^\text{r}\text{ nC}_\text{r}\text{r}^\text{m},&\text{if m}\geq\text{n}\\0,&\text{if m}<\text{n}\end{cases}$
Here, number of elements in A = 4 = m Number of elements in B = 2 = n So, m > n Number of onto functions$=\sum_{\text{r}=1}^2(-1)^\text{r}2\text{C}_\text{r}\text{r}^4$
$= (-1)^12\text{C}_11^4 + (-1)^22\text{C}_22^4$
$= -2 + 16$$= 14$
View full question & answer→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,
$A=\{a, b, c, d\} \text { and } f: A \rightarrow A \text { be given by }$
$f=\{(a, b),(b, d),(c, a),(d, c)\}$
(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 $f: R \rightarrow R$ defined by $f(x)=3 x-4$ is invertible, then write $f^{-1}(x)$.
Answer$\text { Let } f^{-1}(x)=y . . . . .(1) \Rightarrow f(y)=x$
$\Rightarrow 3 y-4=x$
$\Rightarrow 3 y=x+4$
$\Rightarrow y=\frac{x+4}{3}$
$\Rightarrow f^{-1}(x)=\frac{x+4}{3} \text { [from (1)] }$
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 $B=\{a, b\}$
$\Rightarrow n ( A )=3$ and $n ( B )=2$
$\therefore$ Number of functions from $A$ to $B=2^3=8$
View full question & answer→Question 492 Marks
If $f(x)=2 x+5$ and $g(x)=x^2+1$ be two real functions, then describe the following functions:
gof
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: $R \rightarrow R$.
$(g \circ f)(x)=g(f(x))$
$=g(2 x+5)$
$=(2 x+5)^2+1$
$=4 x^2+20 x+26$
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$f: C \rightarrow C$ 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)
$\Rightarrow f ( x )=-1$
$\Rightarrow x^3=-1$
$\Rightarrow x ^3+1=0$
$\Rightarrow(x+1)\left(x^2-x+1\right)=0$
[Using the identity: $\left.a^3+b^3=(a+b)\left(a^2-a b+b^2\right)\right]$
$\Rightarrow( x +1)( x +\omega)\left( x +\omega^2\right)=0$, where $\omega=\frac{1 \pm i \sqrt{3}}{2}$
$\Rightarrow x =-1,-\omega,-\omega^2($ as $x \in C )$
$\Rightarrow f ^{-1}(-1)=\left\{-1,-\omega,-\omega^2\right\}[$ from $1$]
View full question & answer→Question 542 Marks
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, $y \in B$ 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}
View full question & answer→Question 582 Marks
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 $\text{f}=\{1, \cos1, \cos2\}$
View full question & answer→Question 612 Marks
Let $f, g: R \rightarrow R$ be defined by $f(x)=2 x+1$ 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 $f(x)=2 x+1$ and $g(x)=x^2-2$ for all $x \in R$, respectively
Now,
$g \circ f(x)=g(f(x))$
$=g(2 x+1)$
$=(2 x+1)^2-2$
$=4 x^2+4 x+1-2$
$=4 x^2+4 x-1$
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