If (f:R\to R), (f(x)=x^3+5), what is (f^{-1}(x))?
Correct answer: A
Step 1: Write (y=x^3+5). Step 2: Then (x^3=y-5), so (x=\sqrt[3]{y-5}). Step 3: Replacing (y) by (x), (f^{-1}(x)=\sqrt[3]{x-5}).
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Functions
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Medium · Level 20 · inverse function,cubic function,real numbers
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- (\sqrt[3]{x-5})
- (\sqrt[3]{x+5})
- (x^3-5)
- (\sqrt{x-5})
Medium · Level 20 · restricted domain,inverse value,square function
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- (3)
- (-3)
- ({-3,3})
- (9)
Medium · Level 20 · preimage,square function,inverse image
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- ({-2,2})
- ({2})
- ({-2})
- ([1,5])
Medium · Level 20 · zero of function,quadratic function,factorisation
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- (f(2)) and (f(3))
- only (f(4))
- only (f(2))
- all three
Medium · Level 20 · absolute value,range,minimum value
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- ([1,\infty))
- ([0,\infty))
- (R)
- (R-{1})
Medium · Level 20 · onto function,absolute value,range,codomain
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- Because (0) cannot be its image
- Because (1) is its image
- Because it is defined everywhere
- Because it is not a function
Medium · Level 20 · bijective function,finite sets,cardinality
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- (8)
- (7)
- (9)
- Cannot be determined
Medium · Level 20 · inverse function,composition,identity function,bijection
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- (f) is bijective and (g=f^{-1})
- (f) is a constant function
- (f) is not a function
- (g) is only the identity function
Hard · Level 19 · functions,range,quadratic
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- ([2,\infty))
- ([3,\infty))
- ((-\infty,2])
- (\mathbb{R})
Hard · Level 19 · inverse function,linear function,bijection
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- (\frac{x+5}{2})
- (\frac{x-5}{2})
- (2x+5)
- (5-2x)
Hard · Level 19 · number of functions,finite sets,domain codomain
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- (6)
- (8)
- (9)
- (12)
Hard · Level 19 · composition of functions,algebra,functions
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- (x^2+3)
- ((x+3)^2)
- (x^2+6)
- (x+9)
Hard · Level 19 · one-one function,increasing function,real function
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- It is one-one
- It is not one-one
- It is a constant function
- It is not defined
Hard · Level 19 · one-one function,counterexample,square function
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- Because (f(1)=f(-1))
- Because (f(0)) is not defined
- Because every real number has a preimage
- Because its domain is empty
Hard · Level 19 · onto function,not one-one,square function
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- Both one-one and onto
- Onto but not one-one
- One-one but not onto
- Neither one-one nor onto
Hard · Level 19 · onto function,range,codomain
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- Because (0) has no preimage
- Because (1) has no preimage
- Because (2) has no preimage
- Because all positive numbers are missed
Hard · Level 19 · range,rational function,excluded value
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- (1)
- (2)
- (3)
- (5)
Hard · Level 19 · composition,value of function,linear functions
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- (-5)
- (-3)
- (3)
- (7)
Hard · Level 19 · one-one function,finite sets,pigeonhole
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- Such a function is possible
- Such a function is not possible
- Such a function is always onto
- There will be exactly (20) such functions
Hard · Level 19 · bijection,counting functions,finite sets
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- (4)
- (16)
- (24)
- (256)
If (f:[0,\infty)\to[1,\infty)), (f(x)=x^2+1), what is (f^{-1}(10))?
Correct answer: A
Step 1: To find (f^{-1}(10)), solve (x^2+1=10). Step 2: (x^2=9), so (x=3) or (x=-3). Step 3: Since the domain is ([0,\infty)), only (3) is valid.
View question detailsIf (f:R\to R), (f(x)=x^2+1), what is (f^{-1}({5}))?
Correct answer: A
Step 1: For the preimage, solve (x^2+1=5). Step 2: This gives (x^2=4). Step 3: Hence (x=2) or (x=-2), so the preimage is ({-2,2}).
View question detailsIf (f(x)=x^2-5x+6), which of (f(2), f(3), f(4)) are zero?
Correct answer: A
Step 1: (f(x)=x^2-5x+6=(x-2)(x-3)). Step 2: At (x=2) and (x=3), the product becomes zero. Step 3: At (x=4), (f(4)=2), so it is not zero.
View question detailsIf (f:R\to R), (f(x)=|x-4|+1), what is the range?
Correct answer: A
Step 1: (|x-4|\ge0) for every real (x). Step 2: Therefore (|x-4|+1\ge1). Step 3: At (x=4), the minimum value (1) is obtained, so the range is ([1,\infty)).
View question detailsIf (f:R\to R), (f(x)=|x-4|+1), why is it not onto?
Correct answer: A
Step 1: The value of (|x-4|+1) is always at least (1). Step 2: The codomain (R) contains (0), but it cannot be the image of any (x). Step 3: Therefore the function is not onto.
View question detailsIf (f:A\to B) is bijective and (B) has (8) elements, how many elements will (A) have?
