Samacheer Kalvi Class 11 Maths Solutions Chapter 1 Sets, Relations and Functions Exercise 1.4

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets, Relations and Functions Ex 1.4

 

Question 1. For the curve \( y = x^3 \) given in figure below, draw
(i) \( y = -x^3 \)
(ii) \( y = x^3 + 1 \)
(iii) \( y = x^3 - 1 \)
(iv) \( y = (x + 1)^3 \) with the same scale.
Answer:
(i) To graph \( y = -x^3 \), we can first create a table of values:

The graph of \( y = -x^3 \) is a reflection of the graph of \( y = x^3 \) across the x-axis. This transformation flips the curve upside down compared to the original function.
(ii) To graph \( y = x^3 + 1 \), we calculate values for x and y:
Adding 1 to \( x^3 \) shifts the entire graph of \( y = x^3 \) upwards by 1 unit. Each point on the original graph moves vertically up by one unit.
(iii) To graph \( y = x^3 - 1 \), we find corresponding x and y values:
Subtracting 1 from \( x^3 \) shifts the entire graph of \( y = x^3 \) downwards by 1 unit. This is a vertical translation, moving every point one unit lower.
(iv) To graph \( y = (x + 1)^3 \), we create a table of points:
Replacing \( x \) with \( (x + 1) \) shifts the graph of \( y = x^3 \) to the left by 1 unit. This is a horizontal translation, where adding a constant inside the function moves the graph in the negative x-direction.
In simple words: We changed the original graph of \( y = x^3 \) in four ways. For \( y = -x^3 \), we flipped it over the x-axis. For \( y = x^3 + 1 \), we moved it up by 1. For \( y = x^3 - 1 \), we moved it down by 1. For \( y = (x + 1)^3 \), we moved it left by 1.

🎯 Exam Tip: Remember that adding or subtracting a number *outside* the function shifts the graph vertically, while adding or subtracting *inside* the function shifts it horizontally in the opposite direction.

 

Question 2. For the given curve \( y = x^{1/3} \) given in figure draw
(i) \( y = -x^{1/3} \)
(ii) \( y = x^{1/3} + 1 \)
(iii) \( y = x^{1/3} - 1 \)
(iv) \( y = (x + 1)^{1/3} \)
Answer:
(i) For \( y = -x^{1/3} \), we can write this as \( -y = x^{1/3} \). Cubing both sides gives \( (-y)^3 = x \), which simplifies to \( -y^3 = x \). We can find points by choosing y-values and calculating x:
When \( y = 0 \implies -0^3 = x \implies x = 0 \)
When \( y = 1 \implies -1^3 = x \implies x = -1 \)
When \( y = 2 \implies -2^3 = x \implies x = -8 \)
When \( y = 3 \implies -3^3 = x \implies x = -27 \)
When \( y = -1 \implies -(-1)^3 = x \implies x = 1 \)
When \( y = -2 \implies -(-2)^3 = x \implies x = 8 \)
When \( y = -3 \implies -(-3)^3 = x \implies x = 27 \)
Let's organize these points in a table:

