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Graphs to solve linear and non linear equations - class-XI

Description: graphs to solve linear and non linear equations
Number of Questions: 49
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Tags: maths curved graphs sets, relations and functions
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The equation of a line parallel to $x+2y=1$ and passing through the point of intersection of the lines $x-y=4$ and $3x+y=7$ is ?

  1. $x+2y=5$

  2. $4x+8y-1=0$

  3. $4x+8y+1=0$

  4. none of these


Correct Option: B
Explanation:

Let the equation of line parallel to $x+2y=1$ be $y=mx+c$

intersection point of line $x-y=4$ and $3x+y=7$ is given by solving two equation we get $x=\dfrac { 11 }{ 4 } $ and $y=\dfrac { -5 }{ 4 } $
from equation $x+2y=1\ \Rightarrow y=-\dfrac { 1 }{ 2 } x+\dfrac { 1 }{ 2 } $
we get ${ m } _{ 1 }=-\dfrac { 1 }{ 2 } $
since $y=mx+c$ is parallel to $x+2y=1$
$\therefore { m } _{ 2 }=-\dfrac { 1 }{ 2 } $
also, $y=mx+c\ \Rightarrow \dfrac { -5 }{ 4 } =\dfrac { -1 }{ 2 } .\dfrac { 11 }{ 4 } +c\ \Rightarrow c=\dfrac { 1 }{ 8 } $
therefore equation of line parallel to $x+2y=1$ is given as 
$y=mx+c\ \Rightarrow y=\dfrac { -1 }{ 2 } x+\dfrac { 1 }{ 8 } \ \Rightarrow 4x+8y-1=0$

$2x + y = 0$ 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. $\left (1, -2\right) ; 16$


Correct Option: C
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
$(x,y)=\dfrac{3-1}{2}, \dfrac{2-6}{2}$ $=(1,-2)$

The line $y = x$ meets $y = ke^x , k \le 0$ at

  1. no ponits

  2. one point

  3. two points

  4. none of these


Correct Option: B

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. $3bc +2ad =0$


Correct Option: C
Explanation:

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

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

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 $B$.

The least integral value of $a$ for which the graphs of the functions $y = 2ax + 1$ and $\displaystyle y=(a-6)x^{2}-2$ do not intersect is:

  1. -6

  2. -5

  3. 3

  4. 2


Correct Option: B
Explanation:

For no intersection of graphs of functions,  $ y = 2ax + 1$ and $ y = (a-6)x^2 -2$, There should not any common points between two curves.


Putting the value of $y$ from equation of line into equation of given parabola, we get,

$\Rightarrow (2ax + 1) = (a-6)x^2 - 2$

$\Rightarrow (a-6)x^2  - (2a)x -3 = 0$ ...$(1)$

Equation $(1)$ is a quadratic equation in $x$. 

For no intersection of both given functions, the equation $(1)$ must not have any real solutions.

A quadratic equation have no real roots if the value of it's discriminant is less than zero.

Hence $D = b^2 - 4ac < 0 $

$\Rightarrow D = ((-2a)^2) - 4 \times (a-6) \times (-3) < 0$

$\Rightarrow  D = 4a^2 +12a -72 < 0$

$\Rightarrow (a +6)(a -3) <0$

Hence Value of $a$ for the graphs of given functions do not intersect lies between $(-6 ,3)$

So the least integral value will be $(-5)$. Correct answer is $A$.

