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📝 When does inverse function exist (14 MCQs)

📖 From Calculus • 1. Basics before calculus • 14 questions available

What is When does inverse function exist?

Definition:
An inverse function exists if and only if the original function is one-to-one (injective), meaning each output comes from exactly one input, which can be tested via the horizontal line test or by showing f(a)=f(b)a=bf(a)=f(b) \Rightarrow a=b.

Example:
f(x)=x2f(x)=x^2 on all reals is not one-to-one because f(2)=f(2)=4f(2)=f(-2)=4, so no inverse; but on x0x \geq 0, it is one-to-one and has inverse f1(x)=xf^{-1}(x)=\sqrt{x}.

Reason:
This condition ensures that the inverse is a function itself, avoiding ambiguity when reversing the mapping.

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📝 All When does inverse function exist MCQs

Q1. Which of the following best defines a one-to-one (injective) function?

A.Each element of the codomain is paired with at most one element of the domain.
B.Different inputs always produce different outputs. ✅
C.The function’s graph passes the vertical line test.
D.The function has an inverse that is also a function.
💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: A one‑to‑one function never maps two distinct x‑values to the same y‑value. Therefore, whenever x₁≠x₂, we must have f(x₁)≠f(x₂). This property is captured precisely by the statement that different inputs always produce different outputs, which is option B.

Q2. What does the Horizontal Line Test state about a function’s invertibility?

A.If any vertical line intersects the graph more than once, the function is not invertible.
B.If any horizontal line intersects the graph at most once, the function has an inverse. ✅
C.A function is invertible only if its graph is symmetric about the y‑axis.
D.The test determines whether a function is continuous.
💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The Horizontal Line Test says that a function is invertible exactly when no horizontal line cuts its graph more than once. This ensures that each y‑value comes from at most one x‑value, guaranteeing a well‑defined inverse. Option B restates this condition correctly.

Q3. Consider the piecewise function \f(x)=\\begin{cases}x^{2},&x\\le 0\\\\x+2,&x>0\\end{cases}\. Which statement is true regarding its invertibility?

A.f is not invertible because the horizontal line y=4 intersects the graph twice.
B.f is invertible because each piece is one‑to‑one and the pieces do not overlap. ✅
C.f is not invertible because the two pieces share the same output value at x=0.
D.f is invertible only on the interval \(-\\infty,0]\.
💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: For x≤0, f(x)=x² gives non‑negative outputs; for x>0, f(x)=x+2 yields outputs greater than 2. No y‑value is produced by both pieces, and each piece is itself one‑to‑one. Hence the whole function is injective and passes the Horizontal Line Test, making it invertible. Option B is correct.

Q4. If a function f has an inverse and satisfies \f(2)=5\, which of the following must be true?

A.\f^{-1}(5)=2\
B.\f^{-1}(2)=5\
C.\f^{-1}(5)= -2\
D.Both A and B are true.
💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: By definition of an inverse, \f^{-1}(y)\ returns the unique x such that \f(x)=y\. Since \f(2)=5\, the inverse evaluated at 5 must give back 2, i.e., \f^{-1}(5)=2\. The other statements do not follow from the given information. Therefore, option A is correct.

Q5. The function \g(x)=x^{3}-3x\ is defined for all real x. Using the Horizontal Line Test, determine whether g has an inverse.

A.g fails the test because the horizontal line y=0 meets the graph at three points. ✅
B.g passes the test because it is strictly increasing everywhere.
C.g fails the test because it is not continuous.
D.g passes the test because its derivative never changes sign.
💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Setting \g(x)=0\ gives \x(x^{2}-3)=0\, whose solutions are \x=0,\\pm\\sqrt{3}\. Thus the horizontal line y=0 intersects the graph at three distinct points, violating the Horizontal Line Test. Consequently, g is not one‑to‑one and has no inverse. Option A is correct.

Q6. Which statement correctly compares the invertibility of \f(x)=x^{2}\ and \h(x)=x^{3}\ on the entire real line?

A.Both are invertible because they are continuous.
B.Only f is invertible; h is not.
C.Only h is invertible; f is not. ✅
D.Neither is invertible because both fail the Horizontal Line Test.
💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The cubic function \x^{3}\ passes the Horizontal Line Test: each horizontal line meets its graph exactly once, so it is invertible. The quadratic \x^{2}\ fails the test because lines with y>0 intersect twice (e.g., y=4 at x=±2). Hence only h is invertible. Option C is correct.

Q7. Suppose f and g are functions such that the composition \f\\circ g\ is invertible. Which statement must be true?

