Tag: maths

Questions Related to maths

The lines $x+y=\left|\ a\ \right|$ and $ax-y=1$ intersect each other in the first quadrant. Then the set of all possible values of $a$ is the interval :

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

  1. $\left( 0,\infty \right)$

  2. $\left[ 1,\infty \right)$

  3. $\left( -1,\infty \right)$

  4. $\left( -1,1 \right] $

Show answer
Correct Option: B
Explanation:

Given lines are :

$x+y=\left | a\right |$ and $ax-y=1$
Case $1:a>0$
$x+y=a-----(1)$
and $ax-y=1------(2)$
Adding $(1)+(2)$
$\Rightarrow x(1+a)=1+a$
$\Rightarrow x=1$
Hence, $y=a-1$
Since it is in first quadrant,$a-1\ge 0$
$\Rightarrow a\ge 1$

Case $2:a<0$
$x+y=-a$ and $ax-y=1$
Solving For $x,y$
$x=\dfrac{1-a}{1+a}>0$
$\Rightarrow \dfrac{a-1}{a+1}<0$
$\Rightarrow a\epsilon (-1,1)$
also ,$y=-a-\left(  \dfrac{1-a}{1+a}\right )$ which should be $>0$
$\Rightarrow -\dfrac{a^2+1}{a+1}>0$
$\Rightarrow a<-1$
Combining both two cases, we get :
$a\ge 1$
$a\epsilon[1,\infty)$

If the line $y - 1 = m(x -1)$ cuts the circle $x^{2} + y^{2} = 4$ at two real points then the number of possible values of $m$ is:

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

  1. $1$

  2. $2$

  3. Infinite

  4. None of these

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Correct Option: B
Explanation:

Given circle is ${ x }^{ 2 }+{ y }^{ 2 }=4$,

Given that the line $y-1=m(x-1)$ intersects the circle at two different points
If the perpendicular distance from the centre of the circle to the line is less than the radius of the circle then the line intersects at two different real points
$\Longrightarrow \dfrac { \left| m-1 \right|  }{ \sqrt { 1+{ m }^{ 2 } }  } <2\ $ 
Squaring on both sides gives,
${ m }^{ 2 }-2m+1<4+4{ m }^{ 2 }\ \Longrightarrow 3{ m }^{ 2 }+2m+3>0\ $,
Given quadratic equation has complex roots and the co-efficient of ${ x }^{ 2 }$ is positive
$\therefore$ The quadratic equation is always positive
Hence, infinite values of $m$ exist to intersect the line at $2$ diffferent real points.

The set of values of $c$ so that the equations $\displaystyle y=\left | x \right |+c: : and: : x^{2}+y^{2}-8\left | x \right |-9=0 $ have no solution is

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

  1. $\displaystyle \left ( -\infty ,-3 \right )\cup \left ( 3,\infty \right )$

  2. $(-3, 3)$

  3. $\displaystyle \left ( -\infty ,-5\sqrt{2} \right )\cup \left ( 5\sqrt{2},\infty \right )$

  4. $\displaystyle \left ( -\infty ,-4-5\sqrt{2} \right )\cup \left ( 5\sqrt{2}-4,\infty \right )$

Show answer
Correct Option: D
Explanation:

Given equation 

$y=\left |x\right |+c$---(1)
$x^2+y^2-8\left| x\right |-9=0$
From equation (1) and (2)
$x^2+(\left | x\right |+c)^2-8\left |x\right |-9=0$
when $x>0$
$x^2+(x+c)^2-8x-9=0$
$x^2+x^2+c^2+2cx-8x-9=0$
$2x^2+x(2c-8)+c^2-9=0$
For no solution 
$D<0$
$(2c-8)^2-4\times 2 (c^2-9)<0$
$4c^2+64-32c-8c^2+72<0$
$-4c^2-32c+136<0$
$c^2+8c-34>0$
$c=\dfrac{-8\pm\sqrt{64+136}}{2}$

$c=\dfrac{-8\pm\sqrt{200}}{2}$

$c=-4\pm 5\sqrt{2}$
C has root $c=-4\pm5\sqrt{2}$
Hence for no solution c has all value excluding it's roots  
$c\epsilon(-\infty,-4-5\sqrt{2})\cup(5\sqrt{2}-4,\infty)$


The number of points of intersection of the two curves $\mathrm{y}= 2$ sinx and $\mathrm{y}= 5\mathrm{x}^{2}+2\mathrm{x}+3$ is

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

  1. 0

  2. 1

  3. 2

  4. $\infty$

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Correct Option: A
Explanation:

$ 2sinx = 5x^{2} + 2x + 3$
$ 2sinx = 5(x+ \frac{1}{5})^{2} + \frac{4}{5} + 2 > 2 \geq 2sinx$
so no solution

What are the coordinates of the points intersection of the line with equation $y=x+1$ and circle with equation ${x}^{2}+{y}^{2}=5$

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

  1. $-2,0$

  2. $1,2$

  3. $-2,1$

  4. $-2,-1$

  5. $1,3$

Show answer
Correct Option: C
Explanation:

Put $y=x+1$ in the equation of the circle $x^2+y^2=5$ as shown below:

$x^2+y^2=5$
$\Rightarrow x^2+(x+1)^2=5$
$\Rightarrow x^2+x^2+1+2x=5$
$\Rightarrow 2x^2+1+2x-5=0$
$\Rightarrow 2x^2+2x-4=0$ or $x^2+x-2=0$
Factorising the above quadratic equation, we get:
$x^2+x-2=0$
$\Rightarrow x^2+2x-x-2=0$
$\Rightarrow x(x+2)-1(x+2)=0$
$\Rightarrow (x+2)=0$ and $(x-1)=0$ 
$\rightarrow x=-2$ and $x=1$ 
Hence, the coordinates of the points intersection is $-2,1$.