The point of intersection of the two ellipse $x^2+2y^2-6x-12y+23=0$ and $4x^2+2y^2-20x-12y+35=0$

  1. lie on a circle centered at $\displaystyle \left( \frac { 8 }{ 3 } ,3 \right) $ and of radius $\displaystyle \frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 2 }  } $

  2. lie on a circle centered at $\displaystyle \left( -\frac { 8 }{ 3 } ,-3 \right) $ and of radius $\displaystyle \frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 2 }  } $

  3. lie on a circle centered at $\displaystyle \left( 8 ,9 \right) $ and of radius $\displaystyle \frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 3 }  } $

  4. are not concyclic


Correct Option: A
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 ${S} _{2}=0.$

$\Rightarrow \left( \lambda +4 \right) { x }^{ 2 }+2\left( \lambda +1 \right) { y }^{ 2 }-2\left( 3\lambda +10 \right) x-12\left( \lambda +1 \right) y+\left( 23\lambda +35 \right) =0$   ...(1)
For it to be a circle, choose $\lambda$ such that the coefficients of ${x}^{2}$ and ${y}^{2}$ are equal: $\Rightarrow \lambda+4=2\lambda+2$
$\therefore \lambda=2$
This gives the equation of the circle as $6\left( { x }^{ 2 }+{ y }^{ 2 } \right) -32x-36y+81=0$  {(using (1))}
$\displaystyle \Rightarrow { x }^{ 2 }+{ y }^{ 2 }-\frac { 16 }{ 3 } x-6y+\frac { 27 }{ 2 } =0$ 
Its centre is $\displaystyle C\left( \frac { 8 }{ 3 } ,3 \right) $ and radius is $\displaystyle r=\sqrt { \frac { 64 }{ 9 } +9-\frac { 27 }{ 2 }  } =\frac { 1 }{ 3 } \sqrt { \frac { 47 }{ 2 }  } .$

The line $x+y=1$ meets the lines represented by the equation $y^{3}-xy^{2}-14x^{2}y+24x^{3}=0$ at the points $A, B, C$. If $O$ is the origin, then $OA^{2}+OB^{2}+OC^{2}$ is equal to

  1. $\dfrac{22}9$

  2. $\dfrac{85}{72}$

  3. $\dfrac{181}{72}$

  4. $\dfrac{221}{72}$


Correct Option: D
Explanation:

X-coordinate of the points are given by the roots of the equation


$24{ x }^{ 3 }+14{ x }^{ 2 }\left( x-1 \right) -x{ \left( x-1 \right)  }^{ 2 }-{ \left( x-1 \right)  }^{ 3 }=0\ \Rightarrow 36{ x }^{ 3 }-9{ x }^{ 2 }-4x+1=0\ \Rightarrow \left( 3x-1 \right) \left( 3x+1 \right) \left( 4x-1 \right) =0\ \Rightarrow x=\cfrac { 1 }{ 3 } ,-\cfrac { 1 }{ 3 } ,\cfrac { 1 }{ 4 } $

$\Rightarrow A\left( \cfrac { 1 }{ 3 } ,\cfrac { 2 }{ 3 }  \right) ,B\left( -\cfrac { 1 }{ 3 } ,\cfrac { 4 }{ 3 }  \right) $ and $C\left( \cfrac { 1 }{ 4 } ,\cfrac { 3 }{ 4 }  \right) $

Hence,

${ OA }^{ 2 }+{ OB }^{ 2 }+{ OC }^{ 2 }=\cfrac { 1 }{ 9 } +\cfrac { 4 }{ 9 } +\cfrac { 1 }{ 9 } +\cfrac { 16 }{ 9 } +\cfrac { 1 }{ 16 } +\cfrac { 9 }{ 16 } =\cfrac { 221 }{ 72 } $

The points of intersection of the two ellipses $x^{2}+2y^{2}-6x-12y+23=0$ and $4x^{2}+2y^{2}-20x-12y+35=0$.

  1. lie on a circle centred at $\left(\dfrac83, 3\right)$ and of radius $\displaystyle \frac{1}{3}\sqrt{\frac{47}{2}}$

  2. lie on a circle centred at $\left(-\dfrac83, 3\right)$ and of radius $\displaystyle \frac{1}{3}\sqrt{\frac{47}{2}}$

  3. lie on a circle centred at $(8, 9)$ and of radius $\displaystyle \frac{1}{3}\sqrt{\frac{47}{2}}$

  4. are not cyclic.


