Tag: sets, relations and functions

Questions Related to sets, relations and functions

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? 

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  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 $4{ x }^{ 2 }+2{ y }^{ 2 }-20x-12y+35=0$

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  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 $x^2- y^2 = 1$?

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  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 $(0,2)$

If the ellipse $\displaystyle \frac{x^{2}}{4}+\frac{y^{2}}{b^{2}}=1$ meets the ellipse $\displaystyle \frac{x^{2}}{1}+\frac{y^{2}}{a^{2}}=1$ in four distinct points and $\displaystyle a^{2} = b^{2} -4b + 8$, then $b$ lies in

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

  1. $(- \infty ,0)$

  2. $(- \infty ,2)$

  3. $(2,\infty)$

  4. $[2, \infty)$

Show answer
Correct Option: B
Explanation:
Given,
$\displaystyle \dfrac{x^{2}}{1}+\dfrac{y^{2}}{a^{2}}=1$   -- (i)

$\displaystyle \dfrac{x^{2}}{4}+\dfrac{y^{2}}{b^{2}}=1$   -- (ii) 
are the two equations of the ellipses

Eliminating $y^2$ from both the equations we get, 
$ x^2 \left( \dfrac{b^2-4a^2}{4a^2b^2} \right) =  \dfrac{b^2-a^2}{a^2b^2} $

$\dfrac{x^2}{4}  = \dfrac{b^2-a^2}{b^2-4a^2} $

Substituting the value of $a^2$ we get, 
$ \dfrac{x^2}{4} = \dfrac{4b-8}{-3b^2+16b-32} $

The denominator is always negative as the discriminant of the expression is negative and the coefficient of $b^2$ is also negative. 

Hence, $4b-8 < 0$
$\Rightarrow b <2 $

Eliminating $x^2$ from the two equations we get, 

$y^2 \left ( \dfrac{4a^2-b^2}{a^2b^2} \right) = 3 $

$ \dfrac{y^2}{3} = \dfrac {a^2b^2} { 4a^2 - b^2} $

Hence, $4a^2 -b^2 >0 $
$\Rightarrow 3b^2 -16b +32 > 0 $. 

The discriminant of the expression is less than 0 and the coefficient of $b^2$ is greater than 0. Hence, the inequality holds true for all values of $b$. 

Hence the common set of the values of $b$ is $ (-\infty,  2) $. 
Hence, option B is correct

Let $A(z _a), B(z _b), C(z _c)$ are three non-collinear points where $z _a=i, z _b=\dfrac{1}{2}+2i, z _c=1+4i$ and a curve is $z=z _a\cos^4t+2z _b\cos^2t \sin^2t+z _c\sin^4t(t\in R)$
A line bisecting AB and parallel to AC intersects the given curve at

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  1. Two distinct points

  2. Two co-incident points

  3. Only one point

  4. No point

Show answer
Correct Option: C
Explanation:

${ z } _{ a }=i\Rightarrow A\left( 0,1 \right) $
${ z } _{ b }=\cfrac { 1 }{ 2 } +2i\Rightarrow B\left( \cfrac { 1 }{ 2 } ,2 \right) $
${ z } _{ c }=1+4i\Rightarrow C\left( 1,4 \right) $
Let D be the midpoint of AB
$D=\left[ \cfrac { 0+\cfrac { 1 }{ 2 }  }{ 2 } ,\cfrac { 1+2 }{ 2 }  \right] $
$D\left[ \cfrac { 1 }{ 4 } ,\cfrac { 3 }{ 2 }  \right] $
Line bisecting AB at D is parallel to AC
$\therefore $ Slope of AC $=\cfrac { 4-1 }{ 1-0 } =3$
Equation of line bisecting AB is $y-\cfrac { 3 }{ 2 } =3\left( x-\cfrac { 1 }{ 4 }  \right) $

$\Rightarrow \cfrac { 2y-3 }{ 2 } =\cfrac { 12x-3 }{ 4 } $
$\Rightarrow 4y-6=12x-3$
$\Rightarrow 12x-4y+3=0$
$\Rightarrow y=\cfrac { 12x+3 }{ 4 } $
Equation of curve is $y={ \left( x+1 \right)  }^{ 2 }$
$\cfrac { 12x+3 }{ 4 } ={ x }^{ 2 }+2x+1$
$4{ x }^{ 2 }-4x+1=0$
${ \left( 2x-1 \right)  }^{ 2 }=0$
$x=\cfrac { 1 }{ 2 } ,\quad y=\cfrac { 9 }{ 4 } $
$\left( \cfrac { 1 }{ 2 } ,\cfrac { 9 }{ 4 }  \right) \Rightarrow $ given line and curve intersect only at one point

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

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  1. $\dfrac{4}{13}\left ( 3\sqrt{3}-1 \right )$

