This TSA checks functions and relations, domains and ranges, one-one functions, inverse functions, inverse graphs, piecewise functions, periodic functions, composite functions, and self-inverse functions.
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Functions, Domains, Ranges and One-One Functions
1. True or False?
A function assigns exactly one output to each input in its domain, and its domain is part of its definition.
Tested fundamental: Definition of a function.
Explanation: A function gives exactly one output for each input in its domain. The domain is part of the function definition, not an optional extra.
Common wrong choice: False usually comes from thinking that the rule alone is enough to define a function.
2. True or False?
Two functions with the same rule are always equal, even if their domains are different.
Tested fundamental: Equality of functions.
Explanation: Two functions are equal only when they have the same rule and the same domain.
Common wrong choice: True ignores the fact that the domain is part of the function definition.
3. The relation \[ x^2+y^2=4,\qquad -2\le x\le 2 \] is:
Tested fundamental: Distinguishing a function from a relation.
Explanation: The equation represents a circle. For many \(x\)-values, there are two possible \(y\)-values, so it is not a function.
Common wrong choice: C assumes every relation with a stated domain is automatically a function.
4. Find the maximum domain of \[ f(x)=\frac{\sqrt{2x+5}}{x-1}. \]
Tested fundamental: Finding the maximum domain of a function.
Explanation: The square root requires \(2x+5\ge0\), so \(x\ge-\frac52\). Also, the denominator cannot be zero, so \(x\ne1\).
Common wrong choice: A forgets to exclude \(x=1\).
5. The function \[ f(x)=3-2x,\qquad -1\le x<4 \] is defined. Find \(R_f\).
Answer format: Give your answer in interval notation. For example [0,1) or (-inf, 5] etc.
Answer: \((-5,5]\)
Tested fundamental: Finding the range of a decreasing linear function on a restricted domain.
Explanation: Since \(f(x)=3-2x\) is decreasing, \(x=-1\) gives the maximum value \(5\), while \(x\to4^-\) gives \(f(x)\to-5^+\). Hence \(R_f=(-5,5]\).
Common error: Reversing the inclusivity of the endpoints.
6. True or False?
If a function is strictly increasing or strictly decreasing on its whole domain, then it is one-one.
Tested fundamental: Sufficient condition for a one-one function.
Explanation: A strictly increasing or strictly decreasing function cannot take the same output value twice, so it is one-one.
Common wrong choice: False usually comes from confusing one-one with the vertical line test.
7. The function \[ f(x)=x^2-4x+1,\qquad x\ge k \] is to be one-one. Find the least possible value of \(k\).
Answer format: Give your answer as an integer.
Answer: \(2\)
Tested fundamental: Restricting the domain of a quadratic to make it one-one.
Explanation: \(f(x)=(x-2)^2-3\), so the turning point is at \(x=2\). For the function to be one-one on \(x\ge k\), the least possible \(k\) is \(2\).
Common error: Choosing a value less than \(2\), which includes both sides of the turning point.
8. Which pair of distinct \(x\)-values \(a\) and \(b\), where \(a<b\), shows that \[ f(x)=x^2-4x+1,\qquad x\in\mathbb R \] is not one-one?
Tested fundamental: Showing that a function is not one-one by finding two inputs with the same output.
Explanation: \(f(1)=1-4+1=-2\) and \(f(3)=9-12+1=-2\). Since two different inputs give the same output, \(f\) is not one-one.
Common wrong choice: B uses two different inputs, but they do not give the same output.
Inverse Functions and Graphs
9. A function \(f^{-1}\) exists if and only if:
Tested fundamental: Condition for the existence of an inverse function.
Explanation: \(f^{-1}\) exists as a function if and only if \(f\) is one-one.
Common wrong choice: C is not sufficient; a continuous function may still fail the horizontal line test.
10. Given \[ f(x)=2x-3,\qquad x\in\mathbb R, \] which is \(f^{-1}(x)\)?
Tested fundamental: Finding an inverse function algebraically.
Explanation: Let \(y=2x-3\). Then \(x=\frac{y+3}{2}\), so \(f^{-1}(x)=\frac{x+3}{2}\).
Common wrong choice: A subtracts \(3\) instead of undoing \(-3\) by adding \(3\).
11. A function \(f\) has \[ D_f=[1,4),\qquad R_f=(-2,5]. \] Which is correct?
Tested fundamental: Domain and range of an inverse function.
Explanation: The domain of \(f^{-1}\) is the range of \(f\), and the range of \(f^{-1}\) is the domain of \(f\).
Common wrong choice: A forgets to swap the domain and range.