Correct Option: A
Explanation:

Equation of any curve passing through the intersection of the given ellipse is
   $4x^{2}+2y^{2}-20x-12y+35+\lambda \left ( x^{2}+2y^{2}-6x-12y+23 \right )=0$
which represents a circle is
   $4+\lambda =2+2\lambda \Rightarrow \lambda =2$
and the equation of the circle is thus,
   $6x^{2}+6y^{2}-32x-36y+81=0$
$\Rightarrow $   $\displaystyle x^{2}+y^{2}-\left ( \frac{16}{3} \right )x-6y+\frac{81}{6}=0$
centre of the circle is $\left(\dfrac83, 3\right)$
and the radius is $\displaystyle \sqrt{\left ( \frac{8}{3} \right )^{2}+\left ( 3 \right )^{2}-\frac{81}{6}}$
   $\displaystyle =\sqrt{\frac{128+162-243}{18}}=\frac{1}{3}\sqrt{\frac{47}{2}}$

If the points of intersection of curves $\displaystyle C _{1}=\lambda x^{2}+4y^{2}-2xy-9x+3: : and: : C _{2}=2x^{2}+3y^{2}-4xy+3x-1 $ subtends a right angle at origin then the value of $\displaystyle \lambda $ is

  1. $19$

  2. $9$

  3. $-19$

  4. $-9$


Correct Option: C
Explanation:
Given 
$C _{1} : \lambda x^2+4y^2-2xy-9x+3=0$

$C _{2} :  2x^2+3y^2-4xy+3x-1=0\Rightarrow 2x^2+3y^2-4xy-1=-3x$-----(1)

Putting (1) in $C _{1}$

$\lambda x^2+4y^2-2xy+3(2x^2+3y^2-4xy-1)+3=0$

$\lambda x^2+4y^2-2xy+6x^2+9y^2-12xy-3+3=0$

$(\lambda+6 )x^2+13y^2-14xy=0$

Above equation has a condition of perpendicularity 

Hence $\lambda+6+13=0 \Rightarrow \lambda=-19$

If $x^{2}+y^{2}=a^{2}$ touches the line $y=3x+10$, then $a=$ 

  1. $\sqrt{10}$

  2. $10$

  3. $\sqrt{5}$

  4. $\dfrac{10}{2}$


Correct Option: A

Given, $y=3$, $y=ax^2+b$
In the system of equations above, $a$ and $b$ are constants. For which of the following values of $a$ and $b$ does the system of equations have exactly two real solutions?

  1. $a = 2, b = 2$

  2. $a = 2, b = 4$

  3. $a = 2, b = 3$

  4. $a = 4, b = 3$


Correct Option: A
Explanation:

On substituting value of $y=3$ in second equation, we get
$ax^2+b=3$
$\Rightarrow ax^2+b-3=0$
$\Rightarrow ax^2=3-b$
$\Rightarrow x^2=\frac{3-b}a$
Since $x^2$ is positive quantity, therefore just $a=2$ and $b=2$ satisfies this.
Hence, option A is correct.

The system of equations:

$\displaystyle y={ x }^{ 2 }-2x$
$\displaystyle y=2x-1$ has two solutions for ($x,y$). 
Determine the greater value of $x$.

  1. $\displaystyle 2-\sqrt { 3 } $

  2. $\displaystyle \sqrt { 3 } $

  3. $\displaystyle 2+2\sqrt { 3 } $

  4. $\displaystyle 5$


Correct Option: C
Explanation:
Given, $y=x^{2}-2x$
$y=2x-1$
Then $x^{2}-2x=2x-1 $
$\Rightarrow x^{2}-2x-2x+1=0$
$\Rightarrow x^{2}-4x+1=0$
Manipulate this equation $ax^{2}+bx+c=0$
We know $x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}$
$\therefore x=\dfrac{-(-4)\pm \sqrt{(-4)^{2}-4(1)(1)}}{2(1)}$
$\Rightarrow x= \dfrac{4\pm \sqrt{16-4}}{2}$
$\Rightarrow x=\dfrac{4\pm \sqrt{12}}{2}$
$\Rightarrow x=\dfrac{4\pm 2\sqrt{3}}{2}$
The greater of the two possible values for $x$ is $x=2\pm 2\sqrt{3}$
Therefore, the correct answer is (C).

