Composition of Functions and Invertible Functions Quiz - CBSE Class 12 Mathematics
Free practice quiz: Composition of Functions and Invertible Functions (CBSE Class 12 Mathematics). 13 multiple-choice questions with answers and explanations. Play online, no sign-up needed.
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If f(x) = 2x + 3 and g(x) = x - 2, find (f ∘ g)(x).
- 2x - 1
- 2x + 1
- 2x - 5
- 4x - 1
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Answer: A. 2x - 1
The composition (f ∘ g)(x) means f(g(x)). Since g(x) = x - 2, substitute x - 2 into f(x): f(x) = 2(x - 2) + 3 = 2x - 4 + 3 = 2x - 1.
If f(x) = 5x + 7, find the inverse function f⁻¹(x).
- (x - 7)/5
- (x + 7)/5
- 5x + 7
- 5(x - 7)
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Answer: A. (x - 7)/5
To find the inverse, swap x and y in the equation y = 5x + 7, giving x = 5y + 7. Solve for y: y = (x - 7)/5, so f⁻¹(x) = (x - 7)/5.
If h(x) = 4x + 8, what is the value of (h ∘ h⁻¹)(10)?
- 10
- 12
- 8
- 6
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Answer: A. 10
The composition (h ∘ h⁻¹)(x) is equal to x, hence (h ∘ h⁻¹)(10) = 10.
Prove that f(x) = (2x + 3)/(3x + 5) is invertible.
- Yes, it is a bijection.
- No, it is not a bijection.
- Yes, it is injective but not surjective.
- No, it is surjective but not injective.
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Answer: A. Yes, it is a bijection.
To prove that a function is invertible, it must be bijective. f(x) = (2x + 3)/(3x + 5) is both injective and surjective based on its domain, therefore it is a bijection.
Prove f(x) = log(x+1) is injective over the domain x > -1.
- It's injective because f(x) has a unique derivative.
- It is not injective because it is logarithmic.
- It is injective as it's strictly increasing.
- It is not injective because for x = -1
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Answer: C. It is injective as it's strictly increasing.
The function f(x) = log(x+1) is strictly increasing for x > -1, implying that if x1 < x2 -> f(x1) < f(x2), meeting injective criteria definitively over given domain.