OP Malhotra Class 11 Maths Solutions Chapter 2 Relations and Functions Exercise 2 (G)

 

Question 1. \( y = - | x | \)
Answer: The given function is \( y = - | x | \). The domain for this function includes all real numbers. The range for this function includes all negative numbers and zero, meaning \( y \le 0 \). When \( x \) is positive, \( |x| = x \), so \( y = -x \). This forms a straight line passing through the origin with a slope of -1. When \( x \) is negative, \( |x| = -x \), so \( y = -(-x) = x \). This forms a straight line also passing through the origin but with a slope of 1. When \( x=0 \), \( y=0 \). Therefore, the graph forms a V-shape pointing downwards, with its vertex at the origin (0, 0). Each side of the V-shape is a straight line.
In simple words: This function creates a graph that looks like a "V" shape turned upside down, with its pointy end exactly at the center of the graph (origin).

🎯 Exam Tip: When graphing absolute value functions, always consider the cases for positive and negative inputs to correctly draw the V-shape or inverted V-shape.

 

Question 2. \( y = \frac{|x|-x}{2} \)
Answer: The given function is \( y = \frac{|x|-x}{2} \). The domain for this function includes all real numbers. We need to consider two main cases for \( x \). First, if \( x \ge 0 \), then \( |x| = x \). In this case, the function becomes \( y = \frac{x-x}{2} = \frac{0}{2} = 0 \). This means for all non-negative values of \( x \), the graph lies exactly on the x-axis. Second, if \( x < 0 \), then \( |x| = -x \). In this case, the function becomes \( y = \frac{-x-x}{2} = \frac{-2x}{2} = -x \). This forms a straight line with a slope of -1, passing through the origin and extending into the second quadrant for negative \( x \) values. Combining these, the graph is the x-axis for \( x \ge 0 \) and a line with slope -1 for \( x < 0 \).
In simple words: If the number \( x \) is positive or zero, the graph stays flat on the x-axis. If \( x \) is a negative number, the graph goes up diagonally from left to right, like a ramp going down.

🎯 Exam Tip: Always break down absolute value functions into cases based on the sign of the expression inside the absolute value, then graph each piece separately.

 

Question 3. \( y = \frac{1}{|x|} \)
Answer: The given function is \( y = \frac{1}{|x|} \). The domain for this function is all real numbers except for \( x = 0 \), because division by zero is not allowed. The range will always be positive values since \( |x| \) is always positive (or zero, but \( x=0 \) is excluded). When \( x > 0 \), then \( |x| = x \), so \( y = \frac{1}{x} \). This creates a curve in the first quadrant that gets very large as \( x \) approaches 0 from the positive side, and gets very small as \( x \) gets larger. When \( x < 0 \), then \( |x| = -x \), so \( y = \frac{1}{-x} = -\frac{1}{x} \). This creates a curve in the second quadrant that mirrors the first quadrant curve across the y-axis, also getting very large as \( x \) approaches 0 from the negative side. The graph does not touch the x-axis or the y-axis (these are asymptotes).
The table of values shows how \( y \) changes with \( x \):

Also, this graph does not meet coordinate axes for \( x < 0 \). The table for negative \( x \) values:

🎯 Exam Tip: Remember that \( y = \frac{1}{|x|} \) has a vertical asymptote at \( x=0 \) and a horizontal asymptote at \( y=0 \), meaning the graph gets very close to the axes but never touches or crosses them.

 

Question 4. \( y = | 4 - x^2 | \), \( -3 \le x \le 3 \).
Answer: The function given is \( y = | 4 - x^2 | \), defined for \( -3 \le x \le 3 \). The domain is the interval \( [-3, 3] \), and the range is \( [0, \infty) \) because the absolute value ensures all \( y \) values are non-negative. Let's analyze the expression \( 4 - x^2 \). This is a downward-opening parabola with its vertex at (0, 4) and x-intercepts at \( x = \pm 2 \). When \( 4 - x^2 \ge 0 \), which means \( -2 \le x \le 2 \), then \( y = 4 - x^2 \). This is the upper part of the parabola. When \( 4 - x^2 < 0 \), which means \( x < -2 \) or \( x > 2 \), then \( y = - (4 - x^2) = x^2 - 4 \). This takes the parts of the parabola that would normally be below the x-axis and reflects them upwards because of the absolute value. The graph starts from \( y = |4 - (-3)^2| = |4 - 9| = 5 \) at \( x = -3 \). It then follows \( x^2 - 4 \) up to \( x = -2 \) (where \( y = 0 \)). From \( x = -2 \) to \( x = 2 \), it follows the parabola \( 4 - x^2 \), peaking at (0, 4) and returning to \( y = 0 \) at \( x = 2 \). Finally, from \( x = 2 \) to \( x = 3 \), it follows \( x^2 - 4 \) again, reaching \( y = |4 - 3^2| = |4 - 9| = 5 \) at \( x = 3 \). This creates a W-like shape where the portions below the x-axis are folded up.
In simple words: The graph of \( y = 4 - x^2 \) is a hill-shaped curve. Because of the absolute value sign, any part of the hill that would go below the ground (x-axis) is flipped upwards. So, it becomes a "W" shape within the given \( x \) range.

