Tag: sets, relations and functions

Quizzes (13) Questions (60)

Questions Related to sets, relations and functions

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

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. $a = 4, b = 3$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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. $\displaystyle 5$
Reveal answer Fill a bubble to check yourself
C Correct answer
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).
Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

$(x,y)$ 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. $144$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let $x^{2}+y^{2}=153$.......(1)
and $y=-4x$......(2)
Put the value $y=-4x$ in equation (1), we get
$x^{2}+(-4x)^{2}=153$
$\Rightarrow x^{2}+16x^{2}=153$
$\Rightarrow 17x^{2}=153$
$\Rightarrow x^{2}=9$
Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

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. $\displaystyle\frac{1}{6}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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
$\dfrac{9}{4}\left ( x-y \right )=-\dfrac{3}{8}$
$\Rightarrow (x-y)=-\dfrac{3}{8}\times \dfrac{4}{9}$
$\Rightarrow (x-y)=-\dfrac{12}{72}$
$\Rightarrow 2(x-y)=-2\times \dfrac{12}{72}$
$\Rightarrow 2x-2y=-\dfrac{1}{3}$

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

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. $16$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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. $5$
Reveal answer Fill a bubble to check yourself
A Correct answer
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)
$d _{1}=2=d _{2}$

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

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

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

$2(2-1)=2$
Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

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. $\infty$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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. $(5, -1, 1)$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $= (5, -1, 1)$.

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

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. $\underline{+}\sqrt{72}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation


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

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

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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$

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