Tag: functions and their graphs

Questions Related to functions and their graphs

Let $a,b,c$ and $d$ be non-zero numbers. If the point of intersection of the lines $4ax+2ay+c=0$ and $5bx+2by+d=0$ lies in the fourth quadrant and is equidistant from the two axes then

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  1. $2bc-3ad=0$

  2. $2bc+3ad=0$

  3. $3bc-2ad=0$

  4. $3bc+2ad=0$

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Correct Option: C
Explanation:
Since point of intersection lies in the fourth quadrant and is equidistant from coordinate axes,

the $x$ and $y$ co-ordinates will be same 

Hence the coordinates become $(h,-h)$

Passing $4ax+2ay+c=0$ through $(h,-h)$ 

$\Rightarrow 4ah-2ah+c=0$

$\Rightarrow h=-\dfrac{c}{2a}----------(1)$

Also passing the second line $5bx+2by+d=0$ through $(h,-h)$

$\Rightarrow 5bh-2bh+d=0$

$\Rightarrow h=-\dfrac{d}{3b}----(2)$

From eq (1) and (2)

$-\dfrac{c}{2a}=-\dfrac{d}{3b}$

$\Rightarrow 3bc-2ad=0$

The straight line passes through the point of intersection of the straight lines $x+2y-10=0$ and $2x+y+5=0$, is 

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  1. $5x-4y=0$

  2. $5x+4y=0$

  3. $4x-5y=0$

  4. $4x+5y=0$

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Correct Option: B
Explanation:

The line passes through the point of intersection of the equations $x+2y-10=0$  and $2x+y+5=0$

Now,
$\ x+2y-10=0....(i)\ 2x+y+5=0....(ii)\times 2\ =>4x+2y+10=0....(iii)$
Subtracting (iii) and (i), we get,
$-3x-20=0\ =>x=\cfrac { -20 }{ 3 } \ \therefore y=\cfrac { 25 }{ 3 } $
Now the line must pass through $(\cfrac { -20 }{ 3 } ,\cfrac { 25 }{ 3 } )$ 
Therefore, $5x+4y=0$ passes through $(\cfrac { -20 }{ 3 } ,\cfrac { 25 }{ 3 } )$

If the line $y-\sqrt{3}x+3=0$ cuts the curve $y^{2}=x+2$ at $A$ and $B$ and point on the line $P$ is $\left(\sqrt{3},0\right)$ then $\left|PA.PB\right|=$

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  1. $\dfrac{4\left(\sqrt{3}+2\right)}{3}$

  2. $\dfrac{4\left(2-\sqrt{3}\right)}{3}$

  3. $\dfrac{4\sqrt{3}+}{2}$

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

Show answer
Correct Option: A
Explanation:
Line $ = y - \sqrt 3 x + 3 = 0$
$\begin{array}{l} y=\sqrt { 3 } x-3 \\ m=\sqrt { 3 } =\tan  \theta  \\ \theta =\frac { \pi  }{ 3 } =60^{ \circ  }\to \left( i \right)  \end{array}$
Parametric form of line
$\begin{array}{l} =\frac { { x-{ x _{ 1 } } } }{ { \cos  \theta  } } =\frac { { y-{ y _{ 1 } } } }{ { \sin  \theta  } } =r  \\ \therefore x={ x _{ 1 } }+r \cos  \theta  \\ y={ y _{ 1 } }+r \sin  \theta  \end{array}$
As we know $P\left( {\sqrt 3 ,0} \right)$
$\therefore \left. \begin{array}{l} x=\sqrt { 3 } +r \cos  \theta  \\ y=0+r \sin  \theta  \end{array} \right\} lie\, \, on\, \, parabola\, \, { y^{ 2 } }=x+2$
$\begin{array}{l} \therefore { \left( { r\sin  \theta  } \right) ^{ 2 } }=\sqrt { 3 } +r\cos  \theta +2 \\ { r^{ 2 } }{ \sin ^{ 2 }  }\theta =\sqrt { 3 } +r\cos  \theta +2 \\ { r^{ 2 } }{ \sin ^{ 2 }  }\theta -r\cos  \theta -\sqrt { 3 } -2=0\, \, \, \begin{array} { *{ 20 }{ c } }{ { r _{ 1 } }=PA } \\ { { r _{ 2 } }=PB } \end{array} \\ { r _{ 1 } }{ r _{ 2 } }=\frac { { -\sqrt { 3 } -2 } }{ { { { \sin   }^{ 2 } }\theta  } } \to \left( { ii } \right)  \\ from\, \, \left( i \right) \, \, \theta =60^{ \circ  }\, \, \sin  60^{ \circ  }=\frac { { \sqrt { 3 }  } }{ 2 } \, \, { \sin ^{ 2 }  }60=\frac { 3 }{ 4 }  \\ from\, \, \left( { ii } \right) \, \, \frac { { -\sqrt { 3 } -2 } }{ { \frac { 3 }{ 4 }  } } =\frac { { -4\left( { \sqrt { 3 } +2 } \right)  } }{ 3 }  \\ \therefore \left| { PA.PB } \right| =\left| { { r _{ 1 } }{ r _{ 2 } } } \right| =\frac { { 4\left( { \sqrt { 3 } +2 } \right)  } }{ 3 }  \end{array}$
Hence ans is A.

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 :

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  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:

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  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

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  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

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  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

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  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? 

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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)
$x=2-\sqrt{\dfrac{3}{2}}$

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