🎯 Exam Tip: When dealing with absolute value functions like \( y = |f(x)| \), remember to graph \( f(x) \) first, then reflect any part below the x-axis over the x-axis to get the final graph.

 

Question 5. \( y = | x | + x \), \( -2 \le x \le 2 \)
Answer: The given function is \( y = | x | + x \), defined for \( -2 \le x \le 2 \). The domain covers all real numbers. We need to split this into two cases based on the sign of \( x \). Case 1: When \( x \ge 0 \). In this case, \( |x| = x \). So the function becomes \( y = x + x = 2x \). This is a straight line passing through the origin with a slope of 2. Case 2: When \( x < 0 \). In this case, \( |x| = -x \). So the function becomes \( y = -x + x = 0 \). This means for all negative values of \( x \), the graph lies exactly on the x-axis. Combining these, the graph is the x-axis for \( x < 0 \) and a line \( y = 2x \) for \( x \ge 0 \). Within the given range, it starts from \( x = -2 \) (where \( y = 0 \)), continues along the x-axis to \( x = 0 \) (where \( y = 0 \)), and then goes up along the line \( y = 2x \) to \( x = 2 \) (where \( y = 2(2) = 4 \)).
In simple words: For negative numbers, the graph stays flat on the x-axis. For positive numbers and zero, the graph goes up in a straight line, rising twice as fast as \( x \) grows.

🎯 Exam Tip: This function is a classic example of a piecewise function. Always clearly define the rule for each part of the domain to ensure correct graphing.

 

Question 6. \( y = |x + 2| + x \)
Answer: The given function is \( y = |x + 2| + x \). The domain for this function is all real numbers. We need to consider two cases based on the expression inside the absolute value, \( x + 2 \). Case 1: When \( x + 2 \ge 0 \), which means \( x \ge -2 \). In this case, \( |x + 2| = x + 2 \). So, \( y = (x + 2) + x = 2x + 2 \). This is a straight line with a slope of 2 and a y-intercept of 2. Case 2: When \( x + 2 < 0 \), which means \( x < -2 \). In this case, \( |x + 2| = -(x + 2) \). So, \( y = -(x + 2) + x = -x - 2 + x = -2 \). This means for all values of \( x \) less than -2, the graph is a horizontal line at \( y = -2 \). Combining these, the graph is a horizontal line \( y = -2 \) for \( x < -2 \), and a line \( y = 2x + 2 \) for \( x \ge -2 \). The two parts meet at \( x = -2 \), where \( y = 2(-2) + 2 = -4 + 2 = -2 \). The table of values for \( y = 2(x+1) \) (which is \( y=2x+2 \)) for \( x \ge -2 \) is given as:

🎯 Exam Tip: Pay close attention to the point where the behavior of the absolute value changes (the critical point) as this is where the graph will bend or switch forms.

 

Question 7.
(i) Copy and complete this table of values :

(ii) Draw the graph \( y = 3^x \) on squared paper, for \( -2 \le x \le 3 \).
(iii) What features do the graphs of \( y = 2^x \) (drawn on page Ch 2-51) and \( y = 3^x \) have in common?
Answer:
(i) The given function is \( y = 3^x \). Let's complete the table by calculating the values of \( y \) for the given \( x \) values.
When \( x = 1 \), \( y = 3^1 = 3 \).
When \( x = 2 \), \( y = 3^2 = 9 \).
When \( x = 3 \), \( y = 3^3 = 27 \).
The completed table of values is:

(ii) The graph of \( y = 3^x \) shows exponential growth. As \( x \) increases, \( y \) increases rapidly. As \( x \) decreases, \( y \) approaches 0 but never actually reaches it. The graph always passes through the point (0, 1).