The graph of \( y = -x^{1/3} \) is a reflection of the graph of \( y = x^{1/3} \) across the x-axis. This transformation means all y-coordinates change sign.
(ii) For \( y = x^{1/3} + 1 \), we can rearrange to get \( y - 1 = x^{1/3} \). Cubing both sides gives \( (y - 1)^3 = x \). We calculate points by setting y-values:
When \( y = 0 \implies (0 - 1)^3 = x \implies x = -1 \)
When \( y = 1 \implies (1 - 1)^3 = x \implies x = 0 \)
When \( y = 2 \implies (2 - 1)^3 = x \implies x = 1 \)
When \( y = 3 \implies (3 - 1)^3 = x \implies x = 8 \)
When \( y = -1 \implies (-1 - 1)^3 = x \implies x = -8 \)
When \( y = -2 \implies (-2 - 1)^3 = x \implies x = -27 \)
Here is the table of values:
The graph of \( y = x^{1/3} + 1 \) shifts the graph of \( y = x^{1/3} \) upwards by 1 unit. This is a vertical translation.
(iii) For \( y = x^{1/3} - 1 \), we get \( y + 1 = x^{1/3} \). Cubing both sides yields \( (y + 1)^3 = x \). We find points by:
When \( y = 0 \implies (0 + 1)^3 = x \implies x = 1 \)
When \( y = 1 \implies (1 + 1)^3 = x \implies x = 8 \)
When \( y = 2 \implies (2 + 1)^3 = x \implies x = 27 \)
When \( y = -1 \implies (-1 + 1)^3 = x \implies x = 0 \)
When \( y = -2 \implies (-2 + 1)^3 = x \implies x = -1 \)
When \( y = -3 \implies (-3 + 1)^3 = x \implies x = -8 \)
The table for these points is:
The graph of \( y = x^{1/3} - 1 \) shifts the graph of \( y = x^{1/3} \) downwards by 1 unit. This is also a vertical translation, but in the negative direction.
(iv) For \( y = (x + 1)^{1/3} \), we cube both sides to get \( y^3 = x + 1 \). Then \( x = y^3 - 1 \). We find points by setting y-values:
When \( y = 0 \implies 0^3 - 1 = x \implies x = -1 \)
When \( y = 1 \implies 1^3 - 1 = x \implies x = 0 \)
When \( y = 2 \implies 2^3 - 1 = x \implies x = 7 \)
When \( y = 3 \implies 3^3 - 1 = x \implies x = 26 \)
When \( y = -1 \implies (-1)^3 - 1 = x \implies x = -2 \)
When \( y = -2 \implies (-2)^3 - 1 = x \implies x = -9 \)
The table for these points is:
The graph of \( y = (x + 1)^{1/3} \) shifts the graph of \( y = x^{1/3} \) to the left by 1 unit. This is a horizontal shift because the change is applied directly to the x-variable.
In simple words: For \( y = x^{1/3} \), we performed similar transformations. \( y = -x^{1/3} \) flips the graph over the x-axis. \( y = x^{1/3} + 1 \) moves it up. \( y = x^{1/3} - 1 \) moves it down. \( y = (x + 1)^{1/3} \) moves it to the left.

🎯 Exam Tip: When dealing with inverse functions like cube root, it's often easier to convert them into their cubic form (e.g., \( (y-k)^3 = x \)) to find points by choosing y-values.

 

Question 3. Graph the functions \( f(x) = x^3 \) and \( g(x) = \sqrt[3]{x} \) on the same coordinate plane. Find fog and the graph it on the plane as well. Explain your results.
Answer:
Given functions are \( f(x) = x^3 \) and \( g(x) = x^{1/3} \).
First, let's find the composite function \( fog(x) \):
\( fog(x) = f(g(x)) \)
\( \implies fog(x) = f(x^{1/3}) \)
\( \implies fog(x) = (x^{1/3})^3 \)
\( \implies fog(x) = x \)
Now, let's create tables of values for \( f(x) \), \( g(x) \), and \( fog(x) \):
For \( f(x) = x^3 \):

For \( g(x) = x^{1/3} \):
For \( fog(x) = x \):
Since \( fog(x) = x \), this means the graph of \( fog(x) \) is a straight line passing through the origin with a slope of 1. This function is symmetric about the line \( y = x \). This implies that \( g(x) \) is the inverse function of \( f(x) \). In mathematical terms, \( g(x) = f^{-1}(x) \). This makes sense because the cube root function is the inverse of the cube function.
In simple words: We have a cubing function and a cube root function. When we put one into the other (fog), we just get \( x \) back. This means these two functions undo each other. They are inverse functions, and their combined graph is a simple straight line.

🎯 Exam Tip: Always check if \( fog(x) = x \) and \( gof(x) = x \). If both are true, the functions are inverses of each other, and their graphs are reflections across the line \( y = x \).

 

Question 4. Write steps to obtain the graph of the function \( y = 3 (x - 1)^2 + 5 \) from the graph \( y = x^2 \).
Answer:
Here are the steps to transform the graph of \( y = x^2 \) to \( y = 3 (x - 1)^2 + 5 \):
Step 1: Draw the base graph \( y = x^2 \).
We can create a table of values for \( y = x^2 \):

