Mathematics Week 5 Graded Assignment
Graded Assignment
1. Function Identification via Graph (Figure M1W8A-8.1)
Question: A graph is shown in Figure M1W8A-8.1, ◦symbol signifies that the straight line does not touch the point and the - symbol signifies that the line touches the point. Choose the correct option.
- The graph cannot be a function, because it fails the vertical line test.
- The graph cannot be a function, because it passes the horizontal line test but fails the vertical line test.
- The graph can be a function, because it passes the vertical line test.
- The graph cannot be a function, because it passes the vertical line test but fails the horizontal line test.
Solution: To check if the given graph represents a function, use the vertical line test. In Figure M1W8A-8.1, every vertical line crosses the graph only once (including both - and ◦ as per definition). Therefore, the graph represents a function.
- Correct options:
- The graph can be of a function, because it passes the vertical line test.
- The graph fails the horizontal line test.
- The graph represents the graph of neither even function nor odd function.
- The given graph is not invertible in the given domain.
- Incorrect options:
- The graph cannot be of a function, because it passes the vertical line test but fails the horizontal line test.
- The graph cannot be of a function, because it fails the vertical line test.
- The graph cannot be of a function, because it passes the horizontal line test but fails the vertical line test.
- The graph fails the horizontal line test thus it can be an injective function.
- The graph represents the graph of either even function or odd function.1
2. Injectivity of Power Functions (Figures M1W8AS-8.1 and M1W8AS-8.2)
Question: For $ y = x^n $, where $ n $ is a positive integer and $ x \in \mathbb{R} $, which of the following statement is true?
- For all values of n, y is not a one-to-one function.
- For all values of n, y is an injective function.
- y is not a function.
- If n is an even number, then y is not an injective function. If n is an odd number, then y is an injective function.
Solution:
- $ y = x^n $ is a function for all positive integers $ n $.
- If $ n $ is odd, the function is injective (passes the horizontal line test, see Figure M1W8AS-8.1).
- If $ n $ is even, the function is not injective (see Figure M1W8AS-8.2).
- Therefore, the correct option is: If n is even, not injective; if n is odd, injective.1
3. Algebraic Simplification
Question: If $ 4m - n = 0 $, then the value of
$$ \frac{16^m}{2^n} + \frac{27^n}{96^m} $$is
Solution: Given $ 4m - n = 0 $,
$$ \frac{16^m}{2^n} + \frac{27^n}{96^m} = (2^4)^m / 2^n + (3^3)^n / (2^5 \cdot 3)^m = 2^{4m-n} + 3^{3n-6m} = 2^0 + 3^0 = 1 + 1 = 2 $$Answer: 21
4. Radioactive Decay (Half-Life Calculation)
Question: Half-life of an element is the time required for half of a given sample of radioactive element to change to another element. The rate of change of concentration is calculated by the formula $ A(t) = A_0 (1/2)^{t/\gamma} $, where $ \gamma $ is the half-life. If Radium has a half-life of 1600 years and the initial concentration is 100%, calculate the percentage of Radium after 2000 years.
- 35%
- 42%
- 19%
- 21%
Solution:
$$ A(2000) = 100 \times (1/2)^{2000/1600} = 100 \times (1/2)^{1.25} \approx 42\% $$Answer: 42%1
5. Domain of a Composite Function
Question: If $ f(x) = (1 - x)^{1/2} $ and $ g(x) = 1 - x^2 $, find the domain of the composite function $ g \circ f $.
- $ \mathbb{R} $
- $ (-\infty, 1] \cap [-2, \infty) \cup (-\infty, -2) $
- $ [1, \infty) $
- $ \mathbb{R} \setminus (1, \infty) $
Solution:
- Domain of $ f(x) $: $ x \leq 1 $ ($ (-\infty, 1] $)
- Domain of $ g(x) $: $ \mathbb{R} $, but range of $ f(x) $ is $ [0, \infty) $
- So, domain of $ g \circ f $ is $ (-\infty, 1] $ (options 2 and 4 are correct)1
6. Domain of the Inverse Function
Question: Find the domain of the inverse function of $ y = x^3 + 1 $.
- $ \mathbb{R} $
- $ \mathbb{R} \setminus {1} $
- $ [1, \infty) $
- $ \mathbb{R} \setminus [1, \infty) $
Solution: The range of $ y = x^3 + 1 $ is $ \mathbb{R} $, so the domain of its inverse is also $ \mathbb{R} $. Answer: $ \mathbb{R} $1
7. Intersection Points of a Function and Its Inverse
Question: If $ f(x) = x^3 $, then which of the following is the set of points where the graphs of $ f(x) $ and $ f^{-1}(x) $ intersect?