(iii) Both graphs of \( y = 2^x \) and \( y = 3^x \) share common features. Both are exponential growth curves. They both pass through the point (0, 1), which is a key characteristic for any function of the form \( y = a^x \) when \( x=0 \). They both rise steadily from left to right, showing that as \( x \) increases, \( y \) increases. Furthermore, as \( x \) approaches negative infinity (goes far to the left), the y-values for both graphs approach 0, but never actually touch the x-axis. This means the x-axis acts as a horizontal asymptote for both graphs.
When \( x \to \infty \), then \( y \to \infty \).
When \( x \to -\infty \), then \( y \to 0 \).
In simple words: Both graphs start low on the left and go up very fast to the right. They both cross the 'y' line at the same spot (0,1). As you go far left, both lines get very close to the 'x' line but never touch it.

🎯 Exam Tip: For exponential functions of the form \( y = a^x \), remember that all graphs pass through (0,1) and the x-axis is always a horizontal asymptote, meaning the y-value approaches 0 but never reaches it.

 

Question 8.
(i) Copy and complete this table.

(ii) Draw the graphs \( y = 2^x \) and \( y = (\frac{1}{2})^x \) on the same diagram, for \( -3 \le x \le 3 \).
(iii) Which line is the axis of symmetry in the diagram?
Answer:
(i) The given function is \( y = (\frac{1}{2})^x \). Let's complete the table of values.
When \( x = -3 \), \( y = (\frac{1}{2})^{-3} = 2^3 = 8 \).
When \( x = -2 \), \( y = (\frac{1}{2})^{-2} = 2^2 = 4 \).
When \( x = -1 \), \( y = (\frac{1}{2})^{-1} = 2 \).
When \( x = 1 \), \( y = \frac{1}{2} \).
The table of values is completed as:

(ii) The graph of \( y = 2^x \) shows exponential growth, increasing from left to right. The graph of \( y = (\frac{1}{2})^x \) (which can also be written as \( y = 2^{-x} \)) shows exponential decay, decreasing from left to right. Both graphs pass through the point (0, 1) and have the x-axis as a horizontal asymptote.

(iii) Both graphs of \( y = 2^x \) and \( y = (\frac{1}{2})^x \) pass through the point (0, 1). This is a common feature of all exponential functions where the base is positive and not equal to 1. The y-axis (the line \( x = 0 \)) is the axis of symmetry for these two graphs. This is because \( (\frac{1}{2})^x = (2^{-1})^x = 2^{-x} \). So, the graph of \( y = (\frac{1}{2})^x \) is a reflection of the graph of \( y = 2^x \) across the y-axis.
In simple words: Both graphs go through the point (0,1). One graph goes up from left to right, and the other goes down. The 'y' line (the vertical line in the middle) acts like a mirror between the two graphs.

🎯 Exam Tip: Recognize that functions \( y = a^x \) and \( y = (\frac{1}{a})^x \) are reflections of each other across the y-axis, and both will always intersect the y-axis at (0, 1).

 

Question 9. A sketch of the graph \( y = a \log_4 (x + b) \) is shown in Fig. 2.70. Find the values of a and b.
Answer: The given equation of the curve is \( y = a \log_4 (x + b) \). From the provided graph, we can see that the curve passes through two specific points: (-2, 0) and (1, 5). We will use these points to form equations and solve for 'a' and 'b'. Using point (-2, 0):
\( 0 = a \log_4 (-2 + b) \)
Since \( a \ne 0 \) (otherwise \( y \) would always be 0), we must have \( \log_4 (-2 + b) = 0 \).
This means \( -2 + b = 4^0 \)
\( -2 + b = 1 \)
\( b = 1 + 2 \)
\( b = 3 \).
Now, using point (1, 5) and the value of \( b = 3 \):
\( 5 = a \log_4 (1 + 3) \)
\( 5 = a \log_4 (4) \)
We know that \( \log_k k = 1 \), so \( \log_4 4 = 1 \).
\( 5 = a \times 1 \)
\( a = 5 \).
Thus, the values are \( a = 5 \) and \( b = 3 \).
In simple words: We used two points shown on the graph to figure out the hidden numbers 'a' and 'b' in the equation. By plugging in the points, we solved the equations to find that 'a' is 5 and 'b' is 3.

🎯 Exam Tip: When given points on a graph and an equation with unknown constants, substitute the coordinates of each point into the equation to create a system of equations, then solve for the unknowns.