Step 2: Shift the graph horizontally.
The term \( (x - 1)^2 \) means the graph of \( y = x^2 \) shifts to the right by 1 unit. When a constant 'c' is subtracted from 'x' inside the function, the graph \( y = f(x) \) becomes \( y = f(x - c) \), causing a shift to the right by 'c' units.
Table for \( y = (x - 1)^2 \):
Step 3: Apply vertical stretch.
The factor of 3 in \( y = 3(x - 1)^2 \) compresses the graph towards the y-axis, making it appear narrower. This is because the multiplying factor 3 is greater than 1, causing the y-values to be three times larger than for \( (x - 1)^2 \). For a graph \( y = kf(x) \), if \( k > 1 \), the graph moves away from the x-axis, creating a stretch.
Table for \( y = 3(x - 1)^2 \):
Step 4: Shift the graph vertically.
Finally, the \( + 5 \) in \( y = 3(x - 1)^2 + 5 \) shifts the entire graph upwards by 5 units. This is a vertical translation, where adding a constant 'd' to the function \( y = f(x) \) gives \( y = f(x) + d \), moving the graph up by 'd' units.
Table for \( y = 3(x - 1)^2 + 5 \):
By following these four steps, we can accurately plot the final transformed graph starting from the basic \( y = x^2 \) curve.
In simple words: To get the final graph, we start with \( y = x^2 \). First, we move it 1 unit to the right. Then, we make it 3 times taller (stretch it vertically). Lastly, we move the whole graph up by 5 units.

🎯 Exam Tip: Always apply transformations in the correct order: horizontal shifts first (inside the function), then stretches/compressions/reflections, and finally vertical shifts (outside the function).

 

Question 5. From the curve \( y = \sin x \), 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) \( y = \sin\left(\frac{\pi}{2}-x\right) \) which is also \( \cos x \)
Answer:
First, let's consider the base curve \( y = \sin x \). Here is a table of its key values:

(i) For \( y = \sin (-x) \):
We know that \( \sin (-x) = -\sin x \). So, the graph of \( y = \sin (-x) \) is the same as \( y = -\sin x \). This means it is a reflection of the graph of \( y = \sin x \) about the y-axis, or equivalently, a reflection over the x-axis for this specific function. Reflecting over the y-axis means that for any point \( (x, y) \) on \( y = \sin x \), there is a point \( (-x, y) \) on \( y = \sin (-x) \).
Here is the table of values for \( y = -\sin x \):
The graph of \( y = \sin (-x) \) is a reflection of the graph of \( y = \sin x \) about the y-axis. This transformation changes the direction of the wave.
In simple words: We start with the sine wave. For \( y = \sin (-x) \), we flip the wave horizontally across the y-axis. Because sine is an odd function, this is the same as flipping it vertically over the x-axis.

🎯 Exam Tip: Remember that \( \sin (-x) = -\sin x \) and \( \cos (-x) = \cos x \). This property helps simplify transformations involving negative arguments in trigonometric functions.

 

Question 5. From the curve y = sin 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{\pi}{2}-x\right) which is also cos x
Answer:
(i) y = sin (-x)
First, we need the table for the basic function \( y = \sin x \):

The graph for \( y = \sin x \) looks like this:

Now, for \( y = \sin(-x) \), we know that \( \sin(-x) = -\sin x \). So, we can use the same values for \( x \) and multiply the \( y \) values of \( \sin x \) by -1 to get the \( y \) values for \( y = \sin(-x) \). This means reflecting the graph about the x-axis.

The graph of \( y = \sin(-x) \) is a reflection of the graph \( y = \sin x \) about the y-axis. It is the same as the graph of \( y = -\sin x \).

In simple words: The graph of \( y = \sin(-x) \) is like flipping the original \( y = \sin x \) graph across the y-axis. It looks exactly the same as the graph for \( y = -\sin x \), where all the up-and-down values are reversed.

🎯 Exam Tip: Remember that \( \sin(-x) = -\sin x \), which means the graph of \( \sin(-x) \) is a reflection of \( \sin x \) across the y-axis, and also the x-axis (due to its odd function property).

 

(ii) y = -sin(-x)
Answer: We already know that \( \sin(-x) = -\sin x \).
So, \( y = -\sin(-x) \) becomes \( y = -(-\sin x) \).
This simplifies to \( y = \sin x \).
Therefore, the graph of \( y = -\sin(-x) \) is exactly the same as the graph of \( y = \sin x \). This is a helpful identity to recall.

The graph will be identical to the \( y = \sin x \) graph.

In simple words: This function is the same as \( y = \sin x \) because two minus signs cancel each other out. So, the graph is exactly the same as the original sine wave.

🎯 Exam Tip: Simplify the trigonometric expression first (e.g., \( -\sin(-x) = \sin x \)) before attempting to graph, as it makes the task much simpler.

