If (f(x)=x^2-7x+12), which of (f(3), f(4), f(5)) are zero?
Correct answer: A
Step 1: (f(x)=x^2-7x+12=(x-3)(x-4)). Step 2: At (x=3) and (x=4), the product becomes zero. Step 3: At (x=5), (f(5)=2), so it is not zero.
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 · zero of function,quadratic function,factorisation
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- (f(3)) and (f(4))
- only (f(5))
- only (f(3))
- all three
Medium · Level 21 · absolute value,range,minimum value
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- ([2,\infty))
- ([0,\infty))
- (R)
- (R-{2})
Medium · Level 21 · onto function,absolute value,range,codomain
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- Because (1) cannot be its image
- Because (2) is its image
- Because it is defined everywhere
- Because it is not a function
Medium · Level 21 · bijective function,finite sets,cardinality
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- (9)
- (8)
- (10)
- Cannot be determined
Medium · Level 21 · inverse function,composition,identity function,bijection
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- (g=f^{-1})
- (g=f) only
- (g) is a constant function
- (g) is not defined
Medium · Level 21 · composition,domain,radical function,rational function
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- ([-1,\infty)-{3})
- ([-1,\infty))
- (R-{3})
- ((3,\infty))
Medium · Level 8 · sets,linear inequalities,interval notation,real numbers,Introduction to Relations,Relations and Functions,Mathematics,Class 12 MCQ
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- (3, ∞)
- [3, ∞)
- (−∞, 3]
- (−∞, 3)
Easy · Level 23 · relations,cartesian-product,ordered-pairs,relation-definition,Introduction to Relations,Relations and Functions,Mathematics,Class 12 MCQ
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- \(\{(1,3),(2,4)\}\)
- \(\{(1,2),(2,3),(3,4)\}\)
- \(\{(3,1),(4,2)\}\)
- \(\{(1,3),(2,4),(3,5)\}\)
If (f:R\to R), (f(x)=|x+3|+2), what is the range?
Correct answer: A
Step 1: (|x+3|\ge0) for every real (x). Step 2: Therefore (|x+3|+2\ge2). Step 3: At (x=-3), the minimum value (2) is obtained, so the range is ([2,\infty)).
View question detailsIf (f:R\to R), (f(x)=|x+3|+2), why is it not onto?
Correct answer: A
Step 1: The value of (|x+3|+2) is always at least (2). Step 2: The codomain (R) contains (1), 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 (A) has (9) elements, how many elements will (B) 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 (A) has (9) elements, (B) also has (9).
View question detailsIf (f:A\to B) and (g:B\to A) satisfy (g\circ f=I_A) and (f\circ g=I_B), what is correct about (g)?
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 (g) is the inverse of (f), so (g=f^{-1}).
View question detailsIf (f(x)=\sqrt{x+1}) and (g(x)=\frac{1}{x-2}), what is the real domain of ((g\circ f)(x))?
Correct answer: A
Step 1: First, for (f(x)=\sqrt{x+1}), we need (x+1\ge0), so (x\ge-1). Step 2: ((g\circ f)(x)=\frac{1}{\sqrt{x+1}-2}), so the denominator must not be zero. Step 3: (\sqrt{x+1}=2) gives (x=3), hence the domain is ([-1,\infty)-{3}).
View question detailsIf A = {x ∈ R : 2x − 1 ≥ 5}, what is the interval form of A?
Correct answer: B
Solve the defining inequality step by step: 2x − 1 ≥ 5. Adding 1 to both sides gives 2x ≥ 6, and dividing by the positive number 2 gives x ≥ 3. Thus A contains 3 and every real number greater than 3. In interval notation, inclusion of 3 is shown by a square bracket, while infinity always uses a parenthesis. Therefore A = [3,∞), so B is correct.
View question detailsIf \(A=\{1,2,3\}\), \(B=\{1,2,3,4\}\), and \(R=\{(x,y)\in A\times B:y=x+2\}\), what is \(R\)?
Correct answer: A
A pair belongs to \(R\) only when its first component is in \(A\), its second component is in \(B\), and it satisfies \(y=x+2\). For \(x=1\), \(y=3\), producing \((1,3)\). For \(x=2\), \(y=4\), producing \((2,4)\). For \(x=3\), the equation gives \(y=5\), but 5 is not in \(B\), so that pair is excluded. Hence \(R=\{(1,3),(2,4)\}\), option A.
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