If (f(x)=2x+1) and (g(x)=x^2+3), what is the value of ((g\circ f)(1))?
Correct answer: A
Step 1: ((g\circ f)(1)=g(f(1))). Step 2: (f(1)=2\cdot1+1=3). Step 3: (g(3)=3^2+3=12), so the value is (12).
View question detailsMathematics
Introduction to Relations
संबंधों का परिचय
In Class 12 Mathematics, under the chapter Relations and Functions, Introduction to Relations explains a relation as a subset of a Cartesian product. Students learn to form and count relations on sets, and identify important examples such as empty, universal and identity relations. The topic also introduces conditions used to recognise reflexive and symmetric relations.
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Medium · Level 21 · composition of functions,function value,medium
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- (12)
- (10)
- (9)
- (6)
Medium · Level 21 · composition,linear functions,algebra,medium
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- (5x-19)
- (5x-3)
- (x+1)
- (5x+5)
Medium · Level 21 · composition order,functions,non commutative,medium
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- They are generally not equal
- They are always equal
- Both are constant functions
- Both are undefined
Medium · Level 21 · linear function,bijection,one one,onto
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- It is one-one and onto
- It is a constant function
- It is not one-one
- It is not onto
Medium · Level 21 · 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 21 · restricted domain,bijection,linear function,medium
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- One-one and onto
- Not one-one
- Not onto
- Constant function
Medium · Level 21 · onto,not one one,quadratic 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 21 · domain,radical function,inequality,medium
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- ([4,\infty))
- ((4,\infty))
- (R-{4})
- ((-\infty,4])
Medium · Level 21 · domain,radical denominator,real function,medium
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- ((-\infty,5))
- ((-\infty,5])
- (R-{5})
- ([5,\infty))
Medium · Level 21 · domain,rational function,excluded values,medium
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- (R-{-2,2})
- (R-{2})
- (R-{-2})
- (R)
Medium · Level 21 · preimage,absolute value,function inverse image
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- ({2,8})
- ({3})
- ({8})
- ({2})
Medium · Level 21 · 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 21 · onto functions,counting,finite sets,medium
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- (30)
- (32)
- (25)
- (20)
Medium · Level 21 · one one functions,counting,finite sets,permutation
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- (120)
- (216)
- (18)
- (60)
Medium · Level 21 · onto function,finite sets,domain codomain,counting
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- No, because the domain has fewer elements than the codomain
- Yes, there are (12) such functions
- Yes, every function will be onto
- Yes, there are (16) such functions
Medium · Level 21 · inverse function,cubic function,preimage,medium
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- (2)
- (4)
- (8)
- (-2)
Medium · Level 21 · inverse function,linear fractional form,algebra
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- (\frac{2x-1}{5})
- (\frac{2x+1}{5})
- (\frac{5x-1}{2})
- (5x+1)
Medium · Level 21 · inverse function,not one one,quadratic 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 linear
Medium · Level 21 · preimage,absolute value,inverse image,medium
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- ({-10,2})
- ({6})
- ({2})
- ({-10})
Medium · Level 21 · minimum value,quadratic function,range,medium
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- (4)
- (20)
- (0)
- (8)
If (f(x)=5x+1) and (g(x)=x-4), what is ((f\circ g)(x))?
Correct answer: A
Step 1: ((f\circ g)(x)=f(g(x))). Step 2: Put (g(x)=x-4) into (f). Step 3: (f(x-4)=5(x-4)+1=5x-19).
View question detailsIf (f(x)=x^2-1) and (g(x)=x+2), 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+2)=(x+2)^2-1). Step 2: ((g\circ f)(x)=g(x^2-1)=x^2+1). Step 3: The expressions are generally different, so order matters in composition.
View question detailsIf (f:R\to R), (f(x)=9x-4), which statement is correct about (f)?
Correct answer: A
Step 1: (9x-4) gives different outputs for different (x), so it is one-one. Step 2: For any (y\in R), choose (x=\frac{y+4}{9}). Step 3: Hence every real (y) is an image, so the function is onto.
View question detailsWhich statement is correct about (f:R\to R), (f(x)=x^2+2x+2)?
Correct answer: A
Step 1: (f(0)=2) and (f(-2)=2), so the function is not one-one. Step 2: (x^2+2x+2=(x+1)^2+1), so the range is ([1,\infty)). Step 3: With codomain (R), (0) is not an image, so the function is not onto either.
