Exercise 1.4 - Chapter 1 - Sets, Relations and Functions - 11th Maths Guide Samacheer Kalvi Solutions

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EX 1.4
Question 1.
For the curve $y=x^3$ given in Figure, draw
(i) $r=-\mathrm{x}^3$
(ii) $y=x^3+1$
(iii) $y=x^3-1$
(iv) $\mathrm{y}=(\mathrm{x}+1)^3$ with the same scale.

(i)

(ii)

(iii)

(iv)

Solution:
(i) It is the reflection on $y$ axis
(ii) The graph of $y=\mathrm{x}^3+1$ is shifted upward to 1 unit.
(iii) The graph of $y=\mathrm{x}^3-1$ is shifted downward to 1 unit.
(iv) The graph of $y=(x+1)^3$ is shifted to the left for 1 unit.

Question 2.
(i) $y=-x\left(\frac{1}{3}\right)$
(iii) $y=x^{\left(\frac{1}{3}\right)}+1$
(iii) $y=x^{\left(\frac{1}{3}\right)}-1$
(iv) $y=(x+1)^{\left(\frac{1}{3}\right)}$
Solution:

$\text { (i) Let } y=-x^{\left(\frac{1}{3}\right)}$

$
\text { Then } y=-x^{\frac{1}{3}} \text { is the reflection of the graph of } y=x^{\left(\frac{1}{3}\right)} \text { about the } x \text {-axis }
$

$
\text { (ii) Let } y=x^{\left(\frac{1}{3}\right)}+1
$

Then $y=x^{\left(\frac{1}{3}\right)}+1$ is the $x$ graph of $y=x^{\left(\frac{1}{3}\right)}$ shifts to the upward for one unit.

(iii) $y=x^{\left(\frac{1}{3}\right)}-1$
Let $y=-x^{\left(\frac{1}{3}\right)}$

Then $y=x^{\left(\frac{1}{3}\right)}-1$ is the graph of $x^{\frac{1}{3}}$ shifts to the upward for one unit.
(iv) $y=(x+1)^{\left(\frac{1}{3}\right)}$

$y=(x+1)^{\left(\frac{1}{3}\right)}$ causes the graph of $x^{\frac{1}{3}}$ on the same co-ordinate plane. Fine $f o g$ and graph it on the plane as well. Explain your results.
Question 3.
Graph the functions $\mathrm{f}(\mathrm{x})=\mathrm{x}^3$ and $\mathrm{g}(\mathrm{x})=\sqrt[3]{x}$ on the same coordinate plane. Find fog and graph it on the plane as well. Explain your results.
Solution:
$
y=x^3
$

$y=\sqrt[3]{x}$

Now $f \circ g(x)=f[g(x)]=f[\sqrt[3]{x}]=(\sqrt[3]{x})^3=x$
(i.e.) $y=x$
Question 4.
Write the steps to obtain the graph of the function $y=3(x-1)^2+5$ from the graph $y=x^2$.
Solution:
Draw the graph of $y=x^2$
To get $y=(x-1)^2$ we have to shift the curve 1 unit to the right.
Then we have to draw the curve $y=3(x-1)^2$ and finally we have to draw $y=3(x-1)^2+5$
Question 5.
From the curve $\mathrm{y}=\sin \mathrm{x}$, graph the functions
(i) $y=\sin (-x)$
(ii) $y=-\sin (-x)$
(iii) $y=\sin \left(\frac{\pi}{2}+x\right)$ which is $\cos x$
(iv) $y=\sin \left(\frac{2}{2}-x\right)$ which is also $\cos x$ (refer trigonometry)
Solution:
First we have to draw the curve $y=\sin x$
(i) $y=\sin (-x)=-\sin x=f(x)$

$\text { (ii) } y=-\sin (-x)=-[-\sin x]=\sin x$

it is the same as $y=\sin x$ by shifting the graph to $\frac{\pi}{2}$ to the left.
(iii) $y=\sin \left(\frac{\pi}{2}+x\right)$ is the graph which is obtained from $y=\sin x$ by shifting the graph to $\frac{\pi}{2}$ to the left.

$\begin{aligned}
& \text { (iv) } y=\sin \left(\frac{\pi}{2}-x\right) \text { is the graph obtained from } y=\sin x \text { by shifting the graph to } \frac{\pi}{2} \text { to the } \\
& \text { right. }
\end{aligned}$

Question 6.
From the curve $y=x$, draw
(i) $y=-x$
(ii) $y=2 x$
(iii) $\mathrm{y}=\mathrm{x}+1$
(iv) $y=\frac{1}{2} x+1$
(v) $2 x+y+3=0.2$
Solution:
$
y=x
$

$\text { (i) } y=-x$

$\text { (ii) } y=2 x$

$\mathrm{y}=2 \mathrm{x}$ the graph moves away the $\mathrm{x}$-axis, as multiplying factor is 2 which is greater than one.
(iii) $y=2 \mathrm{x}+1$

$\text { (iv) } y=1 / 2 x+1$

$\mathrm{y}=\frac{1}{2} \mathrm{x}$ moves towards $\mathrm{x}-$ axis by a side factor $\frac{1}{2}$ which is less than $\mathrm{y}=\frac{1}{2} \mathrm{x}+1$ upwards by 1 unit.
(v) $y=-2 x-3$

Question 7.
From the curve $y=|x|$, draw
(i) $y=|x-1|+1$
(ii) $\mathrm{y}=|\mathrm{x}+1|-1$
(iii) $y=|x+2|-3$.
Solution:
Given, $y=|x|$
If $y=x$

If $y=-x$

$
\begin{aligned}
& \text { (i) } y=|x-1|+1 \\
& y=x-1+1 \\
& y=-x+1+1=x
\end{aligned}
$

$\text { (For } y=2-x)$

$\begin{aligned}
& \text { (ii) } y=|x+1|-1 \\
& y=x+1-1=x \\
& y=-x-1-1 \\
& y=-x-2 \\
& y=-(x+2)
\end{aligned}$

$\begin{aligned}
&\begin{aligned}
& \text { (iii) } y=|x+2|-3 \\
& \mathrm{y}=\mathrm{x}+2-3 \Rightarrow \mathrm{x}-1 \\
& \mathrm{y}=-\mathrm{x}-2+3=1-\mathrm{x}
\end{aligned}\\
&y=-(x-1)
\end{aligned}$

Question 8.
From the curves $=\sin x$, draw $y=\sin |x|$ (Hint: $\sin (-x)=-\sin x$.)
Solution:
$
\begin{aligned}
& \mathrm{y}=\sin |\mathrm{x}| \\
& \therefore \mathrm{y}=\sin \mathrm{x} \\
& \therefore \mathrm{y}=\sin (-\mathrm{x})=-\sin \mathrm{x} \\
& \mathrm{y}=-\sin \mathrm{x}
\end{aligned}
$