Tag: maths

Questions Related to maths

If $y=2x+3$ and $x < 2$, which of the following represents all the possible values for $y$?

  1. $y < 7$

  2. $y > 7$

  3. $y < 5$

  4. $y > 5$

  5. $5 < y < 7$


Correct Option: A
Explanation:

Given, $y$ $=$ $2x$ $+$ $3$ and $x$ $<$ $2$

To find possible values for $y$,
$\Rightarrow 2x$ $=$ $y$ $-$ $3$
$\Rightarrow x$ $=$ $\dfrac {y \space - \space 3}{2}$
As $x$ $<$ $2$,
$\Rightarrow \dfrac {y \space - \space 3}{2}$  $<$ $2$
$\Rightarrow y$ $-$ $3$ $<$ $4$
$\Rightarrow y$ $<$ $7$
Therefore, possible values for $'y'$ are $'y$ $<$ $7'$.

The graph of equation of the form $ax + by +c=0$ where a, b are non $-$ zero numbers,
represents:

  1. A triangle

  2. A ray

  3. A straight line

  4. a line segment


Correct Option: C
Explanation:

The equation $ax+by+c=0$ represent straight line under one condition i. e, 

$|a|+|b|\neq 0$       or,     $ a\neq b\neq 0$
Also,  here
$y= \dfrac{-a}{b}x$ $  \dfrac{-c}{b}$ represent the slope interspect form where $m= (-a/b), y-m =  -c/b $ 

A right-angled triangle is formed by a straight line : $3x-4y=12$ with both the axis. Then length of perpendicular from the origin to the hypotenuse is :

  1. $3.5$

  2. $2.4$

  3. $4.2$

  4. none of these


Correct Option: B
Explanation:

Given straight line is $3x-4y=12$
$\Rightarrow\frac{x}{4}-\frac{y}{3} = 1$
this line have x intercept y & y- intercept (-3)
so the evaluation of hypotenuse s the given straight line $3x-4y=12$
so, distance of O(0, 0) form the line $3x-4y=12$ is given by
$=\frac{|3\times0-4\times0-12|}{\sqrt{(3)^2+(-4)^2}}$
$=\frac{12}{5}$
$=2.4$

The graph of the function $\displaystyle \cos x.\cos (x+2)-\cos^{2}(x+1)$ is a 

  1. straight line passing through the point $\displaystyle(0,-\sin^{2}1)$ with slope $2$

  2. straight line passing through the origin

  3. parabola with vertex $\displaystyle (1,-\sin^{2}1)$

  4. straight line passing through the point $\displaystyle\left(\dfrac{\pi}2,-\sin^{2}1\right) $ and parallel to the $x-$axis


Correct Option: D
Explanation:

Let $y=\cos { x } \cos { \left( x+2 \right)  } -\cos ^{ 2 }{ \left( x+1 \right)  } \ =\cos { \left( x+1-1 \right)  } \cos { \left( x+1+1 \right)  } -\cos ^{ 2 }{ \left( x+1 \right)  }\$
 Using $\cfrac{1}{2} \left[\cos\left(A+B+A-B\right)+\cos\left(A+B-A+B\right)\right]\
           = \cfrac{1}{2}\left[\cos 2A + \cos 2B\right]
           = \cfrac{1}{2}\left[\cos^{2}A -1 +1 -2 \sin^{2}A\right ]
           =  \cos ^{2}\left(x+1\right)-\sin^{2}1 $
$=\cos ^{ 2 }{ \left( x+1 \right)  } -\sin ^{ 2 }{ 1 } -\cos ^{ 2 }{ \left( x+1 \right)  } \ =-\sin ^{ 2 }{ 1 } $
This is a straight line which is parallel to x-axis, it passes through $\left( \cfrac { \pi  }{ 2 } ,-\sin ^{ 2 }{ 1 }  \right) $

Complete the table, to draw the graph of line $2y=3x+2$.


$x:$ $3$ $\displaystyle \frac{7}{3}$ $-2$
$y:$ $y _1$ $y _2$ $y _3$

  1. $y _1 = \dfrac12, y _2=\dfrac72,y _3 = -2$

  2. $y _1 = \dfrac{11}2, y _2=\dfrac92,y _3 = -2$

  3. $y _1 = \dfrac12, y _2=\dfrac72,y _3 = -3$

  4. $y _1 = \dfrac{11}2, y _2=\dfrac92,y _3 = -3$


Correct Option: B
Explanation:

The equation of line is $2y=3x+2$

Now to complete the given table,
Substitute $x=3$ in the equation $2y=3x+2$ to fill the first column of the table
$\displaystyle 2y _1=3\times 3+2\ \Rightarrow 2y _1=9+2\ \Rightarrow 2y _!=11\ \Rightarrow y _!=\dfrac { 11 }{ 2 }$
 Substitute $x=\dfrac { 7 }{ 3 }$ in the equation $2y=3x+2$ to fill the second column of the table

$2y _2=3\times \dfrac { 7 }{ 3 } +2\\ \Rightarrow 2y _2=7+2\\ \Rightarrow 2y _2=9\\ \Rightarrow y _2=\dfrac { 9 }{ 2 }$
Finally, substitute $x=-2$ in the equation $2y=3x+2$ to fill the first column of the table
$2y _3=3\times -2+2\\ \Rightarrow 2y _3=-6+2\\ \Rightarrow 2y _3=-4\\ \Rightarrow y _3=-2$

Which of the following is true about the three lines
$L _{1}: x - 3y + 7 = 0 , L _{2} : 2x + y - 3 = 0$ and $L _{3} : 7x +\dfrac{7y}{2}-\dfrac{21}{2}=0$

  1. Lines form a triangle

  2. Lines are concurrent

  3. Lines can not bound any region

  4. None of these


Correct Option: C
Explanation:

Given equations of lines as
$L _{1}: x - 3y + 7 = 0 $
Slope of $L _{1}=\displaystyle \frac{1}{3}$
$ L _{2} : 2x + y - 3 = 0$ 
Slope of $L _{2}=-2$
$L _{3} : 7x +\dfrac{7y}{2}-\dfrac{21}{2}=0$
Slope of $L _{3}=-2$
$\Rightarrow L _{2}$ and $L _{3}$ are parallel
Hence, lines cannot bound any region.

The number of circles that touch all the straight
lines $x+y - 4 = 0, x - y+2 = 0$ and $y = 2$ is

  1. 1

  2. 2

  3. 3

  4. 4


Correct Option: D
Explanation:

The given lines form a triangle.
The number of circles are $4$, i.e. incircle and three ex-circles of the triangle. 

The points (4,0), (0,4), (-4,0), and (0, -4) form

  1. a rectangle

  2. a square

  3. a trapezium

  4. none of these


Correct Option: B
Explanation:

From the diagram the points form a square

Find the equation of the straight line passing through the point $ (6,2)  $ and having slope $ -3 .  $

  1. $x-3y-10=0$

  2. $3x+y-20=0$

  3. $x+2y-40=0$

  4. $3x-y-10=0$


Correct Option: B
Explanation:

Equation of any line is $y=mx+c$


here $m$ is slope of line 


so we have $m=-3$

$y=-3x+c$

Also this line passes through $(6, 2)$

$2=-18+c\Rightarrow c=20$

so equation of line will be $y+3x=20$

or $3x+y-20=0$.


A line passing through (2, 2) is perpendicular to the line $3x+y=3$. Its y intercept is _____________.

  1. $\dfrac { 1 }{ 3 } $

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

  3. 1

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


Correct Option: A