Tag: functions and their graphs

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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 ${ x }^{ 2 }+2{ y }^{ 2 }-6x-12y+23=0$ and $4{ x }^{ 2 }+2{ y }^{ 2 }-20x-12y+35=0$

  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

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 ${ S } _{ 2 }=0$

$\Rightarrow \left( \lambda +4 \right) { x }^{ 2 }+2\left( \lambda +1 \right) { y }^{ 2 }-2\left( 3\lambda +10 \right) x-12\left( \lambda +1 \right) y+\left( 23\lambda +35 \right) =0$
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 }  } $

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

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

  1. $0$
  2. $1$
  3. $2$$
  4. $3$
  5. $4$
Reveal answer Fill a bubble to check yourself
E Correct answer
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 $4$.

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

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

  1. $(-2, 0)$ and $(0,-2)$
  2. $(2,0)$ and $(0,2)$
  3. $(3,0)$ and $(0,3)$
  4. $(1,0)$ and $(0,1)$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $(0,2)$
Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

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

  1. $(- \infty ,0)$
  2. $(- \infty ,2)$
  3. $(2,\infty)$
  4. $[2, \infty)$
Reveal answer Fill a bubble to check yourself
B Correct answer
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

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

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

  1. Two distinct points

  2. Two co-incident points

  3. Only one point

  4. No point
Reveal answer Fill a bubble to check yourself
C Correct answer
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

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

If the line $y=x\sqrt{3}$ 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)

  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. $\dfrac{4}{13}\left ( 3\sqrt{3}+1 \right )$
Reveal answer Fill a bubble to check yourself
A Correct answer
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
$\Rightarrow  \displaystyle \frac{r^{3}}{8}+\frac{r^{3}.3\sqrt{3}}{8}+3.\frac{r}{2}.\frac{r\sqrt{3}}{2}+5.\frac{r^{2}}{4}+3.\frac{r^{2}.3}{4}+4.\frac{r}{2}+5.\frac{r\sqrt{3}}{2}-1=0$

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

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

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

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

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

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 $\lambda=$

  1. $3$
  2. $-3$
  3. $1/3$
  4. $-1/3$
Reveal answer Fill a bubble to check yourself
A Correct answer
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)

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

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.

  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})$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $g(x)=x-\dfrac{1}{x}$