MCQ 11 Mark
Let $\text{A}=\{\text{x}\in\text{R}:-1\leq\text{x}\leq1\}=\text{B}.$ Then, the mapping $f : A \rightarrow B$ given by $f(x) = x|x|$ is:
- A
Injective but not surjective.
- B
Surjective but not injective.
- ✓
- D
AnswerGiven function is $\text{A}=\{\text{x}:-1\leq\text{x}\leq1\}$ and $f : A \rightarrow A$ such that $f(x) = x|x|$
For the mod function we have to check three cases as $x < 0, x = 0, x > 0.$
For example,$ x < 0$
$f(x) = x|x| < 0$
$|x| = -x$
$y = -x^2$
$\text{x}=-\sqrt{-\text{y}}$ which is not possible for $x > 0$
Hence, $f$ is onto.
$\Rightarrow f$ is bijection.
View full question & answer→MCQ 21 Mark
The range of the function $\text{f(x)}=^{7-\text{x}}\text{P}_{\text{x}-3}$ is:
- A
$\{1, 2, 3, 4, 5\}$
- B
$\{1, 2, 3, 4, 5, 6\}$
- C
$\{1, 2, 3, 4\}$
- ✓
$\{1, 2, 3\}$
AnswerCorrect option: D. $\{1, 2, 3\}$
We know that
$7-\text{x}>0;\ \text{x}-3\geq0$ and $7-\text{x}\geq\text{x}-3$
$\Rightarrow\ \text{x}<7;\ \text{x}\geq3$ and $2\text{x}\leq10$
$\Rightarrow\ \text{x}<7;\ \text{x}\geq3$ and $\text{x}\leq5$
Therefore, $x = 3, 4, 5$
Range of $\text{f}=\Big\{^{(7-3)}\text{P}_{(3-3)},\ ^{(7-4)}\text{P}_{(4-3)},\ ^{(5-3)}\text{P}_{(7-5)}\Big\}$
$= \{4P_0, 3P_1, 2P_2\}$
$= \{1, 3, 2\}$
$= \{1, 2, 3\}$
View full question & answer→Question 31 Mark
Let $\text{f(x)}=\frac{1}{1-\text{x}}.$ Then, {fo(fof)}(x):
- x for all $\text{x}\in\text{R}$
- x for all $\text{x}\in\text{R}-\{1\}$
- x for all $\text{x}\in\text{R}-\{0,1\}$
- None of these.
Answer
- x for all $\text{x}\in\text{R}-\{0,1\}$
Solution:
Domain of f: $1-\text{x}\neq0$
$\Rightarrow\ \text{x}\neq1$
Domain of f = R - {1}
Range of f: $\text{y}=\frac{1}{1-\text{x}}$
$\Rightarrow\ 1-\text{x}=\frac{1}{\text{y}}$
$\Rightarrow\ \text{x}=1-\frac{1}{\text{y}}$
$\Rightarrow\ \text{y}\neq0$
Range of f = R - {0}
So, f : R - {1} → R - {0} and f : R - {1} → R - {0}
Range of f is not a subset of the domain of f.
Domain (fof) = {x : $\text{x}\in$ domain of f and $\text{f(x)}\in$ domain of f}
Domain (fof) $=\big\{\text{x}:\text{x}\in\text{R}-\{1\}\text{ and }\frac{1}{1-\text{x}}\in\text{R}-\{1\}\big\}$
Domain (fof) $=\big\{\text{x}:\text{x}\neq1\text{ and }\frac{1}{1-\text{x}}\neq1\big\}$
Domain (fof) $=\{\text{x}:\text{x}\neq1\text{ and }1-\text{x}\neq1\}$
Domain (fof) $=\{\text{x}:\text{x}\neq1\text{ and }\text{x}\neq0\}$
Domain (fof) = R - {0, 1}
(fof)(x) = f(f(x))
$=\text{f}\Big(\frac{1}{1-\text{x}}\Big)=\frac{1}{1-\frac{1}{1-\text{x}}}=\frac{1-\text{x}}{1-\text{x}-1}$
$=\frac{1-\text{x}}{-\text{x}}=\frac{\text{x}-1}{\text{x}}$
For range of fof, $\text{x}\neq0$
Now, fof : R → {0, 1} → R - {0} and f : R - {1} → R - {0}
Range of fof is not a subset of domain of f.
Domain (fo(fof)) $=\{\text{x}:\text{x}\in$ domain of fof and (fof)(x) $\in$ domain of f$\}$
Domain (fo(fof)) $=\Big\{\text{x}:\text{x}\in\text{R}-\{0,1\}\text{ and }\frac{\text{x}-1}{\text{x}}\in\text{R}-\{1\}\Big\}$
Domain (fo(fof)) $=\Big\{\text{x}:\text{x}\neq0,1\text{ and }\frac{\text{x}-1}{\text{x}}\neq1\Big\}$
Domain (fo(fof)) $=\{\text{x}:\text{x}\neq0,1\text{ and }\text{x}-1\neq\text{x}\}$
Domain (fo(fof)) $=\{\text{x}:\text{x}\neq0,1\text{ and }\text{x}\in\text{R}\}$
Domain (fo(fof)) = R - {0, 1}
Domain (fo(fof)) = f((fof)(x))
$=\text{f}\Big(\frac{\text{x}-1}{\text{x}}\Big)$
$=\frac{1}{1-\frac{\text{x}-1}{\text{x}}}$
$=\frac{\text{x}}{\text{x}-\text{x}+1}$
$=\text{x}$
So, (fo(fof))(x) = x, where $\text{x}\neq0,1$ View full question & answer→Question 41 Mark
If $\text{g(f(x))}=|\sin\text{x}|$ and $\text{f(g(x))}=(\sin\sqrt{\text{x}})^2,$ then
- $\text{f(x)}=\sin^2\text{x},\ \text{g(x)}=\sqrt{\text{x}}$
- $\text{f(x)}=\sin\text{x},\ \text{g(x)}=|\text{x}|$
- $\text{f(x)}=\text{x}^2,\ \text{g(x)}=\sin\sqrt{\text{x}}$
- $\text{f and g cannot be determined.}$
Answer
- $\text{f(x)}=\sin^2\text{x},\ \text{g(x)}=\sqrt{\text{x}}$
Solution:
If we solve it by the trial-and-error method, we can see that (a) satisfies the given condition.
From (a):
$\text{f(x)}=\sin^2\text{x}$ and $\text{g(x)}=\sqrt{\text{x}}$
$\Rightarrow\ \text{f(g(x))}=\text{f}(\sqrt{\text{x}})=\sin^2\sqrt{\text{x}}$
$=(\sin\sqrt{\text{x}})^2$ View full question & answer→Question 51 Mark
The function $\text{f}:[0,\infty)\rightarrow\ \text{R}$ given by $\text{f(x)}=\frac{\text{x}}{\text{x}+1}$ is:
- One-one and onto.
- One-one but not onto.
- Onto but not one-one.
- Onto but not one-one.