 

Question 10. Diagrams (i) shows the curve \( y = \log_a x \). What is the value of a ?
Answer: The given equation of the curve is \( y = \log_a x \). From the provided graph, we can see that the curve passes through the point (81, 4). We will use this point to find the value of 'a'. Substitute the coordinates \( x = 81 \) and \( y = 4 \) into the equation:
\( 4 = \log_a 81 \)
To solve for 'a', we convert the logarithmic equation to its exponential form. The definition of logarithm states that if \( y = \log_a x \), then \( a^y = x \).
So, \( a^4 = 81 \)
We need to find a number 'a' that, when raised to the power of 4, equals 81. We know that \( 3^4 = 3 \times 3 \times 3 \times 3 = 9 \times 9 = 81 \).
Therefore, \( a^4 = 3^4 \)
Comparing the bases, we find \( a = 3 \).
In simple words: The graph shows that when \( x \) is 81, \( y \) is 4. We use this fact in the equation \( y = \log_a x \) to find the base 'a'. It turns out that 'a' must be 3 because 3 raised to the power of 4 gives 81.

🎯 Exam Tip: Always remember the conversion between logarithmic form \( y = \log_a x \) and exponential form \( a^y = x \); this is crucial for solving such problems.

 

Question 11. Diagram given below (ii) shows the curve \( y = \log_{10} (x + p) \). What is the value of p ?
Answer: The given equation of the curve is \( y = \log_{10} (x + p) \). From the provided graph, we can see that the curve passes through the point (-1, 0). We will use this point to find the value of 'p'. Substitute the coordinates \( x = -1 \) and \( y = 0 \) into the equation:
\( 0 = \log_{10} (-1 + p) \)
To solve for 'p', we convert the logarithmic equation to its exponential form. The definition of logarithm states that if \( y = \log_a x \), then \( a^y = x \).
So, \( 10^0 = -1 + p \)
We know that any non-zero number raised to the power of 0 is 1.
\( 1 = -1 + p \)
\( p = 1 + 1 \)
\( p = 2 \).
Thus, the value of 'p' is 2. The graph shows that the function is defined for \( x+p > 0 \), so \( x+2 > 0 \), which means \( x > -2 \). This aligns with the visual of the curve starting to the right of \( x=-2 \).
In simple words: The graph crosses the 'x' line when \( x \) is -1. Using this point in the equation \( y = \log_{10} (x + p) \), we find that \( p \) must be 2.

🎯 Exam Tip: For logarithmic functions of the form \( y = \log_a (x+p) \), the vertical asymptote is given by \( x = -p \). If a point \( (-1, 0) \) is on the graph, it implies \( \log_{10} (-1+p) = 0 \), which means \( -1+p=1 \).

 

Question 12.
(i) Sketch the graphs \( y = 2 \) and \( y = \log_{10} 2x \) on the same diagram.
(ii) Find the point of intersection of the graphs by solving the equation \( \log_{10} 2x = 2 \).
Answer:
(i) We need to sketch two graphs: \( y = 2 \) and \( y = \log_{10} 2x \).
The graph of \( y = 2 \) is a straight horizontal line that is parallel to the x-axis and passes through the point (0, 2) on the y-axis.
For the graph of \( y = \log_{10} 2x \), we first determine its domain. For the logarithm to be defined, \( 2x > 0 \), which implies \( x > 0 \). The vertical asymptote is at \( x = 0 \) (the y-axis). To sketch the curve, we can find some points. It is helpful to rewrite the logarithmic equation in exponential form: \( 2x = 10^y \), so \( x = \frac{1}{2} \cdot 10^y \).
Let's create a table of values for \( y = \log_{10} 2x \) (or \( x = \frac{1}{2} \cdot 10^y \)):

(ii) To find the point of intersection, we solve the equation \( \log_{10} 2x = 2 \).
Convert the logarithmic equation to exponential form: \( 2x = 10^2 \).
\( 2x = 100 \)
Divide both sides by 2:
\( x = \frac{100}{2} \)
\( x = 50 \).
Since \( y = 2 \) is one of the equations, the y-coordinate of the intersection is 2. Therefore, the graphs intersect at the point (50, 2).
In simple words: We find where the horizontal line and the curved line meet. By using the equation for the curved line and setting it equal to 2, we found that they cross when \( x \) is 50. So the meeting point is (50, 2).

🎯 Exam Tip: When solving logarithmic equations for points of intersection, always convert the logarithm into its exponential form to simplify the algebra and find the exact coordinates.

