Tag: maths

Questions Related to maths

$\displaystyle ax^{2}+2hxy+by^{2}=0$ represents a pair of straight lines through origin & angle between them is given by
$\displaystyle \tan \theta=\frac{2\sqrt{h^{2}-ab}}{a+b}$. If the lines are perpendicular then $\displaystyle a+b=0 $ and the equation of bisectors is given by  $\displaystyle \frac{x^{2}-y^{2}}{a-b}=\frac{xy}{h}$
The general equation of second degree given by
$\displaystyle ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$ represent a pair of straight lines if $\displaystyle \triangle =0 $ or 
$ \displaystyle \begin{vmatrix}a&h  &g \\ h&b  &f \\ g&f  &c \end{vmatrix}=0 $ or $\displaystyle abc+2fgh-af^{2}-bg^{2}-ch^{2}=0$
On the basis of above information answer the following question

If the lines joining origin to the points of intersection of the line $\displaystyle x +y = 1$ with the curve $\displaystyle x^{2}+y^{2}+x-2y-m = 0$ are perpendicular to each other, then value of $m$ is

maths functions and graphs different forms of equation of a line

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

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

  3. $\displaystyle 1 $

  4. $\displaystyle -1 $

Show answer
Correct Option: A
Explanation:

As the line $x+y=1$ intersect the curve,$\displaystyle x^{2}+y^{2}+x-2y-m=0$


$\displaystyle \Rightarrow  x^{2}+y^{2}+\left ( x-2y \right )\left ( 1 \right )-m\left ( 1 \right )^{2}=0$

$\displaystyle \Rightarrow  x^{2}+y^{2}+\left ( x-2y \right )\left ( x+y \right )-m\left ( 1 \right )^{2}=0$

$\displaystyle \Rightarrow  x^{2}+y^{2}+\left ( x-2y \right )\left ( x+y \right )-m\left ( x+y \right )^{2}=0$

$\displaystyle \Rightarrow  x^{2}+y^{2}+\left ( x^{2}-xy-2y^{2} \right )-m\left ( x^{2}+y^{2}+2xy \right )=0$

$\displaystyle \Rightarrow  x^{2}\left ( 2-m \right )+y^{2}\left ( 1-2-m \right )+xy\left ( 1-2m \right )=0$      ......(*)

Now pair of straight lines given bye (*) will be $\perp$er if

$\displaystyle 2-m+1-2-m=0$ (Using $a+b=0$)

$\displaystyle \Rightarrow m=\frac{1}{2}$

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

maths functions and graphs different forms of equation of a line

  1. $m^{2}=1$

  2. $m^{2} = 3$

  3. $m^{2}= 7$

  4. $2m^{2} = 1$

Show answer
Correct Option: C
Explanation:

Joint equation of the lines joining the origin and the point of intersection of the line $y =mx + 2$ and
the curve $x^{2} + y^{2}=1$ is
$ \displaystyle x^{2} +y^{2} =\left( \frac{y-mx}{2}\right)^{2} $
$ x^{2}(4 -m^{2} )+2mxy +3y^{2} = 0$
Since these lines are at right angles
$4 -m^{2} + 3 =0 \Rightarrow m^{2} = 7$

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

maths functions and graphs different forms of equation of a line

  1. is equal to $1$

  2. is equal to $-1$

  3. is equal to $2$

  4. does not exist in the set of real numbers

Show answer
Correct Option: B
Explanation:

Homogenizing the curve with the help of the straight line.


$5x^2+12xy-6y^2+4x(x+ky) -2y(x+ky)+3(x+ky)^2 = 0$

$12x^2 + (10 + 4k + 6k) xy + (3k^2 -2k -6)y^2 = 0$

Lines are equally inclined to the coordinate axes

$\therefore$ coefficient of $xy = 0$

$\Rightarrow 10k + 10 = 0 \Rightarrow k = -1$

The lines joining the origin to the point of intersection of $3x^2 + mxy - 4x + 1 = 0$ and $2x + y - 1= 0$  are at right angles. Then which of the following is/are possible value/s of $m?$

maths functions and graphs different forms of equation of a line

  1. $-4$

  2. $4$

  3. $7$

  4. $3$

Show answer
Correct Option: A,B,C,D
Explanation:

By application of the method of homogenization we get
$3x^2+mxy-4x(2x+y)+1(2x+y)^{2}$
$=3x^2+mxy-8x^2-4xy+4x^2+y^2+4xy$
$=-x^2+mxy+y^2$
$=0$
Hence the equations of the lines is given by $ x^2-mxy-y^2=0$
Hence the lines are perpendicular for all values of $m.$

Find the equation of the lines joining the origin to the points of intersection of the curve $2x^2 + 3xy -4x + 1 = 0$ and the line $3x + y = 1$

maths functions and graphs different forms of equation of a line

  1. $x^2-y^2-5xy=0$

  2. $x^2+y^2-5xy=0$

  3. $x^2+y^2+5xy=0$

  4. $None\ of\ these$

Show answer
Correct Option: A
Explanation:

Given equations of curve and the line are $2x^2 + 3xy -4x + 1 = 0$ and $3x + y = 1$.
Homogenising the curve with the line gives
$2x^2+3xy-4x(3x+y)+(3x+y)^2=0$
$\Rightarrow 2x^2+3xy-12x^2-4xy+9x^2+y^2+6xy=0$
$\Rightarrow y^2-x^2+5xy=0$
$\therefore$ The equation of the lines joining the origin to the points of intersection of the curve and hte line is $x^2-5xy-

y^2=0$
Hence, option A.

