What is the real domain of (f(x)=\frac{x+2}{x-1})?
Correct answer: A
Step 1: The denominator (x-1) must not be zero. Step 2: (x-1=0) gives (x=1). Step 3: So the domain is all real numbers except (1).
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Functions
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Easy · Level 21 · domain,rational function,denominator restriction
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- (R-{1})
- (R-{-2})
- (R)
- ([1,\infty))
Easy · Level 21 · number of functions,finite sets,counting
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- (16)
- (8)
- (6)
- (4)
Easy · Level 21 · counting functions,finite sets,domain codomain
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- (64)
- (12)
- (81)
- (7)
Easy · Level 21 · one one functions,counting,finite sets
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- (6)
- (9)
- (3)
- (2)
Easy · Level 21 · bijective function,finite mapping,one one onto
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- Bijective
- Only many-one
- Not onto
- Not a function
Easy · Level 21 · finite function,not one one,not onto
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- It is neither one-one nor onto
- It is bijective
- It is only onto
- It is not a function
Easy · Level 21 · one one proof,linear function,injective
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- Because (f(a)=f(b)) gives (a=b)
- Because all images are the same
- Because it is constant
- Because its range is ({-2})
Easy · Level 21 · not one one,quadratic function,many one
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- Because (f(3)=f(-3))
- Because (f(0)=-9)
- Because (f(1)=-8)
- Because it is not defined
Easy · Level 21 · onto function,cubic function,real numbers
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- Yes, because for every real (y), (x=\sqrt[3]{y+2}) exists
- No, because a cube is always positive
- No, because (-2) is not an image
- No, because it is constant
Easy · Level 21 · absolute value function,not one one,many one
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- Because (f(-3)=f(-1))
- Because (f(-2)=0)
- Because (f(0)=2)
- Because it is not a function
Easy · Level 21 · absolute value,range,function graph
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- ([0,\infty))
- (R)
- ((-\infty,0])
- ([2,\infty))
Easy · Level 21 · one one function,injective property,basic
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- (f(a_1)\ne f(a_2))
- (f(a_1)=f(a_2))
- (a_1=a_2)
- (f(a_1)) is not defined
Easy · Level 21 · onto function,preimage,codomain,surjective
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- It has at least one preimage in (A)
- It has no preimage
- It always lies in (A) also
- It has exactly two preimages
Easy · Level 21 · identity function,composition,function basics
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- (x)
- (2x)
- (x^2)
- (0)
Easy · Level 21 · composition,inverse functions,identity result
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- (x)
- (x+8)
- (x-8)
- (2x)
Easy · Level 21 · inverse functions,composition,linear functions
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- They are inverses of each other
- Both are constant functions
- They are the same function
- They are not functions
Easy · Level 21 · bijective function,inverse function,domain codomain
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- (f^{-1}:B\to A) is defined
- (f^{-1}:A\to B) only
- The inverse never exists
- The inverse is only a constant function
Medium · Level 19 · inverse function,linear function,preimage,medium
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- (4)
- (8)
- (2)
- (12)
Medium · Level 19 · function difference,quadratic function,algebra,medium
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- (2x+3)
- (2x+1)
- (x+3)
- (x^2+3)
Medium · Level 19 · domain,rational function,denominator,medium
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- (R-{-2})
- (R-{1})
- (R)
- ([-2,\infty))
If (A={1,2,3,4}) and (B={a,b}), how many functions can be formed from (A) to (B)?
Correct answer: A
Step 1: (A) has (4) elements and (B) gives (2) choices. Step 2: Each element has (2) image choices. Step 3: Total functions are (2^4=16).
View question detailsIf (A) has (3) elements and (B) has (4) elements, how many functions are there from (A) to (B)?
Correct answer: A
Step 1: The number of functions is (n^m). Step 2: Here (m=3) and (n=4). Step 3: Therefore the number is (4^3=64).
View question detailsIf (A={1,2}) and (B={a,b,c}), how many one-one functions are there from (A) to (B)?
Correct answer: A
Step 1: In a one-one function, the two inputs must have distinct images. Step 2: The first input has (3) choices and the second has (2) choices. Step 3: Total one-one functions are (3\cdot2=6).
View question detailsIf (A={1,2,3}), (B={a,b,c}), and (f={(1,a),(2,b),(3,c)}), what type of function is (f)?
Correct answer: A
Step 1: The three inputs have distinct images, so the function is one-one. Step 2: All elements of (B) appear as images, so it is onto. Step 3: Since it is both one-one and onto, it is bijective.
