Relations and Functions Quiz - ICSE Class 12 Mathematics
Free practice quiz: Relations and Functions (ICSE Class 12 Mathematics). 57 multiple-choice questions with answers and explanations. Play online, no sign-up needed.
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Let $f: \mathbb{R} \setminus \{0\} o \mathbb{R}$ be defined by $f(x) = \frac{1}{x}$. If $g(x) = f(f(x))$, find the domain of $g(x)$.
- $\mathbb{R}$
- $\mathbb{R} \setminus \{0\}$
- $\mathbb{R} \setminus \{1\}$
- $\mathbb{R} \setminus \{-1\}$
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Answer: B. $\mathbb{R} \setminus \{0\}$
The function $f(x) = \frac{1}{x}$ has a domain of $\mathbb{R} \setminus \{0\}$. For $g(x) = f(f(x))$, the input to the outer function $f$ is $f(x)$. Therefore, $f(x)$ cannot be $0$. Since $f(x) = \frac{1}{x}$, $\frac{1}{x} eq 0$ is always true for any $x \in \mathbb{R} \setminus \{0\}$. The domain of $f(x)$ is $\mathbb{R} \setminus \{0\}$. Thus, the domain of $g(x)$ is also $\mathbb{R} \setminus \{0\}$.
Let $f(x) = 2x + 3$ and $g(x) = x-1$. Find $(f+g)(x)$.
- $3x + 2$
- $2x^2 + 3x - 1$
- $2x - 2$
- $3x + 4$
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Answer: A. $3x + 2$
The sum of two functions $f(x)$ and $g(x)$, denoted by $(f+g)(x)$, is defined as the function obtained by adding the values of $f(x)$ and $g(x)$ for each $x$ in their common domain. In this case, the domain for both $f(x)$ and $g(x)$ is all real numbers ($\mathbb{R}$). $(f+g)(x) = f(x) + g(x) = (2x + 3) + (x - 1)$. Combine like terms: $(f+g)(x) = (2x + x) + (3 - 1) = 3x + 2$.
If $f(x) = x^3 - x$, then $f$ is:
- Even
- Odd
- Neither even nor odd
- Both even and odd
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Answer: B. Odd
A function $f(x)$ is even if $f(-x) = f(x)$ for all $x$ in its domain. A function $f(x)$ is odd if $f(-x) = -f(x)$ for all $x$ in its domain. Let's evaluate $f(-x)$ for $f(x) = x^3 - x$. $f(-x) = (-x)^3 - (-x) = -x^3 + x$. Now let's compare $f(-x)$ with $-f(x)$. $-f(x) = -(x^3 - x) = -x^3 + x$. Since $f(-x) = -f(x)$, the function $f(x) = x^3 - x$ is an odd function.
Let $f(x) = x^2$. Which of the following is not a property of $f(x)$ on $\mathbb{R}$?
- It is not injective.
- It is not surjective.
- It is neither even nor odd.
- Its range is $[0, \infty)$.
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Answer: C. It is neither even nor odd.
Let $f(x) = x^2$ on the domain $\mathbb{R}$. 1. Injectivity: $f(2) = 4$ and $f(-2) = 4$. Since $2 eq -2$, $f$ is not injective. This statement is true. 2. Surjectivity: The range of $f(x) = x^2$ on $\mathbb{R}$ is $[0, \infty)$. The codomain is assumed to be $\mathbb{R}$. Since the range is not equal to the codomain (e.g., -1 is in the codomain but not in the range), $f$ is not surjective onto $\mathbb{R}$. This statement is true. 3. Even/Odd: $f(-x) = (-x)^2 = x^2 = f(x)$. So, $f$ is an even function. This statement claims $f$ is neither even nor odd, which is false. Therefore, this is the statement that is NOT a property of $f(x)$. 4. Range: As established, the range of $f(x) = x^2$ on $\mathbb{R}$ is indeed $[0, \infty)$. This statement is true. The question asks which of the following is NOT a property. The false statement is that $f$ is neither even nor odd.
Let $f(x) = x^3$. Which of the following statements is false?
- $f$ is injective on $\mathbb{R}$.
- $f$ is surjective on $\mathbb{R}$.
- $f$ is an odd function.
- $f$ is a constant function.
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Answer: D. $f$ is a constant function.
Let $f(x) = x^3$. We analyze each statement for $f: \mathbb{R} o \mathbb{R}$. 1. Injectivity: If $f(x_1) = f(x_2)$, then $x_1^3 = x_2^3$, which implies $x_1 = x_2$. So, $f$ is injective. This statement is true. 2. Surjectivity: For any $y \in \mathbb{R}$, there exists $x = \sqrt[3]{y} \in \mathbb{R}$ such that $f(x) = x^3 = (\sqrt[3]{y})^3 = y$. So, $f$ is surjective. This statement is true. 3. Odd function: $f(-x) = (-x)^3 = -x^3 = -f(x)$. So, $f$ is an odd function. This statement is true. 4. Constant function: A constant function has the same output for every input. For $f(x) = x^3$, $f(1) = 1$ and $f(2) = 8$. Since $f(1) eq f(2)$, $f$ is not a constant function. This statement is false. The question asks for the false statement.