Answer
- One-one but not onto.
Solution:
Given function is $\text{f(x)}=\frac{\text{x}}{\text{x}+1}$ on $\text{f}:[0,\infty)\rightarrow\ \text{R}$
If f(x) = f(y)
$\Rightarrow\ \frac{\text{x}}{\text{x}+1}=\frac{\text{y}}{\text{y}+1}$
⇒ xy + x = xy + y
⇒ x = y
Hence, f is one-one.
If y = f(x)
$\text{y}=\frac{\text{x}}{\text{x}+1}$
⇒ xy + y = x
⇒ xy - x = -y
x(y - 1) = -y
$\text{x}=\frac{-\text{y}}{\text{y}-1}\neq\text{f(x)}$
It is not onto. View full question & answer→MCQ 61 Mark
Let $f : R \rightarrow R$ be defined as $\text{f(x)}=\begin{cases}2\text{x},&\text{if x}>3\\\text{x}^2,&\text{if }1<\text{x}\leq3\\3\text{x},&\text{if x}\leq1\end{cases}.$ Then, find $f(-1) + f(2) + f(4):$
AnswerWe have,
$\text{f(x)}=\begin{cases}2\text{x},&\text{if x}>3\\\text{x}^2,&\text{if }1<\text{x}\leq3\\3\text{x},&\text{if x}\leq1\end{cases}$
Now,
$f(-1) + f(2) + f(4)$
$= 3(-1) + 2^2 + 2(4)$
$= -3 + 4 + 8$
$= 9$
View full question & answer→MCQ 71 Mark
Let $f : R \rightarrow R$ be given by $f(x) = [x^2] + [x + 1] - 3$ where $[x]$ denotes the greatest integer less than or equal to $x$. Then$, f(x)$ is:
- A
Many$-$one and onto.
- ✓
Many$-$one and into.
- C
One$-$one and into.
- D
One$-$one and onto.
AnswerCorrect option: B. Many$-$one and into.
$f : R \rightarrow R$
$= [x^2] + [x + 1] - 3$
It is many one function because in this case for two different values of $x$ we would get the same value of $f(x)$.
For $\text{x}=1.1,\ 1.2\in\text{R}$
$f(1.1) = (1.1)^2 + [1.1 + 1] - 3$
$= [1.21] + [2.1] - 3$
$= 1 + 2 + 3 = 0$
$f(1.1) = [1.2]^2 + [1.2 + 1] - 3$
$= [1.44] + [2.2] - 3$
$= 1 + 2 - 3$
$= 0$
It is into function because for the given domain we would only get the integral values of $f(x)$.
But $R$ is the co$-$domain of the given function.
That means, Co$-$domain $\neq$ Range
Hence, the given function is into function.
Therefore$, f(x)$ is many one and into.
View full question & answer→Question 81 Mark
The function f : R → R defined by f(x) = (x - 1)(x - 2)(x - 3) is:
- One-one but not onto.
- Onto but not one-one.
- Both one and onto.
- Neither one-one nor onto.
Answer
- Onto but not one-one.
Solution:
Given function is f(x) = (x - 1)(x - 2)(x - 3)
If f(x) = f(y) then
(x - 1)(x - 2)(x - 3) = (y - 1)(y - 2)(y - 3)
⇒ f(1) = f(2) = f(3) = 0
It is not one-one.
y = f(x)
$\text{x}\in\text{R}$ also $\text{y}\in\text{R}$ hence f is onto. View full question & answer→Question 91 Mark
f : R → R is defined by $\text{f(x)}=\frac{\text{e}^{\text{x}^2}-\text{e}^{-\text{x}^2}}{\text{e}^{\text{x}^2}+\text{e}^{-\text{x}^2}}$ is:
- One-one but not onto.
- Many-one but onto.
- One-one and onto.
- Neither one-one nor onto.
Answer
- Neither one-one nor onto.
Solution:
We have,
$\text{f(x)}=\frac{\text{e}^{\text{x}^2}-\text{e}^{-\text{x}^2}}{\text{e}^{\text{x}^2}+\text{e}^{-\text{x}^2}}$
Here, $-2,2\in\text{R}$
Now, $2\neq-2$
But, f(2) = f(-2)
Therefore, function is not one-one.
And,
The minimum value of the function is 0 and maximum value is 1.
That is range of the function is [0, 1] but the co-domain of the function is given R.
Therefore, function is not onto.
$\therefore$ function is neither one-one nor onto. View full question & answer→Question 101 Mark
Let $\text{A}=\{\text{x}\in\text{R}:\text{x}\geq1\}.$ The inverse of the function f : A → A given by $\text{f(x)}=2^{\text{x}(\text{x}-1)},$ is:
- $\big(\frac{1}{2}\big)^{\text{x}(\text{x}-1)}$
- $\frac{1}{2}\big\{1+\sqrt{1+4\log_2\text{x}}\big\}$
- $\frac{1}{2}\big\{1-\sqrt{1+4\log_2\text{x}}\big\}$
- $\text{Not defined}$
Answer
- $\frac{1}{2}\big\{1+\sqrt{1+4\log_2\text{x}}\big\}$
Solution:
Given function is $\text{A}=\{\text{x}\in\text{R}:\text{x}\geq1\}.$
The inverse of the function f : A → A given by $\text{f(x)}=2^{\text{x}(\text{x}-1)}$
$\text{f(x)}=\text{y}$
$2^{\text{x}(\text{x}-1)}=\text{y}$
$\text{x}(\text{x}-1)=\log_2\text{y}$
$\text{x}^2+\text{x}=\log_2\text{y}$
$\text{x}^2+\text{x}+\frac{1}{4}=\log_2\text{y}+\frac{1}{4}$
$\Big(\text{x}-\frac{1}{2}\Big)^2=\frac{4\log_2\text{y}+1}{4}$
$\text{x}-\frac{1}{2}=\pm\sqrt{\frac{4\log_2\text{y}+1}{4}}$
$\text{x}=\frac{1}{2}\pm\sqrt{\frac{4\log_2\text{y}+1}{4}}$
$\text{x}=\frac{1}{2}+\sqrt{\frac{4\log_2\text{y}+1}{4}}$
$\text{f}^{-1}(\text{x})=\frac{1+\sqrt{4\log_2\text{y}+1}}{2}$ View full question & answer→Question 111 Mark
Which of the following functions form $Z$ to itself are bijections?
- $f(x) = x^3$
- $f(x) = x + 2$
- $f(x) = 2x + 1$
- $f(x) = x^2 + x$
Answer
- $f(x) = x + 2$
$f$ is not because for $\text{y}=3\in$ Co$-$domain $(Z),$ there is no value of $\text{x} \in$ Domain $(Z)$ $\text{x}^3=3$
$\Rightarrow\ \text{x}=\sqrt[3]{3}\notin\text{Z}$
$\Rightarrow f$ is not onto.
So, f is not a bijection.
- Injectivity: Let $x$ and $y$ be two elements of the domain $(Z),$ such that
$x + 2 = y + 2$
$\Rightarrow x = y$
So, f is one$-$one.
