If (f:R\to R), (f(x)=x^2-2x+3), what is the range of (f)?
Correct answer: A
Step 1: Write (x^2-2x+3=(x-1)^2+2). Step 2: Since ((x-1)^2\ge0), we have (f(x)\ge2). Step 3: The minimum value is (2), so the range is ([2,\infty)).
View question detailsMathematics
Functions
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Medium · Level 19 · range,quadratic function,minimum value,medium
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- ([2,\infty))
- ([3,\infty))
- (R)
- ((-\infty,2])
Medium · Level 19 · composition of functions,function value,medium
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- (24)
- (25)
- (15)
- (9)
Medium · Level 19 · composition,linear functions,algebra,medium
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- (2x+2)
- (2x-4)
- (2x+3)
- (x+2)
Medium · Level 19 · composition order,functions,non commutative,medium
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- They are not always equal
- They are always equal
- Both are undefined
- Both are constant functions
Medium · Level 19 · one one,onto,linear function,bijection
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- It is one-one and onto
- It is only constant
- It is not one-one
- It is not onto
Medium · Level 19 · quadratic function,not one one,not onto,range
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- It is neither one-one nor onto
- It is one-one and onto
- It is one-one but not onto
- It is onto but not one-one
Medium · Level 19 · restricted domain,square function,bijection,medium
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- One-one and onto
- Not one-one
- Not onto
- Constant function
Medium · Level 19 · onto,not one one,square function,codomain
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- It is onto but not one-one
- It is one-one and onto
- It is one-one but not onto
- It is not a function
Medium · Level 19 · domain,radical function,inequality,medium
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- ([2,\infty))
- ((2,\infty))
- (R-{2})
- ((-\infty,2])
Medium · Level 19 · domain,radical denominator,real function,medium
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- ((1,\infty))
- ([1,\infty))
- (R-{1})
- ((-\infty,1))
Medium · Level 19 · domain,rational function,excluded values,medium
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- (R-{-3,3})
- (R-{3})
- (R-{-3})
- (R)
Medium · Level 19 · preimage,absolute value,function inverse image
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- ({-1,5})
- ({3})
- ({5})
- ({-1})
Medium · Level 19 · finite function,one one,not onto,codomain
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- One-one but not onto
- Onto but not one-one
- Bijective
- Not a function
Medium · Level 19 · onto functions,counting,finite sets,medium
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- (6)
- (8)
- (4)
- (2)
Medium · Level 19 · one one functions,counting,finite sets,permutation
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- (60)
- (125)
- (15)
- (25)
Medium · Level 19 · onto function,finite sets,counting,domain codomain
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- No, because the domain has fewer elements than the codomain
- Yes, there are (6) such functions
- Yes, there are (9) such functions
- Yes, every function will be onto
Medium · Level 19 · inverse function,cubic function,preimage,medium
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- (2)
- (1)
- (8)
- (3)
Medium · Level 19 · inverse function,linear fractional form,algebra
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- (\frac{3x-1}{2})
- (\frac{x-1}{3})
- (\frac{3x+1}{2})
- (2x+1)
Medium · Level 19 · inverse function,not one one,square function,medium
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- Because it is not one-one
- Because it is not defined everywhere
- Because its value is never positive
- Because it is a constant function
Medium · Level 19 · preimage,absolute value,inverse image,medium
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- ({-2,2})
- ({2})
- ({-2})
- ([0,2])
If (f(x)=2x+1) and (g(x)=x^2-1), what is the value of ((g\circ f)(2))?
Correct answer: A
Step 1: ((g\circ f)(2)=g(f(2))). Step 2: (f(2)=2\cdot2+1=5). Step 3: (g(5)=5^2-1=24), so the answer is (24).
View question detailsIf (f(x)=x+3) and (g(x)=2x-1), what is ((f\circ g)(x))?
Correct answer: A
Step 1: ((f\circ g)(x)=f(g(x))). Step 2: Put (g(x)=2x-1) into (f). Step 3: (f(2x-1)=2x-1+3=2x+2).
View question detailsIf (f(x)=x^2) and (g(x)=x+1), which statement about ((f\circ g)(x)) and ((g\circ f)(x)) is correct?
Correct answer: A
Step 1: ((f\circ g)(x)=f(x+1)=(x+1)^2). Step 2: ((g\circ f)(x)=g(x^2)=x^2+1). Step 3: These expressions are generally not equal, so order matters.
View question detailsIf (f:R\to R), (f(x)=5x-7), which statement is correct about (f)?
Correct answer: A
Step 1: (5x-7) gives different values for different (x), so it is one-one. Step 2: For any real (y), choosing (x=\frac{y+7}{5}) gives (f(x)=y). Step 3: Therefore it is onto as well.
View question detailsWhich statement is correct about (f:R\to R), (f(x)=x^2+1)?