12. The graph of \(y=f(x)\) passes through \((3,5)\) and has vertical asymptote \(x=2\). Which statement about \(y=f^{-1}(x)\) is correct?
Tested fundamental: Reflecting the graph of an inverse function in the line \(y=x\).
Explanation: The point \((3,5)\) becomes \((5,3)\). The vertical asymptote \(x=2\) becomes the horizontal asymptote \(y=2\).
Common wrong choice: C reflects the point but not the asymptote.
13. Given \[ f(x)=\sqrt{x-1}+2,\qquad x\ge 1, \] which is \(f^{-1}(x)\) in similar form?
Tested fundamental: Finding an inverse function and its domain.
Explanation: Let \(y=\sqrt{x-1}+2\). Then \(y-2=\sqrt{x-1}\), so \(x=(y-2)^2+1\). Since \(R_f=[2,\infty)\), \(D_{f^{-1}}=[2,\infty)\).
Common wrong choice: D has the correct domain but the wrong constant term.
14. If the graphs of \(y=f(x)\) and \(y=f^{-1}(x)\) intersect at \((a,b)\), which statement must be true?
Tested fundamental: Intersection of a function and its inverse graph.
Explanation: The graphs of \(y=f(x)\) and \(y=f^{-1}(x)\) are reflections in \(y=x\). Their intersections lie on \(y=x\), so \(a=b\).
Common wrong choice: B confuses reflection in \(y=x\) with reflection in the \(x\)-axis or \(y\)-axis.
15. If \[ f^{-1}(4)=7, \] find \(f(7)\).
Answer format: Give your answer as an integer.
Answer: \(4\)
Tested fundamental: Interpreting inverse-function notation.
Explanation: \(f^{-1}(4)=7\) means that \(f(7)=4\).
Common error: Treating \(f^{-1}(4)\) as \(\frac{1}{f(4)}\).
Piecewise and Periodic Functions
16. The function \(f\) is defined by \[ f(x)= \begin{cases} x^2+1, & x<1,\\ 3x-1, & x\ge 1. \end{cases} \] Find \(f(1)\).
Tested fundamental: Evaluating a piecewise function at a boundary value.
Explanation: Since \(x=1\) falls under \(x\ge1\), use \(3x-1\). Hence \(f(1)=3(1)-1=2\).
Common wrong choice: B uses the first rule even though \(x=1\) is not included in \(x<1\).
17. The function \(f\) is defined by \[ f(x)= \begin{cases} 2x+1, & -1\le x<2,\\ 7-x, & 2\le x\le5. \end{cases} \] Find \(R_f\).
Answer format: Give your answer in interval notation. For example [0,1) or (-inf, 5] etc.
Answer: \([-1,5]\)
Tested fundamental: Finding the range of a piecewise function.
Explanation: For \(-1\le x<2\), \(2x+1\) gives \([-1,5)\). For \(2\le x\le5\), \(7-x\) gives \([2,5]\). Combining these gives \([-1,5]\).
Common error: Missing the endpoint \(5\), which is included by the second piece.
18. The function \[ f(x)= \begin{cases} x+1, & x<0,\\ x^2+1, & x\ge0 \end{cases} \] is one-one because:
Tested fundamental: Checking whether a piecewise function is one-one.
Explanation: The first piece has range \((-\infty,1)\), while the second has range \([1,\infty)\). The ranges do not overlap, and each piece is increasing on its interval.
Common wrong choice: A is false; piecewise functions are not automatically one-one.
19. A function satisfies \[ f(x+4)=f(x) \] for all real \(x\). For \(-1\le x<3\), \[ f(x)=x^2. \] Find \(f(11)\).
Answer format: Give your answer as an integer.
Answer: \(1\)
Tested fundamental: Evaluating a periodic function.
Explanation: Since the period is \(4\), \(f(11)=f(11-12)=f(-1)=(-1)^2=1\).
Common error: Substituting \(x=11\) directly into \(x^2\), even though \(11\) is outside the given base interval.
20. True or False?
If \(f(x+5)=f(x)\) for all real \(x\), then \(5\) must be the least possible period of \(f\).
Tested fundamental: Understanding periodic functions.
Explanation: The equation \(f(x+5)=f(x)\) shows that \(5\) is a period, but it does not prove that \(5\) is the least period.
Common wrong choice: True comes from assuming that any stated period must be the smallest period.
Composite Functions
21. The composite function \(gf\) exists if:
Tested fundamental: Existence condition for a composite function.
Explanation: For \(gf\) to exist, every output of \(f\) must be a valid input for \(g\). Hence \(R_f\subseteq D_g\).
Common wrong choice: B is the condition for \(fg\), not \(gf\).
