Tag: sets, relations and functions

Quizzes (13) Questions (60)

Questions Related to sets, relations and functions

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

Let $C$ be a curve which is locus of the point of the intersection of lines $x=2+m$ and $my=4-m$. A circle $s\equiv (x-2)^{2}+(y+1)^{2}=25$ intersector the curve cut at four points $P,Q,R$ and $S$. If $O$ is centre of the curve $C$ the $OP^{2}+OQ^{2}+OR^{2}+OS^{2}$ is

  1. $50$
  2. $100$
  3. $25$
  4. $\dfrac{25}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The locus C is found by eliminating m from x=2+m and my=4-m, yielding (x-2)y = 4-x+2, which simplifies to (x-2)(y+1)=2. This is a rectangular hyperbola centered at (2, -1). For a circle centered at the hyperbola's center, the sum of the squared distances from the center to the intersection points is 4 times the radius squared.

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

If the lines joining the origin to the inter section 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
A Correct answer
Explanation

Homogenizing the equation of the circle x^2 + y^2 = 1 using the line y = mx + 2 (written as (y-mx)/2 = 1) results in x^2 + y^2 = (y-mx)^2/4. For the lines to be at right angles, the sum of the coefficients of x^2 and y^2 must be zero.

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

If the line $y = \displaystyle \sqrt{3}x$ intersects the curve $\displaystyle x^{3}+y^{3}+3xy+5x^{2}+3y^{2}+4x+5y-1=0$ at the points $A, B, C,$ then the value of $OA.OB.OC$ is equal to: (here O is origin)

  1. $\displaystyle \frac{4}{13}\left ( 3\sqrt{3}+1 \right )$
  2. $\displaystyle \frac{4}{13}\left ( 3\sqrt{3}-1 \right )$
  3. $\displaystyle \frac{1}{26}\left ( 3\sqrt{3}-1 \right )$
  4. $\displaystyle \frac{1}{26}\left ( 3\sqrt{3}+1 \right )$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The lines $y = \sqrt {3x}$ intersects the curve at three points $A$, $B$ and $C$.


The coordinates of these points can be written as ,

$A(x _1, \sqrt{3}x _1)$

$B(x _2, \sqrt{3}x _2)$

$C(x _3, \sqrt{3}x _3)$

If $O (0,0)$ is the origin then $OA = \sqrt { (x _1)^2 + (\sqrt{3x _1})^2 }$

$\Rightarrow OA = 2x _1$

Similarly  $OB = 2x _2$

and $OC = 2x _3$

Hence $OA .OB.OC = 8  \ (x _1.x _2.x _3)$

Now putiing value of $y = \sqrt3$ into equation of given curve, we get,

$ \Rightarrow x^3 + (\sqrt3x)^3 + 3.x.\sqrt3x + 5x^2 + 3 (\sqrt3x)^2 +4x - \sqrt3x -1 = 0$

$\Rightarrow ( 1 + 3\sqrt3)x^3 + (14 +3\sqrt3)x^2 + (4 -\sqrt3)x - 1=0$ ...$(1)$

The equation $(1)$ contains the abscissa of the intersection points of the given line and curve, which are $x _1$ , $x _2$ and $x _3$

From equation $(1)$ we can see that the product of roots is $x _1.x _2.x _3  = - \left ( \dfrac { - 1}{ 1 + 3\sqrt3} \right ) = \dfrac{1}{1 + 3\sqrt3} = \dfrac { 1- 3\sqrt3}{-26}$

Hence $OA.OB.OC = 8(x _1.x _2.x _3) = 8 \times \dfrac {1 - 3\sqrt3}{-26}$

$\Rightarrow OA.OB.OC = \dfrac{4}{13} (3\sqrt3 - 1)$

So correct option is $B$.

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

The least integral value of $a$ for which the graphs of the functions $y = 2ax + 1$ and $\displaystyle y=(a-6)x^{2}-2$ do not intersect is:

  1. -6

  2. -5

  3. 3

  4. 2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For no intersection of graphs of functions,  $ y = 2ax + 1$ and $ y = (a-6)x^2 -2$, There should not any common points between two curves.