$(x,y)$ satisfies the given set of the equations , find the value of ${x}^{2}$.
${x}^{2}+{y}^{2}=153$ and $y=-4x$

  1. $-51$

  2. $3$

  3. $9$

  4. $144$


Correct Option: C
Explanation:
Let $x^{2}+y^{2}=153$.......(1)
and $y=-4x$......(2)
Put the value $y=-4x$ in equation (1), we get
$x^{2}+(-4x)^{2}=153$
$\Rightarrow x^{2}+16x^{2}=153$
$\Rightarrow 17x^{2}=153$
$\Rightarrow x^{2}=9$

If $8x+8y=18$ and $x^2-y^2=-\displaystyle\frac{3}{8}$, calculate the value of $2x-2y$.

  1. $-\displaystyle\frac{1}{3}$

  2. $-\displaystyle\frac{1}{6}$

  3. $\displaystyle\frac{1}{3}$

  4. $\displaystyle\frac{1}{6}$


Correct Option: A
Explanation:

Given, $8x+8y=18$

$\Rightarrow 8(x+y)=18$
$\Rightarrow (x+y)=\dfrac{18}{8}$
$\Rightarrow (x+y)=\dfrac{9}{4}$.......(1)
Also given, $(x^{2}-y^{2})=-\dfrac{3}{8}$
$\Rightarrow (x+y)(x-y)=-\dfrac{3}{8}$.......(2)
Put the value as per equation (1), $(x+y)=\dfrac{9}{4}$, we get
$\dfrac{9}{4}\left ( x-y \right )=-\dfrac{3}{8}$
$\Rightarrow (x-y)=-\dfrac{3}{8}\times \dfrac{4}{9}$
$\Rightarrow (x-y)=-\dfrac{12}{72}$
$\Rightarrow 2(x-y)=-2\times \dfrac{12}{72}$
$\Rightarrow 2x-2y=-\dfrac{1}{3}$

In the xy-plane, the parabola with equation $y = (x - 11)^{2}$ intersects the line with equation $y = 25$ at two points, $A$ and $B$. What is the length of $\overline {AB}$?

  1. $10$

  2. $12$

  3. $14$

  4. $16$


Correct Option: A
Explanation:

Given $y=(x-11)^2$

$\Rightarrow  y=x^2-22x+121$
After substitute $ y=25$. we get,
$x^2-22x+121-25=0$
$\Rightarrow x^2-22x+96=0$
$\Rightarrow (x-16)(x-6)=0$
$\Rightarrow  x=16 , x=6$
$\therefore$ two points are  $A, B$ are $(16,25), (6,25)$.
Distance between A and B is 
$\sqrt {(16-6)^2-(25-25)^2}=10$
Hence, option A is correct.

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 value of $\displaystyle\int^2 _1(g(x)-f(x))dx$ is?

  1. $2$

  2. $3$

  3. $4$

  4. $5$


Correct Option: A
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)
$d _{1}=2=d _{2}$

$\int _{1}^{2} (g(x)-f(x))d(x)$

$\int _{1}^{2} (4-2)d(x)$

$2\int _{1}^{2} d(x)$

$2(2-1)=2$

The number of values of $C$ for which the line $y = 4x + c$ touch the curve $\dfrac {x^{2}}{4} + y^{2} = 1$.

  1. $0$

  2. $1$

  3. $2$

  4. $\infty$


Correct Option: C
Explanation:

The given curve $\dfrac {x^{2}}{4} + y^{2} = 1$ is an ellipse

Here, $a=2$ and $b=1$

And equation of line is slope form is $y=mx+c$
Also, $c = \sqrt {a^{2}m^{2} + b^{2}}$
$\Rightarrow c = \sqrt {4(16) + 1} = \pm \sqrt {65}$
Hence, $c$ has two values.