 

(iii) y = sin\left(\frac{\pi}{2}+x\right)
Answer: We know that \( \sin\left(\frac{\pi}{2}+x\right) = \cos x \). So we will graph \( y = \cos x \).
Let's find some key points for \( y = \cos x \):
When \( x = 0 \), then \( y = \sin\left(\frac{\pi}{2}+0\right) = \sin\left(\frac{\pi}{2}\right) = 1 \)
When \( x = \frac{\pi}{2} \), then \( y = \sin\left(\frac{\pi}{2}+\frac{\pi}{2}\right) = \sin(\pi) = 0 \)
When \( x = \pi \), then \( y = \sin\left(\frac{\pi}{2}+\pi\right) = \sin\left(\frac{3\pi}{2}\right) = -1 \)
When \( x = \frac{3\pi}{2} \), then \( y = \sin\left(\frac{\pi}{2}+\frac{3\pi}{2}\right) = \sin(2\pi) = 0 \)
When \( x = 2\pi \), then \( y = \sin\left(\frac{\pi}{2}+2\pi\right) = \sin\left(\frac{5\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1 \)
When \( x = -\frac{\pi}{2} \), then \( y = \sin\left(\frac{\pi}{2}-\frac{\pi}{2}\right) = \sin(0) = 0 \)
When \( x = -\pi \), then \( y = \sin\left(\frac{\pi}{2}-\pi\right) = \sin\left(-\frac{\pi}{2}\right) = -1 \)
When \( x = -\frac{3\pi}{2} \), then \( y = \sin\left(\frac{\pi}{2}-\frac{3\pi}{2}\right) = \sin(-\pi) = 0 \)
When \( x = -2\pi \), then \( y = \sin\left(\frac{\pi}{2}-2\pi\right) = \sin\left(-\frac{3\pi}{2}\right) = 1 \)

The graph of \( y = \sin\left(\frac{\pi}{2}+x\right) \) is a shift of the graph \( y = \sin x \) to the left by \( \frac{\pi}{2} \) units. This results in the cosine graph.

In simple words: When you add \( \frac{\pi}{2} \) to \( x \) inside a sine function, the whole graph shifts left by that amount. This specific shift turns the sine wave into a cosine wave.

🎯 Exam Tip: Remember the identity \( \sin\left(\frac{\pi}{2}+x\right) = \cos x \). This allows you to directly graph the cosine function, which is a phase shift of the sine function.

 

(iv) y = sin\left(\frac{\pi}{2}-x\right)
Answer: We know that \( \sin\left(\frac{\pi}{2}-x\right) = \cos x \). So we will graph \( y = \cos x \).
Let's find some key points for \( y = \cos x \):
When \( x = 0 \), then \( y = \sin\left(\frac{\pi}{2}-0\right) = \sin\left(\frac{\pi}{2}\right) = 1 \)
When \( x = \frac{\pi}{2} \), then \( y = \sin\left(\frac{\pi}{2}-\frac{\pi}{2}\right) = \sin(0) = 0 \)
When \( x = \pi \), then \( y = \sin\left(\frac{\pi}{2}-\pi\right) = \sin\left(-\frac{\pi}{2}\right) = -1 \)
When \( x = \frac{3\pi}{2} \), then \( y = \sin\left(\frac{\pi}{2}-\frac{3\pi}{2}\right) = \sin(-\pi) = 0 \)
When \( x = 2\pi \), then \( y = \sin\left(\frac{\pi}{2}-2\pi\right) = \sin\left(-\frac{3\pi}{2}\right) = 1 \)
When \( x = -\frac{\pi}{2} \), then \( y = \sin\left(\frac{\pi}{2}-(-\frac{\pi}{2})\right) = \sin(\pi) = 0 \)
When \( x = -\pi \), then \( y = \sin\left(\frac{\pi}{2}-(-\pi)\right) = \sin\left(\frac{3\pi}{2}\right) = -1 \)
When \( x = -\frac{3\pi}{2} \), then \( y = \sin\left(\frac{\pi}{2}-(-\frac{3\pi}{2})\right) = \sin(2\pi) = 0 \)
When \( x = -2\pi \), then \( y = \sin\left(\frac{\pi}{2}-(-2\pi)\right) = \sin\left(\frac{5\pi}{2}\right) = 1 \)

The graph of \( y = \sin\left(\frac{\pi}{2}-x\right) \) is a shift of the graph \( y = \sin x \) to the right by \( \frac{\pi}{2} \) units, and also a reflection. It results in the cosine graph.

