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Questions Related to 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:

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

  1. -6

  2. -5

  3. 3

  4. 2

Show answer
Correct Option: B
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 $A$.

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

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

  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

Show answer
Correct Option: A
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 $\displaystyle r=\sqrt { \frac { 64 }{ 9 } +9-\frac { 27 }{ 2 }  } =\frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 2 }  } .$

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

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

  1. $\dfrac{22}9$

  2. $\dfrac{85}{72}$

  3. $\dfrac{181}{72}$

  4. $\dfrac{221}{72}$

Show answer
Correct Option: D
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 } $

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

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

  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.

Show answer
Correct Option: A
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 $\displaystyle \sqrt{\left ( \frac{8}{3} \right )^{2}+\left ( 3 \right )^{2}-\frac{81}{6}}$
   $\displaystyle =\sqrt{\frac{128+162-243}{18}}=\frac{1}{3}\sqrt{\frac{47}{2}}$

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

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

  1. $19$

  2. $9$

  3. $-19$

  4. $-9$

Show answer
Correct Option: C
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 $\lambda+6+13=0 \Rightarrow \lambda=-19$

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

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

  1. $\sqrt{10}$

  2. $10$

  3. $\sqrt{5}$

  4. $\dfrac{10}{2}$

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Correct Option: A

Given, $y=3$, $y=ax^2+b$
In the system of equations above, $a$ and $b$ are constants. For which of the following values of $a$ and $b$ does the system of equations have exactly two real solutions?

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

  1. $a = 2, b = 2$

  2. $a = 2, b = 4$

  3. $a = 2, b = 3$

  4. $a = 4, b = 3$

Show answer
Correct Option: A
Explanation:

On substituting value of $y=3$ in second equation, we get
$ax^2+b=3$
$\Rightarrow ax^2+b-3=0$
$\Rightarrow ax^2=3-b$
$\Rightarrow x^2=\frac{3-b}a$
Since $x^2$ is positive quantity, therefore just $a=2$ and $b=2$ satisfies this.
Hence, option A is correct.

The system of equations:

$\displaystyle y={ x }^{ 2 }-2x$
$\displaystyle y=2x-1$ has two solutions for ($x,y$). 
Determine the greater value of $x$.

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

  1. $\displaystyle 2-\sqrt { 3 } $

  2. $\displaystyle \sqrt { 3 } $

  3. $\displaystyle 2+2\sqrt { 3 } $

  4. $\displaystyle 5$

Show answer
Correct Option: C
Explanation:
Given, $y=x^{2}-2x$
$y=2x-1$
Then $x^{2}-2x=2x-1 $
$\Rightarrow x^{2}-2x-2x+1=0$
$\Rightarrow x^{2}-4x+1=0$
Manipulate this equation $ax^{2}+bx+c=0$
We know $x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}$
$\therefore x=\dfrac{-(-4)\pm \sqrt{(-4)^{2}-4(1)(1)}}{2(1)}$
$\Rightarrow x= \dfrac{4\pm \sqrt{16-4}}{2}$
$\Rightarrow x=\dfrac{4\pm \sqrt{12}}{2}$
$\Rightarrow x=\dfrac{4\pm 2\sqrt{3}}{2}$
The greater of the two possible values for $x$ is $x=2\pm 2\sqrt{3}$. 
Therefore, the correct answer is (C).

$(x,y)$ satisfies the given set of the equations , find the value of ${x}^{2}$.
${x}^{2}+{y}^{2}=153$ and $y=-4x$

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

  1. $-51$

  2. $3$

  3. $9$

  4. $144$

Show answer
Correct Option: C
Explanation:
Let $x^{2}+y^{2}=153$.......(1)
and $y=-4x$......(2)
Put the value $y=-4x$ in equation (1), we get
$x^{2}+(-4x)^{2}=153$
$\Rightarrow x^{2}+16x^{2}=153$
$\Rightarrow 17x^{2}=153$
$\Rightarrow x^{2}=9$

If $8x+8y=18$ and $x^2-y^2=-\displaystyle\frac{3}{8}$, calculate the value of $2x-2y$.

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

  1. $-\displaystyle\frac{1}{3}$

  2. $-\displaystyle\frac{1}{6}$

  3. $\displaystyle\frac{1}{3}$

  4. $\displaystyle\frac{1}{6}$

Show answer
Correct Option: A
Explanation:

Given, $8x+8y=18$

$\Rightarrow 8(x+y)=18$
$\Rightarrow (x+y)=\dfrac{18}{8}$
$\Rightarrow (x+y)=\dfrac{9}{4}$.......(1)
Also given, $(x^{2}-y^{2})=-\dfrac{3}{8}$
$\Rightarrow (x+y)(x-y)=-\dfrac{3}{8}$.......(2)
Put the value as per equation (1), $(x+y)=\dfrac{9}{4}$, we get
$\dfrac{9}{4}\left ( x-y \right )=-\dfrac{3}{8}$
$\Rightarrow (x-y)=-\dfrac{3}{8}\times \dfrac{4}{9}$
$\Rightarrow (x-y)=-\dfrac{12}{72}$
$\Rightarrow 2(x-y)=-2\times \dfrac{12}{72}$
$\Rightarrow 2x-2y=-\dfrac{1}{3}$