Surjectivity: Let $y$ be an element in the co$-$ domain $(Z),$ such that
$y = f(x)$
$\Rightarrow y = x + 2$
$\Rightarrow\ \text{x}=\text{y}-2\in\text{Z} ($Domain$)$
$\Rightarrow f$ is onto.
So, $f$ is a bijection.
- $f(x) = 2x + 1$ is not onto because if we take $4\in\text{Z} ($co domain$),$ then $4 = f(x)$
$\Rightarrow4 = 2\text{x} + 1$
$\Rightarrow 2\text{x} = 3$
$\Rightarrow\ \text{x}=\frac{3}{2}\notin\text{Z}$
So, $f$ is not a bijection.
- $f(0) = 0^2 + 0 = 0$
$\Rightarrow $ and $f(-1) = (-1)^2 + (-1) = 1 - 1 = 0$
$\Rightarrow 0$ and $-1$ have the same image.
$\Rightarrow f$ is not one$-$one.
So, $f$ is not a bijection. View full question & answer→MCQ 121 Mark
If $f : R \rightarrow R$ is given by $f(x) = x^3 + 3,$ then $f^{-1}(x)$ is equal to$:$
AnswerCorrect option: C. $(\text{x}-3)^\frac{1}{3}$
Let $f^{-1}(x) = y$
$f(y) = x$
$\Rightarrow y^3 + 3 = x$
$\Rightarrow y^3 = x - 3$
$\Rightarrow y = (x - 3)^3$
$\Rightarrow\ \text{y}=(\text{x}-3)^\frac{1}{3}$
View full question & answer→MCQ 131 Mark
If the function $f : R \rightarrow R$ be such that $f(x) = x - [x],$ where $[x]$ denotes the greatest integer less than or equal to $x,$ then $f^{-1}(x)$ is$:$
AnswerGiven function is $f(x) = x - [x]$
$[x]$ is a greatest integer function.
Hence, we will have same values of the function for the different values of $x.$
As we are considering integer only not fraction part.
Hence, it is not defined.
View full question & answer→Question 141 Mark
If the set A contains 5 elements and the set B contains 6 elements, then the number of one-one and onto mappings from A to B is:
- 720
- 120
- 0
- None of these.
Answer
- 0
Solution:
As, the number of bijection from A into B can only be possible when provided $\frac{7}{(\text{A})}>\frac{7}{(\text{B})}$
But here n(A) < n(B)
So, the number of bijection.
i.e. one-one and onto mapping from A to B. View full question & answer→MCQ 151 Mark
If $g(x) = x^2 + x - 2$ and $\frac{1}{2}\text{gof(x)}=2\text{x}^2-5\text{x}+2,$ then $f(x)$ is equal to$:$
AnswerCorrect option: A. $2\text{x}^2-5\text{x}+2$
We will solve this problem by the trial$-$and$-$error method.
Let us check option $(a)$ first.
If $f(x) = 2x - 3$
$\frac{1}{2}(\text{gof})(x)=\text{g(f(x))}$
$=\frac{1}{2}\text{g}(2\text{x}-3)$
$=\frac{1}{2}\big[(2\text{x}-3)^2+(2\text{x}-3)-2\big]$
$=\frac{1}{2}[4\text{x}^2+9-12\text{x}+2\text{x}-3-2]$
$=\frac{1}{2}[4\text{x}^2-10\text{x}+4]$
$=2\text{x}^2-5\text{x}+2$
The given condition is satisfied by $(a).$
View full question & answer→Question 161 Mark
Let g(x) = 1 + x - [x] and $\text{f(x)}=\begin{cases}-1,&\text{x}<0\\0,&\text{x}=0\\1,&\text{x}>0\end{cases}$ where [x] denotes the greatest integer less than or equal to x. Then for all x, f(g(x)) is equal to:
- x
- 1
- f(x)
- g(x)
Answer
- 1
Solution:
When, -1 < x < 0
Then, g(x) = 1 + x - [x]
= 1 + x - (-1) = 2 + x
$\therefore$ f(g(x)) = 1
When, x = 0
Then, g(x) = 1 + x - [x]
= 1 + x - 0 = 1 + x
$\therefore$ f(g(x)) = 1
When, x > 1
Then, g(x) = 1 + x - [x]
= 1 + x - 1 = x
$\therefore$ f(g(x)) = 1
Therefore, for each interval f(g(x)) = 1 View full question & answer→MCQ 171 Mark
If $f : R \rightarrow (-1, 1)$ is defined by $\text{f(x)}=\frac{-\text{x}|\text{x}|}{1+\text{x}^2},$ then $f^{-1}(x)$ equals,
- A
$\sqrt{\frac{|\text{x}|}{1-|\text{x}|}}$
- ✓
$-\text{Sgn (x)}\sqrt{\frac{|\text{x}|}{1-|\text{x}|}}$
- C
$-\sqrt{\frac{\text{x}}{1-\text{x}}}$
- D
$\text{None of these}$
AnswerCorrect option: B. $-\text{Sgn (x)}\sqrt{\frac{|\text{x}|}{1-|\text{x}|}}$
Given function is $f : R \rightarrow (-1, 1)$ is defined by $\text{f(x)}=\frac{-\text{x}|\text{x}|}{1+\text{x}^2}$
Here, for mod function we will have to consider three cases as,
$x < 0, x = 0, x > 0$
$x < 0 \Rightarrow |x| = -x$
$\text{f(|x|)}=\frac{-\text{x}(-\text{x})}{1+\text{x}^2}$
$\text{y}=\frac{\text{x}^2}{1+\text{x}^2}$
$\text{y}(1+\text{x}^2)=\text{x}^2$
$\text{y}+\text{yx}^2=\text{x}^2$
$\text{y}=\text{x}^2-\text{yx}^2$
$\text{y}=(1-\text{y})\text{x}^2$
$\text{x}^2=\frac{\text{y}}{1-\text{y}}$
$\text{x}=-\sqrt{\frac{\text{y}}{1-\text{y}}}$
$\Rightarrow\ \text{x}=-\sqrt{\frac{|\text{y}|}{1-|\text{y}|}}\ \text{x} < 0$
Also you can check for the cases $x = 0$ and $x > 0$
that $\text{x}=-\sqrt{\frac{|\text{y}|}{1-|\text{y}|}}$
$-\text{Sgn (x)}\sqrt{\frac{|\text{x}|}{1-|\text{x}|}}$
View full question & answer→Question 181 Mark
Let $\text{f(x)}=\frac{\alpha\text{x}}{\text{x}+1},\ \text{x}\neq-1.$ Then, for what value of $\alpha$ is f(f(x)) = x?