In simple words: Subtracting \( x \) from \( \frac{\pi}{2} \) inside the sine function also creates the cosine graph. It's like moving the sine wave right by \( \frac{\pi}{2} \) units and then reflecting it.

🎯 Exam Tip: Both \( \sin\left(\frac{\pi}{2}+x\right) \) and \( \sin\left(\frac{\pi}{2}-x\right) \) simplify to \( \cos x \). This property is fundamental for understanding phase shifts and reductions in trigonometry.

 

Question 6. From the curve y = x draw
(i) y = -x
(ii) y = 2x
(iii) y = x + 1
(iv) y = \frac{1}{2}x + 1
(v) 2x + y + 3 = 0
Answer:
First, let's create a table for the base function \( y = x \):

The graph of \( y = x \) looks like this:

(i) y = -x
The graph of \( y = -x \) is a reflection of the graph of \( y = x \) about the x-axis (or y-axis).

The graph of \( y = -x \) looks like this:

In simple words: This graph is a mirror image of \( y = x \), flipped across the x-axis. For example, if \( y = x \) goes up to the right, \( y = -x \) goes down to the right.

🎯 Exam Tip: When \( y = -f(x) \), the graph is a reflection across the x-axis. When \( y = f(-x) \), it's a reflection across the y-axis. For \( y = x \), both reflections yield the same result, \( y = -x \).

 

(ii) y = 2x
Answer: The graph of \( y = 2x \) is a vertical stretch of the graph of \( y = x \) by a factor of 2. It compresses the graph towards the y-axis because the multiplying factor is greater than 1. Each y-value is doubled.

The graph of \( y = 2x \) looks like this:

In simple words: This graph is steeper than \( y = x \). Each 'y' value is twice its 'x' value, so the line climbs faster. This makes the line move away from the x-axis.

🎯 Exam Tip: For \( y = kf(x) \), if \( k > 1 \), the graph is stretched vertically (moves away from the x-axis) compared to \( y = f(x) \). If \( 0 < k < 1 \), it's compressed vertically (moves towards the x-axis).

 

(iii) y = x + 1
Answer: The graph of \( y = x + 1 \) is a vertical shift of the graph of \( y = x \) upward by 1 unit. All the y-values are increased by 1.

The graph of \( y = x + 1 \) looks like this:

In simple words: This graph is the same line as \( y = x \), but it's moved up by one step. Every point on the original line is shifted one unit higher.

🎯 Exam Tip: For \( y = f(x) + d \), where \( d > 0 \), the graph shifts upward by \( d \) units. If \( d < 0 \), it shifts downward. This is a basic vertical translation.

 

(iv) y = \frac{1}{2}x + 1
Answer: The graph of \( y = \frac{1}{2}x + 1 \) involves two transformations. It is a vertical compression of \( y = x \) by a factor of \( \frac{1}{2} \), which means it stretches towards the x-axis, and then a vertical shift upward by 1 unit.
Let's find some key points:
When \( x = 0 \), then \( y = \frac{1}{2} \times 0 + 1 = 1 \)
When \( x = 2 \), then \( y = \frac{1}{2} \times 2 + 1 = 1 + 1 = 2 \)
When \( x = 4 \), then \( y = \frac{1}{2} \times 4 + 1 = 2 + 1 = 3 \)
When \( x = 6 \), then \( y = \frac{1}{2} \times 6 + 1 = 3 + 1 = 4 \)
When \( x = -2 \), then \( y = \frac{1}{2} \times -2 + 1 = -1 + 1 = 0 \)
When \( x = -4 \), then \( y = \frac{1}{2} \times -4 + 1 = -2 + 1 = -1 \)
When \( x = -6 \), then \( y = \frac{1}{2} \times -6 + 1 = -3 + 1 = -2 \)

The graph of \( y = \frac{1}{2}x + 1 \) looks like this:

In simple words: This line is flatter than \( y = x \) because the \( x \) is multiplied by \( \frac{1}{2} \). Then, it's moved up by 1 unit. So it's less steep and starts higher.

🎯 Exam Tip: When graphing \( y = kx + d \), \( k \) determines the slope (steepness), and \( d \) determines the y-intercept (where it crosses the y-axis, indicating vertical shift).

