127 views

Which of the following function(s) is an accurate description of the graph for the range(s) indicated?

1. $y=2x+4 \text{ for } -3 \leq x \leq -1$
2. $y= \mid x-1 \mid \text{ for } -1 \leq x \leq 2$
3. $y= \mid\mid x \mid -1 \mid \text{ for } -1 \leq x \leq 2$
4. $y=1 \text{ for } 2 \leq x \leq 3$
1. i, ii and iii only
2. i, ii and iv only
3. i and iv only
4. ii and iv only

edited | 127 views
Migrated from GO Civil 7 months ago by Arjun

$\bullet$ If we have $2$ points $(x_1,y_1)$ and $(x_2,y_2)$ on $X-Y$ plane,

Then equation of line will be :- $(y - y_1) = \left ( \frac{y_2 - y_1}{x_2 - x_1} \right )(x-x_1)$

$\bullet$  Definition of $y=|x|$ says $y = \left\{\begin{matrix} x & x\geqslant 0 \\ -x & x < 0 \end{matrix}\right.$

Now,

$(i)$ Let, $x_1 = -3, x_2 = -1 , y_1 = -2 , y_2 = 2$

So, Equation of Line will be :- $y - (-2) = \left ( \frac{2-(-2)}{-1 -(-3)} \right )(x-(-3))$

$\Rightarrow y+2 = \frac{4}{2}(x+3)$

$\Rightarrow y= 2x+4$

So, option (i) is correct.

$(ii)$ $y = \left\{\begin{matrix} (x-1), &\; (x-1) \geqslant 0 \; (or) x \geqslant 1 \\ -(x-1)= -x + 1, & \; (x-1) < 0 \; (or) x < 1 & \end{matrix}\right.$

For, $x \geqslant 1,$

Let, $x_1 = 1, x_2 = 2 , y_1 = 0 , y_2 = 1$

So, Equation of Line will be :- $y - 0 = \left ( \frac{1-0}{2 -1} \right )(x-1)$

$\Rightarrow y =(x-1)$

For, $x < 1,$

Let, $x_1 = -1, x_2 = 1 , y_1 = 2 , y_2 = 0$

So, Equation of Line will be :- $y - 2 = \left ( \frac{0-2}{1-(-1)} \right )(x-(-1))$

$\Rightarrow y =-x+1$

Both equations of lines are drawn in the given graph.

So, option (ii) is correct.

$(iii)$ $y = \left\{\begin{matrix} x-1, & when \; x \geqslant 0 \; and \; (x-1) \geqslant 0 \; i.e. \; x \geqslant 1 \\ -(x-1) & when \; x \geqslant 0 \; and \; (x-1) < 0 \; i.e. \; x < 1 \\ x+1 & when \; x < 0 \; and \; (x+1) \geqslant 0 \;\;\; i.e. \; \; x \geqslant -1 \\ -(x+1) & when \; x < 0 \; and \; (x+1) < 0 \; \; i.e. \; x < -1 \end{matrix}\right.$

$\Rightarrow y = \left\{\begin{matrix} x-1, & \; x \geqslant 1 \\ -x+1, & 0 \leqslant x <1 \\ x+1, & -1 \leqslant x < 0 \\ -x-1, & x < -1 \end{matrix}\right.$

Suppose, $x= -1/2$ , $y= 1/2$ but it is not showing correctly in the graph.

So, option (iii) is incorrect.

(iv) is correct as we can see in the given graph. It is constant function $y=1$ between $x=2$ and $x=3$ (both are inclusive)

by Boss (17k points)
selected by
+1
Perfect solution. 👍
+1 vote

Put the values of $x$ and observe the value of $y$ in the graph.

1. $y=2x+4$  for $-3\leq x\leq-1$
• Put the value $x=-3$ we get $y=2(-3)+4=-6+4=-2$
• Put the value $x=-2$ we get $y=2(-2)+4=-4+4=0$
• Put the value $x=-1$ we get $y=2(-1)+4=-2+4=2$

It is true for given range.

ii.$y=\mid x-1 \mid$  for $-1\leq x\leq 2$

•  Put the value $x=-1$ we get $y=\mid -1-1 \mid=\mid -2\mid=2$
•  Put the value $x=0$ we get $y=\mid 0-1 \mid=\mid -1\mid=1$
•  Put the value $x=1$ we get $y=\mid 1-1 \mid=\mid 0\mid=0$
•  Put the value $x=2$ we get $y=\mid 2-1 \mid=\mid 1\mid=1$

It is true for given range.

iii.$y=\bigl\lvert\mid x \mid - 1\bigl\vert$  for $-1\leq x \leq2$

•  Put the value $x=-1$ we get $y=\bigl\lvert\mid -1 \mid - 1\bigl\vert=\mid 1-1\mid =0$

It is ${\color{Red}{\text{false}} }$ because for $x=-1$ value of $y=2.$

iv.$y=1$ for $2\leq x\leq3$

We can see in the graph where $x=2,y=1$ and $x=2,y=1,$ So it is valid.

Option $(B)$ is the right choice.

by Veteran (59.2k points)