Tag: functions and graphs

Quizzes (5) Questions (65)

Questions Related to functions and graphs

Multiple choice maths functions and graphs different forms of equation of a line
$\displaystyle ax^{2}+2hxy+by^{2}=0$ represents a pair of straight lines through origin & angle between them is given by
$\displaystyle \tan \theta=\frac{2\sqrt{h^{2}-ab}}{a+b}$. If the lines are perpendicular then $\displaystyle a+b=0 $ and the equation of bisectors is given by  $\displaystyle \frac{x^{2}-y^{2}}{a-b}=\frac{xy}{h}$
The general equation of second degree given by
$\displaystyle ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$ represent a pair of straight lines if $\displaystyle \triangle =0 $ or 
$ \displaystyle \begin{vmatrix}a&h  &g \\ h&b  &f \\ g&f  &c \end{vmatrix}=0 $ or $\displaystyle abc+2fgh-af^{2}-bg^{2}-ch^{2}=0$
On the basis of above information answer the following question

If the lines joining origin to the points of intersection of the line $\displaystyle x +y = 1$ with the curve $\displaystyle x^{2}+y^{2}+x-2y-m = 0$ are perpendicular to each other, then value of $m$ is

  1. $\displaystyle \frac{1}{2}$
  2. $\displaystyle -\frac{1}{2}$
  3. $\displaystyle 1 $
  4. $\displaystyle -1 $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

As the line $x+y=1$ intersect the curve,$\displaystyle x^{2}+y^{2}+x-2y-m=0$


$\displaystyle \Rightarrow  x^{2}+y^{2}+\left ( x-2y \right )\left ( 1 \right )-m\left ( 1 \right )^{2}=0$

$\displaystyle \Rightarrow  x^{2}+y^{2}+\left ( x-2y \right )\left ( x+y \right )-m\left ( 1 \right )^{2}=0$

$\displaystyle \Rightarrow  x^{2}+y^{2}+\left ( x-2y \right )\left ( x+y \right )-m\left ( x+y \right )^{2}=0$

$\displaystyle \Rightarrow  x^{2}+y^{2}+\left ( x^{2}-xy-2y^{2} \right )-m\left ( x^{2}+y^{2}+2xy \right )=0$

$\displaystyle \Rightarrow  x^{2}\left ( 2-m \right )+y^{2}\left ( 1-2-m \right )+xy\left ( 1-2m \right )=0$      ......(*)

Now pair of straight lines given bye (*) will be $\perp$er if

$\displaystyle 2-m+1-2-m=0$ (Using $a+b=0$)

$\displaystyle \Rightarrow m=\frac{1}{2}$

Multiple choice maths functions and graphs different forms of equation of a line

 If the lines joining the origin to the intersection of the line $y=mx+ 2$ and the curve $x^{2}+ y^{2}= 1$ are at right angles, then

  1. $m^{2}=1$
  2. $m^{2} = 3$
  3. $m^{2}= 7$
  4. $2m^{2} = 1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Joint equation of the lines joining the origin and the point of intersection of the line $y =mx + 2$ and
the curve $x^{2} + y^{2}=1$ is
$ \displaystyle x^{2} +y^{2} =\left( \frac{y-mx}{2}\right)^{2} $
$ x^{2}(4 -m^{2} )+2mxy +3y^{2} = 0$
Since these lines are at right angles
$4 -m^{2} + 3 =0 \Rightarrow m^{2} = 7$

Multiple choice maths functions and graphs different forms of equation of a line

If the straight lines joining the origin and the points of intersection of the curve $5x^2 + 12xy -6y^2 + 4x -2y + 3 = 0$  and $x + ky -1 = 0$ are equally inclined to the co-ordinate axis, then the value of $k$

  1. is equal to $1$
  2. is equal to $-1$
  3. is equal to $2$

  4. does not exist in the set of real numbers
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Homogenizing the curve with the help of the straight line.


$5x^2+12xy-6y^2+4x(x+ky) -2y(x+ky)+3(x+ky)^2 = 0$

$12x^2 + (10 + 4k + 6k) xy + (3k^2 -2k -6)y^2 = 0$

Lines are equally inclined to the coordinate axes

$\therefore$ coefficient of $xy = 0$

$\Rightarrow 10k + 10 = 0 \Rightarrow k = -1$

Multiple choice maths functions and graphs different forms of equation of a line

The lines joining the origin to the point of intersection of $3x^2 + mxy - 4x + 1 = 0$ and $2x + y - 1= 0$  are at right angles. Then which of the following is/are possible value/s of $m?$

  1. $-4$
  2. $4$
  3. $7$
  4. $3$
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

By application of the method of homogenization we get
$3x^2+mxy-4x(2x+y)+1(2x+y)^{2}$
$=3x^2+mxy-8x^2-4xy+4x^2+y^2+4xy$
$=-x^2+mxy+y^2$
$=0$
Hence the equations of the lines is given by $ x^2-mxy-y^2=0$
Hence the lines are perpendicular for all values of $m.$

Multiple choice maths functions and graphs different forms of equation of a line

Find the equation of the lines joining the origin to the points of intersection of the curve $2x^2 + 3xy -4x + 1 = 0$ and the line $3x + y = 1$

  1. $x^2-y^2-5xy=0$
  2. $x^2+y^2-5xy=0$
  3. $x^2+y^2+5xy=0$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given equations of curve and the line are $2x^2 + 3xy -4x + 1 = 0$ and $3x + y = 1$.
Homogenising the curve with the line gives
$2x^2+3xy-4x(3x+y)+(3x+y)^2=0$
$\Rightarrow 2x^2+3xy-12x^2-4xy+9x^2+y^2+6xy=0$
$\Rightarrow y^2-x^2+5xy=0$
$\therefore$ The equation of the lines joining the origin to the points of intersection of the curve and hte line is $x^2-5xy-

y^2=0$
Hence, option A.

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

If f is even function and g is an odd function, then $f _og$ is ............function.

  1. Even

  2. Odd

  3. Neither even nor odd

  4. Either even
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$fog$ function is an even function


Let $f\left(x \right)$ is a even function and $g \left( - x \right)$ is odd function.
So, $f\left( {g\left( { - x} \right)} \right) = f\left( { - g\left( x \right)} \right) = even$

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

State the whether given statement is true or false
If $f\left( x \right) = \dfrac{{x + 1}}{{x - 1}},$ then $f\left( x \right) + f\left( {\dfrac{1}{x}} \right) = 0$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$f\left( x \right) = \dfrac{{x + 1}}{{x - 1}}$


$f\left( \dfrac 1 x \right) = \dfrac{{\dfrac1x + 1}}{{\dfrac1x- 1}}=\dfrac{1+x}{1-x}=-\dfrac{1+x}{x-1}$

Hence, $f(x)+f(\dfrac1x)=\dfrac{{x + 1}}{{x - 1}}-\dfrac{{x + 1}}{{x - 1}}=0$

Multiple choice business maths functions and graphs some functions and their graphs -i graphs of the form y=ax^2+bx+c introduction to sets

Time complexity to check if an edge exists between two vertices would be __________.

  1. O(V*V)

  2. O(V+E)

  3. O(1)

  4. O(E)
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In an adjacency list representation, checking if an edge exists between two vertices u and v requires iterating through the list of neighbors of u, which takes O(degree(u)) time. In the worst case, this is O(E) or O(V). O(E) is the most appropriate choice among the options.