Putting the value of $y$ from equation of line into equation of given parabola, we get,

$\Rightarrow (2ax + 1) = (a-6)x^2 - 2$

$\Rightarrow (a-6)x^2  - (2a)x -3 = 0$ ...$(1)$

Equation $(1)$ is a quadratic equation in $x$. 

For no intersection of both given functions, the equation $(1)$ must not have any real solutions.

A quadratic equation have no real roots if the value of it's discriminant is less than zero.

Hence $D = b^2 - 4ac < 0 $

$\Rightarrow D = ((-2a)^2) - 4 \times (a-6) \times (-3) < 0$

$\Rightarrow  D = 4a^2 +12a -72 < 0$

$\Rightarrow (a +6)(a -3) <0$

Hence Value of $a$ for the graphs of given functions do not intersect lies between $(-6 ,3)$

So the least integral value will be $(-5)$. Correct answer is $A$.

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

The point of intersection of the two ellipse $x^2+2y^2-6x-12y+23=0$ and $4x^2+2y^2-20x-12y+35=0$

  1. lie on a circle centered at $\displaystyle \left( \frac { 8 }{ 3 } ,3 \right) $ and of radius $\displaystyle \frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 2 } } $
  2. lie on a circle centered at $\displaystyle \left( -\frac { 8 }{ 3 } ,-3 \right) $ and of radius $\displaystyle \frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 2 } } $
  3. lie on a circle centered at $\displaystyle \left( 8 ,9 \right) $ and of radius $\displaystyle \frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 3 } } $

  4. are not concyclic
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If ${S} _{1}=0$ and ${S} _{2}=0$ are the equations, then, $\lambda { S } _{ 1 }+{ S } _{ 2 }=0$ is a second degree curve passing through the points of intersection of ${S} _{1}=0$ and ${S} _{2}=0.$

$\Rightarrow \left( \lambda +4 \right) { x }^{ 2 }+2\left( \lambda +1 \right) { y }^{ 2 }-2\left( 3\lambda +10 \right) x-12\left( \lambda +1 \right) y+\left( 23\lambda +35 \right) =0$   ...(1)
For it to be a circle, choose $\lambda$ such that the coefficients of ${x}^{2}$ and ${y}^{2}$ are equal: $\Rightarrow \lambda+4=2\lambda+2$
$\therefore \lambda=2$
This gives the equation of the circle as $6\left( { x }^{ 2 }+{ y }^{ 2 } \right) -32x-36y+81=0$  {(using (1))}
$\displaystyle \Rightarrow { x }^{ 2 }+{ y }^{ 2 }-\frac { 16 }{ 3 } x-6y+\frac { 27 }{ 2 } =0$ 
Its centre is $\displaystyle C\left( \frac { 8 }{ 3 } ,3 \right) $ and radius is $\displaystyle r=\sqrt { \frac { 64 }{ 9 } +9-\frac { 27 }{ 2 }  } =\frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 2 }  } .$

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

The line $x+y=1$ meets the lines represented by the equation $y^{3}-xy^{2}-14x^{2}y+24x^{3}=0$ at the points $A, B, C$. If $O$ is the origin, then $OA^{2}+OB^{2}+OC^{2}$ is equal to

  1. $\dfrac{22}9$
  2. $\dfrac{85}{72}$
  3. $\dfrac{181}{72}$
  4. $\dfrac{221}{72}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

X-coordinate of the points are given by the roots of the equation


$24{ x }^{ 3 }+14{ x }^{ 2 }\left( x-1 \right) -x{ \left( x-1 \right)  }^{ 2 }-{ \left( x-1 \right)  }^{ 3 }=0\ \Rightarrow 36{ x }^{ 3 }-9{ x }^{ 2 }-4x+1=0\ \Rightarrow \left( 3x-1 \right) \left( 3x+1 \right) \left( 4x-1 \right) =0\ \Rightarrow x=\cfrac { 1 }{ 3 } ,-\cfrac { 1 }{ 3 } ,\cfrac { 1 }{ 4 } $