 

Question 13. The sketch shows part of the graph \( y = a \log_2 (x - b) \). Find the values of a and b.
Answer: The given equation of the curve is \( y = a \log_2 (x - b) \). From the provided graph, we can see that the curve passes through two specific points: P(5, 0) and Q(6, 3). We will use these points to form equations and solve for 'a' and 'b'.
Using point P(5, 0):
\( 0 = a \log_2 (5 - b) \)
Since \( a \ne 0 \) (otherwise \( y \) would always be 0), we must have \( \log_2 (5 - b) = 0 \).
This means \( 5 - b = 2^0 \)
\( 5 - b = 1 \)
\( b = 5 - 1 \)
\( b = 4 \).
Now, using point Q(6, 3) and the value of \( b = 4 \):
\( 3 = a \log_2 (6 - 4) \)
\( 3 = a \log_2 (2) \)
We know that \( \log_k k = 1 \), so \( \log_2 2 = 1 \).
\( 3 = a \times 1 \)
\( a = 3 \).
Therefore, the values are \( a = 3 \) and \( b = 4 \).
In simple words: We used two points from the graph, P(5,0) and Q(6,3), to find the missing numbers 'a' and 'b' in the equation. By putting these points into the equation, we found that 'a' is 3 and 'b' is 4.

🎯 Exam Tip: Logarithmic functions of the form \( y = \log_a (x-b) \) have a vertical asymptote at \( x=b \). Always use the x-intercept to find 'b' first, as \( \log_a 1 = 0 \).

 

Question 14.
(i) Sketch the graphs \( y = 4 - x \) and \( y = \log_{10} x \) on the same diagram.
(ii) Write down an equation to find the x-coordinate of the point of intersection of the graphs.
(iii) Show that x satisfies \( 3.4 < x < 3.5 \).
(iv) From your graph find x, correct to 2 decimal places.
Answer:
(i) We need to sketch two graphs: \( y = 4 - x \) and \( y = \log_{10} x \).
For the graph of \( y = 4 - x \): This is a straight line. When \( x = 0 \), \( y = 4 \). When \( y = 0 \), \( 0 = 4 - x \implies x = 4 \). So the line passes through (0, 4) and (4, 0). A table of values for \( y = 4 - x \):

For the graph of \( y = \log_{10} x \): The domain is \( x > 0 \). The vertical asymptote is \( x = 0 \). A table of values for \( y = \log_{10} x \):

(ii) The point of intersection occurs where the y-values of both functions are equal. So, the equation to find the x-coordinate of the intersection is \( 4 - x = \log_{10} x \). This equation cannot be solved algebraically and requires numerical or graphical methods.
In simple words: To find where the two lines cross, we set their equations equal to each other: \( 4 - x = \log_{10} x \).

(iii) To show that \( x \) satisfies \( 3.4 < x < 3.5 \), we can define a function \( f(x) = \log_{10} x + x - 4 \). The root of the equation \( \log_{10} x = 4 - x \) is where \( f(x) = 0 \). If \( f(x) \) changes sign between two values, then a root lies between them.
Let's evaluate \( f(x) \) at \( x = 3.4 \) and \( x = 3.5 \):
\( f(3.4) = \log_{10} (3.4) + 3.4 - 4 \)
\( f(3.4) \approx 0.5314 + 3.4 - 4 \)
\( f(3.4) \approx -0.0686 \)
\( f(3.5) = \log_{10} (3.5) + 3.5 - 4 \)
\( f(3.5) \approx 0.5441 + 3.5 - 4 \)
\( f(3.5) \approx 0.0441 \)
Since \( f(3.4) \) is negative and \( f(3.5) \) is positive, there is a sign change. This means that a root (the point of intersection) lies between \( x = 3.4 \) and \( x = 3.5 \).
In simple words: We tested numbers close to where the lines cross. One number made the equation slightly negative, and the other made it slightly positive. This shows that the crossing point is definitely between these two numbers.

(iv) By carefully examining the graph drawn in part (i), we can estimate the x-coordinate where the line \( y = 4 - x \) and the curve \( y = \log_{10} x \) intersect. From the visual representation, the intersection occurs approximately at \( x = 3.46 \). When estimating from a graph, accuracy depends on the scale and precision of the drawing. Thus, the x-coordinate, correct to 2 decimal places, is \( x = 3.46 \).
In simple words: Looking at the graph, the two lines meet at about \( x = 3.46 \).

🎯 Exam Tip: For graphical solutions, make sure your graph is drawn accurately with a clear scale. When estimating values, read carefully from the grid lines and provide the answer to the requested decimal precision.

