Tag: sets, relations and functions

Quizzes (13) Questions (60)

Questions Related to sets, relations and functions

Multiple choice business maths sets, relations and functions graphs of the form y=ax^2+bx+c some more types of functions functions and their graphs

If $f:\,\left( {3,6} \right) \to \left( {1,3} \right)$ is a function defined by $f\left( x \right) = x - \left[ {\frac{x}{3}} \right],\,then\,{f^{ - 1}}\left( x \right) = $

  1. $x-1$
  2. $x+1$
  3. $x$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given f(x) = x - [x/3] for x in (3, 6). For x in (3, 6), x/3 is in (1, 2), so [x/3] = 1. Thus f(x) = x - 1. Solving y = x - 1 for x gives x = y + 1, so f^-1(x) = x + 1.

Multiple choice business maths sets, relations and functions graphs of the form y=ax^2+bx+c some more types of functions functions and their graphs

The tangents to the graph of the function  $y=f(x)$ at the point with abscissa $x=1$ forms an angle of $\pi/6$ and the point $x=2$ an angle of $\pi/3$ and at the point $x=3$ an angle of $\pi/4$. The value of 
$\displaystyle \int _{1}^{2}{f'(x)f''(x)dx}+\displaystyle \int _{2}^{3}{f''(x)dx}$

  1. $\dfrac{4\sqrt{3}-1}{3\sqrt{3}}$
  2. $\dfrac{3\sqrt{3}-1}{2}$
  3. $\dfrac{4-\sqrt{3}}{3}$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The given integral can be evaluated by substitution. Let u = f'(x), so du = f''(x)dx, turning the first integral into a standard form, while the second integral is directly related to f'(x). Evaluating the trigonometric slopes given by the tangents yields a specific numerical value not matched by A, B, or C.

Multiple choice business maths sets, relations and functions graphs of the form y=ax^2+bx+c some more types of functions functions and their graphs

The graph of the function $\cos x\cos x(x+2)-\cos^{2}(x+1)$ is  

  1. A straight line through $(0, -\sin^{2}1)$ with slope $2$.
  2. A straight line through $(0, 0)$
  3. A parabola with vertex $(1, -\sin^{2}1)$
  4. A straight line through $\left(\dfrac{\pi}{2},-\sin^{2}1\right)$ and parallel to the $x-axis$.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Expanding and simplifying the given expression using trigonometric identities reveals that the function is a constant with respect to x or simplifies to a line parallel to the x-axis passing through the specified point.

Multiple choice business maths sets, relations and functions graphs of the form y=ax^2+bx+c some more types of functions functions and their graphs

If $f(x)=\left | \sin x \right |$, then domain of $f$ for the existence of inverse is  

  1. $[0,\pi ]$
  2. $\left [ 0,\dfrac{\pi }{2} \right ]$
  3. $\left [ -\dfrac{\pi }{4},\dfrac{\pi }{4} \right ]$
  4. $\left [ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right ]$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know that $-1 \leq \sin x \leq 1$ for $x \in \left [ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right] $. 

For $| \sin x |$ to be invertible, the function has to be one-to-one. 
Thus, we need unique values of $x$ that give unique values of $f$ and vice versa.
For $x \in \left [0, \dfrac{\pi}{2} \right]$, $0 \leq \sin x \leq 1 \Rightarrow  0 \leq | \sin x \leq 1$
For $x \in \left [-\dfrac{\pi}{2},0 \right]$, $-1 \leq \sin x \leq 0 \Rightarrow  0 \leq | \sin x \leq 1$.
So, we have
$ \left [0, \dfrac{\pi}{2} \right] \rightarrow\left [0, 1 \right]$
$ \left [- \dfrac{\pi}{2},0 \right] \rightarrow\left [0, 1 \right]$
Since both the domains of $|\sin x|$ map to$\left [0, 1 \right]$, we consider only one of them for $x$ to be unique. 
Here, according to the options, the domain of $f$ must be$\left [0, \dfrac{\pi}{2} \right]$.

Multiple choice maths sets, relations and functions graphs of the form y=ax^2+bx+c some more types of functions functions and their graphs

Two lines $L _{1} :2x+3y-5=0$ and $L _{2} :3x-4y+1=0$ intersect a point $P$ and make an angle $\theta$ with each other. Equation of a line which passes through $P$ and makes an angle $(\pi/2-\theta)$ with the line $L _{1}$ is

  1. $16x+64y+79=0$ and $4x+3y+7=0$
  2. $16x+63y-79=0$ and $4x+3y+7=0$
  3. $16x-63y+79=0$ and $4x-3y+7=0$
  4. $16x-63y-79=0$ and $4x-3y+7=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice maths sets, relations and functions graphs of the form y=ax^2+bx+c some more types of functions functions and their graphs

If the line $3x+4y=\sqrt{7}$ touches the ellipse $3x^{2}+4y^{2}=1$, then the point of contact is 

  1. $(\dfrac{1}{\sqrt{7}},\dfrac{1}{\sqrt{7}})$
  2. $(\dfrac{1}{\sqrt{3}},-\dfrac{1}{\sqrt{3}})$
  3. $(\dfrac{1}{\sqrt{7}},-\dfrac{1}{\sqrt{7}})$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\begin{array}{l} 3{ x^{ 2 } }+4{ y^{ 2 } }=1 \ Equation\, of\, \tan  gent\, to\, ellipse\, at\, \left( { { x _{ 1 } },{ y _{ 1 } } } \right)  \ 3x{ x _{ 1 } }+4y{ y _{ 1 } }=1 \ \frac { { 3x } }{ { \sqrt { 7 }  } } +\frac { { 4y } }{ { \sqrt { 7 }  } } =1 \ { x _{ 1 } }=\frac { 1 }{ { \sqrt { 7 }  } } \, \, \, \, \, { y _{ 1 } }=\frac { 1 }{ { \sqrt { 7 }  } }  \ \left( { \frac { 1 }{ { \sqrt { 7 }  } } ,\frac { 1 }{ { \sqrt { 7 }  } }  } \right)  \ Hence, \ option\, \, A\, is\, correct\, answer. \end{array}$

Multiple choice maths sets, relations and functions graphs of the form y=ax^2+bx+c some more types of functions functions and their graphs

The column sum in an incidence matrix for a simple graph is ________________.

  1. depends on number of edges

  2. always greater than 2

  3. equal to 2

  4. equal to the number of edges
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In an incidence matrix for a simple undirected graph, each column represents an edge. Since an edge connects exactly two vertices, the sum of the entries in each column is 2.

Multiple choice maths sets, relations and functions graphs of the form y=ax^2+bx+c some more types of functions functions and their graphs

Which of the following ways can be used to represent a graph?

  1. Adjacency List and Adjacency Matrix

  2. Incidence Matrix

  3. Adjacency List, Adjacency Matrix as well as Incidence Matrix

  4. None of the mentioned
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Graphs can be represented using adjacency lists, adjacency matrices, and incidence matrices. All these methods are standard ways to store graph data structures.