 

(v) 2x + y + 3 = 0
Answer: First, rewrite the equation in \( y = mx + c \) form: \( y = -2x - 3 \).
This graph is a vertical stretch of \( y = x \) by a factor of 2 (moves away from x-axis), then reflected across the x-axis (due to the negative sign), and finally shifted downward by 3 units.
Let's find some key points:
When \( x = 0 \), then \( y = -2 \times 0 - 3 = -3 \)
When \( x = 1 \), then \( y = -2 \times 1 - 3 = -2 - 3 = -5 \)
When \( x = 2 \), then \( y = -2 \times 2 - 3 = -4 - 3 = -7 \)
When \( x = -1 \), then \( y = -2 \times -1 - 3 = 2 - 3 = -1 \)
When \( x = -2 \), then \( y = -2 \times -2 - 3 = 4 - 3 = 1 \)
When \( x = -3 \), then \( y = -2 \times -3 - 3 = 6 - 3 = 3 \)

The graph of \( y = -2x - 3 \) looks like this:

In simple words: First, we change the equation to \( y = -2x - 3 \). This line is steeper than \( y = x \) (because of 2), slopes downwards (because of -), and is shifted 3 units down.

🎯 Exam Tip: Always rearrange linear equations into \( y = mx + c \) form to clearly identify the slope (\( m \)) and y-intercept (\( c \)), which tell you about stretching/compression/reflection and vertical shifts.

 

Question 7. From the curve y = |x| draw
(ii) y = |x + 1| – 1
(iii) y = |x + 2| – 3
Answer:
First, let's look at the basic absolute value function \( y = |x| \). It creates a V-shaped graph.

The graph of \( y = |x| \) looks like this:

(i) y = |x - 1| + 1 (Note: This sub-part is not explicitly in Q7, but is presented in the source as an answer example after the base function.)
To graph \( y = |x - 1| + 1 \), we first consider \( y = |x - 1| \).
(a) Consider \( y = |x - 1| \)
This means:
\( y = \begin{cases} x - 1 & \text{if } x - 1 \geq 0 \implies x \geq 1 \\ -(x - 1) & \text{if } x - 1 < 0 \implies x < 1 \end{cases} \)
So, \( y = \begin{cases} x - 1 & \text{if } x \geq 1 \\ -x + 1 & \text{if } x < 1 \end{cases} \)
Let's find some key points for \( y = |x - 1| \):
When \( x = 0 \), then \( y = -0 + 1 = 1 \)
When \( x = 1 \), then \( y = 1 - 1 = 0 \)
When \( x = 2 \), then \( y = 2 - 1 = 1 \)
When \( x = 3 \), then \( y = 3 - 1 = 2 \)
When \( x = -1 \), then \( y = -(-1) + 1 = 1 + 1 = 2 \)
When \( x = -2 \), then \( y = -(-2) + 1 = 2 + 1 = 3 \)

The graph of \( y = |x - 1| \) shifts the graph of \( y = |x| \) to the right by 1 unit.

(b) Now consider \( y = |x - 1| + 1 \).
This means:
\( y = \begin{cases} (x - 1) + 1 & \text{if } x - 1 \geq 0 \implies x \geq 1 \\ -(x - 1) + 1 & \text{if } x - 1 < 0 \implies x < 1 \end{cases} \)
So, \( y = \begin{cases} x & \text{if } x \geq 1 \\ -x + 2 & \text{if } x < 1 \end{cases} \)
Let's find some key points for \( y = |x - 1| + 1 \):
When \( x = 0 \), then \( y = -0 + 2 = 2 \)
When \( x = 1 \), then \( y = 1 = 1 \)
When \( x = 2 \), then \( y = 2 = 2 \)
When \( x = 3 \), then \( y = 3 = 3 \)
When \( x = -1 \), then \( y = -(-1) + 2 = 1 + 2 = 3 \)
When \( x = -2 \), then \( y = -(-2) + 2 = 2 + 2 = 4 \)

The graph of \( y = |x - 1| + 1 \) shifts the graph \( y = |x| \) to the right by 1 unit and causes a shift upward by 1 unit.

In simple words: When you subtract a number inside the absolute value, the graph moves to the right. When you add a number outside, it moves up. So, this graph is the V-shape moved 1 unit right and 1 unit up.