- $\sqrt{2}$
- $-\sqrt{2}$
- 1
- -1
Answer
- -1
Solution: Given function is $\text{f(x)}=\frac{\alpha\text{x}}{\text{x}+1},\ \text{x}\neq-1$
Also f(f(x)) = x
$\text{f}\Big(\frac{\alpha\text{x}}{\text{x}+1}\Big)=\text{x}$
$\frac{\alpha\big(\frac{\alpha\text{x}}{\text{x}+1}\big)}{\frac{\alpha\text{x}}{\text{x}+1}+1}=\text{x}$
$\frac{\alpha^2\text{x}}{\alpha\text{x}+\text{x}+1}=\text{x}$
$\alpha^2=\alpha\text{x}+\text{x}+1$
$\alpha^2=(\alpha+1)\text{x}+1$
Comparing on both sides,
$\alpha+1=0\Rightarrow\ \alpha=-1$ View full question & answer→MCQ 191 Mark
Let $A = \{1, 2, ......., n\}$ and $B = \{a, b\}.$ Then the number of subjections from $A$ into $B$ is$:$
- A
$^{\text{n}}\text{P}_2$
- B
$2^\text{n}-2$
- ✓
$2^\text{n}-1$
- D
$^{\text{n}}\text{C}_2$
AnswerCorrect option: C. $2^\text{n}-1$
The number of functions from a set with $n$ number of elements into a set with $2$ number of elements $= 2^n$
But two functions can be many$-$one into function.
View full question & answer→MCQ 201 Mark
The function $f : R \rightarrow R$ defined by $f(x) = 6^x + 6^{|x|}$ is$:$
- A
One$-$one and onto.
- B
- C
One$-$one and into.
- ✓
AnswerGraph of the given function is as follows$:$
A line parallel to $X-$axis is cutting the graph at two different values.
Therefore, for two different values of $x$ we are getting the same value of $y.$
That means it is many one function.
From the given graph we can see that the range is $[2,\infty)$ and $R$ is the co$-$domain of the given function.
Hence, Co$-$dornain $=$ Range
Therefore, the given function is into. View full question & answer→Question 211 Mark
Let f : R → R be a function defined by $\text{f(x)}=\frac{\text{e}^{|\text{x}|}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}.$ Then,
- f is a bijection.
- f is an injection only.
- f is surjection on only.
- f is neither an injection nor a surjection.
Answer
- f is neither an injection nor a surjection.
Solution:
f : R → R
$\text{f(x)}=\frac{\text{e}^{|\text{x}|}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$
For x = -2 and -3 $\in\text{R}$
$\text{f(-2)}=\frac{\text{e}^{|-2|}-\text{e}^2}{\text{e}^{-2}+\text{e}^2}$
$=\frac{\text{e}^2-\text{e}^2}{\text{e}^{-2}+\text{e}^2}$
$=0$
Hence, for different values of x we are getting same values of f(x)
That means, the given function is many one.
Therefore, this function is not injective.
For x < 0
f(x) = 0
For x > 0
$\text{f(x)}=\frac{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$
$=\frac{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{}e^{-\text{x}}}-\frac{2\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$
$=1-\frac{2\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$
The value of $\frac{2\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$ is always positive.
Therefore, the value of f(x) is always less than 1.
Numbers more than 1 are not included in the range but they are included in co-domain.
As the codomain is R.
$\therefore\ \text{Co-domain}\neq\text{Range}$
Hence, the given function is not onto.
Therefore, this function is not surjective. View full question & answer→MCQ 221 Mark
Let $f : R \rightarrow R$ be given by $\text{f(x)}=\tan\text{x}.$ Then$, f^{-1}(1)$ is$:$
AnswerCorrect option: B. $\big\{\text{n}\pi+\frac{\pi}{4}:\text{n}\in\text{Z}\big\}$
We have$, f : R \rightarrow R$ is given by
$\text{f(x)}=\tan\text{x}$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\tan^{-1}\text{x}$
$\therefore\ \text{f}^{-1}(1)=\tan^{-1}1=\big\{\text{n}\pi+\frac{\pi}{4}:\text{n}\in\text{Z}\big\}$
View full question & answer→MCQ 231 Mark
The function $f : R \rightarrow R, f(x) = x^2$ is$:$
- A
Injective but not surjective.
- ✓
Surjective but not injective.
- C
Injective as well as surjective.
- D
Neither injective nor surjective.
AnswerCorrect option: B. Surjective but not injective.
Given function is $f : R \rightarrow R, f(x) = x^2$
If $f(x) = f(y)$ then
$x^2 = y^2$
$\Rightarrow\ \text{x}\pm\text{y}$
Hence, it is not one$-$one or injective.
$f(x) = y$
$y = x^2$
$\text{x}=\pm\sqrt{\text{y}}$
But co$-$domain is $R.$
Hence, it is not onto or surjective.
View full question & answer→MCQ 241 Mark
Which of the following functions from $\text{A}=\{\text{x}:-1\leq\text{x}\leq1\}$ to itself are bijections?
AnswerRange of $\text{f}=\Big[\frac{-1}{2},\frac{1}{2}\Big]\neq\text{A}$
So, $f$ is not a bijection.
Range $=\Big[\sin\Big(\frac{-\pi}{2}\Big),\sin\Big(\frac{\pi}{2}\Big)\Big]=[-1,1]=\text{A}$
So, $g$ is a bijection. $h(-1) = |-1| = 1$
And $h(1) = |1| = 1$
$\Rightarrow -1$ and $1$ have the same images.
So, $h$ is not a bijection.
$k(-1) = (-1)^2 = 1$
And $k(1) = (1)^2 = 1$
$\Rightarrow -1$ and $1$ have the same images.
So, $k$ is not a bijection.
View full question & answer→MCQ 251 Mark
The function $f : R \rightarrow R$ defined by $f(x) = 2^x + 2^{|x| }$ is$:$
- A
One$-$one and onto.
- B
Many$-$one and onto.
- ✓
One$-$one and into.
- D
Many$-$one and into.
AnswerCorrect option: C. One$-$one and into.
The function $f : R \rightarrow R$ defined by $f(x) = 2^x + 2^{|x|}$
Here, for each value of $x$ we will get different values of $f(x).$
Hence, it is one$-$one function.
Also, each element of codomain is mapped to at most one element of the domain.
Function is one$-$one and into.