The point of intersection of line $\dfrac {x - 6}{-1} = \dfrac {y + 1}{0} = \dfrac {z + 3}{4}$ and plane $x + y - z = 3$ is

  1. $(2, 1, 0)$

  2. $(7, -1, -7)$

  3. $(1, 2, -6)$

  4. $(5, -1, 1)$


Correct Option: D
Explanation:

The given line is $\dfrac {x - 6}{-1} = \dfrac {y + 1}{0} = \dfrac {z + 3}{4}=r$(say) ..... $(i)$

And Plane is $x + y - z = 3$ ........ $(ii)$
$\Rightarrow x=-r+6, y=-1, z=4r-3$
Then, the point $(-r + 6, - 1, 4r - 3)$ lies on the line $(i)$. 

It is given that the plane and the line intersects
Thus, the point $(-r + 6, - 1, 4r - 3)$ satisfies the plane
$\Rightarrow (-r + 6) - 1 - (4r - 3) = 3\Rightarrow r = 1$
$\therefore$ Required intersection point $= (5, -1, 1)$.

The value that m can take so that the straight line $y=4x+m$ touches the curve $x^{2}+4y^{2}=4$ is 

  1. $\underline{+}\sqrt{45}$

  2. $\underline{+}\sqrt{60}$

  3. $\underline{+}\sqrt{65}$

  4. $\underline{+}\sqrt{72}$


Correct Option: C
Explanation:


$C=\underline{+}\sqrt{a^{2}\, m^{2}+b^{2}}$

$={+\sqrt{4(16)+1}}=\underline{+}\sqrt{65}$

Find the point of intersection and the inclination of the two lines $Ax+By=A+B$ and $A(x-y)+B(x + y)=2B$.

  1. $(1,1); 45^0$

  2. $(1,2), 60^0$

  3. $(2,1), 75^0$

  4. None of these


Correct Option: A
Explanation:

$Ax+By=A+B$    ........(i)

$  A(x-y)+B(x+y)=2B$

$\Rightarrow  (A+B)x+(B-A)y=2B$    .......(ii)

$\Rightarrow  (A+B)x=2B-(B-A)y\\ \Rightarrow x=\dfrac { 2B-(B-A)y }{ A+B } $

Substituting $x$ in (i), we get

$A\left( \dfrac { 2B-(B-A)y }{ A+B }  \right) +By=A+B$

$\Rightarrow \dfrac { 2AB-ABy+{ A }^{ 2 }y+{ B }^{ 2 }y+ABy }{ A+B } =A+B$

$\Rightarrow 2AB-ABy+{ B }^{ 2 }y+{ A }^{ 2 }y+ABy={ A }^{ 2 }+{ B }^{ 2 }+2AB$ 

$\Rightarrow (A^{2}+{ B }^{ 2 })y=A^{2}+{ B }^{ 2 }$

$\Rightarrow y=1 $

Substituting $y$ in $(i)$

$\Rightarrow  Ax+B\left( 1  \right) =A+B\\ \Rightarrow Ax+B=A+B\\ \Rightarrow Ax=A\\ \Rightarrow x=1 $

So, the point of intersection is $(1,1) $.

Slope of (i), ${ m } _{ 1 }=-\dfrac { A }{ B } $.

Slope of (ii), ${ m } _{ 2 }=-\dfrac { (A+B) }{ B-A } =\dfrac { A+B }{ A-B } $

$\tan { \theta  } =\dfrac { { m } _{ 1 }-{ m } _{ 2 } }{ 1+{ m } _{ 1 }{ m } _{ 2 } } \\ \Rightarrow \tan { \theta  } =\dfrac { -\dfrac{A}{B}-\dfrac { A+B }{ A-B }  }{ 1-\dfrac{A}{B}\times\dfrac { A+B }{ A-B }  } $