If $a, b, c$ form a G,P, with common ratio $r$, the sum of the ordinates of the points of intersection of the line $ax + by + c = 0$ and the curve $x + 2y^{2} =0 $ is

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

  1. $-\dfrac{r^{2}}{2} $

  2. $-\dfrac{r}{2}$

  3. $\dfrac{r}{2}$

  4. $\dfrac{r^2}{2}$

Show answer
Correct Option: C
Explanation:

The equation of the given line is $ax + by +c =0$ $\Rightarrow ax + ary + ar^{2} = 0 $ $\Rightarrow x + ry + r^{2}= 0 $ (i)
(i) intersects the curves  $x + 2y^{2} = 0 $ at the points whose ordinates are given by
$-2y^{2} + ry + r^{2} = 0 $or $2y^{2} -ry -r^{2}= 0 $
Therefore required sum of the ordinates $= r/2$

The equations $(x-2)^2+y^2=3$ and $y=-x+2$ represent a circle and a line that intersects the circle across its diameter. What is the point of intersection of the two equations that lie in quadrant II? 

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

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

  2. $(-4, 2)$

  3. $(2+\sqrt{3}, 2)$

  4. $(2-3\sqrt{2}, 3\sqrt{2})$

Show answer
Correct Option: D
Explanation:
Given equation 
$(x-2)^2+y^2=3----(1)$
$y=-x+2----(2)$
Putting eq (2) in (1)
$(-y)^2+y^2=3$
$2y^2=3$
$y=\sqrt{\dfrac{3}{2}}$(point lies in $II$ quadrant, so $y$ will be positive)
$x=2-\sqrt{\dfrac{3}{2}}$

$\left ( 2-\sqrt{\dfrac{3}{2}},\sqrt{\dfrac{3}{2}} \right )$

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

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

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

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

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

  4. are not cyclic 

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 ${ S } _{ 2 }=0$

$\Rightarrow \left( \lambda +4 \right) { x }^{ 2 }+2\left( \lambda +1 \right) { y }^{ 2 }-2\left( 3\lambda +10 \right) x-12\left( \lambda +1 \right) y+\left( 23\lambda +35 \right) =0$
For it to be a circle, choose $\lambda$ such that the coefficients of ${ x }^{ 2 }$ and ${ y }^{ 2 }$ are equal:
$\Rightarrow \lambda +4=2\lambda +2\Rightarrow \lambda =2$
This gives the equation of the circle as
$\displaystyle 6\left( { x }^{ 2 }+{ y }^{ 2 } \right) -32x-36y+81=0$    (Using (1))
$\displaystyle \Rightarrow { x }^{ 2 }+{ y }^{ 2 }-\frac { 16 }{ 3 } x+6y+\frac { 27 }{ 2 } =0$
Its center 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 }  } $

How many points of intersection are between the graphs of the equations $x^2+ y^2 = 7$ and $x^2- y^2 = 1$?

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

  1. $0$

  2. $1$

  3. $2$$

  4. $3$

  5. $4$

Show answer
Correct Option: E
Explanation:

Given ${x}^{2}+{y}^{2}=7$ and ${x}^{2}-{y}^{2}=1$
Add two equations, we get $2{x}^{2}=8$ , which implies ${x}^{2}=4$
Therefore $x = \pm2$ , we get $y=\pm \sqrt3$
So, number of solutions is $4$.

Find the point(s) of intersection of the circle with equation ${x}^{2}+{y}^{2}=4$ and the circle with equations ${(x-2)}^{2}+{(y-2)}^{2}=4$

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

  1. $(-2, 0)$ and $(0,-2)$

  2. $(2,0)$ and $(0,2)$

  3. $(3,0)$ and $(0,3)$

  4. $(1,0)$ and $(0,1)$

Show answer
Correct Option: B
Explanation:
Let $x^2+y^2=4$    ...........(1)
We first expand the given second equation $(x-2)^2+(y-2)^2=4$ as follows: 
$(x-2)^2+(y-2)^2=4$
$\Rightarrow x^2+4-4x+y^2+4-4y=4$
$\Rightarrow x^2+y^2-4x-4y=-4$       ........(2)
Now subtracting equation (1) from equation (2) we get,
$x^2+y^2-4x-4y-x^2-y^2=-4-4$
$\Rightarrow -4x-4y=-8$
$\Rightarrow 4x+4y=8$
$\Rightarrow x+y=2$
$\Rightarrow x=2-y$
We now substitute $x$ by $2 - y$ in the first equation to obtain 
$(2-y)^2+y^2=4$
$\Rightarrow 4+y^2-4y+y^2=4$
$\Rightarrow 2y^2-4y=4-4$
$\Rightarrow 2y^2-4y=0$
$\Rightarrow 2y(y-2)=0$
$\Rightarrow 2y=0$ and $(y-2)=0$
$\Rightarrow y=0$ and $y=2$
Put $y=0$ in equation (1) that is :
$x^2+(0)^2=4$
$\Rightarrow x^2=4$
$\Rightarrow x=2$
Now put $y=2$ in equation (1) that is :
$x^2+(2)^2=4$
$\Rightarrow x^2+4=4$
$\Rightarrow x^2=4-4$
$\Rightarrow x^2=0$
$\Rightarrow x=0$
The two points of intersection of the two circles are given by, 
$(2,0)$ and $(0,2)$