View full question & answer→MCQ 261 Mark
If $\text{F}:[1,\infty)\rightarrow[2,\infty)$ is given by $\text{f(x)}=\text{x}+\frac{1}{\text{x}},$ then $f^{-1}(x)$ equals$:$
- ✓
$\frac{\text{x}+\sqrt{\text{x}^2-4}}{2}$
- B
$\frac{\text{x}}{1+\text{x}^2}$
- C
$\frac{\text{x}-\sqrt{\text{x}^2-4}}{2}$
- D
$1+\sqrt{\text{x}^2-4}$
AnswerCorrect option: A. $\frac{\text{x}+\sqrt{\text{x}^2-4}}{2}$
Let $f^{-1}(x) = y$
$\Rightarrow\ \text{f(y)} = \text{x}$
$\Rightarrow\ \text{y}+\frac{1}{\text{y}}=\text{x}$
$\Rightarrow\ \text{y}^2 + 1 = \text{xy}$
$\Rightarrow\ \text{y}^2 - \text{xy} + 1 = 0$
$\Rightarrow\ \text{y}^2-2\times\text{y}\times\frac{\text{x}}{2}+\big(\frac{\text{x}}{2}\big)^2-\big(\frac{\text{x}}{2}\big)^2+1=0$
$\Rightarrow\ \text{y}^2-2\times\text{y}\times\frac{\text{x}}{2}+\big(\frac{\text{x}}{2}\big)^2=\frac{\text{x}^2-1}{4}$
$\Rightarrow\ \Big(\text{y}-\frac{\text{x}}{2}\Big)^2=\frac{\text{x}^2-1}{4}$
$\Rightarrow\ \text{y}-\frac{\text{x}}{2}=\frac{\sqrt{\text{x}^2-4}}{2}$
$\Rightarrow\ \text{y}=\frac{\text{x}}{2}+\frac{\sqrt{\text{x}^2-4}}{2}$
$\Rightarrow\ \text{y}=\frac{\text{x}+\sqrt{\text{x}^2-4}}{2}$
$\Rightarrow\ \text{f}^{-1}(\text{x})=\frac{\text{x}+\sqrt{\text{x}^2-4}}{2}$
View full question & answer→Question 271 Mark
f : R → R given by $\text{f(x)}=\text{x}+\sqrt{\text{x}^2}$ is:
- Injective.
- Surjective.
- Bijective.
- None of these.
Answer
- None of these.
Solution:
$\text{f(x)}=\text{x}+\sqrt{\text{x}^2}=\text{x}\pm\text{x}=0\text{ or }2\text{x}$
⇒ Each element of the domain has 2 images.
f is not a function. View full question & answer→MCQ 281 Mark
Let $\text{A}=\{\text{x}:-1\leq\text{x}\leq1\}$ and $f : A \rightarrow A$ such that $\text{f(x)}=\text{x}|\text{x}|,$ then $f$ is$:$
- ✓
- B
Injective but not surjective.
- C
Surjective but not injective.
- D
Neither injective nor surjective.
AnswerGiven function is $\text{A}=\{\text{x}:-1\leq\text{x}\leq1\}$ and $f : A \rightarrow A$ such that $\text{f(x)}=\text{x}|\text{x}|$
For the mod function we have to check three cases as $x < 0, x = 0, x > 0.$
For example,
$x < 0$
$f(x) = x|x| < 0$
$|x| = -x$
$y = -x^2$
$\text{x}=-\sqrt{-\text{y}}$ which is not possible for $x > 0$
Hence, $f$ is onto.
$\Rightarrow f$ is bijection.
View full question & answer→Question 291 Mark
Let f : R - {n} → R be a function defined by $\text{f(x)}=\frac{\text{x}-\text{m}}{\text{x}-\text{n}},$ where $\text{m}\neq\text{n.}$ Then,
- f is one-one onto.
- f is one-one into.
- f is many one onto.
- f is many one into.
Answer
- f is one-one into.
Solution:
Given function f : R - {n} → R be a function defined by $\text{f(x)}=\frac{\text{x}-\text{m}}{\text{x}-\text{n}},\ \text{m}\neq\text{n}$
If f(x) = f(y) then
$\frac{\text{x}-\text{m}}{\text{x}-\text{n}}=\frac{\text{y}-\text{m}}{\text{y}-\text{n}}$
⇒ (x - m)(y - n) = (y - m)(x - n)
After solving this we will get x = y
Hence, it is one-one.
$\text{f(x)}=\frac{\text{x}-\text{m}}{\text{x}-\text{n}},\ \text{m}\neq\text{n}$
$\text{y}=\frac{\text{x}-\text{m}}{\text{x}-\text{n}}$
⇒ y(x - n) = x - m
⇒ yx - yn = x - m
⇒ yx - x = ny - m
⇒ x(y - 1) = ny - m
$\Rightarrow\ \text{x}=\frac{\text{ny}-\text{m}}{\text{y}-1}$
Here, for y = 1 we can not define x.
Hence, it is not onto. View full question & answer→MCQ 301 Mark
Let $f(x) = x^2$ and $g(x) = 2^x.$ Then, the solution set of the equation $fog(x) = gof(x)$ is$:$
AnswerCorrect option: C. ${0, 2}$
Since $(fog)(x) = (gof)(x),$
$f(g(x)) = g(f(x))$
$\Rightarrow\ \text{f}(2^\text{x})=\text{g}(\text{x}^2)$
$\Rightarrow\ \big(2^{\text{x}}\big)^{2}=2^{\text{x}^2}$
$\Rightarrow\ 2^{2\text{x}}=2^{\text{x}^2}$
$\Rightarrow\ \text{x}^2=2\text{x}$
$\Rightarrow\ \text{x}^2-2\text{x}=0$
$\Rightarrow\ \text{x}(\text{x}-2)=0$
$\Rightarrow\ \text{x}=0, 2$
$\Rightarrow\ \text{x}\in\{0,2\}$
View full question & answer→MCQ 311 Mark
Let $f : R \rightarrow R$ be a function defined by $\text{f(x)}=\frac{\text{x}^2-8}{\text{x}^2+2}.$ Then$, f$ is :
AnswerCorrect option: D. Neither one$-$one nor onto.
Injectivity : Let $x$ and $y$ be two elements in the domain $(R),$ such that
$f(x) = f(y)$
$\frac{\text{x}^2-8}{\text{x}^2+2}=\frac{\text{y}^2-8}{\text{y}^2+2}$
$\Rightarrow (x^2 - 8)(y^2 + 2) = (y^2 - 8)(x^2 + 2)$
$\Rightarrow x^2y^2 + 2x^2 - 8y^2 - 16 = x^2y^2 + 2y^2 - 8x^2 - 16$
$\Rightarrow 10x^2 = 10y^2$
$\Rightarrow x^2 = y^2$
$\Rightarrow\ \text{x}=\pm\text{y}$
So$, f$ is not one$-$one.
Surjectivity : $\text{f}(-1)=\frac{(-1)^2-8}{(-1)^2+2}=\frac{1-8}{1+2}=\frac{-7}{3}$
and $\text{f(1)}=\frac{(1)^2-8}{(1)^2+2}=\frac{1-8}{1+2}=\frac{-7}{3}$
$\Rightarrow\ \text{f}(-1)=\text{f}(1)=\frac{-7}{3}$
$\Rightarrow f$ is not onto.