$ \tan { \theta  } =-\dfrac { \left\{ \dfrac { { A }^{ 2 }+AB-AB+{ B }^{ 2 } }{ B(A-B) }  \right\}  }{ \left\{ \dfrac { { -B }^{ 2 }+AB-{ A }^{ 2 }-AB }{ B(A-B) }  \right\}  } \\ \tan { \theta  } =-\dfrac { { A }^{ 2 }+{ B }^{ 2 } }{ { -(A }^{ 2 }+{ B }^{ 2 }) } =1$

$\Rightarrow \theta=45^{o}$

The equation $x-y = 4$ and $x^2 + 4xy + y^2 = 0$ represent the sides of

  1. an equilateral triangle

  2. a right angled triangle

  3. an isosceles triangle

  4. None of these


Correct Option: A
Explanation:

$(y+m _{1}x)(y+m _{2}x)=0$


$y^{2}+(m _{1}+m _{2})xy+m _{2}m _{1}x^2=0$

Comparing coefficients we get

$m _{1}+m _{2}=4$

$m _{1}m _{2}=1$

This implies

$m _{1}^{2}+1=4m _{1}$

$m _{1}^{2}-4m _{1}+1=0$

Therefore $m _{1}=2-\sqrt{3}$ and $m _{1}=2+\sqrt{3}$

$\tan A=2-\sqrt{3}$ and $\tan A=2+\sqrt{3}$

Hence $A=15^0$ and $A=75^0$

Corresponding values of $m _{2}=75^{\circ}$ and $15^{\circ}$

Therefore the angle between the lines is $75^0-15^0$
$=60^0$

The equation of the angle bisectors of the lines is $x=\pm y$

The line $x=-y$ is perpendicular to $x-y=4$

Hence the above is an isosceles triangle with the vertical angle being $60^0$

Hence the triangle is an equilateral triangle.

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. $3bc+2ad=0$


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 

  1. $5x-4y=0$

  2. $5x+4y=0$

  3. $4x-5y=0$

  4. $4x+5y=0$


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 $\left|PA.PB\right|=$

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


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 :

  1. $\left( 0,\infty \right)$

  2. $\left[ 1,\infty \right)$

  3. $\left( -1,\infty \right)$

  4. $\left( -1,1 \right] $


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:

  1. $1$

  2. $2$

  3. Infinite

  4. None of these


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

  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. $\displaystyle \left ( -\infty ,-4-5\sqrt{2} \right )\cup \left ( 5\sqrt{2}-4,\infty \right )$


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

  1. 0

  2. 1

  3. 2

  4. $\infty$


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 ${x}^{2}+{y}^{2}=5$

  1. $-2,0$

  2. $1,2$

  3. $-2,1$

  4. $-2,-1$

  5. $1,3$


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

  1. $-\dfrac{r^{2}}{2} $

  2. $-\dfrac{r}{2}$

  3. $\dfrac{r}{2}$

  4. $\dfrac{r^2}{2}$


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? 

  1. $(-3\sqrt{2}, 3\sqrt{2})$

  2. $(-4, 2)$

  3. $(2+\sqrt{3}, 2)$

  4. $(2-3\sqrt{2}, 3\sqrt{2})$


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

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

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 


Correct Option: A
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 }  } $

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. $4$


Correct Option: E
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 $4$.

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 $(0,1)$


Correct Option: B
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 $(0,2)$

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. $[2, \infty)$


Correct Option: B
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

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


Correct Option: C
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

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. $\dfrac{4}{13}\left ( 3\sqrt{3}+1 \right )$


Correct Option: A
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
$\Rightarrow  \displaystyle \frac{r^{3}}{8}+\frac{r^{3}.3\sqrt{3}}{8}+3.\frac{r}{2}.\frac{r\sqrt{3}}{2}+5.\frac{r^{2}}{4}+3.\frac{r^{2}.3}{4}+4.\frac{r}{2}+5.\frac{r\sqrt{3}}{2}-1=0$

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

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

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

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

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. $-1/3$


Correct Option: A
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)

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})$


Correct Option: A
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 $g(x)=x-\dfrac{1}{x}$
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