Correct answer: A
Step 1: In a bijective function, every element is matched exactly one-to-one. Step 2: Hence the domain and codomain have the same number of elements. Step 3: Since (B) has (8) elements, (A) also has (8).
View question detailsIf (f:A\to B) and (g:B\to A) satisfy (f\circ g=I_B) and (g\circ f=I_A), what is correct about (f)?
Correct answer: A
Step 1: Both compositions give identity functions. Step 2: This means (f) and (g) completely undo each other's action. Step 3: Therefore (f) is bijective and (g) is the inverse of (f).
View question detailsIf (f:\mathbb{R}\to\mathbb{R}) is defined by (f(x)=x^2+2x+3), which statement about the range of (f) is correct?
Correct answer: A
Step 1: Write (x^2+2x+3=(x+1)^2+2). Step 2: Since ((x+1)^2\geq 0), the least value is (2). Step 3: Completing the square is a quick method for finding the range of a quadratic function.
View question detailsIf (f:\mathbb{R}\to\mathbb{R}) is given by (f(x)=2x-5), what is (f^{-1}(x))?
Correct answer: A
Step 1: Put (y=2x-5). Step 2: Solving for (x) gives (x=\frac{y+5}{2}). Step 3: While finding the inverse, replace (y) by (x) at the end.
View question detailsIf (f:A\to B), (A={1,2,3}), and (B={a,b}), how many functions are possible from (A) to (B)?
Correct answer: B
Step 1: Each element of (A) has (2) choices in (B). Step 2: For (3) elements, the total number of functions is (2^3=8). Step 3: In counting functions, the base is the number of elements in the codomain.
View question detailsIf (f:\mathbb{R}\to\mathbb{R}) and (g:\mathbb{R}\to\mathbb{R}) are given by (f(x)=x+3), (g(x)=x^2), what is ((g\circ f)(x))?
Correct answer: B
Step 1: ((g\circ f)(x)) means (g(f(x))). Step 2: Substituting (f(x)=x+3) into (g) gives (g(x+3)=(x+3)^2). Step 3: In composition, changing the order can change the answer.
View question detailsIf (f:\mathbb{R}\to\mathbb{R}) is defined by (f(x)=x^3+x), which statement about (f) is correct?
Correct answer: A
Step 1: If (x_1<x_2), then (x^3+x) increases. Step 2: A strictly increasing function cannot give the same value for two different inputs. Step 3: Monotonic behaviour is useful for checking one-one functions.
View question detailsIf (f:\mathbb{R}\to\mathbb{R}) is given by (f(x)=x^2), why is (f) not one-one?
Correct answer: A
Step 1: A one-one function must give different outputs for different inputs. Step 2: Here (1\neq -1), but (f(1)=1) and (f(-1)=1). Step 3: One counterexample is enough to prove that a function is not one-one.
View question detailsIf (f:\mathbb{R}\to[0,\infty)) is defined by (f(x)=x^2), which statement about (f) is correct?
Correct answer: B
Step 1: For every (y\geq 0), choosing (x=\sqrt{y}) gives (f(x)=y), so it is onto. Step 2: Since (f(1)=f(-1)), it is not one-one. Step 3: The codomain strongly affects onto behaviour.
View question detailsIf (f:\mathbb{R}\to\mathbb{R}) is given by (f(x)=x^2+1), why is (f) not onto?
Correct answer: A
Step 1: (x^2+1\geq 1) for every real (x). Step 2: So (0) is in the codomain but is never attained. Step 3: To disprove onto, show one missing element of the codomain.
View question detailsIf (f:\mathbb{R}\setminus{1}\to\mathbb{R}\setminus{2}) is defined by (f(x)=\frac{2x+3}{x-1}), which value can never occur as (f(x))?
Correct answer: B
Step 1: Let (y=\frac{2x+3}{x-1}). Step 2: Putting (y=2) gives (2x+3=2x-2), which is impossible. Step 3: For rational functions, solve in terms of (y) to find excluded values.
View question detailsIf (f:\mathbb{R}\to\mathbb{R}) and (g:\mathbb{R}\to\mathbb{R}) are given by (f(x)=3x+1), (g(x)=x-4), what is the value of ((f\circ g)(2))?
Correct answer: A
Step 1: First find (g(2)=2-4=-2). Step 2: Then (f(-2)=3(-2)+1=-5). Step 3: In composition, apply the inner function first.
View question detailsIf (f:A\to B) is a one-one function and (|A|=5), (|B|=4), which statement is correct?
Correct answer: B
Step 1: A one-one function sends distinct elements of (A) to distinct elements of (B). Step 2: (A) has (5) elements but (B) has only (4). Step 3: By the pigeonhole idea, such a one-one function cannot exist.
View question detailsIf (A) has (4) elements and (B) has (4) elements, how many functions from (A) to (B) are both one-one and onto?
Correct answer: C
Step 1: For finite sets of equal size, functions that are one-one and onto are permutations. Step 2: The number of permutations of (4) elements is (4!=24). Step 3: Use (n!) directly in such bijection counting questions.
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