View full question & answer→MCQ 321 Mark
If a function $\text{f}:[2,\infty)\rightarrow\ \text{B}$ defined by $f(x) = x^2 - 4x + 5$ is a bijection, then $B =$
- A
$\text{R}$
- ✓
$[1,\infty)$
- C
$[4,\infty)$
- D
$[5,\infty)$
AnswerCorrect option: B. $[1,\infty)$
Since $f$ is a bijection, co$-$domain of $f =$ range of $f$
$\Rightarrow B =$ range of $f$
Given: $f(x) = x^2 - 4x + 5$
Let $f(x) = y$
$\Rightarrow y = x^2 - 4x + 5$
$\Rightarrow x^2 - 4x + (5 - y) = 0$
$\because$ Discrimant, $\text{D}=\text{b}^2-4\text{ac}\geq0,$
$(-4)^2-4\times1\times(5-\text{y})\geq0$
$\Rightarrow\ 16-20+4\text{y}\geq0$
$\Rightarrow\ 4\text{y}\geq4$
$\Rightarrow\ \text{y}\geq1$
$\Rightarrow\ \text{y}\in[1,\infty)$
$\Rightarrow$ Range of $\text{f}=[1,\infty)$
$\Rightarrow\ \text{B}=[1,\infty)$
View full question & answer→Question 331 Mark
Which of the following functions from $\text{A}=\{\text{x}\in\text{R}:-1\leq\text{x}\leq1\}$ to itself are bijections?
- $\text{f(x)}=|\text{x}|$
- $\text{f(x)}=\sin\frac{\pi\text{x}}{2}$
- $\text{f(x)}=\sin\frac{\pi\text{x}}{4}$
- $\text{None of these}$
Answer
- $\text{f(x)}=\sin\frac{\pi\text{x}}{2}$
Solution:
It is clear that f(x) is one-one.
Range of $\text{f}=\Big[\sin\frac{\pi(-1)}{2},\sin\frac{\pi(1)}{2}\Big]=\Big[\sin\frac{-\pi}{2},\sin\frac{\pi}{2}\Big]$
= A = Co-domain of f
⇒ f is onto.
So, f is a bijection. View full question & answer→Question 341 Mark
If the set A contains 7 elements and the set B contains 10 elements, then the number one-one functions from A to B is:
- $^{10}\text{C}_7$
- $^{10}\text{C}_7\times7!$
- $7^{10}$
- $10^7$
Answer
- $^{10}\text{C}_7\times7!$
Solution:
As, the number of one-one functions from A to B with m and n elements, respectively $= \ ^{\text{n}}\text{P}_\text{m}=\ ^{\text{n}}\text{C}_\text{m}\times\text{m}!$
So, the number of one-one functions from A to B with 7 and 10 elements, respectively $=\ ^{10}\text{P}_7=\ ^{10}\text{C}_7\times7!$ View full question & answer→MCQ 351 Mark
If $f(x) = \sin^2x$ and the composite function $\text{g(f(x))} = |\sin\text{x}|,$ then $g(x)$ is equal to$:$
- A
$\sqrt{\text{x}-1}$
- ✓
$\sqrt{\text{x}}$
- C
$\sqrt{\text{x}+1}$
- D
$-\sqrt{\text{x}}$
AnswerCorrect option: B. $\sqrt{\text{x}}$
Given that $\text{f(x)}=\sin^2\text{x}$ and the composite function $\text{g(f(x))}=|\sin\text{x}|$
We will do it using trial and error method.
If we take $\text{g(x)}=-\sqrt{\text{x}}$ and $\text{f(x)}=\sin^2\text{x}$
$\text{g(f(x))}=\text{g}(\sin^2\text{x})$
$=-\sin\text{x}$
Which contradicts to the $\text{g(f(x))}=|\sin\text{x}|$
Hence, we take $\text{g(x)}=\sqrt{\text{x}}$
$\text{g(f(x))}=\text{g}(\sin^2\text{x})$
$=\sqrt{\sin^2\text{x}}=|\sin\text{x}|$
View full question & answer→MCQ 361 Mark
Let $[x]$ denote the greatest integer less than or equal to $x.$ If $f(x) = \sin^{-1}x, g(x) = [x^2]$ and $\text{h(x)}=2\text{x},\frac{1}{2}\leq\text{x}\leq\frac{1}{\sqrt{2}},$ then
- A
$\text{fogoh(x)}=\frac{\pi}{2}$
- B
$\text{fogoh(x)}=\pi$
- ✓
$\text{hofog}=\text{hogof}$
- D
$\text{hofog}\neq\text{hogof}$
AnswerCorrect option: C. $\text{hofog}=\text{hogof}$
$\text{hogof}(x) = h(f(g(x)))$
$= h(f([x]))$
$= h(\sin^{-1}[x])$
$= 2\sin^{-1}[x]$
$= 2 \times 0 = 0$
$f(x) = \sin^{-1}x$
$\text{hogof}(x) = \text{hogo}(x) = 0$
View full question & answer→MCQ 371 Mark
Let $f(x) = x^3$ be a function with domain $\{0, 1, 2, 3\}.$ Then domain of $f^{-1}$ is$:$
- A
${3, 2, 1, 0}$
- B
${0, -1, -2, -3}$
- ✓
${0, 1, 8, 27}$
- D
${0, -1, -8, -27}$
AnswerCorrect option: C. ${0, 1, 8, 27}$
Given function is $f(x) = x^3$ be a function with domain $\{0, 1, 2, 3\}.$
Range $= \{0, 1^3, 2^3, 3^3\} = \{0, 1, 8, 27\}$
$f$ can be written as $\{(0, 0), (1, 1), (2, 8), (3, 27)\}$
Hence, $f^{-1}$ can be written as $\{(0, 0), (1, 1), (8, 2), (27, 3)\}$
Domain of $f^{-1}$ is $\{0, 1, 8, 27\}$
View full question & answer→Question 381 Mark
A function f from the set of natural numbers to integers defined by $\text{f(n)}=\begin{cases}\frac{\text{n}-1}{2},&\text{when n is odd}\\-\frac{\text{n}}{2},&\text{when n is even}\end{cases}$
- Neither one-one nor onto.
- One-one but not onto.
- Onto but not one-one.
- One-one and onto both.
Answer
- One-one and onto both.
Solution:
Injectivity: Let x and y be any two elements in the domain (N).
Case-1: Both x and y are even.
Let f(x) = f(y)
$\Rightarrow\ \frac{-\text{x}}{2}=\frac{-\text{y}}{2}$
$\Rightarrow-\text{x}=-\text{y}$
$\Rightarrow\ \text{x}=\text{y}$
Case-2: Both x and y are odd.
Let f(x) = f(y)
$\Rightarrow\ \frac{\text{x}-1}{2}=\frac{\text{y}-1}{2}$
$\Rightarrow\ \text{x}-1=\text{y}-1$
$\Rightarrow\ \text{x}=\text{y}$
Case-3: Let x be even and y be odd.
Then, $\text{f(x)}=\frac{-\text{x}}{2}$ and $\text{f(y)}=\frac{\text{y}-1}{2}$
Then, clearly
$\text{x}\neq\text{y}$
$\Rightarrow\ \text{f(x)}\neq\text{f(y)}$
From all the cases, f is one-one.