 

Question 15. Sketch the graphs.
(i) \( y = \log_2 x \)
(ii) \( y = \log_2 x + 1 \)
(iii) \( y = \log_2 (x + 1) \)
Answer:
(i) For the function \( y = \log_2 x \), the domain includes all positive real numbers, meaning \( x > 0 \). The graph passes through key points like (1,0), (2,1), and (4,2). As \( x \) gets closer to 0, \( y \) approaches negative infinity, and as \( x \) increases, \( y \) increases slowly.
The table of values for \( y = \log_2 x \) is:

(ii) For the function \( y = \log_2 x + 1 \), the graph is the same as \( y = \log_2 x \) but shifted one unit upwards. Its domain is also all positive real numbers. The equation can be rewritten as \( y - 1 = \log_2 x \), which means \( x = 2^{y-1} \).
The table of values for \( y = \log_2 x + 1 \) is:

(iii) For the function \( y = \log_2 (x + 1) \), the graph is similar to \( y = \log_2 x \) but shifted one unit to the left. The domain requires \( x + 1 > 0 \), so \( x > -1 \). The equation means \( x + 1 = 2^y \), or \( x = 2^y - 1 \).
The table of values for \( y = \log_2 (x + 1) \) is:

In simple words: These are all graphs of logarithmic functions. The first graph is the basic one. The second one is the first graph moved up by one step. The third graph is the first graph moved to the left by one step. Look at the tables to find points for plotting.

🎯 Exam Tip: When sketching logarithmic graphs, always identify the vertical asymptote first (e.g., \( x=0 \) for \( \log_2 x \), or \( x=-1 \) for \( \log_2 (x+1) \)). Key points like where the graph crosses the x-axis (y=0) or passes through (base, 1) are also important.

 

Question 16. Sketch the graphs.
(i) \( \log_4 x \)
(ii) \( 2 \log_4 x \)
(iii) \( 3 \log_4 x \)
Answer:
(i) For the function \( y = \log_4 x \), the domain is all positive real numbers. This means \( x = 4^y \). The graph starts from negative infinity as \( x \) approaches 0 and slowly increases, passing through points like (1,0) and (4,1).
The table of values for \( y = \log_4 x \) is:

(ii) For the function \( y = 2 \log_4 x \), the graph is the same as \( y = \log_4 x \) but stretched vertically by a factor of 2. The domain remains all positive real numbers, and the equation can be written as \( \frac{y}{2} = \log_4 x \), which means \( x = 4^{y/2} \).
The table of values for \( y = 2 \log_4 x \) is:

(iii) For the function \( y = 3 \log_4 x \), the graph is the same as \( y = \log_4 x \) but stretched vertically by a factor of 3. Its domain is also all positive real numbers. The equation can be written as \( \frac{y}{3} = \log_4 x \), which means \( x = 4^{y/3} \).
The table of values for \( y = 3 \log_4 x \) is:

In simple words: These are graphs of logarithmic functions. The graphs stretch upwards more as the number multiplied by \( \log_4 x \) gets bigger. All these graphs have the Y-axis as an asymptote, meaning they get very close to it but never touch it.

🎯 Exam Tip: When a logarithmic function is multiplied by a constant, it stretches or compresses the graph vertically. Understand that \( y = k \log_b x \) has the same vertical asymptote as \( y = \log_b x \) but grows faster if \( k > 1 \).

 

Question 17. For – 2 < x < 1, draw the graph of \( y = 2^x \). (use 1 cm = 1 unit on both axes). Use this graph to solve \( 2^x = 2x \).
Answer:
We are given the equation of the curve \( y = 2^x \) within the range \( -2 < x < 1 \). This is an exponential function. We also need to solve \( 2^x = 2x \), which means finding the intersection points of the graph \( y = 2^x \) and the straight line \( y = 2x \).
The table of values for \( y = 2^x \) is:

From the graph below, we can see the exponential curve \( y = 2^x \) and the straight line \( y = 2x \). They intersect at \( x = 1 \) and \( x = 2 \). Since the question specified the range \( -2 < x < 1 \), only \( x = 1 \) is relevant if we extend the upper bound of the range to include 1. However, the graph itself shows both points. If we consider the graph over a wider range, the solutions are \( x = 1 \) and \( x = 2 \).
In simple words: First, draw the graph of \( y = 2^x \). Then, on the same graph, draw the line \( y = 2x \). The points where these two lines cross each other are the answers to the question. You will find they cross at \( x=1 \) and \( x=2 \).

