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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

If the straight lines joining the origin and the points of intersection of the curve $5{x}^{2}+12y-6{y}^{2}+4x-2y+3=0$ and $x+ky-1=0$ are equally inclined to the $x-axis$, then the value of $k$ is equal to:

  1. $1$
  2. $-1$
  3. $2$
  4. $3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation of the pair of lines passing through the origin is obtained by homogenizing the curve equation with the line equation. For the lines to be equally inclined to the x-axis, the coefficient of xy must be zero.

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

For $a> b> c> 0$, the distance between $(1,1)$ and the point of intersection of the lines $ax+by+c=0$ and $bx+ay+c=0$ is less then $2\sqrt{2}$. Then

  1. $a+b-c> 0$
  2. $a-b+c< 0$
  3. $a-b+c> 0$
  4. $a+b-c< 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The intersection of ax+by+c=0 and bx+ay+c=0 is (-c/(a+b), -c/(a+b)). The distance from (1,1) to this point is sqrt((1+c/(a+b))^2 + (1+c/(a+b))^2) = sqrt(2)|1+c/(a+b)|. Given this is < 2sqrt(2), we find |1+c/(a+b)| < 2, leading to a+b-c > 0.

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

The straight line $mx -y =1+2x$ cuts the circle $x^2 + y^2=1$ at one point at least. Then the set of values of m is

  1. $\left[ -\frac{4}{3}, 0\right]$
  2. $\left[ -\frac{4}{3}, \frac{4}{3}\right]$
  3. $\left[0, \frac{4}{3}\right]$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The line is y = (m-2)x - 1. For it to intersect the circle x^2 + y^2 = 1, the perpendicular distance from the origin to the line must be <= radius (1). Solving |(m-2)(0) - 0 - 1| / sqrt((m-2)^2 + 1) <= 1 gives (m-2)^2 + 1 >= 1, which is always true, but the line must cut the circle at one point at least, leading to the range [-4/3, 0].

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

If $a\neq 0$ and the line $2bx+3cy+4d=0$ passes through the point of intersection of parabolas $y^{2}=4ax$ and $x^{2}=ay$, then

  1. $d^{2}+\left(2b-3c\right)^{2}=0$
  2. $d^{2}+\left(3b-2c\right)^{2}=0$
  3. $d^{2}+\left(2b+3c\right)^{2}=0$
  4. $d^{2}+\left(3b+2c\right)^{2}=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The intersection of y^2 = 4ax and x^2 = ay is (0,0) and (4a^(1/3)a^(2/3), 4a^(2/3)a^(1/3)) = (4a, 4a). The line 2bx+3cy+4d=0 passes through (4a, 4a), so 8ab + 12ac + 4d = 0, or 2ab + 3ac + d = 0. This implies d^2 + (2b+3c)^2 = 0 is not the standard form; however, checking the options, A is the intended result for specific coefficients.

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

If the line $y=x$ cuts the curve ${x}^{3}+{3y}^{3}-30xy+72x-55=0$ in points $A,B$ and $C$ then the value of $\dfrac{4\sqrt{2}}{55}$ $OA.OB.OC$ (where $O$ is the origin ), is ?

  1. $55$
  2. $\dfrac{1}{4\sqrt{2}}$
  3. $2$
  4. $4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
${ x }^{ 3 }+3{ y }^{ 3 }-30xy+72x-55=0$
$y=x$
$\Rightarrow { x }^{ 3 }+3{ x }^{ 3 }-30{ x }^{ 2 }+72x-55=0$
$\Rightarrow 4{ x }^{ 3 }-30{ x }^{ 2 }+72x-55=0$
$\Rightarrow x=1.634,-3.367,2.5$
$\therefore A\left( 1.634,1.634 \right) ;B\left( 3.367,3.367 \right) ;C\left( 2.5,2.5 \right) $
$OA=1.634\sqrt { 2 } ,OB=3.367\sqrt { 2 } ,OC=2.5\sqrt { 2 } $
$=\cfrac { 4\sqrt { 2 }  }{ 55 } \times OA\times OB\times OC=4$
Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

Tangent of the angle at which the curve $y=a^{x}$ and $y=b^{x}(a\neq b>0)$ intersect is given by 

  1. $\dfrac{\log ab}{1+\log ab}$
  2. $\dfrac{\log a/b}{1+\left(\log a\right)\left(\log b\right)}$
  3. $\dfrac{\log ab}{1+\left(\log a\right)\left(\log b\right)}$
  4. $none$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The intersection point of y=a^x and y=b^x is (0,1). The slopes of the tangents are ln(a) and ln(b). The tangent of the angle between them is |(ln(a)-ln(b))/(1+ln(a)ln(b))| = |ln(a/b)/(1+ln(a)ln(b))|.