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

  3. $\dfrac{1}{\sqrt{3}}\left ( 2+7\sqrt{3} \right )$

  4. $\dfrac{4}{13}\left ( 3\sqrt{3}+1 \right )$

Show answer
Correct Option: A
Explanation:

Coordinates of a point on the line $y=x\sqrt{3}$ at a distance r from origin 
is $\left ( r\cos \theta ,r\sin \theta  \right )$
$\therefore \tan \theta =\sqrt{3}$
$\therefore \left ( \dfrac{r}{2},\dfrac{r\sqrt{3}}{2} \right )$ lies on the given curve
$\Rightarrow  \displaystyle \frac{r^{3}}{8}+\frac{r^{3}.3\sqrt{3}}{8}+3.\frac{r}{2}.\frac{r\sqrt{3}}{2}+5.\frac{r^{2}}{4}+3.\frac{r^{2}.3}{4}+4.\frac{r}{2}+5.\frac{r\sqrt{3}}{2}-1=0$

$\Rightarrow \left (\displaystyle  \frac{1+3\sqrt{3}}{8} \right )r^{3}+\dfrac{r^{2}}{4}\left ( 3\sqrt{3}+14 \right )+\dfrac{r}{2}\left ( 5\sqrt{3}+4 \right )-1=0$

$\Rightarrow r _{1}.r _{2}.r _{3}=\dfrac{8}{3\sqrt{3}+1}$

$=\dfrac{8}{27-1}\times \left ( 3\sqrt{3}-1 \right )$

$=\dfrac{4}{13} \left ( 3\sqrt{3}-1 \right )$

The pair of lines $6{ x }^{ 2 }+7xy+\lambda { y }^{ 2 }=0\left( \lambda \neq -6 \right) $ forms a right angled triangle with $x+3y+4=0$ then $\lambda=$

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  1. $3$

  2. $-3$

  3. $1/3$

  4. $-1/3$

Show answer
Correct Option: A
Explanation:

Given line is $L: x+3y+4=0$


$\implies  y=-\dfrac{1}{3}(x+4)$

Slope of this line is $m=\dfrac{-1}{3}$

Now, $6x^2+7xy+\lambda y^2=0$ $(\lambda\neq 6)$

$x^2+\dfrac{7}{6}xy+\dfrac{\lambda}{6}y^2=0$

$\implies (x+ay)(x+by)=0$

$\implies x+ay=0$ and $x+by=0$ are the two equations with 

$a+b=\dfrac{7}{6}$    and $ab=\dfrac{\lambda}{6}$

Slope of these lines are $m _1=\dfrac{-1}{a}$ and $m _2=\dfrac{-1}{b}$

Now, $m _1m _2=\dfrac{1}{ab}=\dfrac{\lambda}{6}\neq -1$  since $\lambda\neq -6$

Hence the lines $x+ay=0$ and $x+by=0$ are not prependicular.

From these two only one is normal to $L$.

Let $x+ay$ is normal to $L$.

$\implies m _1m=-1$

$\implies \dfrac{1}{3a}=-1$

$\implies a=\dfrac{-1}{3}$

Now, $a+b=\dfrac{7}{6}\implies b=\dfrac{7}{6}-\dfrac{-1}{3}$

$\implies b=\dfrac{3}{2}$

Now, $ab=\dfrac{\lambda}{6}$

$\implies \lambda=6ab=6\times \dfrac{-1}{3}\times \dfrac{3}{2}$

$\implies \lambda=-3$

Answer-(B)

Let $y=f(x)$ and $y=g(x)$ be the pair of curves such that
(i) The tangents at point with equal abscissae intersect on y-axis.
(ii) The normal drawn at points with equal abscissae intersect on x-axis and
(iii) curve f(x) passes through $(1, 1)$ and $g(x)$ passes through $(2, 3)$ then: The curve g(x) is given by.

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

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

  2. $x+\displaystyle\frac{2}{x}$

  3. $x^2-\displaystyle\frac{1}{x^2}$

  4. $(x+\displaystyle\frac{1}{x})$$(x+\displaystyle\frac{2}{x})$

Show answer
Correct Option: A
Explanation:
$y=f(x)$ and $y=g(x)$
(i)
$\dfrac{dy}{dy}=c\Rightarrow c _{1}=1$
for second curve 
$\dfrac{dy}{dy}=c _{2}\Rightarrow c _{2}=1$

(ii)
$\dfrac{dy}{dx}=d _{1}---(1)$
for second curve 
$\dfrac{dy}{dx}=d _{2}----(2)$

(iii)
From eq (1) and (2)
$d _{1}=2=d _{2}$

$\int _{1}^{2} (g(x)-f(x))d(x)$

$\int _{1}^{2} (4-2)d(x)$

$2\int _{1}^{2} d(x)$

$2(2-1)=2$

SO $g(x)=x-\dfrac{1}{x}$