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

$2x + y = 0$ is the equation of a diameter of the circle which touches the lines $4x-3y+10=0$ and $4x-3y-30=0$ The center and radius of the circle are ?

  1. $\left (-2, 1\right) ; 4$
  2. $\left (1, -2\right) ; 8$
  3. $\left (1, -2\right) ; 4$
  4. $\left (1, -2\right) ; 16$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given $4x-3y+10=0$ and $4x-3y-30=0$ touches circle implies they are tangent.

Solving the line $2x+y=0$  and  $4x-2y+10=0 $
$x=-1$ and $y=2 $ Point A

Solving the line $2x+y=0$  and  $4x-3y-30=0$
$x=3; y=-6 $ Point B

Distance between the parallel lines is length of diameter
$d=\dfrac{(C _1-C _2)}{\sqrt{(a^2+b^2)}}\\$
$d=\dfrac{(10-(-30)}{\sqrt{(16+9)}}\\$
$d=\dfrac{10+30}{5}$
$d=8$
$r=4$

O is midpoint of AB
$(x,y)=\dfrac{3-1}{2}, \dfrac{2-6}{2}$ $=(1,-2)$
Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

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:

  1. $2bc-3ad =0$
  2. $2bc+3ad =0$
  3. $3bc -2ad =0$
  4. $3bc +2ad =0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If it lies in the fourth quadrant, we get$(x,-x)$

$2ax+c = 0$ and $3bx+d = 0$
$\cfrac{c}{2a} = \cfrac{d}{3b}$
$3bc-2ad = 0$

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))|.

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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.