- {(-1, 1), (0, 0), (1, -1)}
- {(-2, -8), (1, 1), (2, 8)}
- {(-1, -1), (0, 0), (1, 1)}
- {(-2, -8), (0, 0), (2, 8)}
Solution: Solve $ x^3 = x \Rightarrow x(x^2 - 1) = 0 \Rightarrow x = -1, 0, 1 $. So, intersection points are {(-1, -1), (0, 0), (1, 1)}1
8. Population Growth Prediction
Question: In a survey, population growth is given by $ \alpha(T) = \alpha_0 (1 + d/100)^T $. If in 2015, the population of Adyar was 30,000 and the growth rate is 4% per year, what will be the approximate population in 2020?
- 60251
- 71255
- 91000
- 36500
Solution: $ T = 5 $, $ \alpha(5) = 30000 \times (1.04)^5 \approx 36500 $1
9. Reflection of a Function Across $ y = x $ (Figure M1W8AS-8.3)
Question: An ant moves along $ f(x) = x^2 + 1 $ for $ x \in [0, \infty) $. A mirror is placed along $ y = x $. If the reflection moves along $ g(x) $, which is/are correct?
- $ g(x) = f^{-1}(x) $
- $ g(x) = f(x) $
- $ g(x) = \sqrt{x-1} $
- $ g(x) = \sqrt{x+1} $
Solution: The reflection is the inverse function, so $ g(x) = f^{-1}(x) = \sqrt{x-1} $. Correct options: 1 and 31
10. Festival Discount Offers (Applied Math)
Question: A textile shop offers: D1: Shop for more than ₹14,999 and pay only ₹9,999. D2: Avail 30% discount on the total payable amount. If Shalini buys two dresses, each over ₹8,000, and can use both offers, which is/are correct?
- The minimum amount she should pay after applying two offers cannot be determined because the exact values are unknown.
- The minimum amount she should pay after applying both offers is approximately ₹6,999.
- The amount after D2 only is approximately ₹11,199.
- The amount after D1 only is approximately ₹9,999.
- If total is ₹17,999, to pay minimum, avail D1 first, then D2.
- If total is ₹17,999, availing D2 first, then D1 is not possible.
- If total is ₹17,999, to pay minimum, avail D2 first, then D1.
Solution:
- If D1 first, then D2: ₹9,999 × 0.7 = ₹6,999 (minimum).
- If D2 first, amount may fall below ₹14,999, so D1 may not be applicable.
- D1 only: ₹9,999.
- D2 only: ₹17,999 × 0.7 = ₹12,599 (if total is ₹17,999).
- So, correct: 2, 4, 5, 61
11. Injectivity and Function Operations
Question: If $ f(x) = x^2 $ and $ h(x) = x-1 $, which options are incorrect?
- $ f \circ h $ is not injective.
- $ f \circ h $ is injective.
- $ f(f(h(x))) \times h(x) = (x-1)^4 $
- $ f(f(h(x))) \times h(x) = (x-1)^5 $
Solution:
- $ f \circ h = (x-1)^2 $ is not injective.
- $ f(f(h(x))) \times h(x) = ((x-1)^2)^2 \times (x-1) = (x-1)^5 $.
- So, incorrect: 2 and 31
12. Graphical Properties and Inverses (Figure 3)
Question: Let $ f(x), g(x), p(x), q(x) $ be functions defined on $ \mathbb{R} $. Refer Figure 3 (A and B) and choose correct options:
- $ g(x) $ may be the inverse of $ f(x) $.
- $ p(x) $ and $ q(x) $ are even functions but $ f(x) $ and $ g(x) $ are neither even nor odd.
- $ q(x) $ could not be the inverse function of $ p(x) $.
- $ p(x), q(x) $ can be even degree polynomials and $ f(x) $ can be an odd degree polynomial.
Solution:
- $ f(x) $ and $ g(x) $ are symmetric across $ y = x $, so $ g(x) $ may be inverse of $ f(x) $.
- $ p(x) $ and $ q(x) $ are symmetric about the y-axis, so they are even functions.
- $ q(x) $ is not symmetric across $ y = x $, so cannot be inverse of $ p(x) $.
- End behaviors suggest $ p(x), q(x) $ can be even degree polynomials, $ f(x) $ can be odd degree.
- All options (a), (b), (c), (d) are correct1
Note: For all questions involving graphs, the referenced figures (e.g., Figure M1W8A-8.1, M1W8AS-8.1, etc.) are described in the solutions, but the actual images are not included in this text extraction. The reasoning is based on their descriptions in the PDF.1