View question detailsIf (f:[-1,\infty)\to[0,\infty)), (f(x)=x+1), what type of function is (f)?
Correct answer: A
Step 1: For different (x), (x+1) is different, so the function is one-one. Step 2: For every (y\ge0), (x=y-1), which lies in the domain ([-1,\infty)). Step 3: Hence the function is onto as well.
View question detailsIf (f:R\to[2,\infty)), (f(x)=x^2+2), which statement is correct about (f)?
Correct answer: A
Step 1: For every (y\ge2), (x=\sqrt{y-2}) or (x=-\sqrt{y-2}) can be chosen, so it is onto. Step 2: (f(1)=f(-1)=3), so it is not one-one. Step 3: Reading the codomain correctly is very important while checking onto.
View question detailsIf (f(x)=\sqrt{3x-12}), what is the real domain?
Correct answer: A
Step 1: The expression inside the square root, (3x-12), must be non-negative. Step 2: From (3x-12\ge0), we get (x\ge4). Step 3: At (x=4), the value is (0), so (4) is included.
View question detailsIf (f(x)=\frac{1}{\sqrt{5-x}}), what is the real domain?
Correct answer: A
Step 1: For the square root, (5-x\ge0) is needed. Step 2: But this square root is in the denominator, so it cannot be zero. Step 3: Thus (5-x>0), meaning (x<5), so the domain is ((-\infty,5)).
View question detailsIf (f(x)=\frac{x+2}{x^2-4}), what is the real domain?
Correct answer: A
Step 1: The denominator is (x^2-4=(x-2)(x+2)). Step 2: The denominator cannot be zero, so (x\ne2) and (x\ne-2). Step 3: Even if (x+2) cancels, (x=-2) is not valid in the original function.
View question detailsIf (f(x)=|x-5|), what is the set (f^{-1}({3}))?
Correct answer: A
Step 1: To find the preimage, solve (|x-5|=3). Step 2: This gives (x-5=3) or (x-5=-3). Step 3: Hence (x=8) or (x=2), so the preimage is ({2,8}).
View question detailsIf (f:A\to B), (A={1,2,3}), (B={u,v,w,z}), and (f={(1,u),(2,v),(3,w)}), 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: (z) is in the codomain but is not an image. Step 3: Therefore the function is not onto.
View question detailsIf (A={1,2,3,4,5}) and (B={a,b}), how many onto functions are there from (A) to (B)?
Correct answer: A
Step 1: Total functions are (2^5=32). Step 2: There are two non-onto functions, where all elements go only to (a) or only to (b). Step 3: Therefore onto functions are (32-2=30).
View question detailsIf (A) has (3) elements and (B) has (6) 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 (6) choices, the second has (5), and the third has (4). Step 3: Total one-one functions are (6\cdot5\cdot4=120).
View question detailsIf (A={1,2}) and (B={a,b,c,d}), 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 (2) elements and the codomain has (4) elements. Step 3: Two inputs cannot cover four different codomain elements, so no onto function exists.
View question detailsIf (f:R\to R), (f(x)=x^3-4), what is (f^{-1}(4))?
Correct answer: A
Step 1: To find (f^{-1}(4)), solve (x^3-4=4). 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{5x+1}{2}), what is (f^{-1}(x))?
Correct answer: A
Step 1: Write (y=\frac{5x+1}{2}). Step 2: Then (2y=5x+1), so (x=\frac{2y-1}{5}). Step 3: Replacing (y) by (x), (f^{-1}(x)=\frac{2x-1}{5}).
View question detailsIf (f:R\to R), (f(x)=x^2-6x), 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: (f(0)=0) and (f(6)=0), while (0\ne6). 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+4|), what is (f^{-1}({6}))?
Correct answer: A
Step 1: To find the preimage, solve (|x+4|=6). Step 2: This gives (x+4=6) or (x+4=-6). Step 3: Hence (x=2) or (x=-10), so the preimage is ({-10,2}).
View question detailsIf (f:R\to R), (f(x)=x^2-8x+20), what is the minimum value?
Correct answer: A
Step 1: Write (x^2-8x+20=(x-4)^2+4). Step 2: Since ((x-4)^2\ge0), (f(x)\ge4). Step 3: The minimum value is (4), attained at (x=4).
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