22. Let \[ f(x)=2x+1,\qquad 0\le x\le2, \] and \[ g(x)=\sqrt{x},\qquad x\ge0. \] Which statement is correct?
Tested fundamental: Checking whether composite functions exist.
Explanation: \(R_f=[1,5]\subseteq D_g=[0,\infty)\), so \(gf\) exists. But \(R_g=[0,\infty)\) is not a subset of \(D_f=[0,2]\), so \(fg\) does not exist.
Common wrong choice: A assumes that if one composite exists, the other must also exist.
23. Let \[ f(x)=x+2,\qquad x\ge0, \] and \[ g(x)=x^2,\qquad x\in\mathbb R. \] Which is \(gf(x)\) in similar form?
Tested fundamental: Finding the expression and domain of a composite function.
Explanation: \(gf(x)=g(f(x))=g(x+2)=(x+2)^2\). The input \(x\) still comes from \(D_f\), so \(x\ge0\).
Common wrong choice: B gives the correct rule but the wrong domain.
24. Let \[ f(x)=x+1,\qquad 0\le x\le3, \] and \[ g(x)=2x-5,\qquad x\in\mathbb R. \] Find \(R_{gf}\).
Answer format: Give your answer in interval notation. For example [0,1) or (-inf, 5] etc.
Answer: \([-3,3]\)
Tested fundamental: Finding the range of a composite function.
Explanation: \(gf(x)=g(x+1)=2(x+1)-5=2x-3\), where \(0\le x\le3\). Hence \(R_{gf}=[-3,3]\).
Common error: Finding the range of \(f\) only, instead of the range of \(gf\).
25. Let \[ f(x)=\frac{1}{x-2},\qquad x\ne2, \] and \[ g(x)=x+3,\qquad x\in\mathbb R. \] Which is \(fg(x)\)?
Tested fundamental: Understanding the order of composite functions.
Explanation: \(fg(x)=f(g(x))=f(x+3)=\frac{1}{(x+3)-2}=\frac{1}{x+1}\). Also, \(x+3\ne2\), so \(x\ne-1\).
Common wrong choice: D gives the correct expression but the wrong domain restriction.
26. Let \[ f(x)=x^2,\qquad -2\le x\le3, \] and \[ g(x)=\sqrt{x-1},\qquad x\ge1. \] Which is \(D_{gf}\)?
Tested fundamental: Finding the domain of a composite function with a restricted inner function.
Explanation: For \(gf(x)=g(f(x))\), we need \(x\in D_f\) and \(f(x)\in D_g\). Hence \(-2\le x\le3\) and \(x^2\ge1\), giving \([-2,-1]\cup[1,3]\).
Common wrong choice: C ignores the negative values of \(x\) that still give \(x^2\ge1\).
Self-Inverse and Mixed Checks
27. A function \(f\) is self-inverse if:
Tested fundamental: Definition of a self-inverse function.
Explanation: A self-inverse function satisfies \(f^{-1}=f\), so applying \(f\) twice returns the original input: \(f(f(x))=x\).
Common wrong choice: B is only one example of a self-inverse function, not the full definition.
28. A self-inverse function \(f\) satisfies \[ f(7)=2. \] Find \[ f^{2025}(7), \] where \(f^{2025}\) means repeated composition of \(f\) with itself \(2025\) times.
Answer format: Give your answer as an integer.
Answer: \(2\)
Tested fundamental: Repeated composition of a self-inverse function.
Explanation: Since \(f\) is self-inverse, applying \(f\) twice returns the original input. Odd powers behave like \(f\), so \(f^{2025}(7)=f(7)=2\).
Common error: Treating \(f^{2025}(7)\) as \([f(7)]^{2025}\).
29. Which function is self-inverse?
Tested fundamental: Recognising a self-inverse function.
Explanation: If \(f(x)=\frac1x\), then \(f(f(x))=f\left(\frac1x\right)=x\). Hence \(f\) is self-inverse.
Common wrong choice: B has inverse \(x-1\), not itself.
30. Find the range of \[ f(x)=\frac{2}{x-1}+3,\qquad x>1. \]
Answer format: Give your answer in interval notation. For example [0,1) or (-inf, 5] etc.
Answer: \((3,\infty)\)
Tested fundamental: Finding the range of a rational function with a restricted domain.
Explanation: Since \(x>1\), \(\frac{2}{x-1}>0\). Hence \(f(x)>3\). As \(x\to\infty\), \(f(x)\to3^+\), and as \(x\to1^+\), \(f(x)\to\infty\). Therefore \(R_f=(3,\infty)\).
Common error: Including \(3\) in the range even though the graph approaches but never reaches \(y=3\).
Your result is ready.