🎯 Exam Tip: To solve equations graphically, always draw each side of the equation as a separate function. The x-coordinates of the intersection points are the solutions to the equation. Label your curves clearly.

 

Question 18.
(i) Complete the following table for \( y = 4^x \). Enter the values of x and y correct to 1 decimal place.
(ii) Taking 4 cm = 1 unit on both axes, draw the graph of \( y = 4^x \) for \( -2 \le x \le 0.75 \).
(iii) From your graph estimate \( \log_4 1.25 \).
Answer:
(i) We need to complete the table for the function \( y = 4^x \). Let's calculate the values and round them to one decimal place:
When \( x = -2 \): \( y = 4^{-2} = \frac{1}{16} = 0.0625 \approx 0.1 \)
When \( x = -1 \): \( y = 4^{-1} = \frac{1}{4} = 0.25 \approx 0.3 \)
When \( x = 0 \): \( y = 4^0 = 1 \approx 1.0 \)
When \( x = 0.5 \): \( y = 4^{0.5} = \sqrt{4} = 2 \approx 2.0 \)
When \( x = 0.75 \): \( y = 4^{0.75} = 4^{3/4} = (\sqrt[4]{4})^3 = (\sqrt{2})^3 = 2\sqrt{2} \approx 2.828 \approx 2.8 \)
The completed table of values is:

(ii) The graph of \( y = 4^x \) for \( -2 \le x \le 0.75 \) is an exponential curve that rises sharply. It passes through the points calculated in the table. The graph is drawn with the specified scale (4 cm = 1 unit) as shown below.

(iii) To estimate \( \log_4 1.25 \) from the graph, we need to find the value of \( x \) for which \( y = 1.25 \). Locate \( y = 1.25 \) on the y-axis, then draw a horizontal line to intersect the curve \( y = 4^x \). From this intersection point, drop a vertical line to the x-axis. The x-coordinate at this point is the estimated value of \( \log_4 1.25 \). From the graph, it is observed that when \( y = 1.25 \), the corresponding \( x \) value is approximately \( 0.1609 \), which can be rounded to \( 0.16 \) (correct to 2 decimal places).
In simple words: First, fill in the table by finding the values of \( 4^x \) for each \( x \). Then draw the graph of \( y = 4^x \). To find \( \log_4 1.25 \), look for 1.25 on the y-axis, go across to the curve, and then go down to the x-axis to read the answer.

🎯 Exam Tip: When dealing with exponential graphs, remember that \( y=b^x \) and \( x=\log_b y \) are inverse functions. This means you can use the graph of \( y=b^x \) to find \( \log_b y \) by swapping the roles of x and y.

 

Question 19. Copy and complete the table, for the function \( y = \frac{3}{x} \), giving your answer correct to 1 d.p. Then draw and graph.
Answer:
We need to complete the table for the function \( y = \frac{3}{x} \) and round the values to one decimal place:
When \( x = -3 \): \( y = \frac{3}{-3} = -1 \approx -1.0 \)
When \( x = -2.5 \): \( y = \frac{3}{-2.5} = -1.2 \approx -1.2 \)
When \( x = -2 \): \( y = \frac{3}{-2} = -1.5 \approx -1.5 \)
When \( x = -1.5 \): \( y = \frac{3}{-1.5} = -2 \approx -2.0 \)
When \( x = -1 \): \( y = \frac{3}{-1} = -3 \approx -3.0 \)
When \( x = -0.5 \): \( y = \frac{3}{-0.5} = -6 \approx -6.0 \)
When \( x = 0.5 \): \( y = \frac{3}{0.5} = 6 \approx 6.0 \)
When \( x = 1 \): \( y = \frac{3}{1} = 3 \approx 3.0 \)
When \( x = 1.5 \): \( y = \frac{3}{1.5} = 2 \approx 2.0 \)
When \( x = 2 \): \( y = \frac{3}{2} = 1.5 \approx 1.5 \)
When \( x = 2.5 \): \( y = \frac{3}{2.5} = 1.2 \approx 1.2 \)
When \( x = 3 \): \( y = \frac{3}{3} = 1 \approx 1.0 \)
The completed table is:

The graph of \( y = \frac{3}{x} \) is a hyperbola with both the x-axis and y-axis as asymptotes. This means the curve gets infinitely close to these axes but never touches them. The graph has two separate branches, one in the first quadrant (where x and y are positive) and another in the third quadrant (where x and y are negative).
In simple words: Fill the table by dividing 3 by each x-value given. Then plot these points on graph paper. The graph will be a curve that never touches the X or Y axes, forming two separate parts.