A.Both f and g are invertible.
B.f is invertible, but g may fail to be one-to-one.
C.g is invertible, but f may fail to be onto.
D.Both f and g must be one-to-one, and f must be onto the codomain of \f\\circ g\. ✅
💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: For \f\\circ g\ to be bijective, g must be injective (otherwise the composition would map distinct inputs to the same output) and f must be surjective onto the range of the composition. Neither function needs to be invertible on its own, but both must satisfy the stated conditions. Option D captures this requirement.

Q8. If a function is strictly increasing on its entire domain, what can be concluded about its invertibility?

A.It is guaranteed to be invertible. ✅
B.It may still fail the Horizontal Line Test.
C.It is invertible only if its range is all real numbers.
D.Strict monotonicity has no impact on invertibility.
💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: A strictly increasing function never repeats a y‑value; each horizontal line meets its graph at most once. This satisfies the Horizontal Line Test, guaranteeing the existence of a unique inverse function. Hence option A is correct.

Q9. Consider the function \q(x)=x^{2}\ defined on \[-2,2]\. Which domain restriction yields an invertible function?

A.Restrict to \[-2,0]\.
B.Restrict to \[0,2]\. ✅
C.Restrict to \[-2,2]\ (no restriction).
D.Restrict to \[-1,1]\.
💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: On \[0,2]\ the function \x^{2}\ is monotonic increasing, so each y‑value in \[0,4]\ corresponds to a single x‑value. The restriction to the non‑negative side eliminates the duplicate outputs caused by symmetry about the y‑axis, thereby satisfying the Horizontal Line Test. Option B is correct.

Q10. Why does the function \f(x)=|x|\ not have an inverse, while its restriction to \x\\ge 0\ does?

A.Because \|x|\ is not continuous for x<0.
B.Because \|x|\ maps both a positive and a negative x to the same y, violating one‑to‑one. ✅
C.Because the restricted function becomes linear.
D.Because the restricted function has a different codomain.
💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The absolute‑value function yields the same output for x and –x (e.g., \|3|=|-3|=3\), so it is not injective and fails the Horizontal Line Test. When the domain is limited to non‑negative numbers, each y is produced by exactly one x, making the function one‑to‑one and invertible. Option B explains this.

Q11. For the function \f(x)=\\dfrac{x}{1+x^{2}}\, on which intervals is it one‑to‑one?

A.\( -\\infty,-1]\ and \[1,\\infty)\
B.\( -\\infty,0]\ and \[0,\\infty)\
C.\( -\\infty,-\\tfrac{1}{\\sqrt{3}}]\ and \[\\tfrac{1}{\\sqrt{3}},\\infty)\
D.\( -\\infty,-\\tfrac{1}{\\sqrt{2}}]\ and \[\\tfrac{1}{\\sqrt{2}},\\infty)\
💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The derivative is \f&#039;(x)=\\dfrac{1-x^{2}}{(1+x^{2})^{2}}\. It is positive for |x|<1 and negative for |x|>1, so f increases on \(-1,1)\ and decreases on \(-\\infty,-1]\ and \[1,\\infty)\. Each of these monotonic intervals is injective; option A lists two of them correctly.

Q12. If a differentiable function passes the Horizontal Line Test, what can be said about its derivative where it exists?

A.The derivative must be positive everywhere.
B.The derivative cannot be zero at any point.
C.The derivative may be zero, but the function must still be monotonic. ✅
D.The derivative must be negative everywhere.
💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: A function can be monotonic and still have horizontal tangents (derivative zero) at isolated points, as seen with \x^{3}\ at the origin. The Horizontal Line Test ensures injectivity, which only requires the function to be monotonic, not that its derivative be strictly non‑zero. Thus option C is accurate.

Q13. For a continuous function \f:\\mathbb{R}\\to\\mathbb{R}\ that satisfies the Intermediate Value Property, which additional condition guarantees that \f\ is invertible?

A.Being strictly monotonic. ✅
B.Being bounded.
C.Having a derivative equal to zero somewhere.
D.Having a symmetric graph about the y‑axis.
💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Continuity together with the Intermediate Value Property ensures the function takes every value between any two outputs. To be invertible, it must also be one‑to‑one, which is guaranteed if the function is strictly monotonic (always increasing or always decreasing). Hence, strict monotonicity is the required extra condition. Option A is correct.

Q14. Which statement about the function \f(x)=e^{x}+x\ is correct regarding its invertibility on \\\mathbb{R}\?

A.It is not invertible because it is not one‑to‑one.
B.It is invertible because it is strictly increasing. ✅
C.It is invertible only on \(-\\infty,0]\.
D.It fails the Horizontal Line Test at y=1.
💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The derivative of \f\ is \f&#039;(x)=e^{x}+1>0\ for all real x, so the function is strictly increasing everywhere. A strictly increasing continuous function satisfies the Horizontal Line Test, guaranteeing the existence of an inverse on the entire real line. Therefore, option B is correct.

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