🎯 Exam Tip: For transformations of \( y = |x| \), \( y = |x-c| \) shifts right by \( c \) units, \( y = |x+c| \) shifts left by \( c \) units, and \( y = |x| + d \) shifts up by \( d \) units, while \( y = |x| - d \) shifts down by \( d \) units.

 

(ii) y = |x + 1| -1
Answer: To graph \( y = |x + 1| - 1 \), we first consider \( y = |x + 1| \).
(a) Consider \( y = |x + 1| \)
This means:
\( y = \begin{cases} (x + 1) & \text{if } x + 1 \geq 0 \implies x \geq -1 \\ -(x + 1) & \text{if } x + 1 < 0 \implies x < -1 \end{cases} \)
So, \( y = \begin{cases} x + 1 & \text{if } x \geq -1 \\ -x - 1 & \text{if } x < -1 \end{cases} \)
Let's find some key points for \( y = |x + 1| \):
When \( x = 0 \), then \( y = 0 + 1 = 1 \)
When \( x = 1 \), then \( y = 1 + 1 = 2 \)
When \( x = 2 \), then \( y = 2 + 1 = 3 \)
When \( x = 3 \), then \( y = 3 + 1 = 4 \)
When \( x = -1 \), then \( y = -1 + 1 = 0 \)
When \( x = -2 \), then \( y = -(-2) - 1 = 2 - 1 = 1 \)
When \( x = -3 \), then \( y = -(-3) - 1 = 3 - 1 = 2 \)

The graph of \( y = |x + 1| \) shifts the graph of \( y = |x| \) to the left by 1 unit.

(b) Now consider \( y = |x + 1| - 1 \).
This means:
\( y = \begin{cases} (x + 1) - 1 & \text{if } x + 1 \geq 0 \implies x \geq -1 \\ -(x + 1) - 1 & \text{if } x + 1 < 0 \implies x < -1 \end{cases} \)
So, \( y = \begin{cases} x & \text{if } x \geq -1 \\ -x - 2 & \text{if } x < -1 \end{cases} \)
Let's find some key points for \( y = |x + 1| - 1 \):
When \( x = 0 \), then \( y = 0 = 0 \)
When \( x = 1 \), then \( y = 1 = 1 \)
When \( x = 2 \), then \( y = 2 = 2 \)
When \( x = -1 \), then \( y = -1 = -1 \)
When \( x = -2 \), then \( y = -(-2) - 2 = 2 - 2 = 0 \)
When \( x = -3 \), then \( y = -(-3) - 2 = 3 - 2 = 1 \)
When \( x = -4 \), then \( y = -(-4) - 2 = 4 - 2 = 2 \)

The graph of \( y = |x + 1| - 1 \) shifts the graph \( y = |x| \) to the left by 1 unit and causes a shift downward by 1 unit.

In simple words: Adding a number inside absolute value moves the graph left. Subtracting a number outside moves it down. So this V-shape is moved 1 unit left and 1 unit down.

🎯 Exam Tip: When \( x \) is replaced by \( (x+c) \) in \( f(x) \), the graph shifts horizontally: left if \( c > 0 \) and right if \( c < 0 \). The vertical shift applies to \( f(x) \pm d \).

 

(iii) y = |x + 2| - 3
Answer: To graph \( y = |x + 2| - 3 \), we first consider \( y = |x + 2| \).
(a) Consider the curve \( y = |x + 2| \)
This means:
\( y = \begin{cases} (x + 2) & \text{if } x + 2 \geq 0 \implies x \geq -2 \\ -(x + 2) & \text{if } x + 2 < 0 \implies x < -2 \end{cases} \)
So, \( y = \begin{cases} x + 2 & \text{if } x \geq -2 \\ -x - 2 & \text{if } x < -2 \end{cases} \)
Let's find some key points for \( y = |x + 2| \):
When \( x = 0 \), then \( y = 0 + 2 = 2 \)
When \( x = 1 \), then \( y = 1 + 2 = 3 \)
When \( x = 2 \), then \( y = 2 + 2 = 4 \)
When \( x = 3 \), then \( y = 3 + 2 = 5 \)
When \( x = -1 \), then \( y = -1 + 2 = 1 \)
When \( x = -2 \), then \( y = -2 + 2 = 0 \)
When \( x = -3 \), then \( y = -(-3) - 2 = 3 - 2 = 1 \)

The graph of \( y = |x + 2| \) shifts the graph of \( y = |x| \) to the left by 2 units.

