Tag: sets, relations and functions

Questions Related to sets, relations and functions

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?

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  1. $a = 2, b = 2$

  2. $a = 2, b = 4$

  3. $a = 2, b = 3$

  4. $a = 4, b = 3$

Show answer
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 $x$.

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

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

  2. $\displaystyle \sqrt { 3 } $

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

  4. $\displaystyle 5$

Show answer
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 $y=-4x$

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  1. $-51$

  2. $3$

  3. $9$

  4. $144$

Show answer
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 $2x-2y$.

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

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

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

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

Show answer
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 $\overline {AB}$?

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  1. $10$

  2. $12$

  3. $14$

  4. $16$

Show answer
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?

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

  1. $2$

  2. $3$

  3. $4$

  4. $5$

Show answer
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 $\dfrac {x^{2}}{4} + y^{2} = 1$.

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  1. $0$

  2. $1$

  3. $2$

  4. $\infty$

Show answer
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

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

  1. $(2, 1, 0)$

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

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

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

Show answer
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 

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

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

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

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

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

Show answer
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 $A(x-y)+B(x + y)=2B$.

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

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

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

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

  4. None of these

Show answer
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$

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