$\Rightarrow A\left( \cfrac { 1 }{ 3 } ,\cfrac { 2 }{ 3 }  \right) ,B\left( -\cfrac { 1 }{ 3 } ,\cfrac { 4 }{ 3 }  \right) $ and $C\left( \cfrac { 1 }{ 4 } ,\cfrac { 3 }{ 4 }  \right) $

Hence,

${ OA }^{ 2 }+{ OB }^{ 2 }+{ OC }^{ 2 }=\cfrac { 1 }{ 9 } +\cfrac { 4 }{ 9 } +\cfrac { 1 }{ 9 } +\cfrac { 16 }{ 9 } +\cfrac { 1 }{ 16 } +\cfrac { 9 }{ 16 } =\cfrac { 221 }{ 72 } $

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

The points of intersection of the two ellipses $x^{2}+2y^{2}-6x-12y+23=0$ and $4x^{2}+2y^{2}-20x-12y+35=0$.

  1. lie on a circle centred at $\left(\dfrac83, 3\right)$ and of radius $\displaystyle \frac{1}{3}\sqrt{\frac{47}{2}}$
  2. lie on a circle centred at $\left(-\dfrac83, 3\right)$ and of radius $\displaystyle \frac{1}{3}\sqrt{\frac{47}{2}}$
  3. lie on a circle centred at $(8, 9)$ and of radius $\displaystyle \frac{1}{3}\sqrt{\frac{47}{2}}$

  4. are not cyclic.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equation of any curve passing through the intersection of the given ellipse is
   $4x^{2}+2y^{2}-20x-12y+35+\lambda \left ( x^{2}+2y^{2}-6x-12y+23 \right )=0$
which represents a circle is
   $4+\lambda =2+2\lambda \Rightarrow \lambda =2$
and the equation of the circle is thus,
   $6x^{2}+6y^{2}-32x-36y+81=0$
$\Rightarrow $   $\displaystyle x^{2}+y^{2}-\left ( \frac{16}{3} \right )x-6y+\frac{81}{6}=0$
centre of the circle is $\left(\dfrac83, 3\right)$
and the radius is $\displaystyle \sqrt{\left ( \frac{8}{3} \right )^{2}+\left ( 3 \right )^{2}-\frac{81}{6}}$
   $\displaystyle =\sqrt{\frac{128+162-243}{18}}=\frac{1}{3}\sqrt{\frac{47}{2}}$

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

If the points of intersection of curves $\displaystyle C _{1}=\lambda x^{2}+4y^{2}-2xy-9x+3: : and: : C _{2}=2x^{2}+3y^{2}-4xy+3x-1 $ subtends a right angle at origin then the value of $\displaystyle \lambda $ is

  1. $19$
  2. $9$
  3. $-19$
  4. $-9$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given 
$C _{1} : \lambda x^2+4y^2-2xy-9x+3=0$

$C _{2} :  2x^2+3y^2-4xy+3x-1=0\Rightarrow 2x^2+3y^2-4xy-1=-3x$-----(1)

Putting (1) in $C _{1}$

$\lambda x^2+4y^2-2xy+3(2x^2+3y^2-4xy-1)+3=0$

$\lambda x^2+4y^2-2xy+6x^2+9y^2-12xy-3+3=0$

$(\lambda+6 )x^2+13y^2-14xy=0$

Above equation has a condition of perpendicularity 

Hence $\lambda+6+13=0 \Rightarrow \lambda=-19$
Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

If $x^{2}+y^{2}=a^{2}$ touches the line $y=3x+10$, then $a=$ 

  1. $\sqrt{10}$
  2. $10$
  3. $\sqrt{5}$
  4. $\dfrac{10}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a circle x^2 + y^2 = a^2 to touch a line y = mx + c, the perpendicular distance from the center (0,0) to the line must equal the radius a. The distance is |c| / sqrt(m^2 + 1) = |10| / sqrt(3^2 + 1) = 10 / sqrt(10) = sqrt(10).