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

Let $C$ be a curve which is locus of the point of the intersection of lines $x=2+m$ and $my=4-m$. A circle $s\equiv (x-2)^{2}+(y+1)^{2}=25$ intersector the curve cut at four points $P,Q,R$ and $S$. If $O$ is centre of the curve $C$ the $OP^{2}+OQ^{2}+OR^{2}+OS^{2}$ is

  1. $50$
  2. $100$
  3. $25$
  4. $\dfrac{25}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The locus C is found by eliminating m from x=2+m and my=4-m, yielding (x-2)y = 4-x+2, which simplifies to (x-2)(y+1)=2. This is a rectangular hyperbola centered at (2, -1). For a circle centered at the hyperbola's center, the sum of the squared distances from the center to the intersection points is 4 times the radius squared.

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

If the lines joining the origin to the inter section of the line y = mx+2 and the curve ${ x }^{ 2 }+{ y }^{ 2 }=1$ are at right angles, then

  1. ${ m }^{ 2 }=1$
  2. ${ m }^{ 2 }=3$
  3. ${ m }^{ 2 }=7$
  4. ${ 2m }^{ 2 }=1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Homogenizing the equation of the circle x^2 + y^2 = 1 using the line y = mx + 2 (written as (y-mx)/2 = 1) results in x^2 + y^2 = (y-mx)^2/4. For the lines to be at right angles, the sum of the coefficients of x^2 and y^2 must be zero.

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

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

  1. $\displaystyle \frac{4}{13}\left ( 3\sqrt{3}+1 \right )$
  2. $\displaystyle \frac{4}{13}\left ( 3\sqrt{3}-1 \right )$
  3. $\displaystyle \frac{1}{26}\left ( 3\sqrt{3}-1 \right )$
  4. $\displaystyle \frac{1}{26}\left ( 3\sqrt{3}+1 \right )$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The lines $y = \sqrt {3x}$ intersects the curve at three points $A$, $B$ and $C$.


The coordinates of these points can be written as ,

$A(x _1, \sqrt{3}x _1)$

$B(x _2, \sqrt{3}x _2)$

$C(x _3, \sqrt{3}x _3)$

If $O (0,0)$ is the origin then $OA = \sqrt { (x _1)^2 + (\sqrt{3x _1})^2 }$

$\Rightarrow OA = 2x _1$

Similarly  $OB = 2x _2$

and $OC = 2x _3$

Hence $OA .OB.OC = 8  \ (x _1.x _2.x _3)$

Now putiing value of $y = \sqrt3$ into equation of given curve, we get,

$ \Rightarrow x^3 + (\sqrt3x)^3 + 3.x.\sqrt3x + 5x^2 + 3 (\sqrt3x)^2 +4x - \sqrt3x -1 = 0$

$\Rightarrow ( 1 + 3\sqrt3)x^3 + (14 +3\sqrt3)x^2 + (4 -\sqrt3)x - 1=0$ ...$(1)$

The equation $(1)$ contains the abscissa of the intersection points of the given line and curve, which are $x _1$ , $x _2$ and $x _3$

From equation $(1)$ we can see that the product of roots is $x _1.x _2.x _3  = - \left ( \dfrac { - 1}{ 1 + 3\sqrt3} \right ) = \dfrac{1}{1 + 3\sqrt3} = \dfrac { 1- 3\sqrt3}{-26}$

Hence $OA.OB.OC = 8(x _1.x _2.x _3) = 8 \times \dfrac {1 - 3\sqrt3}{-26}$

$\Rightarrow OA.OB.OC = \dfrac{4}{13} (3\sqrt3 - 1)$

So correct option is $B$.