Surjectivity: Co-domain of f = Z = {......, -3, -2, -1, 0, 1, 2, 3, ......}
Range of $\text{f}=\Big\{....,\ \frac{-3-1}{2},\ \frac{-(-2)}{2},\ \frac{-1-1}{2},\ \frac{0}{2},\ \frac{1-1}{2},\ \frac{-2}{2},\ \frac{3-1}{2},\ ....\Big\}$
⇒ Range of f = {....., -2, 1, -1, 0, 0, -1, 1, .....}
⇒ Range of f = {....., -2, -1, 0, 1, 2, ......}
⇒ Co-domain of f = Range of f
⇒ f is onto. View full question & answer→Question 391 Mark
Let M be the set of all 2 × 2 matrices with entries from the set R of real numbers. Then, the function f : M→ R defined by f(A) = |A| for every A ∈ M, is:
- One-one and onto.
- Neither one-one nor onto.
- One-one but-not onto.
- Onto but not one-one.
Answer
- Onto but not one-one.
Solution:
$\text{M}=\begin{Bmatrix}\text{A}=\begin{bmatrix}\text{a}&\text{b}\\\text{c}&\text{d}\end{bmatrix}:\text{a, b, c, d}\in\text{R}\end{Bmatrix}$
f : M → R is given by f(A) = |A|
Injectivity: $\text{f}\begin{pmatrix}\begin{bmatrix}0&0\\0&0\end{bmatrix}\end{pmatrix}=\begin{vmatrix}0&0\\0&0\end{vmatrix}=0$
and $\text{f}\begin{pmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix}\end{pmatrix}=\begin{vmatrix}1&0\\0&0\end{vmatrix}=0$
$\Rightarrow\ \text{f}\begin{pmatrix}\begin{bmatrix}0&0\\0&0\end{bmatrix}\end{pmatrix}=\text{f}\begin{pmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix}\end{pmatrix}=0$
So, f is not one-one.
Surjectivity: Let y be an element of the co-domain, such that
$\text{f(A)}=-\text{y},\ \text{A}=\begin{bmatrix}\text{a} & \text{b} \\\text{c} & \text{d} \end{bmatrix}$
$\Rightarrow\ \begin{vmatrix}\text{a}&\text{b}\\\text{c}&\text{d}\end{vmatrix}=\text{y}$
$\Rightarrow\ \text{ad}-\text{bc}=\text{y}$
$\Rightarrow\ \text{a, b, c, d}\in\text{R}$
$\Rightarrow\ \text{A}=\begin{bmatrix}\text{a} & \text{b} \\\text{c} & \text{d} \end{bmatrix}\in\text{M}$
⇒ f is onto. View full question & answer→Question 401 Mark
A function f from the set of natural numbers to the set of integers defined by $\text{f(n)}\begin{cases}\frac{\text{n}-1}{2},&\text{when n is odd}\\-\frac{\text{n}}{2},&\text{when n is even}\end{cases}$ is:
- Neither one-one nor onto.
- One-one but not onto.
- Onto but not one-one.
- One-one and onto.
Answer
- One-one and onto.
Solution:
Given function is,
$\text{f(n)}=\frac{\text{n}-1}{2}$ for n is odd
$=-\frac{\text{n}}{2}$ for n is even
For n is odd,
If f(n) = f(m) then
$\frac{\text{n}-1}{2}=\frac{\text{m}-1}{2}$
⇒ n = m
Also, for n is even if f(n) = f(m) then n = m
Hence, f is one-one.
Also, each element of y is associated with at least one element of x,
f is onto. View full question & answer→Question 411 Mark
The function $\text{f}:\Big[\frac{-1}{2},\frac{1}{2},\frac{1}{2}\Big]\rightarrow\ \Big[\frac{-\pi}{2},\frac{\pi}{2}\Big],$ defined by $\text{f(x)}=\sin^{-1}(3\text{x}-4\text{x}^3),$ is:
- Bijection.
- Injection but not a surjection.
- Surjection but not an injection.
- Neither an injection nor a surjection.
Answer
- Bijection.
Solution:
$\text{f(x)}=\sin^{-1}(3\text{x}-4\text{x}^3)$
$\Rightarrow\ \text{f(x)}=3\sin^{-1}\text{x}$
Injectivity: Let x and y be two elements in the domain $\Big[\frac{-1}{2},\frac{1}{2}\Big],$ such that
f(x) = f(y)
$\Rightarrow\ 3\sin^{-1}\text{x}=3\sin^{-1}\text{y}$
$\Rightarrow\ \sin^{-1}\text{x}=\sin^{-1}\text{y}$
$\Rightarrow\ \text{x}=\text{y}$
So, f is one-one.
Surjectivity: Let y be any element in the co-domain, such that
f(x) = y
$\Rightarrow\ 3\sin^{-1}\text{x}=\text{y}$
$\Rightarrow\ \sin^{-1}\text{x}=\frac{\text{y}}{3}$
$\Rightarrow\ \text{x}=\sin\frac{\text{y}}{3}\in\Big[\frac{-1}{2},\frac{1}{2}\Big]$
⇒ f is onto.
⇒ f is a bijection. View full question & answer→Question 421 Mark
Let f : Z → Z be given by $\text{f(x)}=\begin{cases}\frac{\text{x}}{2},&\text{if x is even}\\0,&\text{if x is odd}\end{cases}.$ Then, f is:
- Onto but not one-one.
- One-one but not onto.
- One-one and onto.
- Neither one-one nor onto.
Answer
- Onto but not one-one.
Solution:
Given function is
$\text{f(x)}=\frac{\text{x}}{2}$ if x is even
= 0 if x is odd
For f(3) = 0 and f(4) = 0
⇒ f(3) = f(4)
But, $3\neq4$
Hence, it is not one-one.
