Tag: maths

Questions Related to 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. $3$


Correct Option: B

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. $a+b-c< 0$


Correct Option: A

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


Correct Option: A

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. $d^{2}+\left(3b+2c\right)^{2}=0$


Correct Option: A

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. $4$


Correct Option: D
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 } $
$=\cfrac { 4\sqrt { 2 }  }{ 55 } \times OA\times OB\times OC=4$

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. $none$


Correct Option: A

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. $\dfrac{25}{2}$


Correct Option: A

The point of intersection of the tangents drawn to the curve $x^2y=1 -y$ at the point where it is met by the curve xy=1-y is given by 

  1. (0,-1)

  2. (1,1)

  3. (0,1)

  4. $(0,\infty )$


Correct Option: A

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. ${ 2m }^{ 2 }=1$


Correct Option: A

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. $\displaystyle \frac{1}{26}\left ( 3\sqrt{3}+1 \right )$


Correct Option: B
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 $B$.