(b) Now consider \( y = |x + 2| - 3 \).
This means:
\( y = \begin{cases} (x + 2) - 3 & \text{if } x + 2 \geq 0 \implies x \geq -2 \\ -(x + 2) - 3 & \text{if } x + 2 < 0 \implies x < -2 \end{cases} \)
So, \( y = \begin{cases} x - 1 & \text{if } x \geq -2 \\ -x - 5 & \text{if } x < -2 \end{cases} \)
Let's find some key points for \( y = |x + 2| - 3 \):
When \( x = 0 \), then \( y = 0 - 1 = -1 \)
When \( x = 1 \), then \( y = 1 - 1 = 0 \)
When \( x = 2 \), then \( y = 2 - 1 = 1 \)
When \( x = 3 \), then \( y = 3 - 1 = 2 \)
When \( x = -1 \), then \( y = -1 - 1 = -2 \)
When \( x = -2 \), then \( y = -2 - 1 = -3 \)
When \( x = -3 \), then \( y = -(-3) - 5 = 3 - 5 = -2 \)

The graph of \( y = |x + 2| - 3 \) shifts the graph \( y = |x| \) to the left by 2 units and causes a shift downward by 3 units.

In simple words: This graph moves the V-shape from \( y = |x| \) 2 units to the left and 3 units down. The sharp point of the V ends up at (-2, -3).

🎯 Exam Tip: For \( y = |x+c| - d \), the vertex of the V-shape will be at \( (-c, -d) \). This makes it easy to quickly sketch the graph by locating its vertex and drawing the V-shape from there.

 

Question 8. From the curve y = sin x, draw y = sin |x|.
Answer:
(a) For the graph of \( y = \sin x \):

The graph for \( y = \sin x \) is a wave that passes through the origin, reaching a maximum of 1 at \( x = \frac{\pi}{2} \), and a minimum of -1 at \( x = \frac{3\pi}{2} \). It repeats every \( 2\pi \) radians.

(b) To draw \( y = \sin |x| \):
First, we define \( y = \sin |x| \) piecewise:
\( y = \begin{cases} \sin x & \text{if } x \ge 0 \\ \sin(-x) & \text{if } x < 0 \end{cases} \)
We know that \( \sin(-x) = -\sin x \). So, for \( x < 0 \), we are evaluating \( \sin(-x) \).

Let's calculate some values for \( y = \sin |x| \):
When \( x = 0 \), \( y = \sin |0| = \sin 0 = 0 \)
When \( x = \frac{\pi}{2} \), \( y = \sin |\frac{\pi}{2}| = \sin \frac{\pi}{2} = 1 \)
When \( x = \pi \), \( y = \sin |\pi| = \sin \pi = 0 \)
When \( x = \frac{3\pi}{2} \), \( y = \sin |\frac{3\pi}{2}| = \sin \frac{3\pi}{2} = -1 \)
When \( x = 2\pi \), \( y = \sin |2\pi| = \sin 2\pi = 0 \)
For negative values of x:
When \( x = -\frac{\pi}{2} \), \( y = \sin |-\frac{\pi}{2}| = \sin (\frac{\pi}{2}) = 1 \)
When \( x = -\pi \), \( y = \sin |-\pi| = \sin (\pi) = 0 \)
When \( x = -\frac{3\pi}{2} \), \( y = \sin |-\frac{3\pi}{2}| = \sin (\frac{3\pi}{2}) = -1 \)
When \( x = -2\pi \), \( y = \sin |-2\pi| = \sin (2\pi) = 0 \)

The graph of \( y = \sin |x| \) is formed by keeping the part of \( y = \sin x \) for \( x \ge 0 \) exactly as it is. Then, for the part where \( x < 0 \), we reflect the graph from \( x \ge 0 \) over the y-axis. This makes the graph symmetrical about the y-axis, meaning it is an even function. This type of graph looks like a sine wave on the right side and a mirror image of that sine wave on the left side.

In simple words: To draw \( y = \sin |x| \), first draw the regular \( y = \sin x \) graph for all positive x values. Then, for negative x values, simply copy the positive x-side graph by reflecting it over the y-axis. The graph will look the same on both the left and right sides of the y-axis.

🎯 Exam Tip: When drawing \( y = f(|x|) \), always remember that the graph for \( x < 0 \) is a mirror image of the graph for \( x \ge 0 \) across the y-axis.