$\text{x}\in\text{R}\Rightarrow\ \text{y}\in\text{R}$
Here, Domain = range of f
Hence, it is onto. View full question & answer→MCQ 431 Mark
Let $f$ be an injective map with domain $\{x, y, z\}$ and range $\{1, 2, 3\},$ such that exactly one of the following statements is correct and the remaining are false.$\text{f(x)}=1,\ \text{f(y)}\neq1,\ \text{f(z)}\neq2.$ The value of $f^{-1}(1)$ is$:$
AnswerGiven that $f$ be an injective map with domain $\{x, y, z\}$ and range $\{1, 2, 3\}$
$f(x) = 1, \text{f(y)}\neq1,\ \text{f(z)}\neq2$
As $f(x) = 1 $
$\Rightarrow f^{-1}(1) = y$
View full question & answer→MCQ 441 Mark
Let $\text{A}=\{\text{x}\in\text{R}:\text{x}\leq1\}$ and $f : A \rightarrow A$ be defined as $f(x) = x(2 - x).$ Then $f^{-1}(x)$ is$:$
- A
$1+\sqrt{1-\text{x}}$
- ✓
$1-\sqrt{1-\text{x}}$
- C
$\sqrt{1-\text{x}}$
- D
$1\pm\sqrt{1-\text{x}}$
AnswerCorrect option: B. $1-\sqrt{1-\text{x}}$
Let y be the element in the co$-$domain $R$ such that $f^{-1}(x) = y ......(1)$
$\Rightarrow f(y) = x$ and $\text{y}\leq1$
$\Rightarrow y(2 - y) = x$
$\Rightarrow 2y - y^2 = x$
$\Rightarrow y^2 - 2y + x = 0$
$\Rightarrow y^2 - 2y = -x$
$\Rightarrow y^2 - 2y + 1 = 1 - x$
$\Rightarrow\ (\text{y}-1)^2=\sqrt{1-\text{x}}$
$\Rightarrow\ \text{y}-1=\pm\sqrt{1-\text{x}}$
$\Rightarrow\ \text{y}=1\pm\sqrt{1-\text{x}}$
$\Rightarrow\ \text{y}=1-\sqrt{1-\text{x}}$ $(\because\ \text{y}\leq1)$
View full question & answer→MCQ 451 Mark
The distinct linear functions that map $[-1, 1]$ onto $[0, 2]$ are$:$
- ✓
$f(x) = x + 1, g(x) = -x + 1$
- B
$f(x) = x - 1, g(x) = x + 1$
- C
$f(x) = -x - 1, g(x) = x - 1$
- D
AnswerCorrect option: A. $f(x) = x + 1, g(x) = -x + 1$
Since $f$ is invertible, range of $f =$ co$-$domain of $f = x$
So, we need to find the range of $f$ to find $X.$
For finding the range, let $f(x) = y$
$\Rightarrow 4x - x^2 = y$
$\Rightarrow x^2 - 4x = -y$
$\Rightarrow x^2 - 4x + 4 = 4 - y$
$\Rightarrow (x - 2)^2 = 4 - y$
$\Rightarrow \text{x}-2=\pm4-\text{y}$
$\Rightarrow \text{x}=2\pm4-\text{y}$
This is defined only when $4-\text{y}\geq0$
$\Rightarrow \text{y}\leq4,$
$X =$ Range of $f =(-\infty,4]$
View full question & answer→MCQ 461 Mark
Let $f : R \rightarrow R$ be given by $f(x) = x^2 - 3.$ Then, $f^{-1}$ is given by$:$
- ✓
$\sqrt{\text{x}+3}$
- B
$\sqrt{\text{x}}+3$
- C
$\text{x}+\sqrt{3}$
- D
$\text{None of these}$
AnswerCorrect option: A. $\sqrt{\text{x}+3}$
Given function is $f : R \rightarrow R$ be given by $f(x) = x^2 - 3.$
$\text{y} = \text{x}^2 - 3$
$\text{y} + 3 = \text{x}^2$
$\text{x}=\pm\sqrt{\text{y}+3}$
$\Rightarrow\ \text{y}=\pm\sqrt{\text{x}+3}$
View full question & answer→Question 471 Mark
Let the function f : R - {-b} → R - {1} be defined by $\text{f(x)}=\frac{\text{x}+\text{a}}{\text{x}+\text{b}},\ \text{a}\neq\text{b}.$ Then,
- f is one-one but not onto.
- f is onto but not one-one.
- f is both one-one and onto.
- None of these.
Answer
- f is both one-one and onto.
Solution:
Injectivity: Let x and y be two elements in the domain R - {-b}, such that
f(x) = f(y) ⇒ x + ax + b = y + ay + b
⇒ x + ay + b = x + by + a
⇒ xy + bx + ay + ab = xy + ax + by + ab
⇒ bx + ay = ax + by
⇒ a - bx = a - by
⇒ x = y
So, f is one-one.
Surjectivity: Let y be an element in the co-domain of f,
i.e., R - {1}, such that f(x) = y
⇒ x + ax + b = y
⇒ x + a ⇒ x = -a
So, f is onto.
View full question & answer→MCQ 481 Mark
The function $f : A \rightarrow B$ defined by $f(x) = -x^2 + 6x- 8$ is a bijection if,
- ✓
$\text{A}=(-\infty,3]$ and $\text{B}=(-\infty,1]$
- B
$\text{A}=[-3,\infty)$ and $\text{B}=(-\infty,1]$
- C
$\text{A}=(-\infty,3]$ and $\text{B}=[1,\infty)$
- D
$\text{A}=[3,\infty)$ and $\text{B}=[1,\infty)$
AnswerCorrect option: A. $\text{A}=(-\infty,3]$ and $\text{B}=(-\infty,1]$
$f(x) = -x^2 + 6x - 8,$ is a polynomial function and the domain of polynomial function is real number.
$\therefore\ \text{x}\in\text{R}$
$f(x) = -x^2 + 6x - 8$
$= -(x^2 - 6x + 8)$
$= -(x^2 - 6x + 9 - 1)$
$= -(x - 3)^2 + 1$
Maximum value of $-(x - 3)^2$ woud be $0$
$\therefore$ Maximum value of $-(x - 3)^2 + 1$ woud be $1$
$\therefore\ \text{f(x)}\in(-\infty,1]$
We can see from the given graph that function is symmetrical about $x = 3$ and the given function is bijective.
So$, x$ would be either $(-\infty,3]\text{ or }[3,\infty)$
The correct option which satisfy $A$ and $B$ both is :
$\text{A}=(-\infty,3]$ and $\text{B}=(-\infty,1]$ View full question & answer→MCQ 491 Mark
If $f : R \rightarrow R$ is given by $f(x) = 3x - 5,$ then $f^{-1}(x)$
- A
is given by $\frac{1}{3\text{x}-5}$
- ✓
is given by $\frac{\text{x}+5}{3}$
- C
does not exist because $f$ is not one$-$one.
- D
does not exist because $f$ is not onto.
AnswerCorrect option: B. is given by $\frac{\text{x}+5}{3}$
Given function is $f : R \rightarrow R$ is given by $f(x) = 3x - 5$
To find $f^{-1}(x)$
$y = f(x)$
$\Rightarrow y = 3x - 5$
$\Rightarrow y + 5 = 3x$
$\Rightarrow\ \text{y}=\frac{\text{y}+5}{3}$
Hence, $\text{f}^{-1}(\text{x})=\frac{\text{x}+5}{3}$
View full question & answer→Question 501 Mark
If the function f : R → A given by $\text{f(x)}=\frac{\text{x}^2}{\text{x}^2+1}$ is a surjection, then A =
- R
- [0, 1]
- [0, 1)
- [0, 1)
Answer
- [0, 1)
Solution:
As f is surjective, range of f = co-domain of f
⇒ A = range of f
$=\frac{\text{x}^2}{\text{x}^2+1},$
$\text{y}=\frac{\text{x}^2}{\text{x}^2+1}$
$\Rightarrow\ \text{y}(\text{x}^2+1)$
$\Rightarrow\ \text{x}^2=\frac{-\text{y}}{(\text{y}-1)}$
$\Rightarrow\ \text{x}=\sqrt{\frac{\text{y}}{(1-\text{y})}}$
$\Rightarrow\ \frac{\text{y}}{(1-\text{y})}\geq0$
$\Rightarrow\ \text{y}\in[0,1)$
⇒ Range of f = [0, 1)
⇒ A = [0, 1) View full question & answer→