🎯 Exam Tip: When plotting reciprocal functions like \( y = \frac{k}{x} \), remember that x cannot be zero, and y cannot be zero. These are called asymptotes, and the graph will approach them but never cross them. Always check points for both positive and negative x values.

 

Question 20. Sketch the graphs of the following rational functions
(i) \( y = \frac{x+3}{x-2} \)
(ii) \( y = \frac{6}{x-6} \)
(iii) \( y = \frac{5}{2x+1} \)
(iv) \( y = \frac{2x+1}{x-3} \)
(v) \( y = \frac{7-2x}{3x+5} \)
Answer:
(i) For the function \( y = \frac{x+3}{x-2} \):
To find vertical asymptotes, set the denominator to zero: \( x-2 = 0 \implies x = 2 \).
To find horizontal asymptotes, consider the degrees of \( x \) in the numerator and denominator. Since they are equal (both 1), the horizontal asymptote is the ratio of the leading coefficients: \( y = \frac{1}{1} = 1 \).
The function exists for all real numbers except \( y \neq 1 \).
To find the x-intercept (where \( y=0 \)): \( \frac{x+3}{x-2} = 0 \implies x+3 = 0 \implies x = -3 \). So, it meets the x-axis at \( (-3, 0) \).
To find the y-intercept (where \( x=0 \)): \( y = \frac{0+3}{0-2} = -\frac{3}{2} \). So, it meets the y-axis at \( (0, -\frac{3}{2}) \).
The table of values is:

(ii) For the function \( y = \frac{6}{x-6} \):
The vertical asymptote is at \( x-6 = 0 \implies x = 6 \).
The horizontal asymptote is at \( y = 0 \) (since the degree of the numerator is less than the denominator).
The function exists for all real numbers except \( y \neq 0 \).
It does not have an x-intercept because \( y \) can never be 0.
To find the y-intercept (where \( x=0 \)): \( y = \frac{6}{0-6} = -1 \). So, it meets the y-axis at \( (0, -1) \).
The table of values is:

(iii) For the function \( y = \frac{5}{2x+1} \):
The vertical asymptote is at \( 2x+1 = 0 \implies x = -\frac{1}{2} \).
The horizontal asymptote is at \( y = 0 \).
The function exists for all real numbers except \( y \neq 0 \).
It does not have an x-intercept.
To find the y-intercept (where \( x=0 \)): \( y = \frac{5}{2(0)+1} = 5 \). So, it meets the y-axis at \( (0, 5) \).
The table of values is:

(iv) For the function \( y = \frac{2x+1}{x-3} \):
The vertical asymptote is at \( x-3 = 0 \implies x = 3 \).
The horizontal asymptote is at \( y = \frac{2}{1} = 2 \).
The function exists for all real numbers except \( y \neq 2 \).
To find the x-intercept (where \( y=0 \)): \( 2x+1 = 0 \implies x = -\frac{1}{2} \). So, it meets the x-axis at \( (-\frac{1}{2}, 0) \).
To find the y-intercept (where \( x=0 \)): \( y = \frac{2(0)+1}{0-3} = -\frac{1}{3} \). So, it meets the y-axis at \( (0, -\frac{1}{3}) \).
The table of values is:

(v) For the function \( y = \frac{7-2x}{3x+5} \):
The vertical asymptote is at \( 3x+5 = 0 \implies x = -\frac{5}{3} \).
The horizontal asymptote is at \( y = \frac{-2}{3} \).
The function exists for all real numbers except \( y \neq -\frac{2}{3} \).
To find the x-intercept (where \( y=0 \)): \( 7-2x = 0 \implies x = \frac{7}{2} \). So, it meets the x-axis at \( (\frac{7}{2}, 0) \).
To find the y-intercept (where \( x=0 \)): \( y = \frac{7-2(0)}{3(0)+5} = \frac{7}{5} \). So, it meets the y-axis at \( (0, \frac{7}{5}) \).
The table of values is:

In simple words: For each function, first find the vertical and horizontal lines (asymptotes) that the graph will get close to but never touch. Then, find where the graph crosses the X and Y axes. Use these points and a few others from the table to draw the curve.

🎯 Exam Tip: When sketching rational functions, always clearly mark the vertical and horizontal asymptotes. These are critical for understanding the behavior of the graph. Intercepts (where the graph crosses the axes) provide additional key points for accurate drawing.