Find the area of equilateral  triangle inscribed in a circle of unit radius.

maths area of complex plane figures 2d and 3d figures

  1. 3/4

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

  3. 3

  4. $\frac { 3\sqrt { 3 } }{ 2 } $

Show answer
Correct Option: B
Explanation:

The radius of circumcircle  of equilateral triangle is  $\dfrac 23h=1\h=\dfrac 32$

The side of equilateral triangle is given as $\dfrac{4h}{\sqrt 3} \\dfrac{4}{\sqrt 3}\times \dfrac 32=2\sqrt 3$ 
The area of triangle is given as $\dfrac {\sqrt 3}{4}(2\sqrt 3)^2=3\sqrt 3$

A square is inscribed in a circle of radius $7: cm$. Find area of the square.

maths area of complex plane figures 2d and 3d figures

  1. $98 : cm^{2}$

  2. $97 : cm^{2}$

  3. $91 : cm^{2}$

  4. $90 : cm^{2}$

Show answer
Correct Option: A
Explanation:

Given,
Radius of the circle $=7:cm$
Let the side of the square be $a:cm$.
A square when inscribed in a circle then the diameter of the circle must be diagonal of the square.
Therefore,
Diagonal of square $=\sqrt {a^2+a^2}$
                                 $=a\sqrt 2$
Now,
Diameter of the circle $=2\times 7$
                                  $=14:cm$
$=>\sqrt 2 a=14$
$=>a=\dfrac{14}{\sqrt 2}$
$=>a=7\sqrt 2: cm$
Therefore,
Area of square $=a^2$
                       $=(7\sqrt 2 cm)^2$
                       $=(7\sqrt 2 cm)(7\sqrt 2 cm)$
                       $=98: cm^2$

The ratio of areas of square and circle is given n : 1 where n is a natural number. If the ratio of side of square and radius of circle is k :1, where k is a natural number, then n will be multiple of

maths area of complex plane figures 2d and 3d figures

  1. $77$

  2. $22$

  3. $154$

  4. Data insufficient

Show answer
Correct Option: C
Explanation:

Let a be the side of the square & r be the radius of the circle, then $\dfrac {a^2}{\pi r^2}=n$
Now, $\dfrac {a}{r}=k$
$k=\dfrac {a}{r}=\sqrt {\dfrac {22\times n}{7}},n$ has to be multiple of $22\times 7=154$.

A rectangular sheet of acrylic is 50 cm by 25 cm . From it 60 circular buttons, each of diameter 2.8 cm have been cut out. The area of the remaining sheet is

maths area of complex plane figures 2d and 3d figures

  1. 1260.82 $\displaystyle cm^{2}$

  2. 880.4 $\displaystyle cm^{2}$

  3. 630.4 $\displaystyle cm^{2}$

  4. None of these

Show answer
Correct Option: B
Explanation:

Required area
= Area of sheet - 60 $\displaystyle \times $ area of 1 button
$\displaystyle=\left ( 50\times 25-60\times \dfrac{22}{7}\times 1.4\times 1.4 \right )cm^{2}$
$\displaystyle =\left ( 1250-369.6 \right )cm^{2}$
$\displaystyle =880.4cm^{2}$

If one side of a square is 2.4 m. Then what will be the area of the circle inscribed in the square?

maths area of complex plane figures 2d and 3d figures

  1. $1.44 \displaystyle\, m^{2} $

  2. $\displaystyle 1\frac{11}{25}\pi $ $\displaystyle m^{2} $

  3. $\displaystyle \frac{11}{25}\pi $ $\displaystyle m^{2} $

  4. None of these

Show answer
Correct Option: B
Explanation:

The radius of the circle inscribed in the square
of side 2.4m
$\displaystyle r=\dfrac{2.4}{4}m=1.2m$
$\displaystyle \therefore$ Area of the circle $\displaystyle =\pi r^{2}$ square units
$\displaystyle =\pi \times 1.2 m\times 1.2 m$
$\displaystyle =1.44 \pi m^{2}$

$\displaystyle =1\dfrac{11}{25}\pi m^{2}$
$\displaystyle \therefore $ The required area $\displaystyle =1\frac{11}{25}\pi m^{2}$