View question detailsIf (A={1,2,3}), (B={a,b,c}), and (f={(1,a),(2,a),(3,c)}), what is correct about (f)?
Correct answer: A
Step 1: Both (1) and (2) map to (a), so it is not one-one. Step 2: (b) is in the codomain but is not an image, so it is not onto. Step 3: Still, every input has one image, so it is a function.
View question detailsWhy is (f:R\to R), (f(x)=5x-2), one-one?
Correct answer: A
Step 1: Assume (f(a)=f(b)). Step 2: Then (5a-2=5b-2), so (5a=5b). Step 3: This gives (a=b), so the function is one-one.
View question detailsWhy is (f:R\to R), (f(x)=x^2-9), not one-one?
Correct answer: A
Step 1: In a one-one function, different inputs have different images. Step 2: (3\ne -3), but (f(3)=0) and (f(-3)=0). Step 3: Therefore this function is not one-one.
View question detailsIf (f:R\to R), (f(x)=x^3-2), is (f) onto?
Correct answer: A
Step 1: For onto, every real (y) must have a preimage. Step 2: From (x^3-2=y), we get (x=\sqrt[3]{y+2}). Step 3: This is real for every real (y), so the function is onto.
View question detailsWhy is (f:R\to R), (f(x)=|x+2|), not one-one?
Correct answer: A
Step 1: (f(-3)=|-3+2|=1). Step 2: (f(-1)=|-1+2|=1). Step 3: Two different inputs have the same image, so the function is not one-one.
View question detailsIf (f:R\to R), (f(x)=|x+2|), what is its range?
Correct answer: A
Step 1: An absolute value is never negative. Step 2: At (x=-2), (|x+2|=0). Step 3: Therefore the range is ([0,\infty)).
View question detailsIf (f:A\to B) is one-one and (a_1\ne a_2), which statement is true?
Correct answer: A
Step 1: In a one-one function, different inputs have different images. Step 2: Here (a_1\ne a_2) is given. Step 3: Therefore (f(a_1)\ne f(a_2)).
View question detailsIf (f:A\to B) is onto, what is true for every element of (B)?
Correct answer: A
Step 1: In an onto function, no element of the codomain is left out. Step 2: Therefore each element of (B) is reached by at least one element of (A). Step 3: This is the main idea of onto.
View question detailsIf (f:R\to R), (f(x)=x), what is ((f\circ f)(x))?
Correct answer: A
Step 1: (f(x)=x) is the identity function. Step 2: ((f\circ f)(x)=f(f(x))=f(x)). Step 3: Therefore ((f\circ f)(x)=x).
View question detailsIf (f(x)=x+4) and (g(x)=x-4), what is ((g\circ f)(x))?
Correct answer: A
Step 1: ((g\circ f)(x)=g(f(x))). Step 2: Put (f(x)=x+4) into (g). Step 3: (g(x+4)=(x+4)-4=x).
View question detailsIf (f(x)=3x) and (g(x)=\frac{x}{3}), which statement about (f) and (g) is correct?
Correct answer: A
Step 1: (f(g(x))=f(\frac{x}{3})=x). Step 2: (g(f(x))=g(3x)=x). Step 3: Both composites give the identity function, so they are inverses.
View question detailsIf (f:A\to B) is bijective, which statement about the inverse function is correct?
Correct answer: A
Step 1: A bijective function is both one-one and onto. Step 2: Therefore every image has exactly one preimage. Step 3: Hence the inverse function is defined from (B) to (A).
View question detailsIf (f:R\to R) and (f(x)=3x-4), what is the value of (f^{-1}(8))?
Correct answer: A
Step 1: (f^{-1}(8)) means the value of (x) for which (f(x)=8). Step 2: From (3x-4=8), we get (3x=12), so (x=4). Step 3: To find an inverse value, first equate the original function to the given value.
View question detailsIf (f(x)=x^2+2x), what is (f(x+1)-f(x))?
Correct answer: A
Step 1: Write (f(x+1)=(x+1)^2+2(x+1)). Step 2: Simplifying gives (x^2+4x+3), and (f(x)=x^2+2x). Step 3: Subtracting gives (2x+3).
View question detailsIf (f(x)=\frac{x-1}{x+2}), what is the real domain?
Correct answer: A
Step 1: The denominator of the fraction is (x+2). Step 2: For the function to be defined, (x+2\ne0), so (x\ne-2). Step 3: The value that makes the denominator zero is removed from the domain.
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