Correct answer: A
Step 1: (f(1)=f(-1)=2), so it is not one-one. Step 2: Since (x^2+1\ge1), a real number like (0) cannot be an image. Step 3: Therefore it is not onto either.
View question detailsIf (f:[0,\infty)\to[0,\infty)), (f(x)=x^2), what type of function is (f)?
Correct answer: A
Step 1: On ([0,\infty)), (x^2) is increasing, so different (x) values give different images. Step 2: For every (y\ge0), (x=\sqrt{y}) lies in the domain. Step 3: Hence the function is both one-one and onto.
View question detailsIf (f:R\to[0,\infty)), (f(x)=x^2), which statement is correct about (f)?
Correct answer: A
Step 1: For every (y\ge0), (x=\sqrt{y}) or (x=-\sqrt{y}) gives an input, so the range equals the codomain. Step 2: (f(1)=f(-1)=1), so it is not one-one. Step 3: Changing the codomain can change onto status.
View question detailsIf (f(x)=\sqrt{x-2}), what is the real domain?
Correct answer: A
Step 1: The expression inside the square root, (x-2), must be non-negative. Step 2: (x-2\ge0) gives (x\ge2). Step 3: At (x=2), the value is (0), so (2) is included.
View question detailsIf (f(x)=\frac{1}{\sqrt{x-1}}), what is the real domain?
Correct answer: A
Step 1: For the square root, (x-1\ge0) is needed. Step 2: But (\sqrt{x-1}) is in the denominator, so it cannot be zero. Step 3: Thus (x-1>0), so the domain is ((1,\infty)).
View question detailsIf (f(x)=\frac{x+3}{x^2-9}), what is the real domain?
Correct answer: A
Step 1: The denominator is (x^2-9=(x-3)(x+3)). Step 2: The denominator must not be zero, so (x\ne3) and (x\ne-3). Step 3: Even if (x+3) cancels algebraically, (x=-3) is not allowed in the original function.
View question detailsIf (f(x)=|x-2|), what is the set (f^{-1}({3}))?
Correct answer: A
Step 1: For (f^{-1}({3})), solve (|x-2|=3). Step 2: This gives (x-2=3) or (x-2=-3). Step 3: Hence (x=5) or (x=-1), so the preimage is ({-1,5}).
View question detailsIf (f:A\to B), (A={1,2,3}), (B={a,b,c,d}), and (f={(1,a),(2,b),(3,c)}), what type of function is (f)?
Correct answer: A
Step 1: The images of (1,2,3) are distinct, so the function is one-one. Step 2: (d) is in the codomain but is not an image of any element. Step 3: Therefore the function is not onto.
View question detailsIf (A={1,2,3}) and (B={a,b}), how many onto functions are there from (A) to (B)?
Correct answer: A
Step 1: Total functions are (2^3=8). Step 2: Non-onto functions are those sending all elements only to (a) or only to (b), so there are (2). Step 3: Hence onto functions are (8-2=6).
View question detailsIf (A) has (3) elements and (B) has (5) elements, how many one-one functions are there from (A) to (B)?
Correct answer: A
Step 1: In a one-one function, the three inputs must have distinct images. Step 2: The first input has (5) choices, the second has (4), and the third has (3). Step 3: Total functions are (5\cdot4\cdot3=60).
View question detailsIf (A={1,2}) and (B={a,b,c}), can onto functions from (A) to (B) exist?
Correct answer: A
Step 1: In an onto function, every element of the codomain must be an image. Step 2: Here the domain has only (2) elements, while the codomain has (3). Step 3: Two inputs cannot cover three distinct codomain elements, so no onto function exists.
View question detailsIf (f:R\to R), (f(x)=x^3-1), what is (f^{-1}(7))?
Correct answer: A
Step 1: To find (f^{-1}(7)), solve (x^3-1=7). Step 2: (x^3=8), so (x=2). Step 3: For a cubic function, the real cube root can be taken directly.
View question detailsIf (f(x)=\frac{2x+1}{3}), what is (f^{-1}(x))?
Correct answer: A
Step 1: Write (y=\frac{2x+1}{3}). Step 2: Then (3y=2x+1), so (x=\frac{3y-1}{2}). Step 3: Replacing (y) by (x), (f^{-1}(x)=\frac{3x-1}{2}).
View question detailsIf (f(x)=x^2) is considered from (R) to (R), why does an inverse function not exist?
Correct answer: A
Step 1: For an inverse function to exist, the original function must be one-one. Step 2: Here (f(1)=f(-1)=1), while (1\ne-1). Step 3: Since one image has two preimages, the inverse is not a well-defined function.
View question detailsIf (f:R\to R), (f(x)=|x|), what is (f^{-1}({2}))?
Correct answer: A
Step 1: To find the preimage, solve (|x|=2). Step 2: This gives (x=2) or (x=-2). Step 3: Therefore (f^{-1}({2})={-2,2}).
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