Tag: functions and graphs

Questions Related to functions and graphs

$\displaystyle ax^{2}+2hxy+by^{2}=0$ 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

maths functions and graphs different forms of equation of a line

  1. $\displaystyle \frac{1}{2}$

  2. $\displaystyle -\frac{1}{2}$

  3. $\displaystyle 1 $

  4. $\displaystyle -1 $

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Correct Option: A
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 $a+b=0$)

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

 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

maths functions and graphs different forms of equation of a line

  1. $m^{2}=1$

  2. $m^{2} = 3$

  3. $m^{2}= 7$

  4. $2m^{2} = 1$

Show answer
Correct Option: C
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
$4 -m^{2} + 3 =0 \Rightarrow m^{2} = 7$

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 $k$

maths functions and graphs different forms of equation of a line

  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

Show answer
Correct Option: B
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 $xy = 0$

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

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 $m?$

maths functions and graphs different forms of equation of a line

  1. $-4$

  2. $4$

  3. $7$

  4. $3$

Show answer
Correct Option: A,B,C,D
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 $m.$

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 $3x + y = 1$

maths functions and graphs different forms of equation of a line

  1. $x^2-y^2-5xy=0$

  2. $x^2+y^2-5xy=0$

  3. $x^2+y^2+5xy=0$

  4. $None\ of\ these$

Show answer
Correct Option: A
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.

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

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

  1. Even

  2. Odd

  3. Neither even nor odd

  4. Either even

Show answer
Correct Option: A
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, $f\left( {g\left( { - x} \right)} \right) = f\left( { - g\left( x \right)} \right) = even$

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

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

  1. True

  2. False

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Correct Option: A
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, $f(x)+f(\dfrac1x)=\dfrac{{x + 1}}{{x - 1}}-\dfrac{{x + 1}}{{x - 1}}=0$

The identity function on real numbers given by $f(x)=x$ is continuous at every real numbers.

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

  1. True

  2. False

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Correct Option: A

The minimum value of $f\left( x \right) ={ x }^{ 2 }+2x+3 ,x\in R$ is equal to 

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

  1. $2$

  2. $3$

  3. $4$

  4. $1$

Show answer
Correct Option: A
Explanation:

$f(x)=x^2+2x+3$


For minima: 


$f'(x)=0$

$f'(x)=2x+2=0$

$\implies x=-1$

Hence minimum value of $f(x)$ is at $x=-1$, i.e.

$f(-1)=(-1)^2-2+3=2$

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

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

  1. O(V*V)

  2. O(V+E)

  3. O(1)

  4. O(E)

Show answer
Correct Option: D